23
Electric Fields
CHAPTER OUTLINE
23.1
23.2
23.3
23.4
23.5
23.6
23.7
Properties of Electric
Charges
Charging Objects by
Induction
Coulomb’s Law
The Electric Field
Electric Field of a
Continuous Charge
Distribution
Electric Field Lines
Motion of Charged Particles
in a Uniform Electric Field
ANSWERS TO QUESTIONS
Q23.1
A neutral atom is one that has no net charge. This means that it
has the same number of electrons orbiting the nucleus as it has
protons in the nucleus. A negatively charged atom has one or
more excess electrons.
Q23.2
When the comb is nearby, molecules in the paper are polarized,
similar to the molecules in the wall in Figure 23.5a, and the
paper is attracted. During contact, charge from the comb is
transferred to the paper by conduction. Then the paper has the
same charge as the comb, and is repelled.
Q23.3
The clothes dryer rubs dissimilar materials together as it
tumbles the clothes. Electrons are transferred from one kind of
molecule to another. The charges on pieces of cloth, or on
nearby objects charged by induction, can produce strong
electric fields that promote the ionization process in the
surrounding air that is necessary for a spark to occur. Then you
hear or see the sparks.
Q23.4
To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and could
discharge with a spark, possibly causing an explosion of any flammable material in the oxygenenriched atmosphere.
Q23.5
Electrons are less massive and more mobile than protons. Also, they are more easily detached from
atoms than protons.
Q23.6
The electric field due to the charged rod induces charges on near and far sides of the sphere. The
attractive Coulomb force of the rod on the dissimilar charge on the close side of the sphere is larger
than the repulsive Coulomb force of the rod on the like charge on the far side of the sphere. The
result is a net attraction of the sphere to the rod. When the sphere touches the rod, charge is
conducted between the rod and the sphere, leaving both the rod and the sphere like-charged. This
results in a repulsive Coulomb force.
Q23.7
All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite
readily “steal” charge from a charged object, as any physics teacher trying to perform electrostatics
demonstrations in the summer well knows. As a result—it is difficult to accumulate large amounts of
excess charge on an object in a humid climate. During a North American winter, the cold, dry air
allows accumulation of significant excess charge, giving the potential (pun intended) for a shocking
(pun also intended) introduction to static electricity sparks.
1
2
Electric Fields
Q23.8
Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) of
two particles, and inversely proportional to the square of the separation distance. An electrical force
exhibits the same proportionalities, with charge as the intrinsic property.
Differences: The electrical force can either attract or repel, while the gravitational force as
described by Newton’s law can only attract. The electrical force between elementary particles is
vastly stronger than the gravitational force.
Q23.9
No. The balloon induces polarization of the molecules in the wall, so that a layer of positive charge
exists near the balloon. This is just like the situation in Figure 23.5a, except that the signs of the
charges are reversed. The attraction between these charges and the negative charges on the balloon
is stronger than the repulsion between the negative charges on the balloon and the negative charges
in the polarized molecules (because they are farther from the balloon), so that there is a net attractive
force toward the wall. Ionization processes in the air surrounding the balloon provide ions to which
excess electrons in the balloon can transfer, reducing the charge on the balloon and eventually
causing the attractive force to be insufficient to support the weight of the balloon.
Q23.10
The electric field due to the charged rod induces a charge in the aluminum foil. If the rod is brought
towards the aluminum from above, the top of the aluminum will have a negative charge induced on
it, while the parts draping over the pencil can have a positive charge induced on them. These
positive induced charges on the two parts give rise to a repulsive Coulomb force. If the pencil is a
good insulator, the net charge on the aluminum can be zero.
Q23.11
So the electric field created by the test charge does not distort the electric field you are trying to
measure, by moving the charges that create it.
Q23.12
With a very high budget, you could send first a proton and then an electron into an evacuated
region in which the field exists. If the field is gravitational, both particles will experience a force in
the same direction, while they will experience forces in opposite directions if the field is electric.
On a more practical scale, stick identical pith balls on each end of a toothpick. Charge one pith
ball + and the other –, creating a large-scale dipole. Carefully suspend this dipole about its center of
mass so that it can rotate freely. When suspended in the field in question, the dipole will rotate to
align itself with an electric field, while it will not for a gravitational field. If the test device does not
rotate, be sure to insert it into the field in more than one orientation in case it was aligned with the
electric field when you inserted it on the first trial.
Q23.13
The student standing on the insulating platform is held at the same electrical potential as the
generator sphere. Charge will only flow when there is a difference in potential. The student who
unwisely touches the charged sphere is near zero electrical potential when compared to the charged
sphere. When the student comes in contact with the sphere, charge will flow from the sphere to him
or her until they are at the same electrical potential.
Q23.14
An electric field once established by a positive or negative charge extends in all directions from the
charge. Thus, it can exist in empty space if that is what surrounds the charge. There is no material at
point A in Figure 23.23(a), so there is no charge, nor is there a force. There would be a force if a
charge were present at point A, however. A field does exist at point A.
Q23.15
If a charge distribution is small compared to the distance of a field point from it, the charge
distribution can be modeled as a single particle with charge equal to the net charge of the
distribution. Further, if a charge distribution is spherically symmetric, it will create a field at exterior
points just as if all of its charge were a point charge at its center.
Chapter 23
3
Q23.16
The direction of the electric field is the direction in which a positive test charge would feel a force
when placed in the field. A charge will not experience two electrical forces at the same time, but the
vector sum of the two. If electric field lines crossed, then a test charge placed at the point at which
they cross would feel a force in two directions. Furthermore, the path that the test charge would
follow if released at the point where the field lines cross would be indeterminate.
Q23.17
Both figures are drawn correctly. E1 and E 2 are the electric fields separately created by the point
charges q1 and q 2 in Figure 23.14 or q and –q in Figure 23.15, respectively. The net electric field is the
vector sum of E1 and E 2 , shown as E. Figure 23.21 shows only one electric field line at each point
away from the charge. At the point location of an object modeled as a point charge, the direction of
the field is undefined, and so is its magnitude.
Q23.18
The electric forces on the particles have the same magnitude, but are in opposite directions. The
electron will have a much larger acceleration (by a factor of about 2 000) than the proton, due to its
much smaller mass.
Q23.19
The electric field around a point charge approaches infinity as r approaches zero.
Q23.20
Vertically downward.
Q23.21
Four times as many electric field lines start at the surface of the larger charge as end at the smaller
charge. The extra lines extend away from the pair of charges. They may never end, or they may
terminate on more distant negative charges. Figure 23.24 shows the situation for charges +2q and –q.
Q23.22
At a point exactly midway between the two changes.
Q23.23
Linear charge density, λ, is charge per unit length. It is used when trying to determine the electric
field created by a charged rod.
Surface charge density, σ, is charge per unit area. It is used when determining the electric field
above a charged sheet or disk.
Volume charge density, ρ, is charge per unit volume. It is used when determining the electric
field due to a uniformly charged sphere made of insulating material.
Q23.24
Yes, the path would still be parabolic. The electrical force on the electron is in the downward
direction. This is similar to throwing a ball from the roof of a building horizontally or at some angle
with the vertical. In both cases, the acceleration due to gravity is downward, giving a parabolic
trajectory.
Q23.25
No. Life would be no different if electrons were + charged and protons were – charged. Opposite
charges would still attract, and like charges would repel. The naming of + and – charge is merely a
convention.
Q23.26
If the antenna were not grounded, electric charges in the atmosphere during a storm could place the
antenna at a high positive or negative potential. The antenna would then place the television set
inside the house at the high voltage, to make it a shock hazard. The wire to the ground keeps the
antenna, the television set, and even the air around the antenna at close to zero potential.
Q23.27
People are all attracted to the Earth. If the force were electrostatic, people would all carry charge
with the same sign and would repel each other. This repulsion is not observed. When we changed
the charge on a person, as in the chapter-opener photograph, the person’s weight would change
greatly in magnitude or direction. We could levitate an airplane simply by draining away its electric
charge. The failure of such experiments gives evidence that the attraction to the Earth is not due to
electrical forces.
4
Electric Fields
Q23.28
In special orientations the force between two dipoles can be zero or a force of repulsion. In general
each dipole will exert a torque on the other, tending to align its axis with the field created by the first
dipole. After this alignment, each dipole exerts a force of attraction on the other.
SOLUTIONS TO PROBLEMS
Section 23.1
*P23.1
(a)
Properties of Electric Charges
The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces its
mass by a negligible amount, to
e
j
1.007 9 1.660 × 10 −27 kg − 9.11 × 10 −31 kg = 1.67 × 10 −27 kg .
Its charge, due to loss of one electron, is
e
j
0 − 1 −1.60 × 10 −19 C = +1.60 × 10 −19 C .
(b)
By similar logic, charge = +1.60 × 10 −19 C
e
j
mass = 22.99 1.66 × 10 −27 kg − 9.11 × 10 −31 kg = 3.82 × 10 −26 kg
(c)
charge of Cl − = −1.60 × 10 −19 C
e
j
mass = 35.453 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 5.89 × 10 −26 kg
(d)
e
j
charge of Ca ++ = −2 −1.60 × 10 −19 C = +3.20 × 10 −19 C
e
j e
j
mass = 40.078 1.66 × 10 −27 kg − 2 9.11 × 10 −31 kg = 6.65 × 10 −26 kg
(e)
e
e
mass = 14.007 1.66 × 10 −27
(f)
j
kg j + 3e9.11 × 10
charge of N 3 − = 3 −1.60 × 10 −19 C = −4.80 × 10 −19 C
e
−31
j
j
kg = 2.33 × 10 −26 kg
charge of N 4 + = 4 1.60 × 10 −19 C = +6.40 × 10 −19 C
e
j e
j
mass = 14.007 1.66 × 10 −27 kg − 4 9.11 × 10 −31 kg = 2.32 × 10 −26 kg
(g)
We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.
e
j
charge = 7 1.60 × 10 −19 C = 1.12 × 10 −18 C
e
mass = 14.007 1.66 × 10
(h)
−27
j e
j
kg − 7 9.11 × 10 −31 kg = 2.32 × 10 −26 kg
charge = −1.60 × 10 −19 C
b
g
mass = 2 1.007 9 + 15.999 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 2.99 × 10 −26 kg
Chapter 23
P23.2
F 10.0 grams I FG 6.02 × 10
GH 107.87 grams mol JK H
(a)
N=
(b)
# electrons added =
IJ FG 47 electrons IJ =
K H atom K
2.62 × 10 24
Q
1.00 × 10 −3 C
=
= 6.25 × 10 15
e 1.60 × 10 −19 C electron
2.38 electrons for every 10 9 already present .
or
Section 23.2
Charging Objects by Induction
Section 23.3
Coulomb’s Law
P23.3
atoms
mol
23
If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains
N≅
F 70 000 grams I FG 6.02 × 10
GH 18 grams mol JK H
molecules
mol
23
IJ FG 10 protons IJ ≅ 2.3 × 10
K H molecule K
28
protons .
With an excess of 1% electrons over protons, each person has a charge
e
je
j
e3.7 × 10 j
= e9 × 10 j
q = 0.01 1.6 × 10 −19 C 2.3 × 10 28 = 3.7 × 10 7 C .
So
F = ke
q1 q 2
r2
7 2
9
0.6 2
N = 4 × 10 25 N ~ 10 26 N .
This force is almost enough to lift a weight equal to that of the Earth:
e
j
Mg = 6 × 10 24 kg 9.8 m s 2 = 6 × 10 25 N ~ 10 26 N .
*P23.4
The force on one proton is F =
e8.99 × 10
P23.5
(a)
(b)
9
F 1.6 × 10
N ⋅ m C jG
H 2 × 10
Fe =
Fg =
2
k e q1 q 2
r2
r2
r2
−19
−15
e8.99 × 10
=
Gm1 m 2
k e q1 q 2
C
m
9
e6.67 × 10
=
I
JK
away from the other proton. Its magnitude is
2
= 57.5 N .
je
N ⋅ m 2 C 2 1.60 × 10 −19 C
e
3.80 × 10
−11
−10
j
m
2
je
j
2
= 1.59 × 10 −9 N
N ⋅ m 2 C 2 1.67 × 10 −27 kg
e3.80 × 10
−10
j
m
2
j
2
= 1.29 × 10 −45 N
The electric force is larger by 1.24 × 10 36 times .
(c)
If k e
q
=
m
q1 q 2
r
2
=G
m1 m 2
r2
brepulsiong
with q1 = q 2 = q and m1 = m 2 = m , then
6.67 × 10 −11 N ⋅ m 2 kg 2
G
=
= 8.61 × 10 −11 C kg .
ke
8.99 × 10 9 N ⋅ m 2 C 2
5
6
Electric Fields
P23.6
We find the equal-magnitude charges on both spheres:
q1 q 2
F = ke
= ke
r2
q2
r2
a
1.00 × 10 4 N
= 1.05 × 10 −3 C .
8.99 × 10 9 N ⋅ m 2 C 2
f
F
= 1.00 m
ke
q=r
so
The number of electron transferred is then
N xfer =
1.05 × 10 −3 C
= 6.59 × 10 15 electrons .
1.60 × 10 −19 C e −
The whole number of electrons in each sphere is
F 10.0 g I e6.02 × 10
GH 107.87 g mol JK
N tot =
23
je
j
atoms mol 47 e − atom = 2.62 × 10 24 e − .
The fraction transferred is then
f=
P23.7
F
GH
I
JK
N xfer
6.59 × 10 15
=
= 2.51 × 10 −9 = 2.51 charges in every billion.
24
N tot
2.62 × 10
F1 = k e
F2 = k e
e8.99 × 10
=
q1 q 2
r
2
q1 q 2
r2
=
e8.99 × 10
je
je
N ⋅ m 2 C 2 7.00 × 10 −6 C 2.00 × 10 −6 C
9
9
N ⋅ m2
j = 0.503 N
a0.500 mf
C je7.00 × 10 C je 4.00 × 10 C j
= 1.01 N
a0.500 mf
2
−6
2
−6
2
Fx = 0.503 cos 60.0°+1.01 cos 60.0° = 0.755 N
Fy = 0.503 sin 60.0°−1.01 sin 60.0° = −0.436 N
a
f a
f
F = 0.755 N i − 0.436 N j = 0.872 N at an angle of 330°
FIG. P23.7
P23.8
F = ke
P23.9
(a)
q1 q 2
r2
e8.99 × 10
=
9
je
N ⋅ m 2 C 2 1.60 × 10 −19 C
e
j
2 6.37 × 10 6 m
j e6.02 × 10 j
2
23 2
2
= 514 kN
The force is one of attraction . The distance r in Coulomb’s law is the distance between
centers. The magnitude of the force is
F=
(b)
k e q1 q 2
r2
e
= 8.99 × 10 9 N ⋅ m 2 C 2
12.0 × 10 C je18.0 × 10 C j
=
je
a0.300 mf
−9
−9
2
2.16 × 10 −5 N .
The net charge of −6.00 × 10 −9 C will be equally split between the two spheres, or
−3.00 × 10 −9 C on each. The force is one of repulsion , and its magnitude is
F=
k e q1 q 2
r
2
e
= 8.99 × 10 9 N ⋅ m 2 C 2
3.00 × 10 C je3.00 × 10 C j
=
je
a0.300 mf
−9
−9
2
8.99 × 10 −7 N .
Chapter 23
P23.10
x from the left end of the rod. This bead
Let the third bead have charge Q and be located distance
will experience a net force given by
F=
b g
k e 3q Q
x
2
i+
7
b g e− i j .
ad − x f
ke q Q
2
The net force will be zero if
3
1
=
2
x
d−x
a
f
2
, or d − x =
x
3
.
This gives an equilibrium position of the third bead of x = 0.634d .
The equilibrium is stable if the third bead has positive charge .
P23.11
kee2
e
(a)
F=
(b)
We have F =
r2
= 8.99 × 10 N ⋅ m
2
2
−10
C
j
m
2
j
2
= 8.22 × 10 −8 N
mv 2
from which
r
v=
P23.12
e1.60 × 10
C j
e0.529 × 10
−19
9
e
j=
8.22 × 10 −8 N 0.529 × 10 −10 m
Fr
=
m
9.11 × 10
−31
The top charge exerts a force on the negative charge
kg
k e qQ
d 2 + 2
x
2
ch
2.19 × 10 6 m s .
which is directed upward and to the
FG d IJ to the x-axis. The two positive charges together exert force
H 2x K
F 2 k qQ I FG a− xfi IJ
−2 k qQ
d
GG
JJ G
= ma or for x << , a ≈
x.
2
md 8
eH + x j K GH e + x j JJK
left, at an angle of tan −1
e
d2
4
(a)
d2
4
2
12
e
3
The acceleration is equal to a negative constant times the excursion from equilibrium, as in
16 k e qQ
a = −ω 2 x , so we have Simple Harmonic Motion with ω 2 =
.
md 3
T=
(b)
2
2π
ω
=
π
2
md 3
, where m is the mass of the object with charge −Q .
k e qQ
v max = ωA = 4a
k e qQ
md 3
8
Electric Fields
Section 23.4
P23.13
P23.14
The Electric Field
For equilibrium,
Fe = −Fg
or
qE = − mg − j .
Thus,
E=
e j
je
j
9.11 × 10 −31 kg 9.80 m s 2
mg
j=
j = − 5.58 × 10 −11 N C j
−19
q
−1.60 × 10
C
e
(a)
E=
(b)
1.67 × 10 −27 kg 9.80 m s 2
mg
E=
j=
j=
q
1.60 × 10 −19 C
e
e
e
jb
e
j
je
∑ Fy = 0 : QEj + mge− jj = 0
∴m =
P23.15
e
mg
j.
q
j
j
e1.02 × 10
j
−7
j
NC j
g
24.0 × 10 −6 C 610 N C
QE
=
= 1.49 grams
g
9.80 m s 2
The point is designated in the sketch. The magnitudes of the electric fields,
E1 , (due to the −2.50 × 10 −6 C charge) and E 2 (due to the 6.00 × 10 −6 C
charge) are
E1 =
E2 =
ke q
r
2
keq
r
2
e8.99 × 10
=
=
e8.99 × 10
9
je
N ⋅ m 2 C 2 2.50 × 10 −6 C
d
9
j
2
je
ad + 1.00 mf
N ⋅ m 2 C 2 6.00 × 10 −6 C
j
2
(1)
FIG. P23.15
(2)
Equate the right sides of (1) and (2)
to get
ad + 1.00 mf
or
d + 1.00 m = ±1.55d
which yields
d = 1.82 m
or
d = −0.392 m .
2
= 2.40d 2
The negative value for d is unsatisfactory because that locates a point between the charges where
both fields are in the same direction.
Thus,
d = 1.82 m to the left of the − 2.50 µC charge .
9
Chapter 23
P23.16
If we treat the concentrations as point charges,
E+ = ke
E− = ke
q
e
j b1a40000.0 mCgf e− jj = 3.60 × 10
a40.0 Cf e− jj = 3.60 × 10
C j
b1 000 mg
= 8.99 × 10 9 N ⋅ m 2 C 2
r2
q
e
= 8.99 × 10 9 N ⋅ m 2
r2
2
2
2
e ja
f
e ja
f
5
N C − j downward
5
N C − j downward
E = E + + E − = 7.20 × 10 5 N C downward
*P23.17
The first charge creates at the origin field
k eQ
+Q x = 0
to the right.
a2
Suppose the total field at the origin is to the right. Then q must
be negative:
k eQ
a
2
i+
keq
2k Q
− ij =
i
e
a
a3 a f
e
2
2
q
x
FIG. P23.17
q = −9Q .
In the alternative, the total field at the origin is to the left:
k eQ
a
P23.18
2
(a)
i+
ke q
9a
2
e− ij = 2 ka Q e− ij
e
2
q = +27Q .
e8.99 × 10 je7.00 × 10 j = 2.52 × 10
r
a0.500f
k q e8.99 × 10 je 4.00 × 10 j
=
=
= 1.44 × 10
r
a0.500f
E1 =
−6
9
ke q
=
2
2
NC
5
NC
−6
9
E2
5
e
2
2
Ex = E2 − E1 cos 60° = 1. 44 × 10 5 − 2.52 × 10 5 cos 60.0° = 18.0 × 10 3 N C
5
FIG. P23.18
3
Ey = − E1 sin 60.0° = −2.52 × 10 sin 60.0° = −218 × 10 N C
E = 18.0 i − 218 j × 10 3 N C = 18.0 i − 218 j kN C
(b)
P23.19
(a)
e
j
e
j
e8.99 × 10 je3.00 × 10 j e− jj = −e2.70 × 10
− jj =
E =
e
r
a0.100f
k q
e8.99 × 10 je6.00 × 10 j e− ij = −e5.99 × 10
−ij =
E =
e
r
a0.300f
E = E + E = −e5.99 × 10 N C ji − e 2.70 × 10 N C j j
1
2
1
e
−9
9
k e q1
2
2
2
2
2
2
1
e
3
je
j
e−3.00i − 13.5 jj µN
F = qE = 5.00 × 10 −9 C −599 i − 2 700 j N C
e
j
F = −3.00 × 10 −6 i − 13.5 × 10 −6 j N =
e36.0i − 436 jj mN
j
3
NC j
2
NCi
−9
9
2
2
(b)
je
F = qE = 2.00 × 10 −6 C 18.0 i − 218 j × 10 3 N C = 36.0 i − 436 j × 10 −3 N =
j
FIG. P23.19
10
Electric Fields
P23.20
E=
(a)
keq
r2
e8.99 × 10 je2.00 × 10 j = 14 400 N C
a1.12f
and
E = 2b14 400 g sin 26.6° = 1.29 × 10
−6
9
=
2
Ex = 0
y
e
je
j
F = qE = −3.00 × 10 −6 1. 29 × 10 4 j = −3.86 × 10 −2 j N
(a)
E=
r1 +
r12
E = 3.06
ke q
a
2
ke q2
r22
r2 +
i + 5.06
F = qE = 5.91
(b)
P23.22
ke q 2
a2
k e q3
keq
a
2
r32
r3 =
b g i + k b3 qg ei cos 45.0°+ j sin 45.0°j + k b4qg j
ke 2q
e
a2
j = 5.91
keq
a2
e
2a2
a2
at 58.8°
at 58.8°
The electric field at any point x is
E=
a f
ax − af cx − a− afh ex − a j
ke q
2
ke q
−
2
=
k e q 4ax
2
2 2
.
b g
4a k e q
When x is much, much greater than a, we find E ≅
P23.23
FIG. P23.20
(b)
k e q1
NC
E = 1.29 × 10 4 j N C .
so
P23.21
4
(a)
One of the charges creates at P a field E =
the x-axis as shown.
x3
k eQ n
R2 + x2
.
at an angle θ to
When all the charges produce field, for n > 1 , the components
perpendicular to the x-axis add to zero.
The total field is
(b)
P23.24
E=∑
b g cosθ =
nk e Q n i
2
R +x
2
e
k eQx i
2
R + x2
j
3 2
.
FIG. P23.23
A circle of charge corresponds to letting n grow beyond all bounds, but the result does not
depend on n. Smearing the charge around the circle does not change its amount or its
distance from the field point, so it does not change the field .
keq
r2
r=
keq
a2
e− ij + a2kaqf e− ij + a3kaqf e− ij + … = − ka qi FGH1 + 21 + 31 + …IJK =
e
2
e
2
e
2
2
2
−
π 2ke q
i
6a2
Chapter 23
Section 23.5
P23.25
Electric Field of a Continuous Charge Distribution
e
b g
a f a f a f a
je
j
8.99 × 10 9 22.0 × 10 −6
ke Q
keλ
k eQ
=
=
=
E=
0.290 0.140 + 0.290
d +d
d +d
d +d
fa
f
E = 1.59 × 10 6 N C , directed toward the rod.
P23.26
E=
z
k e dq
x2
E = keλ 0
z
∞
x0
P23.27
P23.28
P23.29
E=
ex
, where dq = λ 0 dx
FG 1 IJ
H xK
dx
x
=
x0
keλ 0
x0
The direction is − i or left for λ 0 > 0
e8.99 × 10 je75.0 × 10 jx = 6.74 × 10 x
ex + 0.100 j
ex + 0.010 0j
−6
9
j
+ a2
∞
= keλ 0 −
2
k e xQ
2
FIG. P23.25
3 2
=
5
2 32
2
32
2
(a)
At x = 0.010 0 m ,
E = 6.64 × 10 6 i N C = 6.64i MN C
(b)
At x = 0.050 0 m ,
E = 2.41 × 10 7 i N C = 24.1i MN C
(c)
At x = 0.300 m ,
E = 6.40 × 10 6 i N C = 6. 40 i MN C
(d)
At x = 1.00 m ,
E = 6.64 × 10 5 i N C = 0.664i MN C
z
E = dE =
E=
LM k λ x dxe− ij OP
z MMN x PPQ = −k λ x i z x
∞
e
ex
k e Qx
+ a2
e
3
x0
2
∞
0 0
j
0 0
−3
F
GH
dx = − k e λ 0 x 0 i −
x0
3 2
For a maximum,
dE
= Qk e
dx
LM
MM ex
N
a
x 2 + a 2 − 3 x 2 = 0 or x =
2
1
2
+ a2
j
3 2
−
OP
=0
P
+
x
a
e
j PQ
3x 2
2 52
2
.
Substituting into the expression for E gives
E=
k eQa
2
e aj
3
2
2 32
=
k eQ
3
3
2
a
2
=
2 k eQ
3 3a
2
=
Q
6 3π ∈0 a 2
.
∞
1
2x
2
x0
I=
JK
keλ 0
−i
2 x0
e j
11
12
Electric Fields
P23.30
F
GH
e
E = 2π 8.99 × 10
P23.31
x
E = 2π k eσ 1 −
2
x +R
9
2
I
JK
je7.90 × 10
−3
F
jGG 1 −
H
I
JJ = 4.46 × 10 FG 1 −
H
+ a0.350f K
x
x2
2
(a)
At x = 0.050 0 m ,
E = 3.83 × 10 8 N C = 383 MN C
(b)
At x = 0.100 m ,
E = 3.24 × 10 8 N C = 324 MN C
(c)
At x = 0.500 m ,
E = 8.07 × 10 7 N C = 80.7 MN C
(d)
At x = 2.00 m ,
E = 6.68 × 10 8 N C = 6.68 MN C
(a)
From Example 23.9: E = 2π k eσ 1 −
F
GH
σ=
I
J
+ 0.123 K
x
8
x2
I
JK
x
x2 + R2
Q
= 1.84 × 10 −3 C m 2
πR 2
ja
e
f
E = 1.04 × 10 8 N C 0.900 = 9.36 × 10 7 N C = 93.6 MN C
b
appx: E = 2π k eσ = 104 MN C about 11% high
(b)
F
jGH
e
E = 1.04 × 10 8 N C 1 −
I = 1.04 × 10
J e
cm K
30.0 cm
jb
g
8
N C 0.004 96 = 0.516 MN C
30.0 + 3.00
Q
5.20 × 10 −6
appx: E = k e 2 = 8.99 × 10 9
= 0.519 MN C about 0.6% high
2
r
0.30
e
P23.32
g
2
2
b
j a f
LM
MN
L
= 2π k σ M1 −
MN
The electric field at a distance x is
Ex = 2π k eσ 1 −
This is equivalent to
Ex
For large x,
R2
x2
x2 + R2
OP
PQ
1
e
1+
<< 1 and
x
g
1 + R2 x2
OP
PQ
R2
R2
≈
+
1
x2
2x 2
I
F
e1 + R e2 x j − 1j
1
J
= 2π k σ G 1 −
= 2π k σ
GH 1 + R e2 x j JK
1 + R e2 x j
k Qe1 x j
F RI
=
= k QG x +
H 2 JK
1 + R e2x j
2
Ex
so
Q
Substitute σ =
,
π R2
But for x >> R ,
1
x2 + R2 2
Ex
≈
1
x2
, so
Ex ≈
e
2
2
e
2
k eQ
x2
2
e
2
e
2
2
for a disk at large distances
2
2
2
Chapter 23
P23.33
z
z
Due to symmetry
Ey = dEy = 0 , and Ex = dE sin θ = k e
where
dq = λds = λrdθ ,
so that,
Ex =
z
keλ π
k λ
sin θdθ = e − cos θ
r 0
r
a
f
π
=
0
z
dq sin θ
r2
2k e λ
r
q
L
and r = .
π
L
2
8.99 × 10 9 N ⋅ m 2 C 2 7.50 × 10 −6 C π
2 k e qπ
=
Ex =
.
2
L2
0.140 m
FIG. P23.33
λ=
where
e
Thus,
je
f
a
13
j
7
Ex = 2.16 × 10 N C .
Solving,
e
j
Since the rod has a negative charge, E = −2.16 × 10 7 i N C = −21.6 i MN C .
P23.34
(a)
We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has
Qdx
and produces, at the chosen point, a field
charge
h
kex
Qdx
dE =
i.
3
2
h
x2 + R2
e
j
The total field is
z z
E=
dE =
d+h
e
k Qi ex + R j
E=
2h
b− 1 2g
d
all charge
k eQxdx
h x2 + R
j
2 3 2
d+h
2 −1 2
2
e
(b)
i=
ze
k eQi d + h 2
x + R2
2h x=d
k Qi
= e
h
x=d
LM
MM d
Ne
j
−3 2
2 xdx
OP
−
P
+R j
ead + hf + R j PQ
1
2
1
2 12
2
2
12
Think of the cylinder as a stack of disks, each with thickness dx, charge
per-area σ =
dE =
So,
Qdx
. One disk produces a field
π R2h
2π k eQdx
π R2h
F
GG 1 −
H ex
I
Ji .
+ R j JK
F
2 k Qdx G
1−
R h G
H ex
x
2
2 12
I
JJ i
E=
z dE = z
+R j K
L
O
2 k Qi M
2 k Qi L
1
E=
M z dx − 2 z ex + R j 2xdxPP = R h Mx
R h MN
MN
Q
2 k Qi L
O
d + h − d − ead + hf + R j + ed + R j P
E=
M
R h N
Q
2 k Qi L
O
h + ed + R j − ead + hf + R j P
E=
R h MN
Q
d+h
all charge
e
2
x =d
e
2
d+h
d+h
d
x=d
2
2
2 12
x
2 12
2
2 −1 2
2
e
2
e
2
Qdx
, and chargeh
2
e
2
12
2
2 12
2
2
12
e
2
2
1 x +R
d+h
−
d
2
12
j
1 2 d+h
d
OP
PP
Q
14
Electric Fields
P23.35
(a)
The electric field at point P due to each element of length dx, is
k dq
dE = 2 e 2 and is directed along the line joining the element to
x +y
point P. By symmetry,
Ex = dEx = 0
and since
dq = λdx ,
E = Ey = dEy = dE cos θ
where
cos θ =
z
z
z
ze
2
Therefore,
E = 2 k e λy
0
P23.36
dx
2
x +y
j
2 32
y
2
x + y2
.
FIG. P23.35
2 k e λ sin θ 0
.
y
=
θ 0 = 90°
Ey =
2k e λ
.
y
(b)
For a bar of infinite length,
(a)
The whole surface area of the cylinder is A = 2π r 2 + 2π rL = 2π r r + L .
j b
e
and
a f
g
Q = σA = 15.0 × 10 −9 C m 2 2π 0.025 0 m 0.025 0 m + 0.060 0 m = 2.00 × 10 −10 C
(b)
For the curved lateral surface only, A = 2π rL .
j b
e
f
ga
Q = σA = 15.0 × 10 −9 C m 2 2π 0.025 0 m 0.060 0 m = 1.41 × 10 −10 C
P23.37
j b
e
g b0.060 0 mg =
2
(c)
Q = ρV = ρπ r 2 L = 500 × 10 −9 C m 3 π 0.025 0 m
(a)
Every object has the same volume, V = 8 0.030 0 m
e
je
a
f
3
5.89 × 10 −11 C
= 2.16 × 10 −4 m3 .
j
For each, Q = ρV = 400 × 10 −9 C m 3 2.16 × 10 −4 m3 = 8.64 × 10 −11 C
(b)
We must count the 9.00 cm 2 squares painted with charge:
(i)
6 × 4 = 24 squares
e
j e
j
j e
j
j e
j
j e
j
Q = σA = 15.0 × 10 −9 C m 2 24.0 9.00 × 10 −4 m 2 = 3.24 × 10 −10 C
(ii)
34 squares exposed
e
Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C
(iii)
34 squares
e
Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C
(iv)
32 squares
e
Q = σA = 15.0 × 10 −9 C m 2 32.0 9.00 × 10 −4 m 2 = 4.32 × 10 −10 C
(c)
(i)
total edge length:
e
= e80.0 × 10
Q = λ = 80.0 × 10 −12
(ii)
Q=λ
continued on next page
b
g
C mj24 × b0.030 0 mg =
C mj44 × b0.030 0 mg =
= 24 × 0.030 0 m
−12
5.76 × 10 −11 C
1.06 × 10 −10 C
Chapter 23
Section 23.6
e
j b
g
e
j b
g
(iii)
Q = λ = 80.0 × 10 −12 C m 64 × 0.030 0 m = 1.54 × 10 −10 C
(iv)
Q = λ = 80.0 × 10 −12 C m 40 × 0.030 0 m = 0.960 × 10 −10 C
Electric Field Lines
P23.39
P23.38
FIG. P23.38
P23.40
(a)
(b)
P23.41
(a)
FIG. P23.39
q1 −6
1
=
= −
3
q 2 18
q1 is negative, q 2 is positive
The electric field has the general appearance shown. It is zero
at the center , where (by symmetry) one can see that the three
charges individually produce fields that cancel out.
In addition to the center of the triangle, the electric field lines in the
second figure to the right indicate three other points near the
middle of each leg of the triangle where E = 0 , but they are more
difficult to find mathematically.
(b)
You may need to review vector addition in Chapter Three. The
electric field at point P can be found by adding the electric field
vectors due to each of the two lower point charges: E = E 1 + E 2 .
The electric field from a point charge is E = k e
q
r2
r.
As shown in the solution figure at right,
E1 = k e
E2 = ke
q
a2
q
a2
to the right and upward at 60°
to the left and upward at 60°
E = E1 + E 2 = k e
= 1.73 k e
q
a2
j
q
a
2
ecos 60° i + sin 60° jj + e− cos 60° i + sin 60° jj = k
FIG. P23.41
e
q
a2
e
j
2 sin 60° j
15
16
Electric Fields
Section 23.7
P23.42
Motion of Charged Particles in a Uniform Electric Field
qE
m
F = qE = ma
a=
v f = vi + at
vf =
qEt
m
e1.602 × 10 ja520fe48.0 × 10 j =
=
4.39 × 10 6 m s
e1.602 × 10 ja520fe48.0 × 10 j =
2.39 × 10 3 m s
−19
ve
electron:
−9
−31
9.11 × 10
in a direction opposite to the field
−19
vp =
proton:
P23.43
P23.44
−9
1.67 × 10 −27
in the same direction as the field
a f
qE 1.602 × 10 −19 640
=
= 6.14 × 10 10 m s 2
m
1.67 × 10 −27
(a)
a=
(b)
v f = vi + at
(c)
x f − xi =
(d)
K=
(a)
−19
6.00 × 10 5
qE 1.602 × 10
a=
=
= 5.76 × 10 13 m s so a = −5.76 × 10 13 i m s 2
m
1.67 × 10 −27
(b)
v f = vi + 2 a x f − xi
e
1
vi + v f t
2
d
i
xf =
e
e
je
je
e
d
e
je
j
2
= 1.20 × 10 −15 J
j
j
jb
g
v i = 2.84 × 10 6 i m s
v f = vi + at
e
j
0 = 2.84 × 10 6 + −5.76 × 10 13 t
t = 4.93 × 10 −8 s
The required electric field will be in the direction of motion .
Work done = ∆K
so,
− Fd = −
which becomes
eEd = K
and
E=
j
i
e
P23.45
t = 1.95 × 10 −5 s
1
1.20 × 10 6 1.95 × 10 −5 = 11.7 m
2
1
1
mv 2 = 1.67 × 10 −27 kg 1.20 × 10 6 m s
2
2
0 = vi2 + 2 −5.76 × 10 13 0.070 0
(c)
j
1.20 × 10 6 = 6.14 × 10 10 t
1
mvi2 (since the final velocity = 0 )
2
K
.
ed
Chapter 23
P23.46
17
The acceleration is given by
d
a f
i
v 2f = vi2 + 2 a x f − x i or
v 2f = 0 + 2 a − h .
Solving
a=−
Now
∑ F = ma :
Therefore
qE = −
(a)
v 2f
2h
F
GH
.
mv 2f
2h
− mg j + qE = −
mv 2f j
I
JK
2h
.
+ mg j .
Gravity alone would give the bead downward impact velocity
ja
e
f
2 9.80 m s 2 5.00 m = 9.90 m s .
To change this to 21.0 m/s down, a downward electric field must exert a downward
electric force.
P23.47
F
GH
I
JK
(b)
(a)
t=
(b)
ay =
(c)
I LM b21.0 m sg
JK M 2a5.00 mf
N
OP
PQ
2
− 9.80 m s 2 = 3.43 µC
0.050 0
x
=
= 1.11 × 10 −7 s = 111 ns
v x 4.50 × 10 5
e
je
j
−19
9.60 × 10 3
qE 1.602 × 10
=
= 9.21 × 10 11 m s 2
−27
m
1.67 × 10
e
y f − yi = v yi t +
*P23.48
F
GH
2
1.00 × 10 −3 kg N ⋅ s 2
m vf
−g =
q=
E 2h
1.00 × 10 4 N C kg ⋅ m
j
1
ayt 2 :
2
yf =
1
9.21 × 10 11 1.11 × 10 −7
2
e
je
e
j
2
= 5.68 × 10 −3 m = 5.68 mm
je
j
v yf = v yi + a y t = 9.21 × 10 11 1.11 × 10 −7 = 1.02 × 10 5 m s
v x = 4.50 × 10 5 m s
jb
ge j
e j
∑ F = e 2 × 10 kg ⋅ m s je− jj = 1 × 10 m s − j .
a=
and moves with acceleration:
e
je j
m
2 × 10
kg
Its x-component of velocity is constant at e1.00 × 10 m sj cos 37° = 7.99 × 10 m s . Thus it moves in a
e
The particle feels a constant force: F = qE = 1 × 10 −6 C 2 000 N C − j = 2 × 10 −3 N − j
−3
2
13
−16
5
2
4
parabola opening downward. The maximum height it attains above the bottom plate is described by
d
i
2
2
= v yi
+ 2 a y y f − yi :
v yf
e
0 = 6.02 × 10 4 m s
y f = 1.81 × 10 −4 m .
continued on next page
j − e2 × 10
2
13
jd
m s2 y f − 0
i
18
Electric Fields
Since this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetric
parabola and strikes the bottom plate after a time given by
y f = yi + v yi t +
1
ayt 2
2
e
j
0 = 0 + 6.02 × 10 4 m s t +
1
−1 × 10 13 m s 2 t 2
2
e
j
since t > 0 ,
t = 1.20 × 10 −8 s .
The particle’s range is
x f = xi + v x t = 0 + 7.99 × 10 4 m s 1.20 × 10 −8 s = 9.61 × 10 −4 m .
e
je
j
In sum,
The particle strikes the negative plate after moving in a parabola with a height of 0.181 mm
and a width of 0.961 mm.
P23.49
vi = 9.55 × 10 3 m s
ay =
(a)
e
^
ja f
j
−19
720
eE 1.60 × 10
=
= 6.90 × 10 10 m s 2
−
27
m
1.67 × 10
e
vi2 sin 2θ
= 1. 27 × 10 −3 m so that
ay
R=
e9.55 × 10 j
3 2
6.90 × 10
sin 2θ
10
sin 2θ = 0.961
t=
(b)
FIG. P23.49
= 1.27 × 10 −3
θ = 36.9°
R
R
=
vix vi cos θ
90.0°−θ = 53.1°
If θ = 36.9° , t = 167 ns .
If θ = 53.1° , t = 221 ns .
Additional Problems
*P23.50
The two given charges exert equal-size forces of attraction on each
other. If a third charge, positive or negative, were placed between
them they could not be in equilibrium. If the third charge were at a
point x > 15 cm , it would exert a stronger force on the 45 µ C than
on the −12 µ C , and could not produce equilibrium for both. Thus
the third charge must be at x = − d < 0 . Its equilibrium requires
b
k e q 12 µ C
d
2
g = k qb45 µ Cg
a15 cm + df
e
2
15 cm + d = 1.94d
FG 15 cm + d IJ
H d K
2
=
d
q
x=0
–
–12 µC
15 cm
x
+
45 µ C
FIG. P23.50
45
= 3.75
12
d = 16.0 cm .
The third charge is at x = −16.0 cm . The equilibrium of the −12 µ C requires
b
g = k b45 µ Cg12 µ C
a16.0 cmf
a15 cmf
k e q 12 µ C
2
e
2
q = 51.3 µ C .
All six individual forces are now equal in magnitude, so we have equilibrium as required, and this is
the only solution.
Chapter 23
P23.51
jb
e
while the e − has acceleration
ae =
(a)
e1.60 × 10
jb
C 640 N C
kg
FG
H
IJ
K
1
1
1
a p t 2 + a e t 2 = 1 837 a p t 2 .
2
2
2
1
4.00 cm
2
= 21.8 µm .
d = apt =
2
1 837
g = 1.12 × 10
14
m s 2 = 1 836 a p .
1
a p t 2 ), knowing:
2
The distance from the positive plate to where the meeting occurs equals the distance the
1
sodium ion travels (i.e., d Na = a Na t 2 ). This is found from:
2
eE
eE
1
1
1
1
2
2
t2 +
t2 .
4.00 cm = a Na t + aCl t :
4.00 cm =
2
2
2 22.99 u
2 35. 45 u
1
1
1
This may be written as
4.00 cm = a Na t 2 + 0.649 a Na t 2 = 1.65 a Na t 2
2
2
2
1
4.00 cm
2
= 2.43 cm .
so
d Na = a Na t =
2
1.65
FG
H
(a)
−31
We want to find the distance traveled by the proton (i.e., d =
Thus,
(b)
−19
9.110 × 10
4.00 cm =
P23.52
g
The proton moves with acceleration
−19
C 640 N C
qE 1.60 × 10
=
= 6.13 × 10 10 m s 2
ap =
m
1.673 × 10 −27 kg
IJ
K
b
FG
H
IJ
K
g
FG
H
IJ
K
The field, E1 , due to the 4.00 × 10 −9 C charge is in the –x
direction.
8.99 × 10 9 N ⋅ m 2 C 2 −4.00 × 10 −9 C
keq
E1 = 2 r =
i
2
r
2.50 m
e
a
je
f
j
FIG. P23.52(a)
= −5.75 i N C
Likewise, E 2 and E3 , due to the 5.00 × 10 −9 C charge and the 3.00 × 10 −9 C charge are
E2 =
E3
ke q
r
2
e8.99 × 10
r=
e8.99 × 10
=
9
je
j i = 11.2 N C i
a2.00 mf
N ⋅ m C je3.00 × 10 C j
i = 18.7 N C i
a1.20 mf
9
N ⋅ m 2 C 2 5.00 × 10 −9 C
2
2
−9
2
2
E R = E1 + E 2 + E 3 = 24.2 N C in +x direction.
(b)
E2
E3
ke q
b
ge
j
=
r = b11. 2 N C ge + jj
r
k q
=
r = b5.81 N C ge −0.371i +0.928 jj
r
E1 =
r2
ke q
r = −8.46 N C 0.243 i + 0.970 j
2
e
2
Ex = E1 x + E3 x = −4.21i N C
ER = 9.42 N C
Ey = E1 y + E2 y + E3 y = 8.43 j N C
θ = 63.4° above − x axis
FIG. P23.52(b)
19
20
Electric Fields
*P23.53
(a)
Each ion moves in a quarter circle. The electric force causes the centripetal acceleration.
∑ F = ma
(b)
qE =
mv 2
R
E=
d
For the x-motion,
2
v xf2 = v xi
+ 2 a x x f − xi
0 = v 2 + 2ax R
ax = −
Ex = −
mv 2
qR
i
v 2 Fx qE x
=
=
m
2R m
mv 2
. Similarly for the y-motion,
2 qR
v 2 = 0 + 2ay R
ay = +
v 2 qE y
=
m
2R
Ey =
mv 2
2 qR
The magnitude of the field is
E x2 + E y2 =
P23.54
at 135° counterclockwise from the x -axis .
2 qR
From the free-body diagram shown,
∑ Fy = 0 :
T cos 15.0° = 1.96 × 10 −2 N .
So
T = 2.03 × 10 −2 N .
From
or
P23.55
mv 2
(a)
∑ Fx = 0 , we have
q=
qE = T sin 15.0°
e
j
2.03 × 10 −2 N sin 15.0°
T sin 15.0°
=
= 5.25 × 10 −6 C = 5.25 µC .
E
1.00 × 10 3 N C
FIG. P23.54
Let us sum force components to find
∑ Fx = qEx − T sin θ = 0 , and ∑ Fy = qEy + T cos θ − mg = 0 .
Combining these two equations, we get
q=
e1.00 × 10 ja9.80f
eE cot θ + E j a3.00 cot 37.0°+5.00f × 10
−3
mg
x
=
y
5
= 1.09 × 10 −8 C
= 10.9 nC
(b)
From the two equations for
T=
Free Body Diagram
FIG. P23.55
∑ Fx
and
∑ Fy
qEx
= 5.44 × 10 −3 N = 5.44 mN .
sin 37.0°
we also find
Chapter 23
P23.56
21
This is the general version of the preceding problem. The known quantities are A, B, m, g, and θ. The
unknowns are q and T.
The approach to this problem should be the same as for the last problem, but without
numbers to substitute for the variables. Likewise, we can use the free body diagram given in the
solution to problem 55.
Again, Newton’s second law:
and
qA
, into Eq. (2),
sin θ
Substituting T =
(a)
(b)
∑ Fx = −T sin θ + qA = 0
(1)
∑ Fy = +T cos θ + qB − mg = 0
(2)
qA cos θ
+ qB = mg .
sin θ
mg
A cot θ + B
Isolating q on the left,
q=
a
Substituting this value into Eq. (1),
T=
a A cosθ + B sinθ f
mgA
f
.
.
If we had solved this general problem first, we would only need to substitute the
appropriate values in the equations for q and T to find the numerical results needed for
problem 55. If you find this problem more difficult than problem 55, the little list at the first
step is useful. It shows what symbols to think of as known data, and what to consider
unknown. The list is a guide for deciding what to solve for in the analysis step, and for
recognizing when we have an answer.
P23.57
F=
k e q1 q 2
r
2
tan θ =
:
15.0
60.0
θ = 14.0°
e8.99 × 10 je10.0 × 10 j
F =
a0.150f
e8.99 × 10 je10.0 × 10 j
F =
a0.600f
e8.99 × 10 je10.0 × 10 j
F =
a0.619f
−6 2
9
1
2
−6 2
9
3
2
= 2.50 N
−6 2
9
2
= 40.0 N
2
= 2.35 N
Fx = − F3 − F2 cos 14.0° = −2.50 − 2.35 cos 14.0° = −4.78 N
Fy = − F1 − F2 sin 14.0° = −40.0 − 2.35 sin 14.0° = −40.6 N
Fnet = Fx2 + Fy2 =
tan φ =
Fy
Fx
φ = 263°
=
−40.6
−4.78
a4.78f + a40.6f
2
2
= 40.9 N
FIG. P23.57
22
Electric Fields
P23.58
d cos 30.0° = 15.0 cm,
15.0 cm
d=
cos 30.0°
From Figure A:
or
θ = sin −1
From Figure B:
θ = sin −1
FG d IJ
H 50.0 cm K
F 15.0 cm I = 20.3°
GH 50.0 cmacos 30.0°f JK
Figure A
Fq
or
= tan θ
mg
Fq = mg tan 20.3°
From Figure C:
Fq = 2 F cos 30.0°
(1)
LM k q OP cos 30.0°
MN a0.300 mf PQ
Combining equations (1) and (2),
L k q OP cos 30.0° = mg tan 20.3°
2M
MN a0.300 mf PQ
mg a0.300 mf tan 20.3°
q =
2
e
Fq = 2
e
2
(2)
Figure B
2
2
2
2
2 k e cos 30.0°
e2.00 × 10 kg je9.80 m s ja0.300 mf tan 20.3°
=
2e8.99 × 10 N ⋅ m C j cos 30.0°
−3
q
2
2
2
9
2
2
Figure C
q = 4.20 × 10 −14 C 2 = 2.05 × 10 −7 C = 0.205 µC
P23.59
Charge
FIG. P23.58
Q
resides on each block, which repel as point charges:
2
F=
Q=
Solving for Q,
*P23.60
P23.61
b gb g = kbL − L g .
L
k bL − L g
.
2L
ke Q 2 Q 2
i
2
i
ke
If we place one more charge q at the 29th vertex, the total force on the central charge will add up to
k qQ
k e qQ
toward vertex 29 .
F28 charges =
zero: F28 charges + e 2 away from vertex 29 = 0
a
a2
According to the result of Example 23.7, the left-hand rod creates
this field at a distance d from its right-hand end:
k eQ
E=
d 2a + d
a
dF =
F=
F=
f
k eQQ
dx
2a d d + 2a
a
k eQ
2a
2
+ k eQ
4a
2
FIG. P23.61
f
IJ
z a f FGH
K
F k Q I lnF
FG − ln 2a + b + ln b IJ = k Q ln b
= G
H b
K
b − 2a
ab − 2 afab + 2af H 4a JK GH b
4a
b
k Q2
dx
1
2a + x
= e
− ln
x x + 2a
x
2a
2a
x=b− 2a
2
e
2
2
b
b−2a
2
e
2
2
2
b2
− 4a 2
I
JK
23
Chapter 23
P23.62
b
g
At equilibrium, the distance between the charges is r = 2 0.100 m sin 10.0° = 3.47 × 10 −2 m
Now consider the forces on the sphere with charge +q , and use
∑ Fy = 0 :
∑ Fx = 0 :
∑ Fy = 0 :
mg
cos 10.0°
= F2 − F1 = T sin 10.0°
T cos 10.0° = mg , or T =
(1)
Fnet
(2)
θθ L
–q
+q
r
Fnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1).
mg sin 10.0°
Fnet =
= mg tan 10.0° = 2.00 × 10 −3 kg 9.80 m s 2 tan 10.0° = 3.46 × 10 −3 N
cos 10.0°
e
je
j
Fnet is the resultant of two forces, F1 and F2 . F1 is the attractive force on +q exerted
by −q , and F2 is the force exerted on +q by the external electric field.
FIG. P23.62
Fnet = F2 − F1 or F2 = Fnet + F1
e5.00 × 10 Cje5.00 × 10 Cj = 1.87 × 10
C j
e3.47 × 10 mj
−8
e
9
F1 = 8.99 × 10 N ⋅ m
2
−8
2
−3
2
−2
N
Thus, F2 = Fnet + F1 yields F2 = 3.46 × 10 −3 N + 1.87 × 10 −2 N = 2.21 × 10 −2 N
and F2 = qE , or E =
P23.63
z
Q = λd =
z
F2 2. 21 × 10 −2 N
=
= 4. 43 × 10 5 N C = 443 kN C .
q
5.00 × 10 −8 C
90 .0 °
λ 0 cos θRdθ = λ 0 R sin θ
−90 .0 °
90 .0 °
−90.0 °
a f
= λ 0 R 1 − −1 = 2 λ 0 R
λ = 10.0 µC m
so
b ga f
F b3.00 µCgeλ cos θRdθ j I
1
1 F b3.00 µC gbλd g I
cos θ =
dF =
G
JJ
G
J
4π ∈ G
4π ∈ H
R
R
K
H
K
e3.00 × 10 Cje10.0 × 10 C mj cos θdθ
F = z e8.99 × 10 N ⋅ m C j
a0.600 mf
8.99a30.0f
F =
e10 Nj z FGH 12 + 12 cos 2θ IJK dθ
0.600
F1 1
I = 0.707 N Downward.
F = a0.450 N fG θ + sin 2θ
JK
H2 4
Q = 12.0 µC = 2 λ 0 0.600 m = 12.0 µC
0
2
0
y
2
0
90.0 °
y
2
0
−6
−6
9
2
2
2
−90 .0 °
y
−3
−π 2
−π 2
cosθ
0
–1
π 2
π 2
y
1
1
0
0°
360°
cos2θ
0°
360°
FIG. P23.63
Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0 .
P23.64
At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be
located by determining the angle θ corresponding to equilibrium.
In terms of lengths s,
attractive force
1
a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an
2
es +
continued on next page
k eQq
1
2
a 3
j
2
24
Electric Fields
The other two charges exert equal repulsive forces of magnitude
of the two repulsive forces add, balancing the attractive force,
Fnet = k eQq
1
2
k eQq
r2
. The horizontal components
LM 2 cosθ
MM r − es +
N
2
a
1
2
r=
The equilibrium condition, in terms of θ, is
Fnet =
Thus the equilibrium value of θ satisfies
2 cos θ sin 2 θ
sin θ
2
a
s=
From Figure P23.64
OP
P=0
3j P
Q
1
1
a cot θ
2
FG 4 IJ k QqFG 2 cos θ sin θ −
H a K GH
e
2
2
e
e
3 + cot θ
j
2
I
JJ = 0 .
3 + cot θ j K
1
2
=1.
One method for solving for θ is to tabulate the left side. To three significant figures a value of θ
corresponding to equilibrium is 81.7°.
The distance from the vertical side of the triangle to the equilibrium position is
1
s = a cot 81.7° = 0.072 9 a .
2
θ
2 cos θ sin 2 θ
60°
FIG. P23.64
e
3 + cot θ
j
2
4
70°
2.654
80°
1.226
90°
0
81°
1.091
81.5°
1.024
81.7°
0.997
A second zero-field point is on the negative side of the x-axis, where θ = −9.16° and s = −3.10 a .
P23.65
(a)
(b)
From the 2Q charge we have
Fe − T2 sin θ 2 = 0 and mg − T2 cos θ 2 = 0 .
Combining these we find
Fe
T sin θ 2
= 2
= tan θ 2 .
mg T2 cos θ 2
From the Q charge we have
Fe = T1 sin θ 1 = 0 and mg − T1 cos θ 1 = 0 .
Combining these we find
Fe
T sin θ 1
= 1
= tan θ 1 or θ 2 = θ 1 .
mg T1 cos θ 1
Fe =
k e 2QQ
r
2
=
2 k eQ
FIG. P23.65
2
r2
If we assume θ is small then
tan θ ≈
r 2
.
Substitute expressions for Fe and tan θ into either equation found in part (a) and solve for r.
F I
GH JK
F
GH
Fe
2k Q 2 1
4k eQ 2
r
= tan θ then e 2
≈
and solving for r we find r ≈
mg
2
mg
mg
r
I
JK
13
.
25
Chapter 23
P23.66
(a)
The distance from each corner to the center of the square is
FG L IJ + FG L IJ
H 2K H 2K
2
2
=
L
2
z
+q
.
+q
L/2
L/2
+q
The distance from each positive charge to −Q is then
z2 +
x
–Q
+q
L2
. Each positive charge exerts a force directed
2
FIG. P23.66
k eQq
along the line joining q and −Q , of magnitude
z 2 + L2 2
.
z
The line of force makes an angle with the z-axis whose cosine is
2
z + L2 2
The four charges together exert forces whose x and y components add to zero, while the
4k e Qqz
F= −
k
z-components add to
3 2
2
z + L2 2
e
(b)
For z >> L , the magnitude of this force is Fz = −
4k eQqz
eL 2 j
2
32
F 4a2f k Qq I z = ma
GH L JK
3 2
=−
3
e
j
z
Therefore, the object’s vertical acceleration is of the form a z = −ω 2 z
with ω 2 =
af
42
32
k eQq
3
mL
=
k eQq 128
mL3
.
Since the acceleration of the object is always oppositely directed to its excursion from
equilibrium and in magnitude proportional to it, the object will execute simple harmonic
motion with a period given by
T=
P23.67
(a)
2π
ω
=
2π
a128f
14
mL3
=
k e Qq
mL3
.
k eQq
π
a8 f
14
FG
H
IJ
K
qE
,
m
which is constant and directed downward. Therefore, it behaves like a simple pendulum in
the presence of a modified uniform gravitational field with a period given by:
The total non-contact force on the cork ball is: F = qE + mg = m g +
T = 2π
L
0.500 m
= 2π
g + qE m
9.80 m s 2 + 2.00 × 10 −6 C 1.00 × 10 5 N C 1.00 × 10 −3 kg
e
je
j
= 0.307 s
(b)
Yes . Without gravity in part (a), we get T = 2π
T = 2π
0.500 m
e2.00 × 10 Cje1.00 × 10
−6
5
j
L
qE m
N C 1.00 × 10 −3 kg
= 0.314 s (a 2.28% difference).
26
Electric Fields
P23.68
The bowl exerts a normal force on each bead, directed along the
radius line or at 60.0° above the horizontal. Consider the free-body
diagram of the bead on the left:
∑ Fy = n sin 60.0°− mg = 0 ,
mg
.
sin 60.0°
or
n=
Also,
∑ Fx = − Fe + n cos 60.0° = 0 ,
keq 2
or
R
2
F mg I
RG
H k 3 JK
q=
Thus,
= n cos 60.0° =
n
mg
mg
=
.
tan 60.0°
3
Fe
mg
12
.
FIG. P23.68
e
P23.69
(a)
60.0°
There are 7 terms which contribute:
3 are s away (along sides)
3 are
1 is
1
2s away (face diagonals) and sin θ =
3s away (body diagonal) and sin φ =
= cos θ
2
1
3
.
FIG. P23.69
The component in each direction is the same by symmetry.
F=
P23.70
ke q2
s2
LM1 + 2 + 1 OPei + j + kj =
N 2 2 3 3Q
ke q2
keq 2
s2
a1.90fei + j + kj
(b)
F = Fx2 + Fy2 + Fz2 = 3. 29
(a)
Zero contribution from the same face due to symmetry, opposite
face contributes
FG k q sin φIJ where
Hr K
4
e
2
sin φ =
s
r
r=
s2
away from the origin
FG s IJ + FG s IJ
H 2K H 2K
E=4
2
k e qs
r3
=
2
+ s 2 = 1.5 s = 1.22s
4
a1.22f
3
keq
s2
= 2.18
keq
s2
FIG. P23.70
(b)
The direction is the k direction.
27
Chapter 23
P23.71
k e xQ
E = Ex =
The field on the axis of the ring is calculated in Example 23.8,
The force experienced by a charge −q placed along the axis of the ring is
ex + a j
LM x
F = − k Qq
MM ex + a j
N
F k Qq IJ x
F = −G
Ha K
e
2 3 2
2
e
and when x << a , this becomes
2 3 2
2
OP
PP
Q
3
This expression for the force is in the form of Hooke’s law, with an
k=
effective spring constant of
k
, we have
m
Since ω = 2π f =
P23.72
dE =
F
G
+ a0.150 mf GH
k e dq
x
2
z
E=
dE = k e λ
2
z
0. 400 m
x=0
all charge
f=
I k λe− xi + 0.150 mjjdx
J=
x + a0.150 mf JK
x + a0.150 mf
e− xi + 0.150 mjjdx
x + a0.150 mf
− x i + 0.150 m j
2
2
2
2 3 2
2
2
.
FIG. P23.72
2
2
0
2
ma 3
0. 400 m
2
9
k eQq
1
2π
2 32
2
0. 400 m
2
a3
e
LM
OP
0.150 mf j x
a
+i
E = k λM
+
PP
0
150
m
.
x
+
a
f
a0.150 mf x + a0.150 mf
MN
Q
E = e8.99 × 10 N ⋅ m C je35.0 × 10 C mj ia 2.34 − 6.67f m + ja6.24 − 0 f m
E = e −1.36 i + 1.96 jj × 10 N C = e −1.36 i + 1.96 jj kN C
e
k eQq
0
−9
−1
−1
3
P23.73
The electrostatic forces exerted on the two charges result in a net torque
τ = −2 Fa sin θ = −2Eqa sin θ .
For small θ, sin θ ≈ θ and using p = 2 qa , we have
τ = −Epθ .
The torque produces an angular acceleration given by τ = Iα = I
Combining these two expressions for torque, we have
This equation can be written in the form
d 2θ
dt 2
.
FG IJ
H K
Ep
d 2θ
θ =0.
+
2
I
dt
d 2θ
dt
2
FIG. P23.73
= −ω 2θ where ω 2 =
Ep
.
I
This is the same form as Equation 15.5 and the frequency of oscillation is found by comparison with
Equation 15.11, or
f=
1
2π
pE
1
=
I
2π
2 qaE
.
I
28
Electric Fields
ANSWERS TO EVEN PROBLEMS
P23.2
P23.4
(a) 2.62 × 10 24 ; (b) 2.38 electrons for every
10 9 present
P23.36
(a) 200 pC; (b) 141 pC; (c) 58.9 pC
P23.38
see the solution
P23.40
(a) −
57.5 N
−9
1
; (b) q1 is negative and q 2 is positive
3
P23.6
2.51 × 10
P23.8
514 kN
P23.42
electron: 4.39 Mm s ; proton: 2.39 km s
P23.10
x = 0.634d . The equilibrium is stable if the
third bead has positive charge.
P23.44
(a) −57.6 i Tm s 2 ; (b) 2.84i Mm s; (c) 49.3 ns
P23.46
(a) down; (b) 3.43 µC
P23.48
The particle strikes the negative plate after
moving in a parabola 0.181 mm high and
0.961 mm.
P23.50
Possible only with +51.3 µ C at
x = −16.0 cm
P23.12
md 3
where m is the mass
k e qQ
π
(a) period =
2
k e qQ
of the object with charge −Q ; (b) 4a
md 3
P23.14
1.49 g
P23.16
720 kN C down
P23.52
(a) 24.2 N C at 0°; (b) 9. 42 N C at 117°
P23.18
(a) 18.0 i − 218 j kN C ;
P23.54
5.25 µC
P23.56
(a)
P23.58
0.205 µC
e
j
(b) 36.0 i − 436 j mN
mg
mgA
; (b)
A cot θ + B
A cos θ + B sin θ
P23.20
(a) 12.9 j kN C ; (b) −38.6 j mN
P23.22
see the solution
P23.24
π 2 ke q
−
i
6a2
P23.62
443 i kN C
P23.26
keλ 0
−i
x0
P23.64
0.072 9 a
P23.28
keλ 0
−i
2x0
P23.66
see the solution; the period is
P23.68
R
P23.30
P23.32
P23.34
P23.60
e j
e j
(a) 383 MN C away; (b) 324 MN C away;
(c) 80.7 MN C away; (d) 6.68 MN C away
LMe
j − ead + hf + R j
N
2 k Qi L
(b)
h + ed + R j − ead + hf + R j
R h MN
k eQ i
d 2 + R2
h
e
2
2
a2
toward the 29th vertex
F mg I
GH k 3 JK
12
e
see the solution
(a)
k e qQ
−1 2
2 12
2
2
2
−1 2
2
OP;
Q
OP
Q
12
P23.70
(a) see the solution; (b) k
P23.72
e−1.36i + 1.96 jj kN C
π
81 4
mL3
k eQq
24
Gauss’s Law
CHAPTER OUTLINE
24.1
24.2
24.3
24.4
24.5
Electric Flux
Gauss’s Law
Application of Gauss’s Law
to Various Charge
Distributions
Conductors in Electrostatic
Equilibrium
Formal Derivation of
Gauss‘s Law
ANSWERS TO QUESTIONS
Q24.1
The luminous flux on a given area is less when the sun is low in
the sky, because the angle between the rays of the sun and the
local area vector, dA, is greater than zero. The cosine of this
angle is reduced. The decreased flux results, on the average, in
colder weather.
Q24.2
If the region is just a point, line, or plane, no. Consider two
protons in otherwise empty space. The electric field is zero at
the midpoint of the line joining the protons. If the field-free
region is three-dimensional, then it can contain no charges, but
it might be surrounded by electric charge. Consider the interior
of a metal sphere carrying static charge.
Q24.3
The surface must enclose a positive total charge.
Q24.4
The net flux through any gaussian surface is zero. We can argue it two ways. Any surface contains
zero charge so Gauss’s law says the total flux is zero. The field is uniform, so the field lines entering
one side of the closed surface come out the other side and the net flux is zero.
Q24.5
Gauss’s law cannot tell the different values of the electric field at different points on the surface.
When E is an unknown number, then we can say E cos θdA = E cos θdA . When E x , y , z is an
z
unknown function, then there is no such simplification.
z
b
g
Q24.6
The electric flux through a sphere around a point charge is independent of the size of the sphere. A
sphere of larger radius has a larger area, but a smaller field at its surface, so that the product of field
strength and area is independent of radius. If the surface is not spherical, some parts are closer to the
charge than others. In this case as well, smaller projected areas go with stronger fields, so that the
net flux is unaffected.
Q24.7
Faraday’s visualization of electric field lines lends insight to this question. Consider a section of a
1
field lines pointing out from it horizontally to
vertical sheet carrying charge +1 coulomb. It has
∈0
the right and left, all uniformly spaced. The lines have the same uniform spacing close to the sheet
and far away, showing that the field has the same value at all distances.
29
30
Gauss’s Law
Q24.8
Consider any point, zone, or object where electric field lines begin. Surround it with a close-fitting
gaussian surface. The lines will go outward through the surface to constitute positive net flux. Then
Gauss’s law asserts that positive net charge must be inside the surface: it is where the lines begin.
Similarly, any place where electric field lines end must be just inside a gaussian surface passing net
negative flux, and must be a negative charge.
Q24.9
Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels
every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surface
of the conductor.
Q24.10
If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shell
carries no charge. The person is not harmed by touching this wall. If the person carries a (small)
charge q, the electric field inside the sphere is no longer zero. Charge –q is induced on the inner wall
of the sphere. The person will get a (small) shock when touching the sphere, as all the charge on his
body jumps to the metal.
Q24.11
The electric fields outside are identical. The electric fields inside are very different. We have E = 0
everywhere inside the conducting sphere while E decreases gradually as you go below the surface of
the sphere with uniform volume charge density.
Q24.12
There is zero force. The huge charged sheet creates a uniform field. The field can polarize the
neutral sheet, creating in effect a film of opposite charge on the near face and a film with an equal
amount of like charge on the far face of the neutral sheet. Since the field is uniform, the films of
charge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forces add
to zero.
Q24.13
Gauss’s law predicts, as described in section 24.4, that excess charge on a conductor will reside on
the surface of the conductor. If a car is left charged by a lightning strike, then that charge will remain
on the outside of the car, not harming the occupants. It turns out that during the lightning strike, the
current also remains on the outside of the conductor. Note that it is not necessarily safe to be in a
fiberglass car or a convertible during a thunderstorm.
SOLUTIONS TO PROBLEMS
Section 24.1
P24.1
Electric Flux
ja
e
f
(a)
Φ E = EA cos θ = 3.50 × 10 3 0.350 × 0.700 cos 0° = 858 N ⋅ m 2 C
(b)
θ = 90.0°
(c)
Φ E = 3.50 × 10 3 0.350 × 0.700 cos 40.0° = 657 N ⋅ m 2 C
ΦE = 0
ja
e
e
f
je
j
P24.2
Φ E = EA cos θ = 2.00 × 10 4 N C 18.0 m 2 cos 10.0° = 355 kN ⋅ m 2 C
P24.3
Φ E = EA cos θ
a
a
A = π r 2 = π 0.200
f
5.20 × 10 5 = E 0.126 cos 0°
f
2
= 0.126 m 2
E = 4.14 × 10 6 N C = 4.14 MN C
Chapter 24
P24.4
(a)
a
fa
f
A ′ = 10.0 cm 30.0 cm
30.0 cm
2
A ′ = 300 cm = 0.030 0 m 2
Φ E , A ′ = EA ′ cos θ
jb
e
31
g
Φ E , A ′ = 7.80 × 10 4 0.030 0 cos 180°
10.0 cm
60.0Þ
Φ E , A ′ = −2.34 kN ⋅ m 2 C
(b)
e
ja f
F 10.0 cm IJ = 600 cm = 0.060 0 m
A = a30.0 cmfa wf = a30.0 cmfG
H cos 60.0° K
Φ
= e7.80 × 10 jb0.060 0g cos 60.0° = +2.34 kN ⋅ m C
2
2
2
4
E, A
(c)
FIG. P24.4
Φ E , A = EA cos θ = 7.80 × 10 4 A cos 60.0°
The bottom and the two triangular sides all lie parallel to E, so Φ E = 0 for each of these. Thus,
Φ E, total = −2.34 kN ⋅ m 2 C + 2.34 kN ⋅ m 2 C + 0 + 0 + 0 = 0 .
P24.5
P24.6
j
Φ E = E ⋅ A = a i + bj ⋅ A i = aA
(b)
Φ E = a i + bj ⋅ Aj = bA
(c)
Φ E = a i + bj ⋅ Ak = 0
e
j
e
j
Only the charge inside radius R contributes to the total flux.
ΦE =
P24.7
e
(a)
q
∈0
Φ E = EA cos θ through the base
a fa f
Φ E = 52.0 36.0 cos 180° = −1.87 kN ⋅ m 2 C .
Note the same number of electric field lines go through the base as go through the
pyramid’s surface (not counting the base).
FIG. P24.7
For the slanting surfaces, Φ E = +1.87 kN ⋅ m 2 C .
P24.8
The flux entering the closed surface equals the flux exiting the surface. The flux entering the left side
of the cone is Φ E = E ⋅ dA = ERh . This is the same as the flux that exits the right side of the cone.
z
Note that for a uniform field only the cross sectional area matters, not shape.
32
Gauss’s Law
Section 24.2
P24.9
(a)
Gauss’s Law
ΦE =
b
g
+5.00 µC − 9.00 µC + 27.0 µC − 84.0 µC
qin
=
= −6.89 × 10 6 N ⋅ m 2 C 2
∈0
8.85 × 10 −12 C 2 N ⋅ m 2
Φ E = −6.89 MN ⋅ m 2 C
(b)
P24.10
(a)
Since the net electric flux is negative, more lines enter than leave the surface.
E=
k eQ
r2
8.90 × 10
:
But Q is negative since E points inward.
(b)
P24.11
P24.12
ΦE =
e8.99 × 10 jQ
=
a0.750f
9
2
2
Q = −5.56 × 10 −8 C = −55.6 nC
The negative charge has a spherically symmetric charge distribution.
qin
∈0
Through S1
ΦE =
−2Q + Q
Q
= −
∈0
∈0
Through S 2
ΦE =
+Q − Q
= 0
∈0
Through S3
ΦE =
−2Q + Q − Q
2Q
= −
∈0
∈0
Through S 4
ΦE = 0
(a)
One-half of the total flux created by the charge q goes through the plane. Thus,
q
1
1 q
Φ E , plane = Φ E , total =
=
.
2
2 ∈0
2 ∈0
(b)
The square looks like an infinite plane to a charge very close to the surface. Hence,
q
Φ E , square ≈ Φ E , plane =
.
2 ∈0
(c)
FG IJ
H K
The plane and the square look the same to the charge.
P24.13
The flux through the curved surface is equal to the flux through the flat circle, E0 π r 2 .
P24.14
(a)
Φ E , shell =
(b)
Φ E, half shell =
(c)
qin 12.0 × 10 −6
=
= 1.36 × 10 6 N ⋅ m 2 C = 1.36 MN ⋅ m 2 C
∈0 8.85 × 10 −12
1
1.36 × 10 6 N ⋅ m 2 C = 6.78 × 10 5 N ⋅ m 2 C = 678 kN ⋅ m 2 C
2
e
j
No, the same number of field lines will pass through each surface, no matter how the
radius changes.
Chapter 24
P24.15
With δ very small, all points on the hemisphere are nearly at
a distance R from the charge, so the field everywhere on the
kQ
curved surface is e 2 radially outward (normal to the
R
surface). Therefore, the flux is this field strength times the
area of half a sphere:
(a)
z
δ →0
33
Q
Φ curved = E ⋅ dA = Elocal A hemisphere
FG
H
Φ curved = k e
(b)
Q
R2
IJ FG 1 4π R IJ = 1 Qa2π f =
K H 2 K 4π ∈
2
0
FIG. P24.15
The closed surface encloses zero charge so Gauss’s law gives
Φ curved + Φ flat = 0
*P24.16
+Q
2 ∈0
Φ flat = − Φ curved =
or
−Q
.
2 ∈0
Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.
z
E ⋅ dA =
mf
b+120 N CgA + b−100 N CgA = ρAa100
∈
b20 N Cge8.85 × 10 C N ⋅ m j = 1.77 × 10
ρ=
q
:
∈0
0
−12
2
2
100 m
−12
C m3
The charge is positive , to produce the net outward flux of electric field.
P24.17
The total charge is Q − 6 q . The total outward flux from the cube is
Q−6q
∈0
, of which one-sixth goes
through each face:
bΦ g
=
bΦ g
=
E one face
E one face
P24.18
Q−6q
6 ∈0
Q−6q
6 ∈0
=
a5.00 − 6.00f × 10
−6
6 × 8.85 × 10
C ⋅ N ⋅ m2
−12
C
2
= −18.8 kN ⋅ m 2 C .
The total charge is Q − 6 q . The total outward flux from the cube is
through each face:
bΦ g
E one face
P24.19
=
Q−6q
6 ∈0
.
If R ≤ d , the sphere encloses no charge and Φ E =
qin
= 0 .
∈0
If R > d , the length of line falling within the sphere is 2 R 2 − d 2
so
ΦE =
2λ R 2 − d 2
∈0
.
Q−6q
∈0
, of which one-sixth goes
34
Gauss’s Law
P24.20
Φ E , hole = E ⋅ A hole =
FG k Q IJ eπ r
HR K
e
2
F e8.99 × 10 N ⋅ m C je10.0 × 10 Cj I
JJ π e1.00 × 10
j GG
a0.100 mf
H
K
2
9
2
=
−6
2
2
−3
j
m
2
Φ E, hole = 28.2 N ⋅ m 2 C
P24.21
ΦE =
qin
170 × 10 −6 C
=
= 1.92 × 10 7 N ⋅ m 2 C
∈0 8.85 × 10 −12 C 2 N ⋅ m 2
1.92 × 10 7 N ⋅ m 2 C
1
ΦE =
6
6
(a)
bΦ g
(b)
Φ E = 19.2 MN ⋅ m 2 C
(c)
E one face
=
bΦ g
E one face
= 3.20 MN ⋅ m 2 C
The answer to (a) would change because the flux through each face of the cube would
not be equal with an asymmetric charge distribution. The sides of the cube nearer the
charge would have more flux and the ones further away would have less. The answer
to (b) would remain the same, since the overall flux would remain the same.
P24.22
No charge is inside the cube. The net flux through the cube is zero. Positive flux
comes out through the three faces meeting at g. These three faces together fill
solid angle equal to one-eighth of a sphere as seen from q, and together pass
1 q
. Each face containing a intercepts equal flux going into the cube:
flux
8 ∈0
FG IJ
H K
0 = Φ E , net
Φ E , abcd =
Section 24.3
P24.23
FIG. P24.22
q
= 3 Φ E , abcd +
8 ∈0
−q
24 ∈0
Application of Gauss’s Law to Various Charge Distributions
The charge distributed through the nucleus creates a field at the surface equal to that of a point
k q
charge at its center: E = e2
r
e8.99 × 10 Nm C je82 × 1.60 × 10
E=
a208f 1.20 × 10 m
9
2
13
E = 2.33 × 10 21 N C
2
−15
−19
C
j
2
away from the nucleus
Chapter 24
P24.24
(a)
E=
(b)
E=
k eQr
a3
= 0
e8.99 × 10 je26.0 × 10 ja0.100f = 365 kN C
a
a0.400f
k Q e8.99 × 10 je 26.0 × 10 j
=
= 1.46 MN C
E=
r
a0.400f
k eQr
3
−6
9
=
3
−6
9
e
(c)
E=
(d)
2
2
k eQ
r2
e8.99 × 10 je26.0 × 10 j =
=
a0.600f
−6
9
649 kN C
2
The direction for each electric field is radially outward .
*P24.25
mg = qE = q
0
P24.26
−12
0.01 9.8
Q 2 ∈0 mg 2 8.85 × 10
=
=
= −2.48 µC m 2
A
q
−0.7 × 10 −6
0
jb
e
2 8.99 × 10 9 Q 2.40
2k e λ
4
E=
3.60 × 10 =
0.190
r
Q = +9.13 × 10 −7 C = +913 nC
(a)
g
E= 0
(b)
*P24.27
ja fa f
e
FG σ IJ = qFG Q A IJ
H2∈ K H 2∈ K
The volume of the spherical shell is
a
f − a0.20 mf
4
π 0.25 m
3
3
3
= 3.19 × 10 −2 m3 .
Its charge is
e
je
j
ρV = −1.33 × 10 −6 C m 3 3.19 × 10 −2 m3 = −4.25 × 10 −8 C .
The net charge inside a sphere containing the proton’s path as its equator is
−60 × 10 −9 C − 4. 25 × 10 −8 C = −1.02 × 10 −7 C .
The electric field is radially inward with magnitude
ke q
r
2
=
q
∈0 4π r
2
=
e
8.99 × 10 9 Nm 2 1.02 × 10 −7 C
2
a
f
C 0.25 m
2
j = 1.47 × 10
4
N C.
For the proton
∑ F = ma
F eEr IJ
v=G
HmK
eE =
12
mv 2
r
F 1.60 × 10
=G
GH
−19
e
j
C 1.47 × 10 4 N C 0.25 m
1.67 × 10 −27 kg
I
JJ
K
12
= 5.94 × 10 5 m s .
35
36
Gauss’s Law
P24.28
e
σ = 8.60 × 10 −6 C cm 2
E=
jFGH 100mcm IJK
2
= 8.60 × 10 −2 C m 2
σ
8.60 × 10 −2
=
= 4.86 × 10 9 N C away from the wall
2 ∈0 2 8.85 × 10 −12
e
j
The field is essentially uniform as long as the distance from the center of the wall to the field point is
much less than the dimensions of the wall.
P24.29
If ρ is positive, the field must be radially outward. Choose as the
gaussian surface a cylinder of length L and radius r, contained inside
the charged rod. Its volume is π r 2 L and it encloses charge ρπ r 2 L .
Because the charge distribution is long, no electric flux passes
through the circular end caps; E ⋅ dA = EdA cos 90.0° = 0 . The curved
surface has E ⋅ dA = EdA cos 0° , and E must be the same strength
everywhere over the curved surface.
q
ρπ r 2 L
Gauss’s law, E ⋅ dA =
,
becomes
E
dA =
.
∈0
∈0
Curved
z
FIG. P24.29
z
Surface
Now the lateral surface area of the cylinder is 2π rL :
b g
E 2π r L =
*P24.30
ρπ r 2 L
.
∈0
ρr
radially away from the cylinder axis .
2 ∈0
E=
Thus,
Let ρ represent the charge density. For the field inside the sphere at r1 = 5 cm we have
E1 4π r12 =
q inside 4π r13 ρ
=
∈0
3 ∈0
E1 =
e
je
r1 ρ
3 ∈0
j
−12
C 2 −86 × 10 3 N
3 ∈0 E1 3 8.85 × 10
ρ=
=
= −4.57 × 10 −5 C m 3 .
r1
0.05 m
Nm 2
C
Now for the field outside at r3 = 15 cm
4π r23 ρ
E3 4π r32 =
3 ∈0
a
f e−4.57 × 10 Cj = 8.99 × 10
k 4 π 0.10 m
E3 = 2e
r3 3
3
−5
m
3
9
e
Nm 2 −1.91 × 10 −7 C
a0.15 mf C
2
2
j = −7.64 × 10
4
NC
E 3 = 76. 4 kN C radially inward
P24.31
E= 0
(a)
E=
(b)
P24.32
k eQ
r2
e8.99 × 10 je32.0 × 10 j = 7.19 MN C
=
a0.200f
−6
9
E = 7.19 MN C radially outward
2
The distance between centers is 2 × 5.90 × 10 −15 m . Each produces a field as if it were a point charge
at its center, and each feels a force as if all its charge were a point at its center.
F=
k e q1 q 2
r2
e
a46f e1.60 × 10 Cj
C j
e2 × 5.90 × 10 mj
2
9
= 8.99 × 10 N ⋅ m
2
−19
2
−15
2
2
= 3.50 × 10 3 N = 3.50 kN
Chapter 24
P24.33
Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a
0.05 m separation on the ends of strings making angles of 10° with the vertical.
(a)
mg
∑ Fy = T cos 10°−mg = 0 ⇒ T = cos 10°
∑ Fx = T sin 10°− Fe = 0 ⇒ Fe = T sin 10° , so
Fe =
FG mg IJ sin 10° = mg tan 10° = b0.001 kgge9.8 m s j tan 10°
H cos 10° K
2
Fe ≈ 2 × 10 −3 N ~ 10 −3 N or 1 mN
(b)
Fe =
keq 2
r2
e8.99 × 10 N ⋅ m
N≈
a0.25 mf
9
2 × 10
−3
2
j
C 2 q2
2
q ≈ 1. 2 × 10 −7 C ~ 10 −7 C or 100 nC
*P24.34
37
keq
(c)
E=
(d)
ΦE =
r
≈
2
e8.99 × 10
9
je
N ⋅ m 2 C 2 1.2 × 10 −7 C
a0.25 mf
2
N C ~ 10 kN C
4 3
πa ρ
3
ρ=
3Q
4π a 3
The flux is that created by the enclosed charge within radius r:
ΦE =
(b)
4
q
1.2 × 10 −7 C
≈
= 1. 4 × 10 4 N ⋅ m 2 C ~ 10 kN ⋅ m 2 C
∈0 8.85 × 10 −12 C 2 N ⋅ m 2
The charge density is determined by Q =
(a)
j ≈ 1.7 × 10
ΦE =
q in 4π r 3 ρ
4π r 3 3Q
Qr 3
=
=
=
∈0
3 ∈0
3 ∈0 4π a 3
∈0 a 3
Q
. Note that the answers to parts (a) and (b) agree at r = a .
∈0
(c)
ΦE
Q
∈0
0
0
a
FIG. P24.34(c)
r
FIG. P24.33
38
Gauss’s Law
P24.35
(a)
e
je
j
9
2
2
2.00 × 10 −6 C 7.00 m
2 k e λ 2 8.99 × 10 N ⋅ m C
=
E=
0.100 m
r
E = 51.4 kN C , radially outward
(b)
b
g
Φ E = EA cos θ = E 2π rA cos 0°
j a
e
fb
ga f
Φ E = 5.14 × 10 4 N C 2π 0.100 m 0.020 0 m 1.00 = 646 N ⋅ m 2 C
P24.36
(a)
(b)
ρ=
Q
5.70 × 10 −6
=
= 2.13 × 10 −2 C m 3
3
4π 3
4π
a
0.040 0
3
3
b
g
FG 4 π r IJ = e2.13 × 10
H3 K
F4 I
= ρ G π r J = e 2.13 × 10
H3 K
qin = ρ
3
−2
qin
3
−2
jFGH 34 π IJK b0.020 0g
jFGH 34 π IJK b0.040 0g
3
= 7.13 × 10 −7 C = 713 nC
3
= 5.70 µC
9.00 × 10 −6 C m 2
σ
=
= 508 kN C , upward
2 ∈0 2 8.85 × 10 −12 C 2 N ⋅ m 2
P24.37
E=
P24.38
Note that the electric field in each case is directed radially inward, toward the filament.
j
e
je
j
e
je
j
e
je
j
(a)
E=
−6
9
2
2
2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m
=
= 16.2 MN C
0.100 m
r
(b)
E=
−6
9
2
2
2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m
=
= 8.09 MN C
0.200 m
r
(c)
−6
9
2
2
2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m
=
= 1.62 MN C
E=
1.00 m
r
Section 24.4
P24.39
e
z
Conductors in Electrostatic Equilibrium
b g
EdA = E 2π rl =
qin
∈0
E=
q in l
λ
=
2π ∈0 r 2π ∈0 r
(a)
r = 3.00 cm
E= 0
(b)
r = 10.0 cm
E=
(c)
r = 100 cm
E=
e
30.0 × 10 −9
ja
2π 8.85 × 10 −12 0.100
e
30.0 × 10 −9
f=
=
ja f
2π 8.85 × 10 −12 1.00
5 400 N C , outward
540 N C , outward
Chapter 24
P24.40
σ=
P24.41
EA =
From Gauss’s Law,
ja
Q
∈0
39
f
Q
=∈0 E = 8.85 × 10 −12 −130 = −1.15 × 10 −9 C m 2 = −1.15 nC m 2
A
e
σ conductor
for the field outside the aluminum looks
∈0
The fields are equal. The Equation 24.9 E =
σ insulator
for the field around glass. But its charge will spread out to
2 ∈0
Q
cover both sides of the aluminum plate, so the density is σ conductor =
. The glass carries charge
2A
Q
Q
only on area A, with σ insulator = . The two fields are
the same in magnitude, and both are
A
2 A ∈0
perpendicular to the plates, vertically upward if Q is positive.
different from Equation 24.8 E =
*P24.42
(a)
All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside is
zero .
(b)
The charged sphere creates field at exterior points as if it were a point charge at the center:
E=
P24.43
ke q
r
2
away =
e8.99 × 10
9
je
Nm 2 40 × 10 −9 C
2
a
f
C 0.17 m
(c)
e8.99 × 10
E=
(d)
All three answers would be the same.
(a)
E=
9
je
2
Nm 2 40 × 10 −9 C
2
a
f
C 0.75 m
σ
∈0
2
e
j outward =
j outward =
je
1.24 × 10 4 N C outward
639 N C outward
j
σ = 8.00 × 10 4 8.85 × 10 −12 = 7.08 × 10 −7 C m 2
σ = 708 nC m 2 , positive on one face and negative on the other.
P24.44
e
σ=
(a)
E= 0
k eQ
(b)
E=
(c)
E= 0
(d)
ja
f
Q
2
Q = σA = 7.08 × 10 −7 0.500 C
A
Q = 1.77 × 10 −7 C = 177 nC , positive on one face and negative on the other.
(b)
E=
r2
k eQ
r2
e8.99 × 10 je8.00 × 10 j = 7.99 × 10
=
b0.030 0g
7
NC
E = 79.9 MN C radially outward
e8.99 × 10 je4.00 × 10 j = 7.34 × 10
=
b0.070 0g
6
NC
E = 7.34 MN C radially outward
−6
9
2
−6
9
2
40
Gauss’s Law
P24.45
The charge divides equally between the identical spheres, with charge
like point charges at their centers:
F=
P24.46
b gb g = k Q
aL + R + Rf 4aL + 2Rf
ke Q 2 Q 2
2
e
2
2
=
e
8.99 × 10 9 N ⋅ m 2 60.0 × 10 −6 C
a
f
4 C 2 2.01 m
j
Q
on each. Then they repel
2
2
2
= 2.00 N .
The electric field on the surface of a conductor varies inversely with the radius of curvature of the
surface. Thus, the field is most intense where the radius of curvature is smallest and vice-versa. The
local charge density and the electric field intensity are related by
E=
(a)
σ
σ =∈0 E .
or
∈0
Where the radius of curvature is the greatest,
e
je
j
σ =∈0 Emin = 8.85 × 10 −12 C 2 N ⋅ m 2 2.80 × 10 4 N C = 248 nC m 2 .
(b)
Where the radius of curvature is the smallest,
e
je
j
σ =∈0 Emax = 8.85 × 10 −12 C 2 N ⋅ m 2 5.60 × 10 4 N C = 496 nC m 2 .
P24.47
(a)
(b)
P24.48
(a)
(b)
P24.49
(a)
Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting
shell is zero, the total charge inside the gaussian surface must be zero, so the inside
charge/length = − λ .
0 = λA + qin
so
qin
= −λ
A
Outside surface:
The total charge on the metal cylinder is
2λA = qin + qout
qout = 2λA + λA
so the outside charge/length is
E=
E=
e
r
k eQ
r2
r
e8.99 × 10 je6.40 × 10 j =
=
a0.150f
−6
9
2
2.56 MN C , radially inward
The charge density on each of the surfaces (upper and lower) of the plate is:
E=
FG IJ
H K
e
E=
j
−8
1 q
1 4.00 × 10 C
=
= 8.00 × 10 −8 C m 2 = 80.0 nC m 2 .
2 A
2 0.500 m 2
a
f
FG σ IJ k = F 8.00 × 10 C m I k = b9.04 kN Cgk
H ∈ K GH 8.85 × 10 C N ⋅ m JK
−8
0
(c)
3λ
radially outward
2π ∈0 r
E=0
σ=
(b)
b g = 6k λ =
2 k e 3λ
3λ .
b−9.04 kN Cgk
−12
2
2
2
Chapter 24
P24.50
(a)
The charge +q at the center induces charge −q on the inner surface of the conductor,
where its surface density is:
−q
σa =
.
4π a 2
(b)
The outer surface carries charge Q + q with density
σb =
P24.51
Q+q
41
.
4π b 2
Use Gauss’s Law to evaluate the electric field in each region, recalling that the electric field is zero
everywhere within conducting materials. The results are:
E = 0 inside the sphere and within the material of the shell
P24.52
E = ke
Q
between the sphere and shell, directed radially inward
r2
E = ke
2Q
outside the shell, directed radially outward .
r2
Charge
−Q is on the outer surface of the sphere .
Charge
+Q is on the inner surface of the shell ,
and
+2Q is on the outer surface of the shell.
An approximate sketch is given at the right. Note that the electric field lines
should be perpendicular to the conductor both inside and outside.
FIG. P24.52
Section 24.5
P24.53
(a)
Formal Derivation of Gauss‘s Law
Uniform E, pointing radially outward, so Φ E = EA . The arc length is ds = Rdθ ,
and the circumference is 2π r = 2π R sin θ
z
A = 2π rds =
zb
θ
z
θ
g
a
2π R sin θ Rdθ = 2π R 2 sin θdθ = 2π R 2 − cos θ
0
f
θ
0
0
a
f
a
1
Q
Q
ΦE =
⋅ 2π R 2 1 − cos θ =
1 − cos θ
4π ∈0 R 2
2 ∈0
a
f
[independent of R!]
f
(b)
For θ = 90.0° (hemisphere): Φ E =
Q
Q
1 − cos 90° =
.
2 ∈0
2 ∈0
(c)
For θ = 180° (entire sphere): Φ E =
Q
Q
1 − cos 180° =
2 ∈0
∈0
a
b
= 2π R 2 1 − cos θ
f
[Gauss’s Law].
g
FIG. P24.53
42
Gauss’s Law
Additional Problems
P24.54
E = ay i + bzj + cxk
E = ay i + cxk
In general,
In the xy plane, z = 0 and
z
ze
Φ E = E ⋅ dA =
w
z
(b)
x=0
y =h
x=0
j
x2
Φ E = ch xdx = ch
2
x=0
(a)
y=0
ay i + cxk ⋅ k dA
w
P24.55
z
y
x=w
chw 2
=
2
dA = hdx
x
FIG. P24.54
qin = +3Q − Q = +2Q
The charge distribution is spherically symmetric and qin > 0 . Thus, the field is directed
radially outward .
k e qin
for r ≥ c .
E=
(d)
Since all points within this region are located inside conducting material, E = 0 for
r
2
=
2 k eQ
(c)
r2
b < r < c.
z
(e)
Φ E = E ⋅ dA = 0 ⇒ qin =∈0 Φ E = 0
(f)
qin = +3Q
(g)
E=
(h)
qin = ρV =
(i)
E=
(j)
From part (d), E = 0 for b < r < c . Thus, for a
spherical gaussian surface with b < r < c ,
qin = +3Q + qinner = 0 where qinner is the
charge on the inner surface of the
conducting shell. This yields qinner = −3Q .
(k)
Since the total charge on the conducting
shell is q net = qouter + qinner = −Q , we have
k e qin
r
2
k e qin
r
2
=
3 k eQ
r2
(radially outward) for a ≤ r < b .
F +3Q I FG 4 π r IJ = +3Q r
GH π a JK H 3 K
a
k F
r I
r
=
+3Q J = 3 k Q
(radially outward) for 0 ≤ r ≤ a .
G
r H
a K
a
4
3
e
2
3
3
3
3
3
3
e
3
b g
qouter = −Q − qinner = −Q − −3Q = +2Q .
(l)
This is shown in the figure to the right.
E
a
b
c
FIG. P24.55(l)
r
Chapter 24
43
P24.56
The sphere with large charge creates a strong field to polarize the other sphere. That means it
pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side.
This patch of opposite charge is smaller in amount but located in a stronger external field, so it can
feel a force of attraction that is larger than the repelling force felt by the larger charge in the weaker
field on the other side.
P24.57
(a)
(b)
z
e
j
E ⋅ dA = E 4π r 2 =
qin
∈0
FG 4 π r IJ
H3 K
3
For r < a ,
qin = ρ
so
E=
For a < r < b and c < r ,
qin = Q .
So
Q
E=
.
4π r 2 ∈0
For b ≤ r ≤ c ,
E = 0 , since E = 0 inside a conductor.
ρr
.
3 ∈0
FIG. P24.57
Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the
conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero.
q1 + Q = 0
Therefore,
and
σ1 =
q1
4π b
2
=
−Q
.
4π b 2
Let q 2 = induced charge on the outside surface of the hollow sphere. Since the hollow
sphere is uncharged, we require
q
Q
σ2 = 1 2 =
.
q1 + q 2 = 0
and
4π c
4π c 2
P24.58
z
e
j
E ⋅ dA = E 4π r 2 =
(a)
e−3.60 × 10
3
qin
∈0
j a
f
N C 4π 0.100 m
2
=
Q
8.85 × 10
−12
C 2 N ⋅ m2
aa < r < bf
Q = −4.00 × 10 −9 C = −4.00 nC
(b)
We take Q ′ to be the net charge on the hollow sphere. Outside c,
Q + Q′
2
+2.00 × 10 2 N C 4π 0.500 m =
r>c
8.85 × 10 −12 C 2 N ⋅ m 2
e
j a
f
a f
Q + Q ′ = +5.56 × 10 −9 C , so Q ′ = +9.56 × 10 −9 C = +9.56 nC
(c)
For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner surface of
the hollow sphere. Thus, Q1 = −Q = +4.00 nC .
Then, if Q 2 is the total charge on the outer surface of the hollow sphere,
Q 2 = Q ′ − Q1 = 9.56 nC − 4.0 nC = +5.56 nC .
44
Gauss’s Law
*P24.59
y
The vertical velocity component of the moving charge
increases according to
dv y
m
dv y dx
= qE y .
m
dx dt
= Fy
dt
v
0
q
Now
dx
= v x has the nearly constant value v. So
dt
dv y =
q
Ey dx
mv
z
vy
x
Q
FIG. P24.59
q ∞
Ey dx .
mv −∞
0
vx
d
z
vy
v y = dv y =
θ
The radially outward compnent of the electric field varies along the x axis, but is described by
z
∞
z
∞
Ey dA =
−∞
−∞
z
∞
So
Ey dx =
−∞
tan θ =
P24.60
vy
v
b g
Ey 2π d dx =
=
Q
.
∈0
qQ
Q
and v y =
. The angle of deflection is described by
2π d ∈0
mv 2π d ∈0
qQ
2π ∈0 dmv
θ = tan −1
2
qQ
2π ∈0 dmv 2
.
First, consider the field at distance r < R from the center of a uniform sphere of positive charge
Q = + e with radius R.
b
e
g
j
4π r 2 E =
(a)
qin ρV
=
=
∈0
∈0
F
GH
+e
4π 3
3 R
I
JK
4
3π
r3
∈0
so E =
F e I r directed outward
GH 4π ∈ R JK
0
3
The force exerted on a point charge q = − e located at distance r from the center is then
F = qE = − e
F e I r = −F e I r =
GH 4π ∈ R JK GH 4π ∈ R JK
2
0
3
3
0
− Kr .
k e2
e2
= e3
3
4π ∈0 R
R
(b)
K=
(c)
Fr = m e a r = −
F k e I r , so a
GH R JK
e
2
3
r
=−
F k e I r = −ω r
GH m R JK
e
e
2
3
2
Thus, the motion is simple harmonic with frequency
(d)
f = 2.47 × 10
15
1
Hz =
2π
e8.99 × 10
9
je
f=
ω
1
=
2π
2π
N ⋅ m 2 C 2 1.60 × 10 −19 C
e9.11 × 10
−31
j
j
2
kg R 3
which yields R 3 = 1.05 × 10 −30 m3 , or R = 1.02 × 10 −10 m = 102 pm .
kee2
me R3
.
Chapter 24
P24.61
The field direction is radially outward perpendicular to the axis. The field strength depends on r but
not on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of radius r and length L.
If r < a ,
ΦE =
E=
qin
∈0
and
λ
2π r ∈0
or
b
g
E 2π rL =
E=
b
E=
λ
g
b
E=
r
ar < a f
.
e
j
λL + ρπ r 2 − a 2 L
∈0
e
λ + ρπ r 2 − a 2
2π r ∈0
g
E 2π rL =
If r > b ,
λL
∈0
2π r ∈0
E 2π rL =
If a < r < b ,
P24.62
45
j r
e
aa < r < bf .
j
λL + ρπ b 2 − a 2 L
∈0
e
λ + ρπ b 2 − a 2
2π r ∈0
j r
ar > b f
.
Consider the field due to a single sheet and let E+ and E−
represent the fields due to the positive and negative sheets. The
field at any distance from each sheet has a magnitude given by
Equation 24.8:
E+ = E− =
(a)
σ
.
2 ∈0
To the left of the positive sheet, E+ is directed toward the
left and E− toward the right and the net field over this
region is E = 0 .
(b)
In the region between the sheets, E+ and E− are both
directed toward the right and the net field is
E=
(c)
σ
∈0
to the right .
FIG. P24.62
To the right of the negative sheet, E+ and E− are again oppositely directed and E = 0 .
46
Gauss’s Law
P24.63
The magnitude of the field due to the each sheet given by
Equation 24.8 is
E=
(a)
σ
directed perpendicular to the sheet.
2 ∈0
In the region to the left of the pair of sheets, both fields are
directed toward the left and the net field is
σ
E=
∈0
to the left .
(b)
In the region between the sheets, the fields due to the individual sheets are oppositely
directed and the net field is
E= 0 .
(c)
In the region to the right of the pair of sheets, both are fields are directed toward the right
and the net field is
σ
to the right .
∈0
E=
P24.64
FIG. P24.63
The resultant field within the cavity is the superposition of two
fields, one E + due to a uniform sphere of positive charge of radius
2a, and the other E − due to a sphere of negative charge of radius a
centered within the cavity.
F
GH
I
JK
4 π r 3ρ
= 4π r 2 E+
3 ∈0
–−
F
GH
I
JK
4 π r13 ρ
= 4π r12 E−
3 ∈0
Since r = a + r1 ,
so
E+ =
ρr
ρr
r =
3 ∈0
3 ∈0
so
E− =
ρ r1
−ρ
− r1 =
r1 .
3 ∈0
3 ∈0
a f
−ρ r − a
E− =
3 ∈0
E = E+ + E− =
*P24.65
b g
Thus,
Ex = 0
and
Ey =
ρa
3 ∈0
FIG. P24.64
ρr
ρr
ρa
ρa
ρa −
+
=
= 0 i +
j.
3 ∈0 3 ∈0 3 ∈0 3 ∈0
3 ∈0
at all points within the cavity.
Consider the charge distribution to be an unbroken charged spherical shell with uniform charge
density σ and a circular disk with charge per area −σ . The total field is that due to the whole sphere,
4π R 2σ
σ
σ
σ
Q
=
=
outward plus the field of the disk −
=
radially inward. The total
2
2
∈
∈
∈0
2
2
4π ∈0 R
4πε 0 R
0
0
field is
σ
σ
σ
outward .
−
=
∈0 2 ∈0
2 ∈0
Chapter 24
P24.66
The electric field throughout the region is directed along x; therefore, E will be
perpendicular to dA over the four faces of the surface which are perpendicular
to the yz plane, and E will be parallel to dA over the two faces which are parallel
to the yz plane. Therefore,
e
Φ E = − Ex
x=a
j A + eE
x x=a+c
jA = −e3 + 2 a jab + e3 + 2aa + cf jab = 2abca2 a + cf .
2
2
Substituting the given values for a, b, and c, we find Φ E = 0.269 N ⋅ m 2 C .
FIG. P24.66
Q =∈0 Φ E = 2.38 × 10 −12 C = 2.38 pC
P24.67
z
e
j
E ⋅ dA = E 4π r 2 =
qin
∈0
z
R
e
0
(b)
AR 5
5
j
4π Ar 5
5
AR 5
.
5 ∈0 r 2
and
E=
For r < R ,
qin = Ar 2 4π r 2 dr =
z
r
e
0
Ar 3
.
5 ∈0
E=
and
P24.68
j
qin = Ar 2 4π r 2 dr = 4π
For r > R ,
(a)
The total flux through a surface enclosing the charge Q is
disk is
Q
. The flux through the
∈0
z
Φ disk = E ⋅ dA
where the integration covers the area of the disk. We must evaluate this integral
1Q
to find how b and R are related. In the figure, take dA to be
and set it equal to 4
∈0
the area of an annular ring of radius s and width ds. The flux through dA is
b
FIG. P24.68
g
E ⋅ dA = EdA cos θ = E 2π sds cos θ .
The magnitude of the electric field has the same value at all points within the annular ring,
E=
1 Q
1
Q
=
4π ∈0 r 2 4π ∈0 s 2 + b 2
and
cos θ =
b
b
=
2
r
s + b2
e
j
12
.
Integrate from s = 0 to s = R to get the flux through the entire disk.
Φ E , disk =
Qb
2 ∈0
ze
R
0
sds
s2 + b 2
j
=
32
The flux through the disk equals
This is satisfied if R = 3 b .
LM e
N
Qb
− s2 + b 2
2 ∈0
j OPQ
12
R
=
0
LM
MM e
N
Q
b
1−
2 ∈0
R2 + b2
Q
b
provided that
4 ∈0
R2 + b2
e
j
12
=
OP
j PPQ
12
1
.
2
47
48
Gauss’s Law
P24.69
z
E ⋅ dA =
qin
1
=
∈0 ∈0
z
z
r
0
a
4π r 2 dr
r
4π a r
4π a r 2
E 4π r 2 =
rdr =
∈0 0
∈0 2
E=
a
= constant magnitude
2 ∈0
(The direction is radially outward from center for positive a; radially inward for negative a.)
P24.70
z
z
1
ρdV . We
∈0
use a gaussian surface which is a cylinder of radius r, length A , and is coaxial with the charge
distribution.
In this case the charge density is not uniform, and Gauss’s law is written as
(a)
b
g
When r < R , this becomes E 2π rA =
ρ0
∈0
z FGH
r
a−
0
E ⋅ dA =
IJ
K
r
dV . The element of volume is a cylindrical
b
shell of radius r, length A , and thickness dr so that dV = 2π rAdr .
b
g FGH 2π r∈ Aρ
2
E 2π rA =
(b)
0
b
(a)
I FG a − r IJ so inside the cylinder, E =
JK H 2 3b K
FG
H
ρ 0r
2r
a−
2 ∈0
3b
When r > R , Gauss’s law becomes
g
E 2π rA =
P24.71
0
ρ0
∈0
z FGH a − br IJK b2π rAdrg or outside the cylinder, E =
FG
H
R
IJ
K
.
ρ0R2
2R
a−
2 ∈0 r
3b
0
IJ
K
Consider a cylindrical shaped gaussian surface perpendicular
to the yz plane with one end in the yz plane and the other end
containing the point x:
Use Gauss’s law:
z
E ⋅ dA =
.
y
qin
∈0
gaussian
surface
By symmetry, the electric field is zero in the yz plane and is
perpendicular to dA over the wall of the gaussian cylinder.
Therefore, the only contribution to the integral is over the end
cap containing the point x :
z
a f
(b)
z
ρ Ax
q
E ⋅ dA = in or EA =
∈0
∈0
so that at distance x from the mid-line of the slab, E =
a=
a f
FG
H
x
x
ρx
.
∈0
IJ
K
−e E
ρe
F
=
=−
x
me
me
m e ∈0
FIG. P24.71
The acceleration of the electron is of the form
a = −ω 2 x with ω =
Thus, the motion is simple harmonic with frequency
f=
ω
1
=
2π
2π
ρe
.
m e ∈0
ρe
.
m e ∈0
Chapter 24
P24.72
Consider the gaussian surface described in the solution to problem 71.
(a)
d
,
2
1
E ⋅ dA =
dq
∈0
z
CA
∈0
EA =
E=
(b)
3
Cd
24 ∈0
For −
E=
(a)
dq = ρ dV = ρAdx = CAx 2 dx
For x >
z
P24.73
49
z
d 2
x 2 dx =
0
FG IJ F d I
H K GH 8 JK
3
1 CA
3 ∈0
E=
or
d
d
<x<
2
2
z
Cd 3 d
i for x > ;
24 ∈0
2
E ⋅ dA =
Cx 3 i for x > 0 ;
3 ∈0
E=−
z
E=−
Cd 3 d
i for x < −
24 ∈0
2
z
1
CA x 2
CAx 3
dq =
x dx =
3 ∈0
∈0
∈0 0
Cx 3 i for x < 0
3 ∈0
A point mass m creates a gravitational acceleration
g=−
z
Gm
r2
r at a distance r.
Gm
e
j
4π r 2 = −4π Gm .
r2
Since the r has divided out, we can visualize the field as unbroken field lines. The same flux
would go through any other closed surface around the mass. If there are several or no
masses inside a closed surface, each creates field to make its own contribution to the net flux
according to
The flux of this field through a sphere is
z
(b)
g ⋅ dA = −
g ⋅ dA = −4π Gm in .
Take a spherical gaussian surface of radius r. The field is inward so
g ⋅ dA = g 4π r 2 cos 180° = − g 4π r 2
z
and
Then,
Or, since
4
−4π Gm in = −4π G π r 3 ρ .
3
4
4 3
2
− g 4π r = −4π G π r ρ and g = π rρG .
3
3
M EGr
ME
M EGr
, g=
or g =
inward .
ρ= 4
3
RE3
RE3
3 π RE
ANSWERS TO EVEN PROBLEMS
P24.2
355 kN ⋅ m 2 C
P24.10
(a) −55.6 nC ; (b) The negative charge has a
spherically symmetric distribution.
P24.4
(a) −2.34 kN ⋅ m 2 C ; (b) +2.34 kN ⋅ m 2 C ;
(c) 0
P24.12
(a)
P24.14
(a) 1.36 MN ⋅ m 2 C ; (b) 678 kN ⋅ m 2 C ;
(c) No; see the solution.
P24.6
q
∈0
P24.8
ERh
q
q
; (b)
; (c) Plane and square
2 ∈0
2 ∈0
both subtend a solid angle of a hemisphere
at the charge.
50
Gauss’s Law
P24.16
P24.18
1.77 pC m3 positive
P24.46
(a) 248 nC m 2 ; (b) 496 nC m 2
Q−6q
P24.48
(a) 2.56 MN C radially inward; (b) 0
P24.50
(a)
P24.52
see the solution
P24.54
chw 2
2
P24.56
see the solution
P24.58
(a) −4.00 nC; (b) +9.56 nC ; (c) +4.00 nC
and +5.56 nC
P24.60
(a, b) see the solution; (c)
6 ∈0
P24.20
28. 2 N ⋅ m 2 C
P24.22
−q
24 ∈0
P24.24
(a) 0; (b) 365 kN C ; (c) 1.46 MN C;
(d) 649 kN C
P24.26
(a) 913 nC ; (b) 0
P24.28
4.86 GN C away from the wall. It is
constant close to the wall
P24.30
76.4 kN C radially inward
P24.32
3.50 kN
−q
4π a
2
; (b)
Q+q
4π b 2
1
2π
ke e2
me R3
;
(d) 102 pm
P24.34
P24.36
P24.38
(a)
3
Qr
Q
; (c) see the solution
; (b)
∈0
∈0 a 3
713 nC ; (b) 5.70 µC
(a) 16.2 MN C toward the filament;
(b) 8.09 MN C toward the filament;
(c) 1.62 MN C toward the filament
2
P24.40
−1.15 nC m
P24.42
(a) 0; (b) 12.4 kN C radially outward;
(c) 639 N C radially outward; (d) Nothing
would change.
P24.44
(a) 0; (b) 79.9 MN C radially outward;
(c) 0; (d) 7.34 MN C radially outward
σ
to the right; (c) 0
∈0
P24.62
(a) 0; (b)
P24.64
see the solution
P24.66
0.269 N ⋅ m 2 C ; 2.38 pC
P24.68
see the solution
P24.70
(a)
P24.72
(a) E =
FG
H
IJ
K
FG
H
ρ 0r
ρ R2
2r
2R
; (b) 0
a−
a−
2 ∈0
3b
2 ∈0 r
3b
IJ
K
Cd 3 d
i for x > ;
24 ∈0
2
3
Cd d
E=−
i for x < − ;
24 ∈0
2
3
Cx Cx 3 i for x > 0 ; E = −
i for x < 0
(b) E =
3 ∈0
3 ∈0
25
Electric Potential
CHAPTER OUTLINE
25.1
25.2
25.3
25.4
25.5
25.6
25.7
25.8
Potential Difference and
Electric Potential
Potential Difference in a
Uniform Electric Field
Electric Potential and
Potential Energy Due to
Point Charges
Obtaining the Value of the
Electric Field from the
Electric Potential
Electric Potential Due to
Continuous Charge
Distributions
Electric Potential Due to a
Charged Conductor
The Milliken Oil Drop
Experiment
Application of Electrostatistics
ANSWERS TO QUESTIONS
Q25.1
When one object B with electric charge is immersed in the
electric field of another charge or charges A, the system
possesses electric potential energy. The energy can be measured
by seeing how much work the field does on the charge B as it
moves to a reference location. We choose not to visualize A’s
effect on B as an action-at-a-distance, but as the result of a twostep process: Charge A creates electric potential throughout the
surrounding space. Then the potential acts on B to inject the
system with energy.
Q25.2
The potential energy increases. When an outside agent makes it
move in the direction of the field, the charge moves to a region
of lower electric potential. Then the product of its negative
charge with a lower number of volts gives a higher number of
joules. Keep in mind that a negative charge feels an electric force
in the opposite direction to the field, while the potential is the
work done on the charge to move it in a field per unit charge.
Q25.3
To move like charges together from an infinite separation, at which the potential energy of the
system of two charges is zero, requires work to be done on the system by an outside agent. Hence
energy is stored, and potential energy is positive. As charges with opposite signs move together
from an infinite separation, energy is released, and the potential energy of the set of charges
becomes negative.
Q25.4
The charge can be moved along any path parallel to the y-z plane, namely perpendicular to the field.
Q25.5
The electric field always points in the direction of the greatest change in electric potential. This is
∂V
∂V
∂V
, Ey = −
and Ez = −
.
implied by the relationships Ex = −
∂x
∂y
∂z
Q25.6
(a)
The equipotential surfaces are nesting coaxial cylinders around an infinite line of charge.
(b)
The equipotential surfaces are nesting concentric spheres around a uniformly charged
sphere.
Q25.7
If there were a potential difference between two points on the conductor, the free electrons in the
conductor would move until the potential difference disappears.
51
52
Electric Potential
Q25.8
No. The uniformly charged sphere, whether hollow or solid metal, is an equipotential volume. Since
there is no electric field, this means that there is no change in electrical potential. The potential at
every point inside is the same as the value of the potential at the surface.
Q25.9
Infinitely far away from a line of charge, the line will not look like a point. In fact, without any
distinguishing features, it is not possible to tell the distance from an infinitely long line of charge.
Another way of stating the answer: The potential would diverge to infinity at any finite distance, if it
were zero infinitely far away.
Q25.10
The smaller sphere will. In the solution to the example referred to, equation 1 states that each will
q
have the same ratio of charge to radius, . In this case, the charge density is a surface charge
r
q
, so the smaller-radius sphere will have the greater charge density.
density,
4π r 2
Q25.11
The main factor is the radius of the dome. One often overlooked aspect is also the humidity of the
air—drier air has a larger dielectric breakdown strength, resulting in a higher attainable electric
potential. If other grounded objects are nearby, the maximum potential might be reduced.
Q25.12
The intense—often oscillating—electric fields around high voltage lines is large enough to ionize the
air surrounding the cables. When the molecules recapture their electrons, they release that energy in
the form of light.
Q25.13
A sharp point in a charged conductor would imply a large electric field in that region. An electric
discharge could most easily take place at that sharp point.
Q25.14
Use a conductive box to shield the equipment. Any stray electric field will cause charges on the outer
surface of the conductor to rearrange and cancel the stray field inside the volume it encloses.
Q25.15
No charge stays on the inner sphere in equilibrium. If there were any, it would create an electric
field in the wire to push more charge to the outer sphere. All of the charge is on the outer sphere.
Therefore, zero charge is on the inner sphere and 10.0 µC is on the outer sphere.
Q25.16
The grounding wire can be touched equally well to any point on the sphere. Electrons will drain
away into the ground and the sphere will be left positively charged. The ground, wire, and sphere
are all conducting. They together form an equipotential volume at zero volts during the contact.
However close the grounding wire is to the negative charge, electrons have no difficulty in moving
within the metal through the grounding wire to ground. The ground can act as an infinite source or
sink of electrons. In this case, it is an electron sink.
SOLUTIONS TO PROBLEMS
Section 25.1
P25.1
Potential Difference and Electric Potential
∆V = −14.0 V
∆V =
W
,
Q
e
je
j
jb
g
and
Q = − N A e = − 6.02 × 10 23 1.60 × 10 −19 = −9.63 × 10 4 C
so
W = Q∆V = −9.63 × 10 4 C −14.0 J C = 1.35 MJ
e
Chapter 25
P25.2
53
a f
7.37 × 10 −17 = q 115
∆K = q ∆V
q = 6. 41 × 10 −19 C
P25.3
(a)
Energy of the proton-field system is conserved as the proton moves from high to low
potential, which can be defined for this problem as moving from 120 V down to 0 V.
K i + Ui + ∆Emech = K f + U f
0 + qV + 0 =
1
mv p2 + 0
2
e1.60 × 10
C 120 V
−19
ja
fFGH 1 V1 J⋅ C IJK = 12 e1.67 × 10
−27
j
kg v p2
v p = 1.52 × 10 5 m s
(b)
The electron will gain speed in moving the other way,
from Vi = 0 to V f = 120 V :
K i + Ui + ∆Emech = K f + U f
0+0+0=
0=
1
mv e2 + qV
2
jb
1
9.11 × 10 −31 kg v e2 + −1.60 × 10 −19 C 120 J C
2
e
j e
g
v e = 6.49 × 10 6 m s
P25.4
W = ∆K = − q∆V
0−
1
9.11 × 10 −31 kg 4.20 × 10 5 m s
2
e
je
j
2
e
j
= − −1.60 × 10 −19 C ∆V
From which, ∆V = −0.502 V .
Section 25.2
P25.5
(a)
Potential Difference in a Uniform Electric Field
We follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm).
∆U = − (work done)
∆U = − (work from origin to (20.0 cm, 0)) – (work from (20.0 cm, 0) to (20.0 cm, 50.0 cm))
Note that the last term is equal to 0 because the force is perpendicular to the displacement.
b g
e
jb
ga
f
∆U = − qEx ∆x = − 12.0 × 10 −6 C 250 V m 0.200 m = −6.00 × 10 −4 J
(b)
P25.6
E=
∆V =
∆U
6.00 × 10 −4 J
=−
= −50.0 J C = −50.0 V
q
12.0 × 10 −6 C
∆V 25.0 × 10 3 J C
=
= 1.67 × 10 6 N C = 1.67 MN C
d
1.50 × 10 −2 m
54
Electric Potential
P25.7
∆U = −
jLNMe1.40 × 10 m sj − e3.70 × 10
= e−1.60 × 10 j∆V
1
1
m v 2f − vi2 = − 9.11 × 10 −31 kg
2
2
e
j
e
+6.23 × 10 −18
∆U = q∆V :
5
2
6
ms
j OQP = 6.23 × 10
2
−18
J
−19
∆V = −38.9 V. The origin is at highest potential.
P25.8
jb
e
g
(a)
∆V = Ed = 5.90 × 10 3 V m 0.010 0 m = 59.0 V
(b)
1
mv 2f = q∆V :
2
ja f
1
9.11 × 10 −31 v 2f = 1.60 × 10 −19 59.0
2
e
j
e
v f = 4.55 × 10 6 m s
P25.9
z
z
z
B
C
B
A
A
C
VB − VA = − E ⋅ ds = − E ⋅ ds − E ⋅ ds
a
f z dy − aE cos 90.0°f
= a325fa0.800 f = +260 V
VB − VA = − E cos 180°
VB − VA
z
0 .500
0. 400
−0.300
−0 . 200
dx
FIG. P25.9
*P25.10
Assume the opposite. Then at some point A on some equipotential surface the electric field has a
nonzero component Ep in the plane of the surface. Let a test charge start from point A and move
z
B
some distance on the surface in the direction of the field component. Then ∆V = − E ⋅ ds is nonzero.
A
The electric potential charges across the surface and it is not an equipotential surface. The
contradiction shows that our assumption is false, that Ep = 0 , and that the field is perpendicular to
the equipotential surface.
P25.11
(a)
Arbitrarily choose V = 0 at 0. Then at other points
V = − Ex
and
U e = QV = −QEx .
Between the endpoints of the motion,
bK + U
s
+ Ue
g = bK + U
i
s
+ Ue
g
f
1 2
2QE
0 + 0 + 0 = 0 + kx max
− QEx max so x max =
.
2
k
(b)
At equilibrium,
∑ Fx = − Fs + Fe = 0 or
kx = QE .
So the equilibrium position is at x =
continued on next page
QE
.
k
FIG. P25.11
Chapter 25
(c)
d2x
∑ Fx = − kx + QE = m dt 2
The block’s equation of motion is
.
QE
QE
,
, or x = x ′ +
k
k
so the equation of motion becomes:
d 2 x + QE k
QE
d 2 x′
k
=−
−k x′ +
+ QE = m
x′ .
, or
2
k
m
dt
dt 2
x′ = x −
Let
FG
H
b
IJ
K
g
FG IJ
H K
This is the equation for simple harmonic motion a x′ = −ω 2 x ′
(d)
with
ω=
The period of the motion is then
T=
bK + U
g
b
+ U e i + ∆Emech = K + U s + U e
s
0 + 0 + 0 − µ k mgx max = 0 +
b
2 QE − µ k mg
x max =
P25.12
g
g
k
.
m
2π
ω
= 2π
m
.
k
f
1 2
kx max − QEx max
2
k
1
ayt 2
2
y f − yi = v yi t +
For the entire motion,
1
ayt 2
2
2mvi
− mg − qE = −
t
m 2 vi
−g
E=
q t
0 − 0 = vi t +
∑ Fy = ma y :
FG
H
d
IJ
K
2
2
= v yi
+ 2 a y y f − yi
v yf
For the upward flight:
FG
H
0 = vi2 + 2 −
2 vi
t
IJ by
K
so
ay = −
and
E=−
and
y max =
i
max
−0
g
FG
IJ
FG
IJ FG 1 v tIJ
z
H
K
H
KH 4 K
I L1
2.00 kg F 2b 20.1 m sg
O
∆V =
− 9.80 m s J M b 20.1 m sga 4.10 sfP =
G
4.10 s
5.00 × 10 C H
Q
KN4
∆V = −
FG
H
IJ
K
m 2 vi
− g j.
q t
1
vi t
4
y
ymax
E ⋅ dy = +
0
max
m 2 vi
m 2 vi
−g y
=
−g
q t
q t
0
i
2
−6
P25.13
2 vi
t
40.2 kV
Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length,
V = − Ed and U e = − λLEd .
(a)
aK + U f = a K + U f
i
0+0=
v=
(b)
f
1
µLv 2 − λLEd
2
2 λEd
µ
The same.
=
e
jb
ga
f=
2 40.0 × 10 −6 C m 100 N C 2.00 m
b0.100 kg mg
0.400 m s
55
56
Electric Potential
P25.14
Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the original position of
the charge is − E ⋅ s = − EL cos θ . At the final point a, V = − EL . Suppose the table is frictionless:
K +U i = K +U f
a
f a
f
0 − qEL cos θ =
v=
Section 25.3
P25.15
1
mv 2 − qEL
2
a
f
2 qEL 1 − cos θ
=
m
jb
e
fa
ga
2 2.00 × 10 −6 C 300 N C 1.50 m 1 − cos 60.0°
0.010 0 kg
f=
0.300 m s
Electric Potential and Potential Energy Due to Point Charges
e
je
j
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
q
=
= 1.44 × 10 −7 V .
r
1.00 × 10 −2 m
(a)
The potential at 1.00 cm is V1 = k e
(b)
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
q
The potential at 2.00 cm is V2 = k e =
= 0.719 × 10 −7 V .
r
2.00 × 10 −2 m
Thus, the difference in potential between the two points is ∆V = V2 − V1 = −7.19 × 10 −8 V .
(c)
The approach is the same as above except the charge is −1.60 × 10 −19 C . This changes the
sign of each answer, with its magnitude remaining the same.
That is, the potential at 1.00 cm is −1.44 × 10 −7 V .
The potential at 2.00 cm is −0.719 × 10 −7 V , so ∆V = V2 − V1 = 7.19 × 10 −8 V .
P25.16
(a)
Since the charges are equal and placed symmetrically, F = 0 .
(b)
Since F = qE = 0 , E = 0 .
(c)
q
V = 2 k e = 2 8.99 × 10 9 N ⋅ m 2 C 2
r
e
× 10 C I
jFGH 2.000.800
m JK
−6
V = 4.50 × 10 4 V = 45.0 kV
P25.17
(a)
(b)
E=
Q
4π ∈0 r 2
V=
Q
4π ∈0 r
r=
V
3 000 V
=
= 6.00 m
E 500 V m
V = −3 000 V =
Q=
Q
4π ∈0 6.00 m
a
−3 000 V
e8.99 × 10
9
V ⋅m C
f
j
a6.00 mf =
−2.00 µC
FIG. P25.16
Chapter 25
P25.18
Ex =
(a)
k e q1
x2
+
ke q2
ax − 2.00f
2
=0
Ex = k e
becomes
a
Dividing by k e ,
2 qx 2 = q x − 2.00
Therefore E = 0
when
f
2
F + q + −2 q I = 0 .
GH x ax − 2.00f JK
2
2
x 2 + 4.00 x − 4.00 = 0 .
x=
−4.00 ± 16.0 + 16.0
= −4.83 m .
2
(Note that the positive root does not correspond to a physically valid situation.)
V=
(b)
k e q1
k q
+ e 2 =0
x
2.00 − x
FG + q − 2 q IJ = 0 .
H x 2.00 − x K
2 qx = qa 2.00 − xf .
or
V = ke
when
x = 0.667 m
For x < 0
x = −2.00 m .
Again solving for x,
For 0 ≤ x ≤ 2.00 V = 0
and
P25.19
V = ∑k
i
−2 q
q
=
.
x
2−x
qi
ri
e
je
V = 8.99 × 10 9 7.00 × 10 −6
1
1 O
−1
−
+
jLMN 0.010
P
0 0.010 0 0.038 7 Q
V = −1.10 × 10 7 V = −11.0 MV
FIG. P25.19
P25.20
(a)
e
je
je
f
j
5.00 × 10 −9 C −3.00 × 10 −9 C 8.99 × 10 9 V ⋅ m C
qQ
U=
=
= −3.86 × 10 −7 J
4π ∈0 r
0.350 m
a
The minus sign means it takes 3.86 × 10 −7 J to pull the two charges apart from 35 cm to a
much larger separation.
(b)
V=
Q1
Q2
+
4π ∈0 r1 4π ∈0 r2
e5.00 × 10 Cje8.99 × 10
=
−9
0.175 m
V = 103 V
9
V ⋅m C
57
j + e−3.00 × 10 Cje8.99 × 10
−9
0.175 m
9
V ⋅m C
j
58
Electric Potential
P25.21
U e = q 4V1 + q 4V2 + q 4V3 = q 4
e
U e = 10.0 × 10 −6 C
je
2
FG 1 IJ FG q
H 4π ∈ K H r
0
1
+
1
8.99 × 10 9 N ⋅ m 2 C 2
q 2 q3
+
r2 r3
IJ
K
F 1
jGG 0.600 m + 0.1501 m + 0.600 m 1+ 0.150 m
a
f a
f
H
2
2
I
JJ
K
U e = 8.95 J
P25.22
V=
(a)
FG IJ
H K
k e q1 k e q 2
k q
+
=2 e
r1
r2
r
F e8.99 × 10 N ⋅ m C je2.00 × 10 Cj I
GG
JJ
1
00
+
0
500
.
m
.
m
a
a
f
f
H
K
9
V=2
2
−6
2
2
2
V = 3.22 × 10 4 V = 32.2 kV
e
je
j
U = qV = −3.00 × 10 −6 C 3.22 × 10 4 J C = −9.65 × 10 −2 J
(b)
P25.23
FIG. P25.22
U = U1 + U 2 + U 3 + U 4
b
g b
U = 0 + U12 + U 13 + U 23 + U 14 + U 24 + U 34
U =0+
U=
k eQ 2 k e Q 2
+
s
s
k eQ
s
2
FG 4 + 2 IJ =
H 2K
FG 1 + 1IJ + k Q FG 1 +
H 2 K s H
e
5.41
k eQ
s
2
g
1
2
IJ
K
+1
2
FIG. P25.23
FG
H
An alternate way to get the term 4 +
IJ is to recognize that there are 4 side pairs and 2 face
2K
2
diagonal pairs.
P25.24
Each charge creates equal potential at the center. The total potential is:
V =5
P25.25
(a)
LM k b− qg OP =
MN R PQ
e
−
5keq
.
R
Each charge separately creates positive potential everywhere. The total potential produced
by the three charges together is then the sum of three positive terms. There is no point
located at a finite distance from the charges, at which this total potential is zero.
(b)
V=
2k e q
ke q keq
+
=
a
a
a
Chapter 25
P25.26
59
Consider the two spheres as a system.
(a)
e j
m1 v1
m2
Conservation of momentum:
0 = m1 v1 i + m 2 v 2 − i or v 2 =
By conservation of energy,
0=
and
k e q1 q 2 k e q1 q 2 1
1 m12 v12
−
= m1 v12 +
2
2 m2
r1 + r2
d
v1 =
d
=
b g
k e − q1 q 2
1
1
m1 v12 + m 2 v 22 +
r1 + r2
2
2
FG 1 − 1 IJ
b
g H r + r dK
2b0.700 kg ge8.99 × 10 N ⋅ m C je 2 × 10 C je3 × 10 C j F
1
G
H
8 × 10
b0.100 kg gb0.800 kg g
2 m 2 k e q1 q 2
m1 m1 + m 2
1
2
9
v1 =
b g
k e − q1 q 2
2
−6
2
−6
−3
m
−
1
1.00 m
IJ
K
= 10.8 m s
v2 =
(b)
b
g
m1 v1 0.100 kg 10.8 m s
=
= 1.55 m s
0.700 kg
m2
If the spheres are metal, electrons will move around on them with negligible energy loss to
place the centers of excess charge on the insides of the spheres. Then just before they touch,
the effective distance between charges will be less than r1 + r2 and the spheres will really be
moving faster than calculated in (a) .
P25.27
Consider the two spheres as a system.
(a)
e j
Conservation of momentum:
0 = m 1 v1 i + m 2 v 2 − i
or
v2 =
By conservation of energy,
0=
and
k e q1 q 2 k e q1 q 2 1
1 m12 v12
−
= m1 v12 +
.
2
2 m2
r1 + r2
d
b g
k e − q1 q 2
v1 =
v2 =
(b)
m1 v1
.
m2
d
=
b g
k e − q1 q 2
1
1
m1 v12 + m 2 v 22 +
r1 + r2
2
2
IJ
K
2m k q q F 1
G
m bm + m g H r + r
2 m 2 k e q1 q 2
m 1 m1 + m 2
b
FG m IJ v
Hm K
1
2
1
=
FG 1
gHr +r
1
−
2
1
d
1 e 1 2
2
1
2
1
2
−
1
d
IJ
K
If the spheres are metal, electrons will move around on them with negligible energy loss to
place the centers of excess charge on the insides of the spheres. Then just before they touch,
the effective distance between charges will be less than r1 + r2 and the spheres will really be
moving faster than calculated in (a) .
60
Electric Potential
*P25.28
(a)
In an empty universe, the 20-nC charge can be placed at its location with no energy
investment. At a distance of 4 cm, it creates a potential
V1 =
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 20 × 10 −9 C
k e q1
=
= 4.50 kV .
0.04 m
r
To place the 10-nC charge there we must put in energy
e
je
j
U12 = q 2 V1 = 10 × 10 −9 C 4.5 × 10 3 V = 4.50 × 10 −5 J .
Next, to bring up the –20-nC charge requires energy
b
U 23 + U 13 = q3 V2 + q3 V1 = q 3 V2 + V1
g
e
= −20 × 10 −9 C 8.99 × 10 9 N ⋅ m 2 C 2
× 10 C 20 × 10 C I
+
jFGH 100.04
m
0.08 m JK
−9
−9
= −4.50 × 10 −5 J − 4.50 × 10 −5 J
The total energy of the three charges is
U12 + U 23 + U13 = −4.50 × 10 −5 J .
(b)
The three fixed charges create this potential at the location where the fourth is released:
e
V = V1 + V2 + V3 = 8.99 × 10 9 N ⋅ m 2 C 2
F
jGH
20 × 10 −9
0.04 2 + 0.03 2
+
10 × 10 −9 20 × 10 −9
−
0.03
0.05
I Cm
JK
V = 3.00 × 10 3 V
Energy of the system of four charged objects is conserved as the fourth charge flies away:
FG 1 mv + qV IJ = FG 1 mv + qV IJ
H2
K H2
K
1
0 + e 40 × 10 C je3.00 × 10 V j = e 2.00 × 10
2
2e1.20 × 10 Jj
= 3.46 × 10 m s
v=
2
2
i
f
−9
3
−13
j
kg v 2 + 0
−4
4
2 × 10 −13 kg
*P25.29
The original electrical potential energy is
U e = qV = q
ke q
.
d
In the final configuration we have mechanical equilibrium. The spring and electrostatic forces on
k q2
k q
each charge are − k 2d + q e 2 = 0 . Then k = e 3 . In the final configuration the total potential
18d
3d
a f a f
1 k q
+ qV =
a 2d f
2 18d
keq 4 keq2
=
. The missing energy must have become internal
3
3d 9 d
k q 2 4k q 2
energy, as the system is isolated: e = e + ∆Eint
d
9d
energy is
∆Eint =
1 2
kx
2
5 keq2
.
9 d
e
2
2
+q
Chapter 25
P25.30
af
V x =
(a)
b g
k +Q
k e Q1 k e Q 2
+
= e
+
r1
r2
x2 + a2
af
2 k eQ
kQ
= e
2
2
a
x +a
V x =
af
2
b k Q ag = b x ag
F 2
GG
H b x ag
2
b g
+ a − af
k e +Q
x2
I
JJ
+1K
2
V x
e
2
+1
FIG. P25.30(a)
bg
V y =
(b)
b g
b g
k e Q 1 k e Q 2 k e +Q k e −Q
+
=
+
r1
r2
y−a
y+a
b g k aQ FGH y a1− 1 − y a1+ 1 IJK
F 1 − 1 I
V b yg
= G
bk Q ag H y a − 1 y a + 1 JK
e
V y =
e
FIG. P25.30(b)
P25.31
V=
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 8.00 × 10 −9 C
kQ
k eQ
72.0 V ⋅ m
=
so r = e =
.
V
V
V
r
For V = 100 V , 50.0 V, and 25.0 V, r = 0.720 m, 1.44 m, and 2.88 m .
The radii are inversely proportional to the potential.
P25.32
Using conservation of energy for the alpha particle-nucleus system,
we have
K f + U f = K i + Ui .
But
Ui =
and
ri ≈ ∞.
Thus,
Ui = 0 .
Also
K f = 0 ( v f = 0 at turning point),
so
U f = Ki
ri
k e qα qgold
or
rmin =
k e qα qgold
rmin
2 k e qα qgold
mα vα2
=
=
1
mα vα2
2
e
ja fa fe
kg je 2.00 × 10
2 8.99 × 10 9 N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C
e6.64 × 10
−27
7
ms
j
2
j
2
= 2.74 × 10 −14 m = 27.4 fm .
61
62
Electric Potential
P25.33
P25.34
Using conservation of energy
k e eQ k e qQ 1
=
+ mv 2
we have:
2
r1
r2
2 k e eQ 1 1
−
m
r1 r2
FG
H
IJ
K
a2fe8.99 × 10
N ⋅ m 2 C 2 −1.60 × 10 −19 C 10 −9 C
which gives:
v=
or
v=
Thus,
v = 7.26 × 10 6 m s .
k e qi q j
9
je
9.11 × 10
e
, summed over all pairs of i , j where i ≠ j .
rij
2
2
2
FIG. P25.34
−6 2
Each charge moves off on its diagonal line. All charges have equal speeds.
∑ K +U i = ∑ K +U f
a
0+
f
a
2
2
4k e q
2k q
+ e
L
2L
f
F 1 I 4k q
= 4G mv J +
H 2 K 2L
2
FG 2 + 1 IJ k q = 2mv
H 2K L
F 1 IJ k q
v = G1 +
H 8 K mL
e
2
e
e
2
+
2k e q 2
2 2L
2
2
A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 × 6 = 12 face
diagonal pairs separated by 2s and 4 interior diagonal pairs separated 3s .
U=
Section 25.4
P25.37
2
2
9
P25.36
kg
jF 1 − 1 I .
GH 0.030 0 m 0.020 0 m JK
b g
L qb−2 qg + b−2 qgb3qg + b2 qgb3qg + qb2 qg + qb3qg + 2 qb−2 qg OP
U=k M
a
b
a
MN b
a +b
a + b PQ
L −2 − 6 + 6 + 2 + 3 − 4 OP
U=k q M
N 0.400 0.200 0.400 0.200 0.447 0.447 Q
L 4 − 4 − 1 OP = −3.96 J
U = e8.99 × 10 je6.00 × 10 j M
N 0.400 0.200 0.447 Q
U=∑
e
P25.35
je
−31
LM
N
OP
Q
keq2
k q2
12
4
+
= 22.8 e
12 +
s
s
2
3
Obtaining the Value of the Electric Field from the Electric Potential
b
g
V = a + bx = 10.0 V + −7.00 V m x
(a)
(b)
At x = 0 ,
V = 10.0 V
At x = 3.00 m ,
V = −11.0 V
At x = 6.00 m ,
V = −32.0 V
E=−
b
g
dV
= − b = − −7.00 V m = 7.00 N C in the + x direction
dx
Chapter 25
P25.38
k eQ
R
dV
= 0
Er = −
dr
(a)
For r < R
V=
(b)
For r ≥ R
V=
k eQ
r
kQ
kQ
dV
= − − e2 = e2
Er = −
dr
r
r
FG
H
P25.39
IJ
K
V = 5 x − 3 x 2 y + 2 yz 2
b
Evaluate E at 1, 0 , − 2
g
a fa f
af a f
−4yz = −4a0fa −2f = 0
+ E = a −5 f + a −5 f + 0 =
∂V
= −5 + 6 xy = −5 + 6 1 0 = −5
∂x
∂V
2
2
= +3 x 2 − 2 z 2 = 3 1 − 2 −2 = −5
Ey = −
∂y
Ex = −
Ez = −
∂V
=
∂z
E = E x2 + E y2
P25.40
2
2
z
2
7.07 N C
∆V
∆s
E A > EB since E =
(a)
2
a f
6−2 V
∆V
=−
= 200 N C down
2 cm
∆s
(b)
EB = −
(c)
The figure is shown to the right, with sample field lines
sketched in.
FIG. P25.40
P25.41
Ey = −
Ey =
Section 25.5
P25.42
LM
MMN
∂V
∂ k eQ
=−
ln
∂y
∂y
LM
MN
k eQ
1−
y
F
GG
H
2
+
y
y2
2
+ y2 +
+ y2
2
+ y2
OP
PQ =
I OP
JJ P
K PQ
k eQ
y
2
+ y2
Electric Potential Due to Continuous Charge Distributions
∆V = V2 R − V0 =
k eQ
a f
R2 + 2R
2
−
k eQ k eQ
=
R
R
FG 1 − 1IJ =
H 5 K
−0.553
k eQ
R
63
64
Electric Potential
P25.43
LM λ OP = C ⋅ FG 1 IJ =
N x Q m H mK
(a)
α =
(b)
V = ke
z
C
m2
z
LM
N
z
FG
H
L
dq
λdx
xdx
L
= ke
= k eα
= k eα L − d ln 1 +
r
r
d+x
d
0
IJ OP
KQ
FIG. P25.43
P25.44
V=
z
k e dq
= ke
r
z
αxdx
b
b2 + L 2 − x
g
2
L
−x.
2
Let z =
Then x =
V = k eα
L
− z , and dx = − dz
2
zb
ga f = − k αL
L 2 − z − dz
2
b +z
e
2
2
z
dz
2
b +z
2
+ k eα
z
zdz
2
b +z
2
=−
FH
LMF L I F L I OP
GMH 2 − xJK + GH 2 − xJK + b P + k α FGH L2 − xIJK + b
N
Q
L
O L FL I
k αL M L 2 − L + b L 2 g + b P
F LI
V=−
ln M
+ k α M G − LJ + b − G J
P
H
K
H 2K
2
2
M
MN L 2 + bL 2g + b PQ N
L
O
k αL M b + eL 4j − L 2 P
V= −
ln M
2
MN b + eL 4j + L 2 PPQ
L
2
k αL
V = − e ln
2
2
L
2
2
e
0
0
2
2
e
2
2
2
2
2
2
2
2
e
IK
k eαL
ln z + z 2 + b 2 + k eα z 2 + b 2
2
2
+ b2
OP
PQ
e
P25.45
z
1
4π ∈0
V = dV =
z
dq
r
All bits of charge are at the same distance from O.
So V =
P25.46
dV =
FG IJ e
H K
1
Q
= 8.99 × 10 9 N ⋅ m 2 C 2
4π ∈0 R
k e dq
z
b
a
−6
−1.51 MV .
where dq = σdA = σ 2π rdr
r 2 + x2
V = 2πσk e
.50 × 10 C I
=
jFGH −70.140
m π JK
rdr
2
r + x2
= 2π k eσ
LM
N
x2 + b2 − x2 + a2
OP
Q
FIG. P25.46
Chapter 25
P25.47
V = ke
z
z
a f
V = − k e λ ln − x
V = k e ln
Section 25.6
P25.48
z
z
3R
−R
dq
λdx
λds
λdx
= ke
+ ke
+ ke
−x
r
R
x
R
all charge
semicircle
−3 R
−R
−3 R
+
keλ
3R
π R + k e λ ln x R
R
a
3R
+ k e λ π + k e ln 3 = k e λ π + 2 ln 3
R
f
Electric Potential Due to a Charged Conductor
Substituting given values into V =
ke q
r
7.50 × 10 3 V =
e8.99 × 10
9
j
N ⋅ m2 C 2 q
0.300 m
.
Substituting q = 2.50 × 10 −7 C ,
N=
P25.49
(a)
2.50 × 10 −7 C
= 1.56 × 10 12 electrons .
1.60 × 10 −19 C e −
E= 0 ;
e
je
j
8.99 × 10 9 26.0 × 10 −6
ke q
=
= 1.67 MV
V=
0.140
R
(b)
e8.99 × 10 je26.0 × 10 j = 5.84 MN C
r
a0.200f
k q e8.99 × 10 je 26.0 × 10 j
=
= 1.17 MV
V=
E=
keq
2
−6
9
=
2
away
−6
9
e
(c)
E=
V=
R
0.200
keq
e8.99 × 10 je26.0 × 10 j =
=
a0.140f
R2
−6
9
ke q
= 1.67 MV
R
2
11.9 MN C away
65
66
Electric Potential
*P25.50
(a)
Both spheres must be at the same potential according to
where also
q1 + q 2 = 1.20 × 10 −6 C .
Then
q1 =
k e q1 k e q 2
=
r1
r2
q 2 r1
r2
q 2 r1
+ q 2 = 1.20 × 10 −6 C
r2
1.20 × 10 −6 C
= 0.300 × 10 −6 C on the smaller sphere
1 + 6 cm 2 cm
q2 =
q1 = 1.20 × 10 −6 C − 0.300 × 10 −6 C = 0.900 × 10 −6 C
V=
(b)
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 0.900 × 10 −6 C
k e q1
=
= 1.35 × 10 5 V
r1
6 × 10 −2 m
Outside the larger sphere,
E1 =
k e q1
r12
r=
V1
1.35 × 10 5 V
r=
r = 2. 25 × 10 6 V m away .
0.06 m
r1
Outside the smaller sphere,
1.35 × 10 5 V
r = 6.74 × 10 6 V m away .
0.02 m
E2 =
The smaller sphere carries less charge but creates a much stronger electric field than the
larger sphere.
Section 25.7
The Milliken Oil Drop Experiment
Section 25.8
Application of Electrostatistics
P25.51
(a)
Emax = 3.00 × 10 6 V m =
6
k eQ
a
r
2
=
f
FG IJ
HK
FG IJ
HK
k eQ 1
1
= Vmax
r r
r
Vmax = Emax r = 3.00 × 10 0.150 = 450 kV
(b)
P25.52
V=
k eQmax
r2
= Emax
RSor k Q
T r
e
max
= Vmax
UV
W
ke q
k q
V
and E = e2 . Since E = ,
r
r
r
(b)
r=
6.00 × 10 5 V
V
=
= 0.200 m and
E 3.00 × 10 6 V m
(a)
q=
Vr
= 13.3 µC
ke
Q max =
a
6
Emax r 2 3.00 × 10 0.150
=
ke
8.99 × 10 9
f
2
= 7.51 µC
Chapter 25
67
Additional Problems
P25.53
q q
U = qV = k e 1 2 = 8.99 × 10 9
r12
P25.54
(a)
e
a38fa54f 1.60 × 10
j a5.50 +e6.20f × 10 j
−19 2
−15
= 4.04 × 10 −11 J = 253 MeV
To make a spark 5 mm long in dry air between flat metal plates requires potential difference
e
je
j
∆V = Ed = 3 × 10 6 V m 5 × 10 −3 m = 1.5 × 10 4 V ~ 10 4 V .
The area of your skin is perhaps 1.5 m 2 , so model your body as a sphere with this surface
area. Its radius is given by 1.5 m 2 = 4π r 2 , r = 0.35 m . We require that you are at the
potential found in part (a):
(b)
ke q
r
V=
q=
a
f FG
H
1.5 × 10 4 V 0.35 m
Vr
J
=
k e 8.99 × 10 9 N ⋅ m 2 C 2 V ⋅ C
IJ FG N ⋅ m IJ
KH J K
q = 5.8 × 10 −7 C ~ 10 −6 C .
P25.55
P25.56
e
je
j
2
e
je
j
2
(a)
−19
9
k e q1 q 2 − 8.99 × 10 1.60 × 10
=
U=
r
0.052 9 × 10 −9
(b)
−19
9
k e q1 q 2 − 8.99 × 10 1.60 × 10
U=
=
r
2 2 0.052 9 × 10 −9
(c)
U=
e
j
= −4.35 × 10 −18 J = −27.2 eV
= −6.80 eV
k e q1 q 2 − k e e 2
=
= 0
r
∞
From Example 25.5, the potential created by the ring at the electron’s starting point is
Vi =
k eQ
x i2
+a
2
=
b
g
xi2
2
k e 2πλa
+a
while at the center, it is V f = 2π k e λ . From conservation of energy,
b g
v 2f
1
m e v 2f + − eV f
2
i
I
4π ek λ F
2e
a
1−
=
V −V i=
G
d
GH x + a JJK
m
m
4π e1.60 × 10 je8.99 × 10 je1.00 × 10 j F
0.200
GG1 −
=
9.11 × 10
H a0.100f + a0.200f
0 + − eVi =
e
f
d
e
i
2
i
e
−19
v 2f
v f = 1.45 × 10 7 m s
9
−31
2
−7
2
2
I
JJ
K
68
Electric Potential
*P25.57
b g = 2V
V0 − −V0
0
.
d
d
Assume the ball swings a small distance x to the right. It moves to a place where the voltage created
2V
by the plates is lower by − Ex = − 0 x . Its ground connection maintains it at V = 0 by allowing
d
2V x k q
2V xR
. Then the ball
charge q to flow from ground onto the ball, where − 0 + e = 0
q= 0
d
R
ked
The plates create uniform electric field to the right in the picture, with magnitude
feels electric force F = qE =
4V02 xR
ked 2
to the right. For equilibrium this must be balanced by the
horizontal component of string tension according to T cos θ = mg
tan θ =
4V02 xR
=
2
k e d mg
F
GH
k d 2 mg
x
for small x. Then V0 = e
L
4RL
I
JK
T sin θ =
12
4V02 xR
ked 2
.
If V0 is less than this value, the only equilibrium position of the ball is hanging straight down. If V0
exceeds this value the ball will swing over to one plate or the other.
P25.58
(a)
Take the origin at the point where we will find the potential. One ring, of width dx, has
Qdx
charge
and, according to Example 25.5, creates potential
h
k eQdx
dV =
.
h x2 + R2
The whole stack of rings creates potential
z
V=
dV =
all charge
(b)
z
d+h
d
k eQdx
h x2 + R2
=
FH
k eQ
ln x + x 2 + R 2
h
A disk of thickness dx has charge
IK
d+h
=
d
F
GG
H
a f
2
d + h + d + h + R2
k eQ
ln
h
d + d2 + R2
Qdx
Qdx
and charge-per-area
. According to
h
π R2h
Example 25.6, it creates potential
Qdx
π R2h
Integrating,
dV = 2π k e
V=
2 k eQ
2
z
W = Vdq
0
where V =
ke q
.
R
Therefore, W =
2
2
k eQ 2
.
2R
2
e
2
2
e
2
Q
P25.59
IK
x2 + R2 − x .
z R h FH x + R dx − xdxIK = 2Rk Qh LMN 12 x x + R
L
kQM
ad + hf ad + hf + R − d d + R − 2dh − h
R hM
NM
d+h
d
V=
FH
2
2
2
2
+
FH
IK OP
Q
F d + h + ad + hf + R I OP
lnG
GH d + d + R JJK PPQ
R2
x2
ln x + x 2 + R 2 −
2
2
2
2
+R
2
2
d+h
d
2
2
I
JJ
K
.
Chapter 25
P25.60
The positive plate by itself creates a field E =
69
36.0 × 10 −9 C m 2
σ
= 2.03 kN C away
=
2 ∈0 2 8.85 × 10 −12 C 2 N ⋅ m 2
e
j
from the + plate. The negative plate by itself creates the same size field and between the plates it is
in the same direction. Together the plates create a uniform field 4.07 kN C in the space between.
(a)
Take V = 0 at the negative plate. The potential at the positive plate is then
V −0=−
zb
12.0 cm
g
−4.07 kN C dx .
0
e
ja
f
The potential difference between the plates is V = 4.07 × 10 3 N C 0.120 m = 488 V .
(b)
FG 1 mv + qV IJ = FG 1 mv + qV IJ
H2
K H2
K
1
qV = e1.60 × 10
C ja 488 V f = mv
2
2
2
i
f
−19
(c)
v f = 306 km s
(d)
v 2f = vi2 + 2 a x f − xi
d
e3.06 × 10
5
ms
j
2
2
f
= 7.81 × 10 −17 J
i
a
f
= 0 + 2 a 0.120 m
a = 3.90 × 10 11 m s 2
P25.61
(e)
∑ F = ma = e1.67 × 10 −27
je
j
(f)
E=
(a)
VB − VA = − E ⋅ ds and the field at distance r from a uniformly
kg 3.90 × 10 11 m s 2 = 6.51 × 10 −16 N
F 6.51 × 10 −16 N
=
= 4.07 kN C
q 1.60 × 10 −19 C
z
B
A
charged rod (where r > radius of charged rod) is
E=
λ
2π ∈0 r
2keλ
.
r
=
In this case, the field between the central wire and the coaxial
cylinder is directed perpendicular to the line of charge so that
VB − VA = −
z
rb
ra
or
FG IJ
H K
r
2keλ
dr = 2 k e λ ln a ,
r
rb
FG r IJ
Hr K
∆V = 2 k e λ ln
continued on next page
a
b
.
FIG. P25.61
70
Electric Potential
(b)
From part (a), when the outer cylinder is considered to be at zero potential, the potential at a
distance r from the axis is
V = 2 k e λ ln
FG r IJ .
HrK
a
The field at r is given by
E=−
FG
H
r
∂V
= −2 k e λ
ra
∂r
b
∆V F 1 I
E=
GJ
lnbr r g H r K
a
P25.62
(a)
a
2
e
∆V
.
ln ra rb
But, from part (a), 2k e λ =
Therefore,
IJ FG − r IJ = 2k λ .
KH r K r
g
.
b
From Problem 61,
E=
∆V 1
.
ln ra rb r
b
g
We require just outside the central wire
5.50 × 10 6 V m =
50.0 × 10 3 V
ln 0.850 m rb
b
FG 1 IJ
gHr K
b
e110 m jr lnFGH 0.850r m IJK = 1 .
−1
or
b
b
We solve by homing in on the required value
a f
jr lnFGH 0.850r m IJK
rb m
e110 m
−1
b
0.0100
0.00100
0.00150
0.00145
0.00143
0.00142
4.89
0.740
1.05
1.017
1.005
0.999
b
Thus, to three significant figures,
rb = 1.42 mm .
(b)
At ra ,
E=
P25.63
z
r2
b
V2 − V1 = − E ⋅ dr = −
r1
V2 − V1 =
FG 1 IJ =
g H 0.850 m K
50.0 kV
ln 0.850 m 0.001 42 m
z
r2
r1
λ
2π ∈0 r
FG IJ
H K
r
−λ
ln 2
r1
2π ∈0
dr
9.20 kV m .
71
Chapter 25
*P25.64
Take the illustration presented with the problem as an initial picture.
No external horizontal forces act on the set of four balls, so its center of
mass stays fixed at the location of the center of the square. As the
charged balls 1 and 2 swing out and away from each other, balls 3 and 4
move up with equal y-components of velocity. The maximum-kineticenergy point is illustrated. System energy is conserved:
v
1
+
P25.65
CM
3
v
2
+
4
v
FIG. P25.64
keq2 keq 2 1
1
1
1
=
+ mv 2 + mv 2 + mv 2 + mv 2
3a
2
2
2
2
a
2k e q 2
= 2mv 2
3a
v
keq2
3 am
v=
g k rbqg + k br−2 qg
b
e
V x , y, z =
For the given charge distribution,
e
1
2
ax + Rf + y + z and r = x + y + z .
V b x , y , zg = 0
F1 2I
k qG − J = 0 , or 2r = r .
Hr r K
4a x + R f + 4y + 4z = x + y + z
F8 I
F4 I
x + y + z + G RJ x + a0fy + a0fz + G R J = 0 .
H3 K
H3 K
2
r1 =
where
The surface on which
is given by
e
1
2
which may be written in the form:
2
2
2
2
2
2
1
2
2
This gives:
2
2
2
2
2
2
2
2
2
b
[1]
g
The general equation for a sphere of radius a centered at x 0 , y 0 , z0 is:
bx − x g + by − y g + bz − z g − a = 0
x + y + z + b −2 x gx + b −2 y gy + b −2 z gz + e x
or
2
2
2
0
2
0
2
0
0
2
2
0
0
2
0
j
+ y 02 + z 02 − a 2 = 0 .
[2]
Comparing equations [1] and [2], it is seen that the equipotential surface for which V = 0 is indeed a
sphere and that:
−2 x 0 =
Thus, x 0 = −
4
8
R ; −2 y 0 = 0 ; −2 z 0 = 0 ; x 02 + y 02 + z 02 − a 2 = R 2 .
3
3
FG
H
IJ
K
4
16 4 2 4 2
R , y 0 = z 0 = 0 , and a 2 =
− R = R .
3
9 3
9
The equipotential surface is therefore a sphere centered at
FG − 4 R, 0 , 0IJ
H 3
K
, having a radius
2
R .
3
72
Electric Potential
P25.66
(a)
E A = 0 (no charge within)
From Gauss’s law,
EB = k e
e1.00 × 10 j = FG 89.9 IJ V m
e
j
Hr K
r
r
bq + q g = e8.99 × 10 j e−5.00 × 10 j = FG − 45.0 IJ V m
H r K
r
r
−8
qA
= 8.99 × 10 9
2
2
2
−9
EC = k e
VC = k e
(b)
A
bq
A
B
+ qB
r
∴ At r2 , V = −
9
2
g = e8.99 × 10 j e−5.00 × 10 j = FG − 45.0 IJ V
H r K
r
−9
9
45.0
= −150 V
0.300
Inside r2 , VB = −150 V +
∴ At r1 , V = −450 +
P25.67
2
FG
H
z
r
IJ FG −450 + 89.9 IJ V
K H
r K
89.9
1
1
=
dr = −150 + 89.9 −
2
r
0
.
300
r2 r
89.9
= +150 V so VA = +150 V .
0.150
From Example 25.5, the potential at the center of the ring is
kQ
Vi = e and the potential at an infinite distance from the ring is
R
V f = 0 . Thus, the initial and final potential energies of the point
charge-ring system are:
U i = QVi =
k eQ 2
R
FIG. P25.67
U f = QV f = 0 .
and
From conservation of energy,
K f + U f = K i + Ui
P25.68
or
k Q2
1
Mv 2f + 0 = 0 + e
R
2
giving
vf =
V = ke
z
a+L
a
λdx
x2 + b2
2 k eQ 2
MR
.
= k λ ln LM x + e x
N
e
2
+b
2
j OPQ
=
a
L a + L + aa + Lf
k λ ln M
MMN a + a + b
2
a +L
e
2
2
+ b2
OP
PPQ
Chapter 25
*P25.69
(a)
(b)
V=
keq ke q keq
−
=
r2 − r1
r1
r2
r1 r2
b
g
From the figure, for r >> a ,
r2 − r1 ≅ 2 a cos θ .
Then
V≅
Er = −
73
ke q
k p cos θ
2 a cos θ ≅ e 2
.
r1 r2
r
2 k e p cos θ
∂V
=
∂r
r3
FG IJ
H K
1 ∂
In spherical coordinates, the θ component of the gradient is
.
r ∂θ
Eθ = −
Therefore,
FIG. P25.69
FG IJ
H K
k p sin θ
1 ∂V
.
= e 3
r ∂θ
r
a f 2rk p
E a90°f = 0 ,
E a0°f = 0
k p
E a90°f =
.
r
For r >> a
E r 0° =
and
e
3
r
θ
and
θ
e
3
These results are reasonable for r >> a . Their directions are as shown in Figure 25.13 (c).
af
However, for r → 0 , E 0 → ∞. This is unreasonable, since r is not much greater than a if it
is 0.
(c)
V=
and
ex
k e py
2
+ y2
j
Ex = −
3 2
3 k e pxy
∂V
=
52
2
∂x
x + y2
e
j
e
2
2
∂V k e p 2 y − x
=
Ey = −
5 2
∂y
x2 + y2
e
j
j
74
Electric Potential
P25.70
Inside the sphere, Ex = Ey = Ez = 0 .
j IK
L
O
F 3I
So
E = − M0 + 0 + E a z G − J e x + y + z j
2 x fP = 3E a xze x
a
K
H
2
N
Q
∂V
∂
I
= − F V − E z + E a ze x + y + z j
E =−
K
∂y
∂y H
F 3I
E = − E a zG − J e x + y + z j
2 y = 3E a yze x + y + z j
H 2K
∂V
F 3I
E =−
= E − E a zG − J e x + y + z j
a 2 zf − E a e x + y + z j
H 2K
∂z
E = E + E a e 2 z − x − y je x + y + z j
Ex = −
Outside,
F
H
∂V
∂
=−
V0 − E0 z + E0 a 3 z x 2 + y 2 + z 2
∂x
∂x
x
0
y
0
y
3
0
z
0
z
P25.71
0
2
2
0
3
0
0
e
3
0
2
3
2
3
2
2
2
2
0
3
3
2
+ y2 + z2
j
−5 2
2 −5 2
2 −5 2
2
0
3
2
2
2 −3 2
2 −5 2
2
k e dq
For an element of area which is a ring of radius r and width dr, dV =
b
2
2 −3 2
0
2
2
2 −5 2
2
2
2 −5 2
−3 2
.
r 2 + x2
g
dq = σdA = Cr 2π rdr and
b
V = C 2π k e
gz
R
0
P25.72
L
b gMM
N
r 2 dr
= C π k e R R 2 + x 2 + x 2 ln
r 2 + x2
dU = Vdq where the potential V =
F
GH R +
x
R2 + x2
I OP
JK P
Q
.
ke q
.
r
e
j
The element of charge in a shell is dq = ρ (volume element) or dq = ρ 4π r 2 dr and the charge q in a
sphere of radius r is
z
r
q = 4πρ r 2 dr = ρ
0
F 4π r I .
GH 3 JK
3
Substituting this into the expression for dU, we have
FG k q IJ dq = k ρFG 4π r IJ FG 1 IJ ρe4π r dr j = k FG 16π IJ ρ r dr
HrK
H 3 K
H 3 KH r K
F 16π I ρ r dr = k F 16π I ρ R
U = z dU = k G
GH 15 JK
H 3 JK z
dU =
3
e
e
2
e
2
R
2
2
4
e
2
e
2
2 4
5
0
3 k eQ 2
4
But the total charge, Q = ρ π R 3 . Therefore, U =
.
5 R
3
Chapter 25
*P25.73
(a)
The whole charge on the cube is
3
q = 100 × 10 −6 C m3 0.1 m = 10 −7 C . Divide up the cube into
ja
e
f
64 or more elements. The little cube labeled a creates at P
ke q
. The others in the
potential
2
64 6.25 + 1.25 2 + 1.25 2 10 −2 m
horizontal row behind it contribute
e
keq
64 10 −2
F
G
mj H
1
8.75 2 + 3.125
+
1
11.25 2 + 3.125
1
+
13.75 2
I.
J
+ 3.125 K
d
c
b
a
P
The little cubes in the rows containing b and c add
e
2k e q
64 10
−2
Le6.25 + 1.25 + 3.75 j + e8.75 + 15.625j
mj MN
OP
+ e11.25 + 15.625j
+ e13.75 + 15.625 j
Q
2
2 −1 2
2
−1 2
2
2
e
keq
64 10 −2
LMe6.25
mj N
2
+ 28.125
The whole potential at P is
j
−1 2
−1 2
1.25 cm
−1 2
2
and the bits in row d make potential at P
e
+ … + 13.75 2 + 28.125
j OQP .
−1 2
8.987 6 × 10 9 Nm 2 × 10 −7 C
e
FIG. P25.73
j
C 2 64 10 −2 m
b1.580 190g4 =
8 876 V . If we use
more subdivisions of the large cube, we get the same answer to four digits.
(b)
75
A sphere centered at the same point would create potential
k e q 8.987 6 × 10 9 Nm 2 × 10 −7 C
=
= 8 988 V , larger by 112 V .
r
C2
10 −1 m
ANSWERS TO EVEN PROBLEMS
P25.2
6.41 × 10 −19 C
P25.4
−0.502 V
P25.6
1.67 MN C
P25.8
(a) 59.0 V ; (b) 4.55 Mm s
P25.10
see the solution
P25.12
40.2 kV
P25.14
0.300 m s
P25.16
(a) 0; (b) 0; (c) 45.0 kV
P25.18
(a) −4.83 m ; (b) 0.667 m and −2.00 m
P25.20
(a) −386 nJ ; (b) 103 V
P25.22
(a) 32.2 kV ; (b) −96.5 mJ
P25.24
−
P25.26
(a) 10.8 m s and 1.55 m s ; (b) greater
P25.28
(a) −45.0 µ J ; (b) 34.6 km s
P25.30
see the solution
P25.32
27.4 fm
P25.34
−3.96 J
P25.36
22.8
5k e q
R
keq 2
s
76
Electric Potential
k eQ
P25.60
P25.38
(a) 0; (b)
P25.40
(a) larger at A; (b) 200 N C down;
(c) see the solution
P25.42
−0.553
P25.44
L
k αL M
−
ln M
2
MN
r
radially outward
2
(d) 390 Gm s 2 toward the negative plate;
(e) 6.51 × 10 −16 N toward the negative plate;
(f) 4.07 kN C toward the negative plate
k eQ
R
e
LM
N
b
2
b2
+ eL 4j − L 2 OP
P
+ e L 4j + L 2 P
Q
2
P25.62
(a) 1.42 mm ; (b) 9. 20 kV m
P25.64
Fk q I
GH 3am JK
2
x2 + b 2 − x2 + a2
P25.66
OP
Q
P25.46
2π k eσ
P25.48
1.56 × 10 12 electrons
P25.50
(a) 135 kV ; (b) 2.25 MV m away from the
P25.54
(a) ~ 10 V ; (b) ~ 10
P25.56
14.5 Mm s
−6
C
P25.70
e
2
2
2
B
2
2
2
2
2
OP
I PP
JJ P
K PQ
2
e
j ;
E = 3E a yze x + y + z j
;
E a e2z − x − y j
E =E +
outside and
ex + y + z j
z
0
3
2
0
E = 0 inside
P25.72
2
Ex = 3E0 a 3 xz x 2 + y 2 + z 2
0
2
2
2
2
2
e
y
d + h + d + h + R2
k eQ
ln
h
d + d 2 + R2
2
FG 89.9 IJ V m radially
Hr K
F 45.0 IJ V m radially
outward; E = G −
H r K
outward;
89.9 I
F
(b) V = 150 V ; V = G −450 +
J V;
H
r K
F 45.0 IJ V
V = G−
H r K
L a + L + aa + Lf + b OP
k λ ln M
MMN a + a + b PPQ
(a) E A = 0 ; E B =
C
2
(a)
12
A
P25.68
F
a f IJ ;
GG
JK
H
LMad + hf ad + hf + R − d d + R
kQ M
F d + h + a d + hf + R
(b)
R h MM−2dh − h + R lnG
GH d + d + R
MN
P25.58
2
2
(a) 13.3 µC ; (b) 0.200 m
4
e
C
large sphere and 6.74 MV m away from
the small sphere
P25.52
(a) 488 V ; (b) 7.81 × 10 −17 J ; (c) 306 km s ;
3 k eQ 2
5 R
−5 2
2 −5 2
2
3
2
2
2
2
2 5 2
2
26
Capacitance and Dielectrics
CHAPTER OUTLINE
26.1
26.2
26.3
26.4
26.5
26.6
26.7
Q26.4
ANSWERS TO QUESTIONS
Definition of Capacitance
Calculating Capacitance
Combinations of
Capacitors
Energy Stored in a Charged
Capacitor
Capacitors with Dielectrics
Electric Dipole in an Electric
Field
An Atomic Description of
Dielectrics
Q26.1
Nothing happens to the charge if the wires are disconnected. If
the wires are connected to each other, charges in the single
conductor which now exists move between the wires and the
plates until the entire conductor is at a single potential and the
capacitor is discharged.
Q26.2
336 km. The plate area would need to be
Q26.3
The parallel-connected capacitors store more energy, since they
have higher equivalent capacitance.
1
m2 .
∈0
Seventeen combinations:
Individual
C1 , C 2 , C 3
Parallel
C 1 + C 2 + C 3 , C 1 + C 2 , C1 + C 3 , C 2 + C 3
FG 1 + 1 IJ + C , FG 1 + 1 IJ + C , FG 1 + 1 IJ + C
HC C K
HC C K
HC C K
FG 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ
HC +C C K HC +C C K HC +C C K
FG 1 + 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ
HC C C K HC C K HC C K HC C K
−1
Series-Parallel
1
−1
3
2
1
−1
2
3
−1
1
2
3
1
2
3
1
3
−1
1
3
−1
2
−1
Series
2
2
3
−1
1
2
1
−1
2
3
1
−1
3
Q26.5
This arrangement would decrease the potential difference between the plates of any individual
capacitor by a factor of 2, thus decreasing the possibility of dielectric breakdown. Depending on the
application, this could be the difference between the life or death of some other (most likely more
expensive) electrical component connected to the capacitors.
Q26.6
No—not just using rules about capacitors in series or in parallel. See Problem 72 for an example. If
connections can be made to a combination of capacitors at more than two points, the combination
may be irreducible.
77
78
Capacitance and Dielectrics
Q26.7
A capacitor stores energy in the electric field between the plates. This is most easily seen when using
a “dissectable” capacitor. If the capacitor is charged, carefully pull it apart into its component pieces.
One will find that very little residual charge remains on each plate. When reassembled, the capacitor
is suddenly “recharged”—by induction—due to the electric field set up and “stored” in the
dielectric. This proves to be an instructive classroom demonstration, especially when you ask a
student to reconstruct the capacitor without supplying him/her with any rubber gloves or other
insulating material. (Of course, this is after they sign a liability waiver).
Q26.8
The work you do to pull the plates apart becomes additional electric potential energy stored in the
capacitor. The charge is constant and the capacitance decreases but the potential difference increases
1
to drive up the potential energy Q∆V . The electric field between the plates is constant in strength
2
but fills more volume as you pull the plates apart.
Q26.9
A capacitor stores energy in the electric field inside the dielectric. Once the external voltage source is
removed—provided that there is no external resistance through which the capacitor can
discharge—the capacitor can hold onto this energy for a very long time. To make the capacitor safe
to handle, you can discharge the capacitor through a conductor, such as a screwdriver, provided that
you only touch the insulating handle. If the capacitor is a large one, it is best to use an external
resistor to discharge the capacitor more slowly to prevent damage to the dielectric, or welding of the
screwdriver to the terminals of the capacitor.
Q26.10
The work done, W = Q∆V , is the work done by an external agent, like a battery, to move a charge
through a potential difference, ∆V . To determine the energy in a charged capacitor, we must add
the work done to move bits of charge from one plate to the other. Initially, there is no potential
difference between the plates of an uncharged capacitor. As more charge is transferred from one
plate to the other, the potential difference increases as shown in Figure 26.12, meaning that more
work is needed to transfer each additional bit of charge. The total work is the area under the curve of
1
Figure 26.12, and thus W = Q∆V .
2
Q26.11
Energy is proportional to voltage squared. It gets four times larger.
Q26.12
Let C = the capacitance of an individual capacitor, and C s represent the equivalent capacitance of the
group in series. While being charged in parallel, each capacitor receives charge
e
ja
f
Q = C∆Vcharge = 500 × 10 −4 F 800 V = 0.400 C .
While being discharged in series,
(or 10 times the original voltage).
∆Vdischarge =
Q
Q
0. 400 C
=
=
= 8.00 kV
C s C 10 5.00 × 10 −5 F
Q26.13
Put a material with higher dielectric strength between the plates, or evacuate the space between the
plates. At very high voltages, you may want to cool off the plates or choose to make them of a
different chemically stable material, because atoms in the plates themselves can ionize, showing
thermionic emission under high electric fields.
Q26.14
The potential difference must decrease. Since there is no external power supply, the charge on the
capacitor, Q, will remain constant—that is assuming that the resistance of the meter is sufficiently
large. Adding a dielectric increases the capacitance, which must therefore decrease the potential
difference between the plates.
Q26.15
Each polar molecule acts like an electric “compass” needle, aligning itself with the external electric
field set up by the charged plates. The contribution of these electric dipoles pointing in the same
direction reduces the net electric field. As each dipole falls into a configuration of lower potential
energy it can contribute to increasing the internal energy of the material.
Chapter 26
79
Q26.16
The material of the dielectric may be able to support a larger electric field than air, without breaking
down to pass a spark between the capacitor plates.
Q26.17
The dielectric strength is a measure of the potential difference per unit length that a dielectric can
withstand without having individual molecules ionized, leaving in its wake a conducting path from
plate to plate. For example, dry air has a dielectric strength of about 3 MV/m. The dielectric constant
in effect describes the contribution of the electric dipoles of the polar molecules in the dielectric to
the electric field once aligned.
Q26.18
In water, the oxygen atom and one hydrogen atom considered alone have an electric dipole moment
that points from the hydrogen to the oxygen. The other O-H pair has its own dipole moment that
points again toward the oxygen. Due to the geometry of the molecule, these dipole moments add to
have a non-zero component along the axis of symmetry and pointing toward the oxygen.
A non-polarized molecule could either have no intrinsic dipole moments, or have dipole
moments that add to zero. An example of the latter case is CO 2 . The molecule is structured so that
each CO pair has a dipole moment, but since both dipole moments have the same magnitude and
opposite direction—due to the linear geometry of the molecule—the entire molecule has no dipole
moment.
Q26.19
Heating a dielectric will decrease its dielectric constant, decreasing the capacitance of a capacitor.
When you heat a material, the average kinetic energy per molecule increases. If you refer back to the
answer to Question 26.15, each polar molecule will no longer be nicely aligned with the applied
electric field, but will begin to “dither”—rock back and forth—effectively decreasing its contribution
to the overall field.
Q26.20
The primary choice would be the dielectric. You would want to chose a dielectric that has a large
dielectric constant and dielectric strength, such as strontium titanate, where κ ≈ 233 (Table 26.1). A
convenient choice could be thick plastic or mylar. Secondly, geometry would be a factor. To
maximize capacitance, one would want the individual plates as close as possible, since the
capacitance is proportional to the inverse of the plate separation—hence the need for a dielectric
with a high dielectric strength. Also, one would want to build, instead of a single parallel plate
capacitor, several capacitors in parallel. This could be achieved through “stacking” the plates of the
capacitor. For example, you can alternately lay down sheets of a conducting material, such as
aluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none of
the conducting sheets are in contact with their next neighbors, connect every other plate together.
Figure Q26.20 illustrates this idea.
Dielectric
Conductor
FIG. Q26.20
This technique is often used when “home-brewing” signal capacitors for radio applications, as
they can withstand huge potential differences without flashover (without either discharge between
plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich
together flexible materials such as aluminum roof flashing and thick plastic, so the whole product
can be rolled up into a “capacitor burrito” and placed in an insulating tube, such as a PVC pipe, and
then filled with motor oil (again to prevent flashover).
80
Capacitance and Dielectrics
SOLUTIONS TO PROBLEMS
Section 26.1
P26.1
P26.2
P26.4
e
ja
f
e
ja
f
(a)
Q = C∆V = 4.00 × 10 −6 F 12.0 V = 4.80 × 10 −5 C = 48.0 µC
(b)
Q = C∆V = 4.00 × 10 −6 F 1.50 V = 6.00 × 10 −6 C = 6.00 µC
(a)
C=
(b)
∆V =
Section 26.2
P26.3
Definition of Capacitance
E=
Q 10.0 × 10 −6 C
=
= 1.00 × 10 −6 F = 1.00 µF
∆V
10.0 V
Q 100 × 10 −6 C
=
= 100 V
C 1.00 × 10 −6 F
Calculating Capacitance
keq
r2
e4.90 × 10 N Cja0.210 mf
e8.99 × 10 N ⋅ m C j
4
q=
:
2
9
2
= 0.240 µC
q 0.240 × 10 −6
=
= 1.33 µC m 2
A 4π 0.120 2
(a)
σ=
(b)
C = 4π ∈0 r = 4π 8.85 × 10 −12 0.120 = 13.3 pF
(a)
C = 4π ∈0 R
C
R=
= k eC = 8.99 × 10 9 N ⋅ m 2 C 2 1.00 × 10 −12 F = 8.99 mm
4π ∈0
a
f
ja
e
f
e
P26.5
2
je
e
je
j
j=
4π 8.85 × 10 −12 C 2 2.00 × 10 −3 m
(b)
C = 4π ∈0 R =
(c)
Q = CV = 2.22 × 10 −13 F 100 V = 2.22 × 10 −11 C
(a)
Q1 R1
=
Q 2 R2
ja
e
FG
H
Q1 + Q 2 = 1 +
Q 2 = 2.00 µC
(b)
N ⋅m
V1 = V2 =
2
0. 222 pF
f
IJ
K
R1
Q 2 = 3.50Q 2 = 7.00 µC
R2
Q1 = 5.00 µC
Q1 Q 2
5.00 µC
=
=
= 8.99 × 10 4 V = 89.9 kV
−1
9
C1 C 2
8.99 × 10 m F
0.500 m
e
j a
f
Chapter 26
P26.6
a fe
je
j
−12
3
2
κ ∈0 A 1.00 8.85 × 10 C 1.00 × 10 m
C=
=
d
N ⋅ m 2 800 m
a
f
2
= 11.1 nF
The potential between ground and cloud is
ja
e
f
∆V = Ed = 3.00 × 10 6 N C 800 m = 2.40 × 10 9 V
a f e
je
j
Q = C ∆V = 11.1 × 10 −9 C V 2.40 × 10 9 V = 26.6 C
P26.7
∆V = Ed
(a)
E=
E=
(b)
20.0 V
1.80 × 10 −3 m
= 11.1 kV m
σ
∈0
e
je
j
σ = 1.11 × 10 4 N C 8.85 × 10 −12 C 2 N ⋅ m 2 = 98.3 nC m 2
e
jb
je
g
(c)
8.85 × 10 −12 C 2 N ⋅ m 2 7.60 cm 2 1.00 m 100 cm
∈0 A
=
C=
d
1.80 × 10 −3 m
(d)
∆V =
Q
C
a
fe
j
Q = 20.0 V 3.74 × 10 −12 F = 74.7 pC
P26.8
C=
κ ∈0 A
= 60.0 × 10 −15 F
d
a fe
je
−12
21.0 × 10 −12
κ ∈0 A 1 8.85 × 10
=
C
60.0 × 10 −15
d = 3.10 × 10 −9 m = 3.10 nm
d=
P26.9
Q=
d=
a f
∈0 A
∆V
d
a f=
∈0 ∆V
σ
j
a f
∈ ∆V
Q
=σ = 0
A
d
e8.85 × 10
e30.0 × 10
−9
−12
ja
C 2 N ⋅ m 2 150 V
je
f
C cm 2 1.00 × 10 4 cm 2 m 2
j
= 4.42 µm
2
= 3.74 pF
81
82
Capacitance and Dielectrics
P26.10
With θ = π , the plates are out of mesh and the overlap area is zero. With
π R2
. By proportion, the
2
π −θ R2
effective area of a single sheet of charge is
2
When there are two plates in each comb, the number of adjoining
sheets of positive and negative charge is 3, as shown in the sketch. When
there are N plates on each comb, the number of parallel capacitors is 2 N − 1
and the total capacitance is
θ = 0 , the overlap area is that of a semi-circle,
a
a2 N − 1f ∈ aπ − θ f R
A
=
f ∈distance
d 2
a
C = 2N − 1
P26.11
(a)
(b)
C=
0
0
effective
A
2 k e ln
f
2
50.0
c h 2e8.99 × 10 j lnc h =
F bI
∆V = 2 k λ lnG J
Method 1:
H aK
b
a
=
9
7. 27
2.58
2
a2 N − 1f ∈ aπ − θ fR
0
=
FIG. P26.10
2
.
d
2.68 nF
e
q 8.10 × 10 −6 C
=
= 1.62 × 10 −7 C m
A
50.0 m
7.27
= 3.02 kV
∆V = 2 8.99 × 10 9 1.62 × 10 −7 ln
2.58
λ=
e
Method 2:
P26.12
∆V =
j FGH
je
IJ
K
Q 8.10 × 10 −6
=
= 3.02 kV
C 2.68 × 10 −9
Let the radii be b and a with b = 2 a . Put charge Q on the inner conductor and – Q on the outer.
Electric field exists only in the volume between them. The potential of the inner sphere is Va =
that of the outer is Vb =
k eQ
. Then
b
FG
H
Va − Vb =
k e Q k eQ
Q
b−a
−
=
a
b
4π ∈0 ab
Here C =
4π ∈0 2 a 2
= 8π ∈0 a
a
IJ and C = Q
K
V −V
a
a=
4π ∈0 ab
.
b−a
C
.
8π ∈0
Volume =
The intervening volume is
b
=
Volume =
FG
H
b
ga
f
e
7 20.0 × 10 −6
384π
2
e8.85 × 10
(a)
C=
0.070 0 0.140
ab
=
= 15.6 pF
ke b − a
8.99 × 10 9 0.140 − 0.070 0
(b)
C=
Q
∆V
a f e
∆V =
jb
IJ FG IJ
K H K
C N ⋅ mj
= 2.13 × 10
C N ⋅m j
C3
4 3 4 3
4
4
7C 3
π b − π a = 7 π a3 = 7 π 3 3 3 =
3
3
3
3
8 π ∈0 384π 2 ∈30
The outer sphere is 360 km in diameter.
P26.13
k eQ
;
a
g
Q 4.00 × 10 −6 C
=
= 256 kV
C 15.6 × 10 −12 F
−12
3
2
2
2 3
16
m3 .
Chapter 26
P26.14
P26.15
∑ Fy = 0 :
T cos θ − mg = 0
∑ Fx = 0 :
T sin θ − Eq = 0
Dividing,
tan θ =
so
E=
and
∆V = Ed =
mg
tan θ
q
mgd tan θ
.
q
e
je
j
C = 4π ∈0 R = 4π 8.85 × 10 −12 C N ⋅ m 2 6.37 × 10 6 m = 7.08 × 10 −4 F
Section 26.3
P26.16
Eq
mg
(a)
Combinations of Capacitors
Capacitors in parallel add. Thus, the equivalent capacitor has a value of
C eq = C1 + C 2 = 5.00 µF + 12.0 µF = 17.0 µF .
(b)
The potential difference across each branch is the same and equal to the voltage of the
battery.
∆V = 9.00 V
(c)
b
f
ga
Q5 = C∆V = 5.00 µF 9.00 V = 45.0 µC
b
ga
f
and Q12 = C∆V = 12.0 µF 9.00 V = 108 µC
P26.17
(a)
In series capacitors add as
1
1
1
1
1
=
+
=
+
C eq C 1 C 2 5.00 µF 12.0 µF
(c)
and
C eq = 3.53 µF .
The charge on the equivalent capacitor is
Q eq = C eq ∆V = 3.53 µF 9.00 V = 31.8 µC .
b
ga
f
Each of the series capacitors has this same charge on it.
(b)
So
Q1 = Q 2 = 31.8 µC .
The potential difference across each is
∆V1 =
Q1 31.8 µC
=
= 6.35 V
C1 5.00 µF
and
∆V2 =
Q 2 31.8 µC
=
= 2.65 V .
C 2 12.0 µF
83
84
Capacitance and Dielectrics
P26.18
The circuit reduces first according to the rule for capacitors in
series, as shown in the figure, then according to the rule for
capacitors in parallel, shown below.
FG
H
C eq = C 1 +
P26.19
IJ
K
1 1
11
+
= C = 1.83C
2 3
6
FIG. P26.18
C p = C1 + C 2
1
1
1
=
+
C s C1 C 2
Substitute C 2 = C p − C1
C p − C1 + C1
1
1
1
=
+
=
.
C s C1 C p − C1 C1 C p − C1
Simplifying,
C 12 − C 1C p + C pC s = 0 .
C1 =
⇒
C p ± C p2 − 4C p C s
2
e
=
j
1
1 2
Cp ±
C p − C pCs
2
4
We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus sign,
we would get the same two answers with their names interchanged.)
C1 =
C 2 = C p − C1 =
P26.20
b
g
b
1
1 2
1
1
Cp +
C p − C p C s = 9.00 pF +
9.00 pF
2
4
2
4
b
g − b9.00 pFgb2.00 pFg =
2
6.00 pF
g
1
1 2
1
Cp −
C p − C p C s = 9.00 pF − 1.50 pF = 3.00 pF
2
4
2
C p = C1 + C 2
and
1
1
1
=
+
.
C s C1 C 2
Substitute
C 2 = C p − C1 :
Simplifying,
and
C p − C1 + C1
1
1
1
=
+
=
.
C s C1 C p − C1 C1 C p − C1
e
j
C 12 − C 1C p + C pC s = 0
C1 =
C p ± C p2 − 4C pC s
2
=
1
1 2
Cp +
C p − C pCs
2
4
where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values for
the capacitances, with the names reversed).
Then, from
C 2 = C p − C1
C2 =
1
1 2
Cp −
C p − C pC s .
2
4
Chapter 26
P26.21
(a)
85
1
1
1
=
+
C s 15.0 3.00
C s = 2.50 µF
C p = 2.50 + 6.00 = 8.50 µF
C eq =
(b)
FG 1 + 1 IJ
H 8.50 µF 20.0 µF K
b
−1
ga
= 5.96 µF
f
Q = C∆V = 5.96 µF 15.0 V = 89.5 µC on 20.0 µF
Q 89.5 µC
=
= 4.47 V
C 20.0 µF
15.0 − 4.47 = 10.53 V
∆V =
b
f
ga
Q = C∆V = 6.00 µF 10.53 V = 63.2 µC on 6.00 µF
89.5 − 63.2 = 26.3 µC on 15.0 µF and 3.00 µF
*P26.22
FIG. P26.21
(a)
Capacitors 2 and 3 are in parallel and present equivalent capacitance 6C. This is in series
−1
1
1
= 2C .
with capacitor 1, so the battery sees capacitance
+
3C 6C
(b)
If they were initially unchanged, C 1 stores the same charge as C 2 and C 3 together. With
LM
N
OP
Q
greater capacitance, C 3 stores more charge than C 2 . Then Q1 > Q3 > Q 2 .
b
g
(c)
The C 2||C 3 equivalent capacitor stores the same charge as C 1 . Since it has greater
Q
implies that it has smaller potential difference across it than C 1 . In
capacitance, ∆V =
C
parallel with each other, C 2 and C 3 have equal voltages: ∆V1 > ∆V2 = ∆V3 .
(d)
If C 3 is increased, the overall equivalent capacitance increases. More charge moves through
the battery and Q increases. As ∆V1 increases, ∆V2 must decrease so Q 2 decreases. Then
Q 3 must increase even more: Q3 and Q1 increase; Q 2 decreases .
P26.23
Q
so
∆V
Q = 120 µC
and
6.00 × 10 −6 =
C=
and
or
Q1 = 120 µC − Q 2
Q
∆V = :
C
120 − Q 2
Q
= 2
6.00
3.00
a3.00fb120 − Q g = a6.00fQ
2
360
= 40.0 µC
Q2 =
9.00
Q
20.0
120 − Q 2 Q 2
=
C1
C2
2
Q1 = 120 µC − 40.0 µC = 80.0 µC
FIG. P26.23
86
Capacitance and Dielectrics
P26.24
In series , to reduce the effective capacitance:
(a)
1
1
1
=
+
µ
µ
32.0 F 34.8 F C s
1
= 398 µF
Cs =
2.51 × 10 −3 µF
In parallel , to increase the total capacitance:
(b)
29.8 µF + C p = 32.0 µF
C p = 2.20 µF
P26.25
100
100
=
1
+ C +" n C
nC =
1
1
C + C
n capacitors
nC =
*P26.26
100C
so n 2 = 100 and n = 10
n
For C 1 connected by itself, C 1 ∆V = 30.8 µC where ∆V is the battery voltage: ∆V =
For C 1 and C 2 in series:
F
GH 1 C
30.8 µC
.
C1
I
JK
1
∆V = 23.1 µC
1 + 1 C2
substituting,
30.8 µC 23.1 µC 23.1 µC
=
+
C1
C1
C2
C 1 = 0.333C 2 .
For C 1 and C 3 in series:
F
GH 1 C
I
JK
1
∆V = 25.2 µC
+
1 1 C3
30.8 µC 25. 2 µC 25.2 µC
=
+
C1
C1
C3
For all three:
Q=
F
GH 1 C
C 1 = 0.222C 3 .
I
JK
C 1 ∆V
1
30.8 µC
∆V =
=
= 19.8 µC .
1
1
1
1
0
C
C
C
C
C
C
.333 + 0.222
+
+
+
+
+
1
2
3
1
2
1
3
This is the charge on each one of the three.
P26.27
FG 1 + 1 IJ = 3.33 µF
H 5.00 10.0 K
= 2a3.33f + 2.00 = 8.66 µF
= 2a10.0f = 20.0 µF
F 1 + 1 IJ = 6.04 µF
=G
H 8.66 20.0 K
Cs =
C p1
C p2
−1
−1
C eq
FIG. P26.27
Chapter 26
P26.28
a f e
ja
f
Q eq = C eq ∆V = 6.04 × 10 −6 F 60.0 V = 3.62 × 10 −4 C
Q eq
Q p1 = Q eq , so ∆Vp1 =
e
C p1
=
3.62 × 10 −4 C
= 41.8 V
8.66 × 10 −6 F
ja
j e
f
Q 3 = C 3 ∆Vp1 = 2.00 × 10 −6 F 41.8 V = 83.6 µC
P26.29
Cs =
FG 1 + 1 IJ
H 5.00 7.00 K
−1
= 2.92 µF
C p = 2.92 + 4.00 + 6.00 = 12.9 µF
FIG. P26.29
*P26.30
According to the suggestion, the combination of
capacitors shown is equivalent to
Then
1
1
1
1 C + C0 + C0 + C + C0
=
+
+
=
C C0 C + C0 C0
C0 C + C0
b
+ C 02
C 0C
g
2
= 2C + 3C 0 C
2
2C + 2C 0 C − C 02 = 0
C=
FIG. P26.30
e j
−2C 0 ± 4C 02 + 4 2C 02
4
Only the positive root is physical
C=
Section 26.4
P26.31
P26.32
C0
2
e
j
3 −1
Energy Stored in a Charged Capacitor
a f
2
=
1
3.00 µF 12.0 V
2
a f
2
=
1
3.00 µF 6.00 V
2
(a)
U=
1
C ∆V
2
(b)
U=
1
C ∆V
2
U=
b
ga
f
2
= 216 µJ
b
ga
f
2
= 54.0 µJ
1
C∆V 2
2
∆V =
2U
=
C
a
2 300 J
f
30 × 10 −6 C V
= 4. 47 × 10 3 V
87
88
Capacitance and Dielectrics
P26.33
U=
a f
1
C ∆V
2
2
The circuit diagram is shown at the right.
C p = C1 + C 2 = 25.0 µF + 5.00 µF = 30.0 µF
(a)
U=
ja f
1
30.0 × 10 −6 100
2
e
2
= 0.150 J
F 1 + 1 IJ = FG 1 + 1 IJ
C =G
H C C K H 25.0 µF 5.00 µF K
1
U = C a ∆V f
2
2a0.150f
2U
=
= 268 V
∆V =
−1
(b)
s
1
−1
= 4.17 µF
2
2
4.17 × 10 −6
C
P26.34
Use U =
∈ A
1 Q2
and C = 0 .
2 C
d
If d 2 = 2d1 , C 2 =
*P26.35
e
je
j
(a)
Q = C∆V = 150 × 10 −12 F 10 × 10 3 V = 1.50 × 10 −6 C
(b)
U=
a f
1
C ∆V
2
u=
2
2U
=
C
∆V =
P26.36
1
C 1 . Therefore, the stored energy doubles .
2
e
2 250 × 10 −6 J
150 × 10
−12
F
j=
U 1
= ∈0 E 2
V 2
1.00 × 10 −7 1
= 8.85 × 10 −12 3 000
V
2
jb
e
g
2
e
V = 2.51 × 10 −3 m 3 = 2.51 × 10 −3 m 3
P26.37
1.83 × 10 3 V
z
LI
J=
jFGH 1 000
m K
3
W = U = Fdx
so F =
F I
GH JK
F
GH
I
JK
dU d Q 2
d
Q2x
Q2
=
=
=
2 ∈0 A
dx dx 2C
dx 2 ∈0 A
2.51 L
FIG. P26.33
Chapter 26
P26.38
With switch closed, distance d ′ = 0.500d and capacitance C ′ =
a f
a f e
ja
f
(a)
Q = C ′ ∆V = 2C ∆V = 2 2.00 × 10 −6 F 100 V = 400 µC
(b)
The force stretching out one spring is
a f
4C 2 ∆V
Q2
F=
=
2 ∈0 A
2 ∈0 A
a f = 2Ca∆V f
=
∈
A
b dgd d
2
2C 2 ∆ V
2
k=
a f FG 4 IJ = 8Ca∆V f
H dK d
F 2C ∆V
=
x
d
2
2
=
2
H ET =
The energy transferred is
2
.
0
One spring stretches by distance x =
P26.39
∈0 A 2 ∈0 A
=
= 2C .
d′
d
d
, so
4
ja
e
8 2.00 × 10 −6 F 100 V
e
8.00 × 10 −3 m
a
j
2
fe
f
2
= 2.50 kN m .
1
1
Q∆V = 50.0 C 1.00 × 10 8 V = 2.50 × 10 9 J
2
2
j
and 1% of this (or ∆Eint = 2.50 × 10 7 J ) is absorbed by the tree. If m is the amount of water boiled
away,
*P26.40
b
f e
ga
∆Eint = m 4 186 J kg⋅° C 100° C − 30.0° C + m 2.26 × 10 6 J kg = 2.50 × 10 7 J
giving
m = 9.79 kg .
a f
1
C ∆V
2
2
+
a f
1
C ∆V
2
2
a f
= C ∆V
2
(a)
U=
(b)
The altered capacitor has capacitance C ′ =
a f a f a f C2 a∆V ′f
C ∆V + C ∆V = C ∆V ′ +
FG
H
1
4 ∆V
U′ = C
2
3
(c)
(d)
P26.41
j
then
IJ
K
2
FG
H
11
4∆V
+
C
22
3
IJ
K
2
=
C
. The total charge is the same as before:
2
∆V ′ =
a ∆V f
4C
4 ∆V
.
3
2
3
The extra energy comes from work put into the system by the agent pulling the capacitor
plates apart.
a f
U=
1
C ∆V
2
U=
1 R
2 ke
FG
H
2
where C = 4π ∈0 R =
IJ FG k Q IJ
KH R K
e
2
=
k eQ 2
2R
kQ
kQ
R
and ∆V = e − 0 = e
R
R
ke
89
90
Capacitance and Dielectrics
*P26.42
(a)
b
g
2
q12
1 q12 1 q 22 1
1 Q − q1
The total energy is U = U 1 + U 2 =
+
=
+
.
2 C 1 2 C 2 2 4π ∈0 R1 2 4π ∈0 R 2
dU
= 0:
dq1
For a minimum we set
b
ga f
1 2 q1
1 2 Q − q1
+
−1 = 0
2 4π ∈0 R1 2 4π ∈0 R 2
R 2 q1 = R1Q − R1 q1
Then q 2 = Q − q1 =
(b)
q1 =
R1Q
R1 + R 2
R 2Q
= q2 .
R1 + R 2
V1 =
k e q1
k e R1Q
k eQ
=
=
R1
R1 R1 + R 2
R1 + R 2
V2 =
ke q2
k e R2 Q
k eQ
=
=
R2
R 2 R1 + R 2
R1 + R 2
b
g
b
g
and V1 − V2 = 0 .
Section 26.5
P26.43
P26.44
Capacitors with Dielectrics
e
je
j
(a)
−12
F m 1.75 × 10 −4 m 2
κ ∈0 A 2.10 8.85 × 10
C=
=
= 8.13 × 10 −11 F = 81.3 pF
d
4.00 × 10 −5 m
(b)
∆Vmax = Emax d = 60.0 × 10 6 V m 4.00 × 10 −5 m = 2.40 kV
e
je
j
Q max = C∆Vmax ,
∆Vmax = Emax d .
but
κ ∈0 A
.
d
Also,
C=
Thus,
Q max =
(a)
κ ∈0 A
Emax d = κ ∈0 AEmax .
d
b
g
With air between the plates, κ = 1.00
and
Emax = 3.00 × 10 6 V m .
Therefore,
e
je
je
j
Q max = κ ∈0 AEmax = 8.85 × 10 −12 F m 5.00 × 10 −4 m 2 3.00 × 10 6 V m = 13.3 nC .
(b)
With polystyrene between the plates, κ = 2.56 and Emax = 24.0 × 10 6 V m .
e
je
je
j
Q max = κ ∈0 AEmax = 2.56 8.85 × 10 −12 F m 5.00 × 10 −4 m 2 24.0 × 10 6 V m = 272 nC
Chapter 26
P26.45
C=
or
κ ∈0 A
d
95.0 × 10
−9
=
jb
e
g
3.70 8.85 × 10 −12 0.070 0 A
0.025 0 × 10 −3
A = 1.04 m
P26.46
Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between
2.54 cm
. Then,
them. Suppose the plastic has κ ≅ 3 , Emax ~ 10 7 V m and thickness 1 mil =
1 000
e
je
j
−12
C 2 N ⋅ m 2 0. 4 m 2
κ ∈0 A 3 8.85 × 10
~ 10 −6 F
~
C=
−5
d
2.54 × 10 m
e
je
j
∆Vmax = Emax d ~ 10 7 V m 2.54 × 10 −5 m ~ 10 2 V
P26.47
C=
Originally,
(a)
a f
e8.85 × 10
−12
je
a f
∈0 A ∆V
i
ja
C 2 N ⋅ m 2 25.0 × 10 −4 m 2 250 V
e1.50 × 10
−2
.
d
j
m
f=
369 pC
Finally,
Cf =
κ ∈0 A
Q
=
d
∆V
a f
Cf =
e
−2
f
=
Qd
κ ∈0 A
a f
Originally,
Ui =
1
C ∆V
2
Finally,
Uf =
1
C f ∆V
2
e8.85 × 10
∆U = −
2
i
−12
C
2
0
a f
2
f
i
κ ∈0 Ad
=
∆U = U f − U i =
So,
je
e1.50 × 10 mj
∈ Aa ∆V f d a ∆V f
=
=
80.0 8.85 × 10 −12 C 2 N ⋅ m 2 25.0 × 10 −4 m 2
f
a ∆V f
(c)
.
i
The charge is the same before and after immersion, with value Q =
Q=
(b)
∈0 A
Q
=
∆V
d
a f
∈0 A ∆V
2
i
2d
=
κ
−2
=
a f
κ ∈0 A ∆V
2dκ 2
2
i
=
a f
∈0 A ∆V
2 dκ
2
i
.
a f aκ − 1f
− ∈0 A ∆V
2 dκ
2
118 pF
250 V
= 3.12 V .
80.0
.
2
i
je25.0 × 10 m ja250 V f a79.0f =
2e1.50 × 10 mja80.0f
N⋅m
−4
2
i
j=
2
−45.5 nJ .
91
92
Capacitance and Dielectrics
P26.48
a fe
je
j
−12
1.00 × 10 −4 m 2
κ ∈0 A 173 8.85 × 10
=
= 1.53 nF
d
0.100 × 10 −3 m
(a)
C = κC 0 =
(b)
The battery delivers the free charge
a f e
ja
f
Q = C ∆V = 1.53 × 10 −9 F 12.0 V = 18.4 nC .
(c)
The surface density of free charge is
Q
18.4 × 10 −9 C
σ= =
= 1.84 × 10 −4 C m 2 .
A 1.00 × 10 −4 m 2
The surface density of polarization charge is
FG
H
σp =σ 1−
(d)
We have E =
E=
P26.49
IJ FG
K H
IJ
K
1
1
=σ 1−
= 1.83 × 10 −4 C m 2 .
κ
173
E0
κ
and E0 =
∆V
; hence,
d
∆V
12.0 V
=
= 694 V m .
κd
173 1.00 × 10 −4 m
a fe
j
The given combination of capacitors is equivalent to the circuit diagram
shown to the right.
Put charge Q on point A. Then,
b
g
b
g
b
g
Q = 40.0 µF ∆VAB = 10.0 µF ∆VBC = 40.0 µF ∆VCD .
FIG. P26.49
So, ∆VBC = 4∆VAB = 4∆VCD , and the center capacitor will break down first, at ∆VBC = 15.0 V . When
this occurs,
∆VAB = ∆VCD =
b
g
1
∆VBC = 3.75 V
4
and VAD = VAB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V .
Section 26.6
P26.50
(a)
Electric Dipole in an Electric Field
The displacement from negative to positive charge is
e
j
e
j
e
j
2 a = −1.20 i + 1.10 j mm − 1.40 i − 1.30 j mm = −2.60 i + 2.40 j × 10 −3 m.
The electric dipole moment is
e
je
j
p = 2aq = 3.50 × 10 −9 C −2.60 i + 2.40 j × 10 −3 m =
(b)
e
j
e−9.10 i + 8.40jj × 10
e
j
τ = p × E = −9.10 i + 8.40 j × 10 −12 C ⋅ m × 7.80 i − 4.90 j × 10 3 N C
e
j
τ = +44.6k − 65.5k × 10 −9 N ⋅ m = −2.09 × 10 −8 N ⋅ mk
continued on next page
−12
C⋅m .
Chapter 26
(c)
e
j
e
93
j
U = − p ⋅ E = − −9.10 i + 8.40 j × 10 −12 C ⋅ m ⋅ 7.80 i − 4.90 j × 10 3 N C
a
f
U = 71.0 + 41.2 × 10 −9 J = 112 nJ
a9.10f + a8.40f
a7.80f + a4.90f
p=
(d)
E=
2
2
× 10 −12 C ⋅ m = 12.4 × 10 −12 C ⋅ m
2
2
× 10 3 N C = 9.21 × 10 3 N C
U max = p E = 114 nJ,
U min = −114 nJ
U max − U min = 228 nJ
P26.51
(a)
Let x represent the coordinate of the negative charge.
Then x + 2 a cos θ is the coordinate of the positive
charge. The force on the negative charge is
F− = −qE x i . The force on the positive charge is
F+
(b)
F-
af
dE
= + qEa x + 2 a cos θ fi ≈ qEa xfi + q
a2 a cosθ fi .
dx
The force on the dipole is altogether
F = F− + F+ = q
The balloon creates field along the x-axis of
Thus,
a f
−2 k e q
dE
=
.
dx
x3
a fe
F+
p
θ
E
FIG. P26.51(a)
a
f
dE
dE
2 a cos θ i = p
cos θ i .
dx
dx
ke q i.
x2
je
a f
j
−2 8.99 × 10 9 2.00 × 10 −6
dE
=
= −8.78 MN C ⋅ m
3
dx
0.160
At x = 16.0 cm ,
e
je
j
F = 6.30 × 10 −9 C ⋅ m −8.78 × 10 6 N C ⋅ m cos 0° i = −55.3 i mN
Section 26.7
P26.52
An Atomic Description of Dielectrics
2π rAE =
so
qin
∈0
E=
λ
2π r ∈0
z
λ max
z
r2
r2
r1
r1
∆V = − E ⋅ d r =
FG IJ
H K
r
λ
λ
dr =
ln 1
r2
2π r ∈0
2π ∈0
= Emax rinner
2π ∈0
e
je
j FGH 025.200.0 IJK
∆V = 1. 20 × 10 6 V m 0.100 × 10 −3 m ln
∆Vmax = 579 V
FIG. P26.52
94
Capacitance and Dielectrics
P26.53
(a)
Consider a gaussian surface in the form of a cylindrical pillbox with ends of area A ′ << A
parallel to the sheet. The side wall of the cylinder passes no flux of electric field since this
surface is everywhere parallel to the field. Gauss’s law becomes
EA ′ + EA ′ =
(b)
Q
away from the positive and
2 ∈A
toward the negative sheet. Together, they create a field of
Q
.
∈A
Assume that the field is in the positive x-direction. Then, the potential of the positive plate
relative to the negative plate is
z
+ plate
∆V = −
z
+ plate
E ⋅ ds = −
− plate
(d)
directed away from the positive sheet.
In the space between the sheets, each creates field
E=
(c)
Q
Q
A ′ , so E =
2 ∈A
∈A
Q Qd
i ⋅ − idx = +
.
A
∈
∈
A
− plate
e
Capacitance is defined by: C =
j
Q
Q
∈ A κ ∈0 A
=
=
=
.
d
d
∆V Qd ∈ A
Additional Problems
P26.54
(a)
C=
(c)
Q ac
LM 1 + 1 OP + LM 1 + 1 OP = 3.33 µF
N 3.00 6.00 Q N 2.00 4.00 Q
= C b ∆V g = b 2.00 µFga90.0 V f = 180 µC
−1
ac
−1
ac
Therefore, Q3 = Q6 = 180 µC
i b
d
ga
f
Q df = C df ∆Vdf = 1.33 µF 90.0 V = 120 µC
(b)
(d)
Q3
C3
Q
∆V6 = 6
C6
Q
∆V2 = 2
C2
Q
∆V4 = 4
C4
∆V3 =
UT =
180 µC
=
3.00 µF
180 µC
=
=
6.00 µF
120 µC
=
=
2.00 µF
120 µC
=
=
4.00 µF
=
a f
1
C eq ∆V
2
2
=
60.0 V
30.0 V
60.0 V
30.0 V
ja
1
3.33 × 10 −6 90.0 V
2
e
f
2
= 13.4 mJ
FIG. P26.54
Chapter 26
*P26.55
Each face of P2 carries charge, so the three-plate system is equivalent to
(a)
P2
P3
P2
P1
Each capacitor by itself has capacitance
C=
e
j
−12
−4
2
2
κ ∈0 A 1 8.85 × 10 C 7.5 × 10 m
=
= 5.58 pF .
d
N ⋅ m 2 1.19 × 10 −3 m
Then equivalent capacitance = 5.58 + 5.58 = 11.2 pF .
a
f
(b)
Q = C∆V + C∆V = 11.2 × 10 −12 F 12 V = 134 pC
(c)
Now P3 has charge on two surfaces and in effect three capacitors are in parallel:
b
g
C = 3 5.58 pF = 16.7 pF .
Only one face of P4 carries charge:
(d)
a f
Q = C∆V = 5.58 × 10 −12 F 12 V = 66.9 pC .
*P26.56
From the example about a cylindrical capacitor,
Vb − Va = −2 k e λ ln
b
a
e
= −2a8.99fe1.4 × 10
je
j
Vb − 345 kV = −2 8.99 × 10 9 Nm 2 C 2 1.40 × 10 −6 C m ln
3
j
12 m
0.024 m
J C ln 500
5
= −1.564 3 × 10 V
Vb = 3. 45 × 10 5 V − 1.56 × 10 5 V = 1.89 × 10 5 V
*P26.57
Imagine the center plate is split along its midplane and pulled apart.
We have two capacitors in parallel, supporting the same ∆V and
∈ A
and the
carrying total charge Q. The upper has capacitance C 1 = 0
d
∈ A
lower C 2 = 0 . Charge flows from ground onto each of the outside
2d
∆V1 = ∆V2 = ∆V .
plates so that Q1 + Q 2 = Q
Then
(a)
(b)
Q 1 Q 2 Q1 d Q 2 2 d
=
=
=
C 1 C 2 ∈0 A ∈0 A
Q1 = 2Q 2
2Q 2 + Q 2 = Q .
Q2 =
Q
Q
. On the lower plate the charge is − .
3
3
Q1 =
2Q
2Q
. On the upper plate the charge is −
.
3
3
∆V =
Q1
2Qd
=
C1
3 ∈0 A
d
2d
FIG. P26.57
95
96
Capacitance and Dielectrics
P26.58
(a)
We use Equation 26.11 to find the potential energy of the capacitor. As we will see, the
potential difference ∆V changes as the dielectric is withdrawn. The initial and final
energies are U i =
F I
GH JK
1 Q2
2 Ci
Uf =
and
F I
GH JK
1 Q2
.
2 Cf
F I
GH JK
1 Q2
But the initial capacitance (with the dielectric) is Ci = κC f . Therefore, U f = κ
.
2
Ci
Since the work done by the external force in removing the dielectric equals the change in
F I F I F Ia f
GH JK GH JK GH JK
1 Q2
1 Q2
1 Q2
−
=
κ −1 .
potential energy, we have W = U f − U i = κ
2
2 Ci
2 Ci
Ci
b g
To express this relation in terms of potential difference ∆Vi , we substitute Q = Ci ∆Vi , and
b ga f e
ja
fa
f
1
1
2
2
C i ∆Vi κ − 1 = 2.00 × 10 −9 F 100 V 5.00 − 1.00 = 4.00 × 10 −5 J .
2
2
The positive result confirms that the final energy of the capacitor is greater than the initial
energy. The extra energy comes from the work done on the system by the external force that
pulled out the dielectric.
evaluate: W =
The final potential difference across the capacitor is ∆V f =
(b)
Ci
Q
.
Cf
a
b g
f
and Q = Ci ∆Vi gives ∆V f = κ∆Vi = 5.00 100 V = 500 V .
κ
Even though the capacitor is isolated and its charge remains constant, the potential
difference across the plates does increase in this case.
Substituting C f =
P26.59
κ = 3.00 , Emax = 2.00 × 10 8 V m =
For
C=
κ ∈0 A
= 0.250 × 10 −6 F
d
∆Vmax
d
jb
je
e
e
g
0. 250 × 10 −6 4 000
C∆Vmax
Cd
A=
= 0.188 m 2
=
=
κ ∈0 κ ∈0 Emax 3.00 8.85 × 10 −12 2.00 × 10 8
*P26.60
j
The original kinetic energy of the particle is
K=
1
1
mv 2 = 2 × 10 −16 kg 2 × 10 6 m s
2
2
e
je
j
2
= 4.00 × 10 −4 J .
Q 1 000 µC
=
= 100 V .
10 µF
C
For the particle to reach the negative plate, the particle-capacitor system would need energy
The potential difference across the capacitor is ∆V =
ja
e
f
U = q∆V = −3 × 10 6 C −100 V = 3.00 × 10 −4 J .
Since its original kinetic energy is greater than this, the particle will reach the negative plate .
As the particle moves, the system keeps constant total energy
aK + U f
at + plate
a
= K +U
f
at − plate
:
ja
e
f 12 e2 × 10 jv
4.00 × 10 −4 J + −3 × 10 −6 C +100 V =
vf =
e
j=
2 1.00 × 10 −4 J
2 × 10
−16
kg
1.00 × 10 6 m s .
−16
2
f
+0
Chapter 26
P26.61
(a)
C1 =
FG 1
HC
2
κ 1 ∈0 A 2
κ ∈ A2
κ ∈ A2
; C2 = 2 0
; C3 = 3 0
d
d 2
d 2
+
1
C3
C = C1 +
(b)
*P26.62
97
IJ
K
−1
FG 1
HC
2
FG
IJ
H
K
κ κ I
∈ A Fκ
+
G
J
d H 2 κ +κ K
=
∈ A κ 2κ 3
C 2C 3
= 0
C 2 + C3
d κ 2 +κ 3
+
1
C3
IJ
K
−1
0
=
1
2
2
3
FIG. P26.61
3
C total = 1.76 × 10 −12 F = 1.76 pF .
Using the given values we find:
The initial charge on the larger capacitor is
a f
Q = C∆V = 10 µF 15 V = 150 µC .
An additional charge q is pushed through the 50-V battery, giving the smaller capacitor charge q and
the larger charge 150 µC + q .
150 µC + q
5 µF
10 µF
500 µC = 2 q + 150 µC + q
50 V =
Then
q
+
q = 117 µC
P26.63
So across the 5- µF capacitor
∆V =
q 117 µC
=
= 23.3 V .
5 µF
C
Across the 10- µF capacitor
∆V =
150 µC + 117 µC
= 26.7 V .
10 µF
(a)
(b)
Put charge Q on the sphere of radius a and – Q on the other sphere. Relative to V = 0 at
infinity,
the potential at the surface of a is
Va =
k eQ k eQ
−
a
d
and the potential of b is
Vb =
− k e Q k eQ
+
.
b
d
The difference in potential is
Va − Vb =
and
C=
As d → ∞ ,
C=
k e Q k eQ k eQ k eQ
+
−
−
a
b
d
d
Q
=
Va − Vb
F 4π ∈
I
GH b1 ag + b1 bg − b2 dg JK
1
1
becomes negligible compared to . Then,
d
a
4π ∈0
1
1
1
+
and =
4π ∈0 a 4π ∈0 b
1 a+1 b
C
as for two spheres in series.
0
.
98
Capacitance and Dielectrics
P26.64
a f
a f
(a)
C=
∈0
∈
A − x A + κ Ax = 0 A 2 + A x κ − 1
d
d
(b)
U=
1
C ∆V
2
(c)
F=−
(d)
F=
a f
2
FG dU IJ i =
H dx K
F a f IA
GH
JK
1 ∈0 ∆V
2
d
=
2
a f
+ Ax κ − 1
a f Aaκ − 1f to the left
∈0 ∆V
2d
2
b2 000g e8.85 × 10 jb0.050 0ga4.50 − 1f =
2e 2.00 × 10 j
2
P26.65
2
(out of the capacitor)
−12
−3
1.55 × 10 −3 N
The portion of the capacitor nearly filled by metal has
a f
capacitance
κ ∈0 Ax
→∞
d
and stored energy
Q2
→0.
2C
The unfilled portion has
a f
∈0 A A − x
.
d
capacitance
Q=
The charge on this portion is
(a)
aA − xfQ
0
.
A
The stored energy is
a f
a f
a f
2
A − x Q0 A
Q02 A − x d
Q2
=
U=
=
.
2C 2 ∈0 A A − x d
2 ∈0 A 3
(b)
F=−
F=
F a fI
GH
JK
2
Q02 d
dU
d Q0 A − x d
=−
=
+
dx
dx
2 ∈0 A 3
2 ∈0 A 3
Q02 d
2 ∈0 A 3
to the right (into the capacitor)
Q02
F
=
Ad
2 ∈0 A 4
(c)
Stress =
(d)
σ
1
1
u = ∈0 E 2 = ∈0
∈0
2
2
FG IJ
H K
2
F
GH
Q0
1
= ∈0
2
∈0 A 2
I
JK
2
=
Q02
2 ∈0 A 4
P26.66
.00 m I
= 5.24 × 10
b126 000 Btu galgb1 054 J BtugFGH 3.7861.00× 10gal m IJK FGH 1670
kg JK
b12.0 J Cgb100 C sgb3 600 sg = 2.70 × 10 J kg
3
Gasoline:
−3
3
Chapter 26
7
99
J kg
5
Battery:
16.0 kg
Capacitor:
1
2
a0.100 Ffa12.0 Vf
2
0.100 kg
= 72.0 J kg
Gasoline has 194 times the specific energy content of the battery and 727 000 times that of the
capacitor.
P26.67
Call the unknown capacitance C u
b g b gd i
C d ∆V i
b10.0 µFga30.0 Vf =
=
=
b∆V g − d∆V i a100 V − 30.0 V f
Q = C u ∆Vi = Cu + C ∆V f
Cu
f
i
*P26.68
4.29 µF
f
She can clip together a series combination of parallel combinations of two
100- µF capacitors. The equivalent capacitance is
1
= 100 µF . When 90 V is connected across the
−1
−1
200 µF + 200 µF
b
g b
g
combination, only 45 V appears across each individual capacitor.
P26.69
(a)
C0 =
FIG. P26.68
∈0 A Q0
=
∆V0
d
When the dielectric is inserted at constant voltage,
C = κC 0 =
U0 =
U=
and
Q
;
∆V0
b g
C 0 ∆V0
2
2
b g
C ∆V0
2
U
=κ .
U0
2
=
e j
κC 0 ∆V02
2
The extra energy comes from (part of the) electrical work done by the battery in separating
the extra charge.
(b)
Q 0 = C 0 ∆V0
and
so
Q = C∆V0 = κC 0 ∆V0
Q
=κ .
Q0
100
P26.70
Capacitance and Dielectrics
The vertical orientation sets up two capacitors in parallel, with equivalent capacitance
Cp =
b g + κ ∈ b A 2g = FG κ + 1 IJ ∈ A
H 2 K d
d
d
∈0 A 2
0
0
where A is the area of either plate and d is the separation of the plates. The horizontal orientation
produces two capacitors in series. If f is the fraction of the horizontal capacitor filled with dielectric,
the equivalent capacitance is
b g OP d , or C = LM κ OP ∈ A .
PQ ∈ A
MN f + κ b1 − f g PQ d
κ +1
κ
=
gives
, or aκ + 1f f + κ b1 − f g = 2κ .
2
f + κ b1 − f g
b g LM
MN
1− f d
f +κ 1 − f
fd
1
=
+
=
∈0 A
κ
C s κ ∈0 A
Requiring that C p = C s
s
0
0
a f
For κ = 2.00 , this yields 3.00 2.00 − 1.00 f = 4.00 , with the solution f =
P26.71
2
.
3
Initially (capacitors charged in parallel),
a f b
ga f
= C a ∆V f = b 2.00 µF ga 250 V f = 500 µC .
q1 = C1 ∆V = 6.00 µF 250 V = 1 500 µC
q2
2
After reconnection (positive plate to negative plate),
qtotal
′ = q1 − q 2 = 1 000 µC and ∆V ′ =
qtotal
1 000 µC
′
=
= 125 V .
8.00 µF
C total
Therefore,
a f b
ga f
q ′ = C a ∆V ′f = b 2.00 µF ga125 V f =
q1′ = C1 ∆V ′ = 6.00 µF 125 V = 750 µC
2
P26.72
2
250 µC .
Assume a potential difference across a and b, and notice that the potential difference across the
8.00 µF capacitor must be zero by symmetry. Then the equivalent capacitance can be determined
from the following circuit:
⇒
⇒
FIG. P26.72
C ab = 3.00 µF .
Chapter 26
P26.73
Emax occurs at the inner conductor’s surface.
Emax =
2keλ
from Equation 24.7.
a
∆V = 2 k e λ ln
Emax =
FG b IJ from Example 26.2
H aK
∆V
a ln b a
b g
FG b IJ = e18.0 × 10 V mje0.800 × 10
H aK
F bI
∆V = 2κλ lnG J
H aK
F bI
a lnG J
H aK
LMlnFG b IJ + aF 1 I FG − b IJ OP = 0
MN H a K GH b a JK H a K PQ
6
∆Vmax = Emax a ln
P26.74
E=
2κλ
;
a
∆Vmax = Emax
dVmax
= Emax
da
ln
P26.75
−3
.00 I
JK =
j FGH 03.800
m ln
19.0 kV .
2
FG b IJ = 1 or b = e
H aK
a
1
so a =
b
e
By symmetry, the potential difference across 3C is zero, so the circuit reduces to
C eq =
FG 1 + 1 IJ
H 2C 4C K
−1
=
8
4
C=
C .
6
3
FIG. P26.75
101
102
P26.76
Capacitance and Dielectrics
The electric field due to the charge on the positive wire is perpendicular to the wire, radial, and of
magnitude
E+ =
λ
.
2π ∈0 r
The potential difference between wires due to the presence of this charge is
∆V1 = −
z
+ wire
E ⋅ dr = −
− wire
λ
2π ∈0
FG
H
z
d
IJ
K
dr
D−d
λ
ln
.
=
π
r
d
2
∈
0
D−d
The presence of the linear charge density − λ on the negative wire makes an identical contribution
to the potential difference between the wires. Therefore, the total potential difference is
b g
∆V = 2 ∆V1 =
FG
H
λ
D−d
ln
π ∈0
d
IJ
K
and the capacitance of this system of two wires, each of length A , is
C=
π ∈0 A
λA
λA
Q
=
=
=
.
∆V ∆V
λ π ∈0 ln D − d d ln D − d d
b
g a
f
The capacitance per unit length is:
*P26.77
a
f
π ∈0
C
=
A ln D − d d
a
f
.
The condition that we are testing is that the capacitance increases by less than 10%, or,
C′
< 1.10 .
C
Substituting the expressions for C and C ′ from Example 26.2, we have,
A
b
k
2
ln
C′
e
1.10 a
=
A
C
2 k e ln ba
c h = lnc h < 1.10 .
lnc
h
ch
b
a
b
1.10 a
This becomes,
ln
FG b IJ < 1.10 lnFG b IJ = 1.10 lnFG b IJ + 1.10 lnFG 1 IJ = 1.10 lnFG b IJ − 1.10 lna1.10f .
H aK
H 1.10a K
H aK
H 1.10 K
H aK
We can rewrite this as,
FG b IJ < −1.10 lna1.10f
H aK
F bI
lnG J > 11.0 lna1.10f = lna1.10f
H aK
−0.10 ln
11.0
where we have reversed the direction of the inequality because we multiplied the whole expression
by –1 to remove the negative signs. Comparing the arguments of the logarithms on both sides of the
inequality, we see that,
a f
b
> 1.10
a
11.0
= 2.85 .
Thus, if b > 2.85 a , the increase in capacitance is less than 10% and it is more effective to increase A .
Chapter 26
103
ANSWERS TO EVEN PROBLEMS
P26.2
(a) 1.00 µF ; (b) 100 V
P26.4
(a) 8.99 mm ; (b) 0.222 pF ; (c) 22.2 pC
P26.6
11.1 nF ; 26.6 C
P26.8
3.10 nm
P26.10
a2 N − 1f ∈ aπ − θ fR
0
2.13 × 10 16 m3
P26.14
mgd tan θ
q
P26.18
P26.20
P26.22
2
R1Q
R 2Q
and q 2 =
;
R1 + R 2
R1 + R 2
(b) see the solution
P26.44
(a) 13.3 nC ; (b) 272 nC
P26.46
~ 10 −6 F and ~ 10 2 V for two 40 cm by
100 cm sheets of aluminum foil
sandwiching a thin sheet of plastic.
P26.48
(a) 1.53 nF ; (b) 18.4 nC ; (c) 184 µC m 2
P26.50
e
j
(a) −9.10 i + 8. 40 j pC ⋅ m ;
(b) −20.9 nN ⋅ mk ; (c) 112 nJ ; (d) 228 nJ
C p2
− C p C s and
4
Cp
2
−
C p2
4
− C pC s
P26.52
579 V
P26.54
(a) 3.33 µF ;
(b) ∆V3 = 60.0 V ; ∆V6 = 30.0 V ;
∆V2 = 60.0 V ; ∆V4 = 30.0 V ;
(c) Q 3 = Q6 = 180 µC ; Q 2 = Q 4 = 120 µC ;
(d) 13.4 mJ
(d) Q 3 and Q1 increase and Q 2 decreases
P26.24
(a) 398 µF in series; (b) 2.20 µF in parallel
P26.56
189 kV
P26.26
19.8 µC
P26.58
(a) 40.0 µJ ; (b) 500 V
P26.28
83.6 µC
P26.60
yes; 1.00 Mm s
P26.30
e
P26.62
23.3 V ; 26.7 V
P26.64
(a)
P26.32
P26.34
;
(a) q1 =
(a) 2C ; (b) Q1 > Q3 > Q 2 ;
(c) ∆V1 > ∆V2 = ∆V3 ;
3 −1
2
P26.42
1.83C
+
2
free; 183 µC m 2 induced; (d) 694 V m
(a) 17.0 µF ; (b) 9.00 V ;
(c) 45.0 µC and 108 µC
Cp
a ∆V f
; (c) 4C
a f ; (b) 4∆V
3
3
(a) C ∆V
(d) Positive work is done on the system by
the agent pulling the plates apart.
2
d
P26.12
P26.16
P26.40
j C2
0
4.47 kV
energy doubles
P26.36
2.51 × 10 −3 m3 = 2.51 L
P26.38
(a) 400 µC ; (b) 2.50 kN m
(b)
a f
∈0 A 2 + Ax κ − 1
a f
∈0 ∆V
d
2
a f;
A 2 + Ax κ − 1
2d
a f a f
2
;
∈0 ∆V A κ − 1
to the left ;
2d
(d) 1.55 mN left
(c)
104
Capacitance and Dielectrics
Gasoline has 194 times the specific energy
content of the battery, and 727 000 times
that of the capacitor.
P26.72
3.00 µF
P26.74
see the solution
P26.68
see the solution; 45 V
P26.76
see the solution
P26.70
2
3
P26.66
27
Current and Resistance
CHAPTER OUTLINE
27.1
27.2
27.3
27.4
27.5
27.6
Electric Current
Resistance
A Model for Electrical
Conduction
Resistance and
Temperature
Superconductors
Electric Power
ANSWERS TO QUESTIONS
Q27.1
Individual vehicles—cars, trucks and motorcycles—would
correspond to charge. The number of vehicles that pass a
certain point in a given time would correspond to the current.
Q27.2
Voltage is a measure of potential difference, not of current.
“Surge” implies a flow—and only charge, in coulombs, can flow
through a system. It would also be correct to say that the victim
carried a certain current, in amperes.
Q27.3
Geometry and resistivity. In turn, the resistivity of the material
depends on the temperature.
Q27.4
Resistance is a physical property of the conductor based on the
material of which it is made and its size and shape, including
the locations where current is put in and taken out. Resistivity
is a physical property only of the material of which the resistor
is made.
Q27.5
The radius of wire B is 3 times the radius of wire A, to make its cross–sectional area 3 times larger.
Q27.6
Not all conductors obey Ohm’s law at all times. For example, consider an experiment in which a
variable potential difference is applied across an incandescent light bulb, and the current is
measured. At very low voltages, the filament follows Ohm’s law nicely. But then long before the
∆V
becomes non-linear, because the resistivity is temperaturefilament begins to glow, the plot of
I
dependent.
Q27.7
A conductor is not in electrostatic equilibrium when it is carrying a current, duh! If charges are
placed on an isolated conductor, the electric fields established in the conductor by the charges will
cause the charges to move until they are in positions such that there is zero electric field throughout
the conductor. A conductor carrying a steady current is not an isolated conductor—its ends must be
connected to a source of emf, such as a battery. The battery maintains a potential difference across
the conductor and, therefore, an electric field in the conductor. The steady current is due to the
response of the electrons in the conductor due to this constant electric field.
105
106
Current and Resistance
Q27.8
The bottom of the rods on the Jacob’s Ladder are close enough so that the supplied voltage is
sufficient to produce dielectric breakdown of the air. The initial spark at the bottom includes a tube
of ionized air molecules. Since this tube containing ions is warmer than the air around it, it is buoyed
up by the surrounding air and begins to rise. The ions themselves significantly decrease the
resistivity of the air. They significantly lower the dielectric strength of the air, marking longer sparks
possible. Internal resistance in the power supply will typically make its terminal voltage drop, so
that it cannot produce a spark across the bottom ends of the rods. A single “continuous” spark,
therefore will rise up, becoming longer and longer, until the potential difference is not large enough
to sustain dielectric breakdown of the air. Once the initial spark stops, another one will form at the
bottom, where again, the supplied potential difference is sufficient to break down the air.
Q27.9
The conductor does not follow Ohm’s law, and must have a resistivity that is current-dependent, or
more likely temperature-dependent.
Q27.10
A power supply would correspond to a water pump; a resistor corresponds to a pipe of a certain
diameter, and thus resistance to flow; charge corresponds to the water itself; potential difference
corresponds to difference in height between the ends of a pipe or the ports of a water pump.
Q27.11
The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electrons
more efficiently.
Q27.12
In a metal, the conduction electrons are not strongly bound to individual ion cores. They can move
in response to an applied electric field to constitute an electric current. Each metal ion in the lattice
of a microcrystal exerts Coulomb forces on its neighbors. When one ion is vibrating rapidly, it can set
its neighbors into vibration. This process represents energy moving though the material by heat.
Q27.13
The resistance of copper increases with temperature, while the resistance of silicon decreases with
increasing temperature. The conduction electrons are scattered more by vibrating atoms when
copper heats up. Silicon’s charge carrier density increases as temperature increases and more atomic
electrons are promoted to become conduction electrons.
Q27.14
A current will continue to exist in a superconductor without voltage because there is no resistance
loss.
Q27.15
Superconductors have no resistance when they are below a certain critical temperature. For most
superconducting materials, this critical temperature is close to absolute zero. It requires expensive
refrigeration, often using liquid helium. Liquid nitrogen at 77 K is much less expensive. Recent
discoveries of materials that have higher critical temperatures suggest the possibility of developing
superconductors that do not require expensive cooling systems.
Q27.16
In a normal metal, suppose that we could proceed to a limit of zero resistance by lengthening the
average time between collisions. The classical model of conduction then suggests that a constant
applied voltage would cause constant acceleration of the free electrons, and a current steadily
increasing in time.
On the other hand, we can actually switch to zero resistance by substituting a
superconducting wire for the normal metal. In this case, the drift velocity of electrons is established
by vibrations of atoms in the crystal lattice; the maximum current is limited; and it becomes
impossible to establish a potential difference across the superconductor.
Q27.17
Because there are so many electrons in a conductor (approximately 10 28 electrons m 3 ) the average
velocity of charges is very slow. When you connect a wire to a potential difference, you establish an
electric field everywhere in the wire nearly instantaneously, to make electrons start drifting
everywhere all at once.
Chapter 27
107
Q27.18
Current moving through a wire is analogous to a longitudinal wave moving through the electrons of
the atoms. The wave speed depends on the speed at which the disturbance in the electric field can
be communicated between neighboring atoms, not on the drift velocities of the electrons
themselves. If you leave a direct-current light bulb on for a reasonably short time, it is likely that no
single electron will enter one end of the filament and leave at the other end.
Q27.19
More power is delivered to the resistor with the smaller resistance, since P =
Q27.20
The 25 W bulb has a higher resistance. The 100 W bulb carries more current.
Q27.21
One ampere–hour is 3 600 coulombs. The ampere–hour rating is the quantity of charge that the
battery can lift though its nominal potential difference.
Q27.22
Choose the voltage of the power supply you will use to drive the heater. Next calculate the required
∆V 2
resistance R as
. Knowing the resistivity ρ of the material, choose a combination of wire length
P
and cross–sectional area to make
∆V 2
.
R
FG A IJ = FG R IJ . You will have to pay for less material if you make both
H AK H ρ K
A and A smaller, but if you go too far the wire will have too little surface area to radiate away the
energy; then the resistor will melt.
SOLUTIONS TO PROBLEMS
Section 27.1
P27.1
I=
Electric Current
∆Q
∆t
N=
P27.2
ja
e
f
∆Q = I∆t = 30.0 × 10 −6 A 40.0 s = 1.20 × 10 −3 C
1.20 × 10 −3 C
Q
=
= 7.50 × 10 15 electrons
e 1.60 × 10 −19 C electron
The molar mass of silver = 107.9 g mole and the volume V is
a fa
f e
je
j
V = area thickness = 700 × 10 −4 m 2 0.133 × 10 −3 m = 9.31 × 10 −6 m 3 .
e
je
j
The mass of silver deposited is m Ag = ρV = 10.5 × 10 3 kg m3 9.31 × 10 −6 m 3 = 9.78 × 10 −2 kg .
And the number of silver atoms deposited is
e
N = 9.78 × 10 −2 kg
I=
10 atoms I F 1 000 g I
jFGH 6.02 ×107.9
JK GH 1 kg JK = 5.45 × 10
g
23
∆V 12.0 V
=
= 6.67 A = 6.67 C s
1.80 Ω
R
∆t =
e
je
j
23
atoms
5.45 × 10 23 1.60 × 10 −19 C
∆Q Ne
=
=
= 1.31 × 10 4 s = 3.64 h
I
I
6.67 C s
108
P27.3
Current and Resistance
af
z
t
e
Q t = Idt = I 0τ 1 − e − t τ
0
P27.4
af
j
j a0.632fI τ
e
(a)
Q τ = I 0 τ 1 − e −1 =
(b)
Q 10τ = I 0τ 1 − e −10 =
(c)
Q ∞ = I 0τ 1 − e −∞ = I 0τ
af
e
(a)
Using
kee2
(b)
The time for the electron to revolve around the proton once is:
a f
j b0.999 95gI τ
e
r2
0
0
j
kee2
= 2.19 × 10 6 m s .
mr
mv 2
, we get: v =
r
=
e
j
−11
m
2π r 2π 5.29 × 10
t=
=
= 1.52 × 10 −16 s .
v
2.19 × 10 6 m s
e
j
The total charge flow in this time is 1.60 × 10 −19 C , so the current is
I=
P27.5
1.60 × 10 −19 C
1.52 × 10 −16 s
The period of revolution for the sphere is T =
revolving charge is I =
P27.6
2π
ω
, and the average current represented by this
q
qω
=
.
2π
T
q = 4t 3 + 5t + 6
e
A = 2.00 cm 2
a
J=
(b)
I=
1.00 m I
J
jFGH 100
cm K
f
I 1.00 s =
(a)
P27.7
= 1.05 × 10 −3 A = 1.05 mA .
dq
dt
2
= 2.00 × 10 −4 m 2
e
= 12t 2 + 5
t = 1.00 s
j
t =1.00 s
= 17.0 A
I
17.0 A
=
= 85.0 kA m 2
A 2.00 × 10 −4 m 2
dq
dt
a100 Af sinFGH 120sπ t IJK dt
−100 C L F π I
O +100 C = 0.265 C
q=
cosG J − cos 0 P =
M
H
K
120π N
2
Q 120π
z z
q = dq = Idt =
z
1 240 s
0
Chapter 27
P27.8
5.00 A
I
=
A π 4.00 × 10 −3 m
= 99.5 kA m 2
(a)
J=
(b)
J2 =
1
I
1 I
=
J1 ;
4
A 2 4 A1
A1 =
1
A 2 so π 4.00 × 10 −3
4
e
j
2
e
e
j
2
=
1
π r22
4
j
r2 = 2 4.00 × 10 −3 = 8.00 × 10 −3 m = 8.00 mm
P27.9
8.00 × 10 −6 A
I
=
A π 1.00 × 10 −3 m
= 2.55 A m 2
(a)
J=
(b)
From J = nev d , we have
n=
(c)
∆Q
, we have
From I =
∆t
6.02 × 10 23 1.60 × 10 −19 C
∆Q N A e
=
=
= 1. 20 × 10 10 s .
∆t =
I
I
8.00 × 10 −6 A
e
j
2
2.55 A m 2
J
=
= 5.31 × 10 10 m −3 .
ev d
1.60 × 10 −19 C 3.00 × 10 8 m s
e
je
e
j
je
j
(This is about 382 years!)
P27.10
(a)
K=
The speed of each deuteron is given by
e2.00 × 10 je1.60 × 10 Jj = 12 e2 × 1.67 × 10
6
−19
−27
j
kg v 2 and
1
mv 2
2
v = 1.38 × 10 7 m s .
q
t
The time between deuterons passing a stationary point is t in
I=
10.0 × 10 −6 C s = 1.60 × 10 −19 C t or
t = 1.60 × 10 −14 s .
e
je
j
So the distance between them is vt = 1.38 × 10 7 m s 1.60 × 10 −14 s = 2.21 × 10 −7 m .
(b)
One nucleus will put its nearest neighbor at potential
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
keq
=
= 6.49 × 10 −3 V .
V=
r
2.21 × 10 −7 m
This is very small compared to the 2 MV accelerating potential, so repulsion within the
beam is a small effect.
109
110
P27.11
Current and Resistance
We use I = nqAv d n is the number of charge carriers per unit volume, and is identical to the number
of atoms per unit volume. We assume a contribution of 1 free electron per atom in the relationship
above. For aluminum, which has a molar mass of 27, we know that Avogadro’s number of atoms,
N A , has a mass of 27.0 g. Thus, the mass per atom is
27.0 g
27.0 g
=
= 4.49 × 10 −23 g atom .
NA
6.02 × 10 23
n=
Thus,
density of aluminum
2.70 g cm 3
=
mass per atom
4.49 × 10 −23 g atom
n = 6.02 × 10 22 atoms cm3 = 6.02 × 10 28 atoms m3 .
vd =
or,
v d = 0.130 mm s .
Section 27.2
*P27.12
I=
P27.14
(a)
e
je
je
j
Resistance
E
J = σE =
P27.13
ρ
=
0.740 V m
2.44 × 10 −8 Ω ⋅ m
FG 1 Ω ⋅ A IJ =
H 1V K
3.03 × 10 7 A m 2
∆V 120 V
=
= 0.500 A = 500 mA
R
240 Ω
Applying its definition, we find the resistance of the rod,
R=
(b)
∆V
15.0 V
=
= 3 750 Ω = 3.75 kΩ .
I
4.00 × 10 −3 A
ρA
. Solving for A
A
and substituting numerical values for R, A, and the value of ρ given for carbon in Table 27.1,
we obtain
The length of the rod is determined from the definition of resistivity: R =
A=
P27.15
5.00 A
I
=
= 1.30 × 10 −4 m s
28
3
−
nqA
6.02 × 10 m
1.60 × 10 −19 C 4.00 × 10 −6 m 2
Therefore,
RA
ρ
e3.75 × 10 Ωje5.00 × 10
=
e3.50 × 10 Ω ⋅ mj
−6
3
−5
m2
j=
536 m .
∆V = IR
and
ρA
R=
:
A
IρA
∆V =
:
A
F 1.00 m I = 6.00 × 10
A = a0.600 mmf G
H 1 000 mm JK
∆VA a0.900 V fe6.00 × 10 m j
I=
=
ρA
e5.60 × 10 Ω ⋅ mja1.50 mf
2
2
−7
−8
I = 6.43 A
2
−7
m2
Chapter 27
P27.16
J=
I
πr
2
= σE =
a
f
σ = 55.3 Ω ⋅ m
P27.17
(a)
b
3.00 A
−1
ρ=
2
1
σ
b
= σ 120 N C
M = ρ d V = ρ d AA
we obtain:
A=
V=
M
.
ρdA
MR
A=
g
= 0.018 1 Ω ⋅ m
Given
Thus,
(b)
g
π 0.012 0 m
ρr ρd
where
ρ d ≡ mass density,
Taking ρ r ≡ resistivity,
R=
e1.00 × 10 ja0.500f
e1.70 × 10 je8.92 × 10 j
=
−8
3
M
π ρdA
=
e
1.00 × 10 −3
π 8.92 × 10
3
ja1.82f
diameter = 280 µm .
The volume of the gram of gold is given by ρ =
V=
m
ρ
=
10 −3 kg
m
V
e
j
= 5.18 × 10 −8 m3 = A 2.40 × 10 3 m
19.3 × 10 3 kg m3
A = 2.16 × 10 −11 m 2
R=
P27.19
(a)
e
j
−8
3
ρ A 2.44 × 10 Ω ⋅ m 2.4 × 10 m
=
= 2.71 × 10 6 Ω
A
2.16 × 10 −11 m 2
Suppose the rubber is 10 cm long and 1 mm in diameter.
e
je
j
−1
13
ρA 4ρA 4 10 Ω ⋅ m 10 m
R=
~
=
= ~ 10 18 Ω
2
2
−
3
A πd
π 10 m
e
e
j
je
j
(b)
4 1.7 × 10 −8 Ω ⋅ m 10 −3 m
4ρA
R=
~
~ 10 −7 Ω
2
−2
π d2
π 2 × 10 m
(c)
I=
e
I~
∆V
10 2 V
~ 18
~ 10 −16 A
R
10 Ω
10 2 V
10 −7 Ω
~ 10 9 A
j
M
.
ρd
r = 1.40 × 10 −4 m .
The diameter is twice this distance:
*P27.18
A = 1.82 m .
π r 2A =
or
r=
ρrA
ρrA
ρ ρ A2
=
= r d .
A
M ρdA
M
−3
M
,
ρd
Thus,
111
112
P27.20
Current and Resistance
F 90.0 g I
The distance between opposite faces of the cube is A = G
H 10.5 g cm JK
R=
ρA ρA ρ 1.59 × 10 −8 Ω ⋅ m
=
= =
= 7.77 × 10 −7 Ω = 777 nΩ
−2
A A2 A
2.05 × 10 m
∆V 1.00 × 10 −5 V
=
= 12.9 A
R
7.77 × 10 −7 Ω
10.5 g cm 3
6.02 × 10 23 electrons mol
n=
107.87 g mol
I=
(b)
e
e
n = 5.86 × 10 22 electrons cm 3
I = nqvA and v =
P27.22
Originally, R =
ρ Al A
b g
π rAl
2
=
b g
π rCu
P27.23
J = σE
P27.24
R=
ρ=
3
6
3
e
28
m3
jb
je
Finally, R f =
b g = ρA =
ρA3
3A
9A
g
2
= 3.28 µm s
R
.
9
2
σ=
so
−13
A m2
J 6.00 × 10
=
= 6.00 × 10 −15 Ω ⋅ m
E
100 V m
a
ρ 1 A1 ρ 2 A 2 ρ 1 A1 + ρ 2 A 2
+
=
A1
A2
d2
e4.00 × 10
R=
Section 27.3
jFGH 1.001×.0010m cm IJK = 5.86 × 10
12.9 C s
I
=
nqA
5.86 × 10 28 m3 1.60 × 10 −19 C 0.020 5 m
ρA
.
A
ρ Cu A
j
ρ Al
2.82 × 10 −8
=
= 1.29
ρ Cu
1.70 × 10 −8
rAl
=
rCu
P27.25
= 2.05 cm .
3
(a)
P27.21
13
−3
ja
f e
ja
f=
j
je
j
Ω ⋅ m 0.250 m + 6.00 × 10 −3 Ω ⋅ m 0.400 m
e3.00 × 10
−3
j
m
2
f
−1
378 Ω
A Model for Electrical Conduction
m
nq 2 τ
so
e
je
9.11 × 10 −31
m
τ=
=
= 2.47 × 10 −14 s
ρnq 2
1.70 × 10 −8 8. 49 × 10 28 1.60 × 10 −19
e
vd =
qE
τ
m
e1.60 × 10 jEe2.47 × 10 j
−19
so
7.84 × 10 −4 =
Therefore,
E = 0.181 V m .
9.11 × 10 −31
−14
Chapter 27
P27.26
113
n is unaffected
(a)
J =
(b)
I
∝I
A
so it doubles .
J = nev d
(c)
so v d
τ=
(d)
doubles .
mσ
is unchanged as long as σ does not change due to a temperature change in the
nq 2
conductor.
P27.27
From Equation 27.17,
τ=
me
9.11 × 10 −31
=
e8.49 × 10 je1.60 × 10 j e1.70 × 10 j
A = vτ = e8.60 × 10 m sje 2.47 × 10
sj = 2.12 × 10
2
nq ρ
28
P27.28
−8
−14
5
Section 27.4
−19 2
= 2. 47 × 10 −14 s
−8
m = 21.2 nm
Resistance and Temperature
At the low temperature TC we write
RC =
∆V
= R0 1 + α TC − T0
IC
At the high temperature Th ,
Rh =
∆V ∆V
=
= R0 1 + α Th − T0 .
Ih
1A
b
g
where T0 = 20.0° C .
b
a∆V f a1.00 Af = 1 + e3.90 × 10
a∆V f I 1 + e3.90 × 10
F 1.15 IJ = 1.98 A
I = a1.00 A fG
H 0.579 K
g
ja38.0f
ja−108f
−3
Then
−3
C
and
P27.29
C
a f
R = R 0 1 + α ∆T
gives
Solving,
And, the final temperature is
a
f e
j
.
140 Ω = 19.0 Ω 1 + 4.50 × 10 −3 ° C ∆T .
∆T = 1.42 × 10 3 ° C = T − 20.0° C .
T = 1.44 × 10 3 ° C .
114
P27.30
Current and Resistance
b
g
b
R = R c + Rn = R c 1 + α c T − T0 + Rn 1 + α n T − T0
b
g
b
g
0 = Rc α c T − T0 + Rnα n T − T0 so R c = − Rn
R = − Rn
αn
+ Rn
αc
F α IJ
F α IJ
= RG 1 −
R = RG 1 −
H αK
H αK
L e0.400 × 10 ° Cj OP
= 10.0 kΩ M1 −
MN e−0.500 × 10 ° Cj PQ
−1
Rn
n
n
−3
Rn = 5.56 kΩ
P27.31
−1
−1
−3
Rn
αn
αc
c
c
c
g
R c = 4.44 kΩ
and
b
g e
a
j
f
(a)
ρ = ρ 0 1 + α T − T0 = 2.82 × 10 −8 Ω ⋅ m 1 + 3.90 × 10 −3 30.0° = 3.15 × 10 −8 Ω ⋅ m
(b)
J=
(c)
F π d I = 6.35 × 10
I = JA = J G
H 4 JK e
(d)
n=
E
ρ
=
0.200 V m
3.15 × 10 −8 Ω ⋅ m
2
= 6.35 × 10 6 A m 2
LM π e1.00 × 10
A m j
MM
4
N
2
6
6.02 × 10 23 electrons
e
6
26.98 g 2.70 × 10 g m
3
j
e
−4
j OP =
PP
Q
m
2
49.9 mA
= 6.02 × 10 28 electrons m 3
j
6.35 × 10 6 A m 2
J
=
= 659 µm s
vd =
ne
6.02 × 10 28 electrons m3 1.60 × 10 −19 C
e
(e)
P27.32
b
je
j
f
ga
∆V = EA = 0.200 V m 2.00 m = 0.400 V
For aluminum,
R=
b
ga
α E = 3.90 × 10 −3 ° C −1
(Table 27.1)
α = 24.0 × 10 −6 ° C −1
(Table 19.1)
f
b
a
g a
f
fFGH
I
JK
1 + α E ∆T
ρA ρ 0 1 + α E ∆T A 1 + α∆T
1.39
=
= R0
= 1.71 Ω
= 1.234 Ω
2
A
T
.
α
+
1
∆
1
002 4
A 1 + α ∆T
a
f
Chapter 27
P27.33
R = R 0 1 + αT
R − R0 = R0α∆T
R − R0
= α∆T = 5.00 × 10 −3 25.0 = 0.125
R0
e
P27.34
j
a
Assuming linear change of resistance with temperature, R = R0 1 + α∆T
a
ja
f e
f
f
R77 K = 1.00 Ω 1 + 3.92 × 10 −3 −216° C = 0.153 Ω .
P27.35
a
f
1
FG ρ
Hρ
W
IJ
K
ρ = ρ 0 1 + α∆T or
∆TW =
Require that ρ W = 4ρ 0 Cu so that
∆TW =
Therefore,
TW = 47.6° C + T0 = 67.6° C .
Section 27.5
αW
−1
0W
F 1
GH 4.50 × 10
I FG 4e1.70 × 10 j − 1IJ = 47.6° C .
JK
° C JK GH 5.60 × 10
−8
−3
Superconductors
Problem 48 in Chapter 43 can be assigned with this section.
Section 27.6
P27.36
I=
Electric Power
P
∆V
=
and R =
*P27.37
P27.38
600 W
= 5.00 A
120 V
∆V 120 V
=
= 24.0 Ω .
I
5.00 A
e
b
gb
g
P = 0.800 1 500 hp 746 W hp = 8.95 × 10 5 W
b
8.95 × 10 5 = I 2 000
P = I∆V
P27.39
j
P = I∆V = 500 × 10 −6 A 15 × 10 3 V = 7.50 W
g
I = 448 A
The heat that must be added to the water is
b
gb
f
ga
Q = mc∆T = 1.50 kg 4 186 J kg ° C 40.0° C = 2.51 × 10 5 J .
Thus, the power supplied by the heater is
P=
W
Q 2.51 × 10 5 J
=
=
= 419 W
∆t ∆t
600 s
a∆V f = a110 V f
2
and the resistance is R =
P
2
419 W
= 28.9 Ω .
−8
115
116
*P27.40
Current and Resistance
The battery takes in energy by electric transmission
a fa f
e
FG 3 600 s IJ = 469 J .
H 1h K
j
P∆t = ∆V I ∆t = 2.3 J C 13.5 × 10 −3 C s 4.2 h
It puts out energy by electric transmission
a∆V fIa∆tf = 1.6 J C e18 × 10
−3
FG 3 600 s IJ = 249 J.
H 1h K
j
C s 2. 4 h
useful output 249 J
=
= 0.530
total input
469 J
(a)
efficiency =
(b)
The only place for the missing energy to go is into internal energy:
469 J = 249 J + ∆Eint
∆Eint = 221 J
(c)
We imagine toasting the battery over a fire with 221 J of heat input:
Q = mc∆T
∆T =
P27.41
221 J
Q
=
mc 0.015 kg
kg ° C
= 15.1° C
975 J
a f R = FG ∆V IJ = FG 140 IJ = 1.361
b g R H ∆V K H 120 K
F P − P IJ a100%f = FG P − 1IJ a100%f = a1.361 − 1f100% =
∆% = G
H P K
HP K
∆V
P
=
P0
∆V0
2
2
2
2
0
0
0
36.1%
0
a f a∆VR f = 500 W
a110 V f = 24.2 Ω
R=
a500 Wf
2
P27.42
P = I ∆V =
2
a24.2 Ωfπ e2.50 × 10
=
−4
m
j
2
(a)
ρ
R= A
A
(b)
R = R0 1 + α∆T = 24.2 Ω 1 + 0.400 × 10 −3 1 180 = 35.6 Ω
A=
so
RA
ρ
e
a∆V f = a110f
2
P=
R
35.6
2
= 340 W
1.50 × 10 −6 Ω ⋅ m
jb
g
= 3.17 m
Chapter 27
P27.43
R=
e
117
j
−6
ρA 1.50 × 10 Ω ⋅ m 25.0 m
=
= 298 Ω
2
A
π 0.200 × 10 −3 m
e
j
a
fa
f
∆V = IR = 0.500 A 298 Ω = 149 V
∆V 149 V
=
= 5.97 V m
25.0 m
A
(a)
E=
(b)
P = ∆V I = 149 V 0.500 A = 74.6 W
(c)
R = R0 1 + α T − T0 = 298 Ω 1 + 0.400 × 10 −3 ° C 320° C = 337 Ω
I=
a f a
fa
b
g
a
a
f
f
f
e
149 V
∆V
=
= 0.443 A
R
337 Ω
a f a
fa
j
f
P = ∆V I = 149 V 0.443 A = 66.1 W
P27.44
P27.45
a f a f a
∆U = q ∆V = It ∆V = 55.0 A ⋅ h 12.0 V
(b)
Cost = 0.660 kWh
FG $0.060 0 IJ =
H 1 kWh K
fFGH 11AC⋅ s IJK FGH 1 V1 J⋅ C IJK FGH 1 W1 J⋅ s IJK = 660 W ⋅ h =
0.660 kWh
3.96¢
a f ∆V = IR
a∆V f = a10.0f = 0.833 W
P=
P = I ∆V
2
R
*P27.46
fa
(a)
(a)
2
120
The resistance of 1 m of 12-gauge copper wire is
ρA
ρA
=
R=
A π d 2
b g
2
=
4ρ A
πd
2
=
e
j
4 1.7 × 10 −8 Ω ⋅ m 1 m
e
π 0.205 3 × 10
−2
j
m
2
= 5.14 × 10 −3 Ω .
a
f
2
The rate of internal energy production is P = I∆V = I 2 R = 20 A 5.14 × 10 −3 Ω = 2.05 W .
(b)
PAl = I 2 R =
PAl ρ Al
=
PCu ρ Cu
I 2 4ρ Al A
π d2
PAl =
2.82 × 10 −8 Ω ⋅ m
2.05 W = 3.41 W
1.7 × 10 −8 Ω ⋅ m
Aluminum of the same diameter will get hotter than copper.
118
*P27.47
Current and Resistance
The energy taken in by electric transmission for the fluorescent lamp is
a
sI
fFGH 3 600
J = 3.96 × 10 J
1h K
F $0.08 IJ FG k IJ FG W ⋅ s IJ FG h IJ = $0.088
JG
H kWh K H 1 000 K H J K H 3 600 s K
6
P∆t = 11 J s 100 h
cost = 3.96 × 10 6
For the incandescent bulb,
a
sI
fFGH 3 600
J = 1.44 × 10
1h K
F $0.08 IJ = $0.32
JG
H 3.6 × 10 J K
P∆t = 40 W 100 h
cost = 1.44 × 10 7
7
J
6
saving = $0.32 − $0.088 = $0.232
P27.48
The total clock power is
e270 × 10
jFGH
6
clocks 2.50
Js
clock
IJ FG 3 600 s IJ = 2.43 × 10
KH 1 h K
12
J h.
Wout
, the power input to the generating plants must be:
Qin
From e =
Q in Wout ∆ t 2.43 × 10 12 J h
=
=
= 9.72 × 10 12 J h
∆t
e
0.250
and the rate of coal consumption is
e
Rate = 9.72 × 10 12 J h
P27.49
a f a
fa
.00 kg coal I
jFGH 133.0
J = 2.95 × 10
× 10 J K
6
5
kg coal h = 295 metric ton h .
f
P = I ∆V = 1.70 A 110 V = 187 W
a
fa
f
Energy used in a 24-hour day = 0.187 kW 24.0 h = 4.49 kWh
FG $0.060 0 IJ = $0.269 = 26.9¢
H kWh K
P = I∆V = a 2.00 A fa120 V f = 240 W
∆E = b0.500 kg gb 4 186 J kg⋅° C ga77.0° C f = 161 kJ
∴ cost = 4.49 kWh
P27.50
int
∆t =
∆Eint
P
=
1.61 × 10 5 J
= 672 s
240 W
Chapter 27
P27.51
At operating temperature,
a
fa
f
(a)
P = I∆V = 1.53 A 120 V = 184 W
(b)
Use the change in resistance to find the final operating temperature of the toaster.
a
R = R0 1 + α∆T
f
120 120
1 + 0.400 × 10 −3 ∆T
=
1.53 1.80
e
∆T = 441° C
*P27.52
j
T = 20.0° C + 441° C = 461° C
You pay the electric company for energy transferred in the amount E = P ∆t
(a)
(b)
(c)
P27.53
119
7 d I F 86 400 s I F 1 J I
fFGH 1 week
JK GH 1 d JK GH 1 W ⋅ s JK = 48.4 MJ
F 7 d IJ FG 24 h IJ FG k IJ = 13.4 kWh
P ∆t = 40 Wa 2 weeksfG
H 1 week K H 1 d K H 1 000 K
F 7 d IJ FG 24 h IJ FG k IJ FG 0.12 $ IJ = $1.61
P ∆t = 40 Wa 2 weeksfG
H 1 week K H 1 d K H 1 000 K H kWh K
a
P ∆t = 40 W 2 weeks
h I F k I F 0.12 $ I
fFGH 601min
JK GH 1 000 JK GH kWh JK = $0.005 82
F 1 h IJ FG k IJ FG 0.12 $ IJ = $0.416
P ∆t = 5 200 Wa 40 minfG
H 60 min K H 1 000 K H kWh K
a
P ∆t = 970 W 3 min
= 0.582¢
Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to the
dryer is
b
gb
FG 1 kWh IJ ≈ 20 kWh .
H 3.6 × 10 J K
f
ga
P ∆t = 400 J s 600 s d 365 d ≈ 9 × 10 7 J
6
We suppose that electrically transmitted energy costs on the order of ten cents per kilowatt-hour.
Then the cost of using the dryer for a year is on the order of
a
fb
g
Cost ≅ 20 kWh $0.10 kWh = $2 ~ $1 .
120
Current and Resistance
Additional Problems
P27.54
(a)
I=
R=
(b)
(c)
(d)
∆V
R
∆V
a ∆V f
R
∆V f
a
a120 V f
R=
=
2
P = I∆V =
so
a f = a120 V f
2
2
25.0 W
P
2
= 576 Ω
and
100 W
P
2
= 144 Ω
P
25.0 W
Q 1.00 C
=
= 0.208 A =
=
∆V
120 V
∆t
∆t
1.00 C
∆t =
= 4.80 s
0.208 A
The bulb takes in charge at high potential and puts out the same amount of charge at low
potential.
I=
∆U 1.00 J
1.00 J
=
∆t =
= 0.040 0 s
∆t
∆t
25.0 W
The bulb takes in energy by electrical transmission and puts out the same amount of energy
by heat and light.
P = 25.0 W =
b
gb
f
ga
∆U = P∆t = 25.0 J s 86 400 s d 30.0 d = 64.8 × 10 8 J
The electric company sells energy .
FG $0.070 0 IJ FG k IJ FG W ⋅ s IJ FG h IJ = $1.26
H kWh K H 1 000 K H J K H 3 600 s K
$0.070 0 F
kWh I
Cost per joule =
G
J = $1.94 × 10 J
kWh H 3.60 × 10 J K
Cost = 64.8 × 10 6 J
−8
6
*P27.55
The original stored energy is Ui =
1
1 Q2
.
Q∆Vi =
2
2 C
(a)
When the switch is closed, charge Q distributes itself over the plates of C and 3C in parallel,
Q
presenting equivalent capacitance 4C. Then the final potential difference is ∆V f =
for
4C
both.
(b)
The smaller capacitor then carries charge C∆V f =
charge 3C
(c)
Q
3Q
.
=
4C
4
The smaller capacitor stores final energy
capacitor possesses energy
(d)
Q
Q
C=
. The larger capacitor carries
4C
4
FG IJ
H K
1
Q
3C
2
4C
2
=
1
C ∆V f
2
d i
2
=
FG IJ
H K
1
Q
C
2
4C
2
=
Q2
. The larger
32C
3Q 2
.
32C
Q 2 3Q 2 Q 2
+
=
. The loss of potential energy is the energy
32C 32C 8C
3Q 2
Q2 Q2
appearing as internal energy in the resistor:
=
+ ∆Eint
∆Eint =
.
8C
2C 8C
The total final energy is
Chapter 27
P27.56
vd =
v=
P27.57
I = nqv d A = nqv d π r 2
We find the drift velocity from
I
nqπ r
x
t
2
t=
=
1 000 A
8.49 × 10
28
m
−3
e1.60 × 10
−19
je
C π 10
m
= 2.34 × 10 −4 m s
1 dρ
.
ρ dT
We begin with the differential equation
α=
(a)
z z
Separating variables,
ln
2
200 × 10 3 m
x
=
= 8.54 × 10 8 s = 27.0 yr
v 2.34 × 10 −4 m s
ρ
dρ
ρ
ρ0
FG ρ IJ = α bT − T g and
Hρ K
T
= αdT
T0
ρ = ρ 0 eα bT −T0 g .
0
0
a
f
From the series expansion e x ≅ 1 + x , x << 1 ,
(b)
b
ρ ≅ ρ 0 1 + α T − T0
P27.58
j
−2
g
The resistance of one wire is
.
FG 0.500 Ω IJ a100 mif = 50.0 Ω .
H mi K
The whole wire is at nominal 700 kV away from ground potential, but the potential difference
between its two ends is
b
f
ga
IR = 1 000 A 50.0 Ω = 50.0 kV .
a f e
jb
g
Then it radiates as heat power P = ∆V I = 50.0 × 10 3 V 1 000 A = 50.0 MW .
P27.59
ρ=
a f
∆V A
RA
=
I A
A
A (m)
0.540
R ( Ω)
10.4
ρ ( Ω ⋅ m)
1.41 × 10 −6
1.028
21.1
1.50 × 10 −6
1.543
31.8
1.50 × 10 −6
ρ = 1.47 × 10 −6 Ω ⋅ m (in agreement with tabulated value of 1.50 × 10 −6 Ω ⋅ m in Table 27.1)
P27.60
2 wires → A = 100 m
R=
a
f
0.108 Ω
100 m = 0.036 0 Ω
300 m
(a)
a∆V f
(b)
P = I ∆V = 110 A 116 V = 12.8 kW
(c)
Pwires = I 2 R = 110 A
home
a f
= ∆V
a f a
line
a fb
fa
a
g
− IR = 120 − 110 0.036 0 = 116 V
f
f b0.036 0 Ωg =
2
436 W
121
122
P27.61
P27.62
P27.63
Current and Resistance
a
a
E=−
(b)
−8
ρA 4.00 × 10 Ω ⋅ m 0.500 m
=
= 0.637 Ω
R=
2
A
π 1.00 × 10 −4 m
(c)
I=
4.00 V
∆V
=
= 6.28 A
R
0.637 Ω
(d)
J=
I 6. 28 A
i=
A
π 1.00 × 10 −4 m
(e)
ρJ = 4.00 × 10 −8 Ω ⋅ m 2.00 × 10 8 i A m 2 = 8.00 i V m = E
(a)
E=−
(b)
R=
4 ρL
ρA
=
A
π d2
(c)
I=
Vπ d 2
∆V
=
R
4ρ L
(d)
J=
I V i=
i
ρL
A
(e)
ρJ=
f
e
j
e
j
e
2
= 2.00 × 10 8 i A m 2 = 200 i MA m 2
je
j
af
dV x V
i=
i
dx
L
V
i= E
L
b
g
so
I0
,
10
so
R = R0 1 + α T − T0
R=
ja
e
In this case, I =
P27.64
f
f
0 − 4.00 V
dV i=−
= 8.00 i V m
0.500 − 0 m
dx
(a)
LM OP
LM OP
N Q
N Q
1
9
T = T + a9f = 20°+
= 2 020° C
0.004 50 ° C
α
T = T0 +
1 R
1 I0
−1 .
− 1 = T0 +
α R0
α I
0
6.00
∆V 12.0
=
=
thus 12.0 I − 36.0 = 6.00 I and I = 6.00 A .
I
I
I − 3.00
Therefore, R =
a
f
12.0 V
= 2.00 Ω .
6.00 A
.
Chapter 27
P27.65
(a)
P = I∆V
so I =
(b)
123
P
8.00 × 10 3 W
=
= 667 A .
∆V
12.0 V
∆U
∆t =
P
=
2.00 × 10 7 J
3
8.00 × 10 W
= 2.50 × 10 3 s
b
ge
j
and ∆x = v∆t = 20.0 m s 2.50 × 10 3 s = 50.0 km .
P27.66
(a)
We begin with
R=
which reduces to
R=
b
g
b
b g
1 + α bT − T g 1 + α ′bT − T g
1 + 2α ′bT − T g
g
ρA ρ 0 1 + α T − T0 A 0 1 + α ′ T − T0
=
,
A
A0 1 + 2α ′ T − T0
R0
0
0
.
0
(b)
ρ 0 = 1.70 × 10 −8 Ω ⋅ m , α = 3.90 × 10 −3 ° C −1 , and α ′ = 17.0 × 10 −6 ° C −1
For copper:
R0 =
ja f
j
e
1.70 × 10 −8 2.00
ρ 0A0
=
= 1.08 Ω .
2
A0
π 0.100 × 10 −3
e
The simple formula for R gives:
a
ja
f e
f
R = 1.08 Ω 1 + 3.90 × 10 −3 ° C −1 100° C − 20.0° C = 1.420 Ω
while the more complicated formula gives:
R=
P27.67
a1.08 Ωf 1 + e3.90 × 10
−3
ja
ja
f e
° C ja80.0° C f
° C −1 80.0° C 1 + 17.0 × 10 −6 ° C −1 80.0° C
e
1 + 2 17.0 × 10 −6
f
= 1.418 Ω .
−1
Let α be the temperature coefficient at 20.0°C, and α ′ be the temperature coefficient at 0 °C. Then
a
f
a
ρ = ρ 0 1 + α T − 20.0° C , and ρ = ρ ′ 1 + α ′ T − 0° C
temperature T. That is, we must have:
f
must both give the correct resistivity at any
a
a
f
f
ρ 0 1 + α T − 20.0° C = ρ ′ 1 + α ′ T − 0° C .
a f
= ρ ′ 1 + α ′a 20.0° C f .
Setting T = 0 in equation (1) yields:
ρ ′ = ρ 0 1 − α 20.0° C ,
and setting T = 20.0° C in equation (1) gives:
ρ0
Put ρ ′ from the first of these results into the second to obtain:
a
f
a
f
ρ 0 = ρ 0 1 − α 20.0° C 1 + α ′ 20.0° C .
continued on next page
(1)
124
Current and Resistance
a
f
Therefore
1 + α ′ 20.0° C =
which simplifies to
α′ =
1
1 − α 20.0° C
a
f
α
.
1 − α 20.0° C
a
f
From this, the temperature coefficient, based on a reference temperature of 0°C, may be computed
for any material. For example, using this, Table 27.1 becomes at 0° C :
Material
Silver
Temp Coefficients at 0°C
4.1 × 10 −3 ° C
Copper
4.2 × 10 −3 ° C
Gold
3.6 × 10 −3 ° C
Aluminum
4.2 × 10 −3 ° C
Tungsten
4.9 × 10 −3 ° C
Iron
5.6 × 10 −3 ° C
Platinum
4.25 × 10 −3 ° C
Lead
4.2 × 10 −3 ° C
Nichrome
0.4 × 10 −3 ° C
Carbon
P27.68
−0.5 × 10 −3 ° C
Germanium
−24 × 10 −3 ° C
Silicon
−30 × 10 −3 ° C
(a)
A thin cylindrical shell of radius r, thickness dr, and length L contributes resistance
dR =
F
b g GH
I
JK
ρdA
ρdr
ρ dr
=
=
.
A
2π L r
2π r L
The resistance of the whole annulus is the series summation of the contributions of the thin
shells:
z
r
FG IJ
H K
r
ρ b dr
ρ
=
R=
ln b
ra
2π L r r
2π L
a
(b)
In this equation
we solve for
.
FG IJ
H K
r
∆V
ρ
=
ln b
I
ra
2π L
ρ=
2π L∆V
I ln rb ra
b
g
.
Chapter 27
P27.69
125
Each speaker receives 60.0 W of power. Using P = I 2 R , we then have
P
60.0 W
=
= 3.87 A .
4.00 Ω
R
I=
The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less .
P27.70
∆V = − E ⋅ A or dV = − E ⋅ dx
∆V = − IR = − E ⋅ A
I=
dq E ⋅ A A
A
dV
dV
=
=
= σA
E ⋅ A = E = −σA
dt
R
dx
dx
ρA
ρ
Current flows in the direction of decreasing voltage. Energy flows as heat in the direction of
decreasing temperature.
P27.71
R=
z
z
y − y1
ρdx
ρdx
=
where y = y1 + 2
x
A
wy
L
z
LM
g N
y − y1
ρL
ρL
dx
=
R=
ln y1 + 2
x
w 0 y1 + y 2 − y 1 L x w y 2 − y 1
L
R=
P27.72
b
b
g
F I
g GH JK
OP
Q
L
0
FIG. P27.71
y
ρL
ln 2
y1
w y 2 − y1
b
From the geometry of the longitudinal section of the resistor shown in the figure,
we see that
b−r
b−a
=
.
y
h
a f a f
From this, the radius at a distance y from the base is
For a disk-shaped element of volume dR =
Using the integral formula
*P27.73
za
du
au + b
f
2
=−
ρdy
:
π r2
R=
1
,
a au + b
a
f
ρ
π
R=
za
h
0
dy
fb g
a−b y h +b
ρ h
.
π ab
ρA d
=
.
A σA
κ ∈0 A
.
The capacitance of the capacitor is C =
d
d κ ∈0 A κ ∈0
=
Then RC =
is a characteristic of the material only.
σA d
σ
(a)
The resistance of the dielectric block is R =
(b)
R=
a f
FIG. P27.72
a f yh + b .
r = a−b
κ ∈0 ρκ ∈0 75 × 10 16 Ω ⋅ m 3.78 8.85 × 10 −12 C 2
=
=
= 1.79 × 10 15 Ω
C
σC
14 × 10 −9 F
N ⋅ m2
2
.
126
P27.74
Current and Resistance
LM F e∆V I − 1OP and R = ∆V
I
MN GH k T JK PQ
I = I 0 exp
B
with I 0 = 1.00 × 10 −9 A , e = 1.60 × 10 −19 C , and k B = 1.38 × 10 −23 J K .
The following includes a partial table of calculated values and a graph for each of the specified
temperatures.
(i)
For T = 280 K :
af
∆V V
af
a f
RΩ
I A
0. 400
0.015 6 25.6
0. 440
0.081 8
5.38
0. 480
0.429
1.12
2.25
0.232
0.520
0.560
11.8
0.047 6
0.600
61.6
0.009 7
FIG. P27.74(i)
(ii)
For T = 300 K :
af
∆V V
af
I A
af
RΩ
0. 400
0.005
77.3
0. 440
0.024
18.1
0. 480
0.114
4.22
0.520
0.534
0.973
0.560
2.51
0.223
0.600
11.8
0.051
FIG. P27.74(ii)
(iii)
For T = 320 K :
a f IaAf
af
∆V V
RΩ
0. 400
0.002 0 203
0. 440
0.008 4
52.5
0. 480
0.035 7
13.4
0.520
0.152
3.42
0.560
0.648
0.864
0.600
2.76
0.217
FIG. P27.74(iii)
Chapter 27
*P27.75
(a)
(b)
127
Think of the device as two capacitors in parallel. The one on the left has κ 1 = 1 ,
A
+ x A. The equivalent capacitance is
A1 =
2
∈ A
κ 1 ∈0 A1 κ 2 ∈0 A 2 ∈0 A A
κ ∈0 A A
− x = 0 A + 2 x + κA − 2κx .
+
=
+x +
d
d
d 2
d
2
2d
FG
H
IJ
K
FG
H
IJ
K
FG
H
IJ
K
a
f
The charge on the capacitor is Q = C∆V
∈ A∆ V
A + 2 x + κA − 2κx .
Q= 0
2d
The current is
∈ A∆Vv
dQ dQ dx ∈0 A∆V
=
=
κ −1 .
I=
0 + 2 + 0 − 2κ v = − 0
dt
dx dt
d
2d
The negative value indicates that the current drains charge from the capacitor. Positive
∈ A∆Vv
current is clockwise 0
κ −1 .
d
a
f
a
f
a f
a f
ANSWERS TO EVEN PROBLEMS
P27.2
3.64 h
P27.32
1.71 Ω
P27.4
(a) see the solution; (b) 1.05 mA
P27.34
0.153 Ω
P27.6
(a) 17.0 A ; (b) 85.0 kA m 2
P27.36
5.00 A , 24.0 Ω
P27.8
(a) 99.5 kA m 2 ; (b) 8.00 mm
P27.38
448 A
P27.10
(a) 221 nm ; (b) no; see the solution
P27.40
(a) 0.530; (b) 221 J; (c) 15.1°C
P27.12
30.3 MA m 2
P27.42
(a) 3.17 m ; (b) 340 W
P27.14
(a) 3.75 kΩ ; (b) 536 m
P27.44
(a) 0.660 kWh ; (b) 3.96¢
P27.16
0.018 1 Ω ⋅ m
P27.46
(a) 2.05 W; (b) 3.41 W; no
P27.18
2.71 MΩ
P27.48
295 metric ton h
P27.20
(a) 777 nΩ ; (b) 3. 28 µm s
P27.50
672 s
P27.22
rAl
= 1.29
rCu
P27.52
(a) $1.61; (b) $0.005 82; (c) $0.416
P27.54
(a) 576 Ω and 144 Ω ;
(b) 4.80 s; The charge is the same. The
charge-field system is in a lower-energy
configuration.
(c) 0.040 0 s; The energy enters by electric
transmission and exits by heat and
electromagnetic radiation;
(d) $1.26; energy; 1.94 × 10 −8 $ J
P27.24
378 Ω
P27.26
(a) nothing; (b) doubles; (c) doubles;
(d) nothing
P27.28
1.98 A
P27.30
carbon, 4.44 kΩ ; nichrome, 5.56 kΩ
128
Current and Resistance
P27.56
27.0 yr
P27.58
50.0 MW
P27.60
(a) 116 V ; (b) 12.8 kW ; (c) 436 W
P27.62
(a) E =
Vπ d
V i
4 ρL
;
; (c) I =
; (b) R =
2
L
4ρ L
πd
(d) J =
V i
; (e) see the solution
ρL
P27.64
2.00 Ω
2
P27.66
(a) see the solution;
(b) 1.418 Ω nearly agrees with 1.420 Ω
P27.68
(a) R =
P27.70
see the solution
P27.72
see the solution
P27.74
see the solution
2π L∆V
r
ρ
ln b ; (b) ρ =
I ln rb ra
2π L ra
b
g
28
Direct Current Circuits
CHAPTER OUTLINE
28.1
28.2
28.3
28.4
28.5
28.6
Electromotive Force
Resistors in Series and
Parallel
Kirchhoff’s Rules
RC Circuits
Electrical Meters
Household Wiring and
Electrical Safety
ANSWERS TO QUESTIONS
Q28.1
The load resistance in a circuit is the effective resistance of all of
the circuit elements excluding the emf source. In energy terms,
it can be used to determine the energy delivered to the load by
electrical transmission and there appearing as internal energy
to raise the temperature of the resistor. The internal resistance
of a battery represents the limitation on the efficiency of the
chemical reaction that takes place in the battery to supply
current to the load. The emf of the battery represents its
conversion of chemical energy into energy which it puts out by
electric transmission; the battery also creates internal energy
within itself, in an amount that can be computed from its
internal resistance. We model the internal resistance as
constant for a given battery, but it may increase greatly as the
battery ages. It may increase somewhat with increasing current
demand by the load. For a load described by Ohm’s law, the
load resistance is a precisely fixed value.
Q28.2
The potential difference between the terminals of a battery will equal the emf of the battery when
there is no current in the battery. At this time, the current though, and hence the potential drop
across the internal resistance is zero. This only happens when there is no load placed on the
battery—that includes measuring the potential difference with a voltmeter! The terminal voltage
will exceed the emf of the battery when current is driven backward through the battery, in at its
positive terminal and out at its negative terminal.
Q28.3
No. If there is one battery in a circuit, the current inside it will be from its negative terminal to its
positive terminal. Whenever a battery is delivering energy to a circuit, it will carry current in this
direction. On the other hand, when another source of emf is charging the battery in question, it will
have a current pushed through it from its positive terminal to its negative terminal.
Q28.4
Connect the resistors in series. Resistors of 5.0 kΩ, 7.5 kΩ and 2.2 kΩ connected in series present
equivalent resistance 14.7 kΩ.
Q28.5
Connect the resistors in parallel. Resistors of 5.0 kΩ, 7.5 kΩ and 2.2 kΩ connected in parallel present
equivalent resistance 1.3 kΩ.
129
130
Direct Current Circuits
Q28.6
Q28.7
In series, the current is the same through each resistor. Without knowing individual resistances,
nothing can be determined about potential differences or power.
Q28.8
In parallel, the potential difference is the same across each resistor. Without knowing individual
resistances, nothing can be determined about current or power.
Q28.9
In this configuration, the power delivered to one individual resistor is significantly less than if only
one equivalent resistor were used. This decreases the possibility of component failure, and possible
electrical disaster to some more expensive circuit component than a resistor.
Q28.10
Each of the two conductors in the extension cord itself has a small resistance. The longer the
extension cord, the larger the resistance. Taken into account in the circuit, the extension cord will
reduce the current from the power supply, and also will absorb energy itself in the form of internal
energy, leaving less power available to the light bulb.
Q28.11
The whole wire is very nearly at one uniform potential. There is essentially zero difference in
potential between the bird’s feet. Then negligible current goes through the bird. The resistance
through the bird’s body between its feet is much larger than the resistance through the wire
between the same two points.
Q28.12
The potential difference across a resistor is positive when it is measured against the direction of the
current in the resistor.
Q28.13
The bulb will light up for a while immediately after the switch is closed. As the capacitor charges, the
bulb gets progressively dimmer. When the capacitor is fully charged the current in the circuit is zero
and the bulb does not glow at all. If the value of RC is small, this whole process might occupy a very
short time interval.
Q28.14
An ideal ammeter has zero resistance. An ideal voltmeter has infinite resistance. Real meters cannot
attain these values, but do approach these values to the degree that they do not alter the current or
potential difference that is being measured within the accuracy of the meter. Hooray for
experimental uncertainty!
Q28.15
A short circuit can develop when the last bit of insulation frays away between the two conductors in
a lamp cord. Then the two conductors touch each other, opening a low-resistance branch in parallel
with the lamp. The lamp will immediately go out, carrying no current and presenting no danger. A
very large current exists in the power supply, the house wiring, and the rest of the lamp cord up to
the contact point. Before it blows the fuse or pops the circuit breaker, the large current can quickly
raise the temperature in the short-circuit path.
Chapter 28
131
Q28.16
A wire or cable in a transmission line is thick and made of material with very low resistivity. Only
when its length is very large does its resistance become significant. To transmit power over a long
distance it is most efficient to use low current at high voltage, minimizing the I 2 R power loss in the
transmission line. Alternating current, as opposed to the direct current we study first, can be stepped
up in voltage and then down again, with high-efficiency transformers at both ends of the power
line.
Q28.17
Car headlights are in parallel. If they were in series, both would go out when the filament of one
failed. An important safety factor would be lost.
Q28.18
Kirchhoff’s junction rule expresses conservation of electric charge. If the total current into a point
were different from the total current out, then charge would be continuously created or annihilated
at that point.
Kirchhoff’s loop rule expresses conservation of energy. For a single-loop circle with two
resistors, the loop rule reads +ε − IR1 − IR 2 = 0 . This is algebraically equivalent to qε = qIR1 + qIR 2 ,
where q = I∆t is the charge passing a point in the loop in the time interval ∆t . The equivalent
equation states that the power supply injects energy into the circuit equal in amount to that which
the resistors degrade into internal energy.
Q28.19
∆V 2
, the bulbs present resistances
R
2
2
2
120 V
120 V
120 V
∆V 2
= 190 Ω , and
= 72 Ω . The nominal 60 W lamp
R=
=
= 240 Ω , and
P
60 W
75 W
200 W
has greatest resistance. When they are connected in series, they all carry the same small current.
Here the highest-resistance bulb glows most brightly and the one with lowest resistance is faintest.
This is just the reverse of their order of intensity if they were connected in parallel, as they are
designed to be.
At their normal operating temperatures, from P =
a
f
a
f
a
f
Q28.20
Answer their question with a challenge. If the student is just looking at a diagram, provide the
materials to build the circuit. If you are looking at a circuit where the second bulb really is fainter, get
the student to unscrew them both and interchange them. But check that the student’s
understanding of potential has not been impaired: if you patch past the first bulb to short it out, the
second gets brighter.
Q28.21
Series, because the circuit breaker trips and opens the circuit when the current in that circuit loop
exceeds a certain preset value. The circuit breaker must be in series to sense the appropriate current
(see Fig. 28.30).
Q28.22
The hospital maintenance worker is right. A hospital room is full of electrical grounds, including the
bed frame. If your grandmother touched the faulty knob and the bed frame at the same time, she
could receive quite a jolt, as there would be a potential difference of 120 V across her. If the 120 V is
DC, the shock could send her into ventricular fibrillation, and the hospital staff could use the
defibrillator you read about in Section 26.4. If the 120 V is AC, which is most likely, the current could
produce external and internal burns along the path of conduction. Likely no one got a shock from
the radio back at home because her bedroom contained no electrical grounds—no conductors
connected to zero volts. Just like the bird in Question 28.11, granny could touch the “hot” knob
without getting a shock so long as there was no path to ground to supply a potential difference
across her. A new appliance in the bedroom or a flood could make the radio lethal. Repair it or
discard it. Enjoy the news from Lake Wobegon on the new plastic radio.
132
Direct Current Circuits
Q28.23
So long as you only grab one wire, and you do not touch anything that is grounded, you are safe
(see Question 28.11). If the wire breaks, let go! If you continue to hold on to the wire, there will be a
large—and rather lethal—potential difference between the wire and your feet when you hit the
ground. Since your body can have a resistance of about 10 kΩ, the current in you would be sufficient
to ruin your day.
Q28.24
Both 120-V and 240-V lines can deliver injurious or lethal shocks, but there is a somewhat better
safety factor with the lower voltage. To say it a different way, the insulation on a 120-V line can be
thinner. On the other hand, a 240-V device carries less current to operate a device with the same
power, so the conductor itself can be thinner. Finally, as we will see in Chapter 33, the last stepdown transformer can also be somewhat smaller if it has to go down only to 240 volts from the high
voltage of the main power line.
Q28.25
As Luigi Galvani showed with his experiment with frogs’ legs, muscles contract when electric
current exists in them. If an electrician contacts a “live” wire, the muscles in his hands and fingers
will contract, making his hand clench. If he touches the wire with the front of his hand, his hand will
clench around the wire, and he may not be able to let go. Also, the back of his hand may be drier
than his palm, so an actual shock may be much weaker.
Q28.26
Grab an insulator, like a stick or baseball bat, and bat for a home run. Hit the wire away from the
person or hit them away from the wire. If you grab the person, you will learn very quickly about
electrical circuits by becoming part of one.
Q28.27
A high voltage can lead to a high current when placed in a circuit. A device cannot supply a high
current—or any current—unless connected to a load. A more accurate sign saying potentially high
current would just confuse the poor physics student who already has problems distinguishing
between electrical potential and current.
Q28.28
The two greatest factors are the potential difference between the wire and your feet, and the
conductivity of the kite string. This is why Ben Franklin’s experiment with lightning and flying a
kite was so dangerous. Several scientists died trying to reproduce Franklin’s results.
Q28.29
Suppose ε = 12 V and each lamp has R = 2 Ω. Before the switch is closed the current is
12 V
= 2 A.
6Ω
The potential difference across each lamp is 2 A 2 Ω = 4 V . The power of each lamp is
2 A 4 V = 8 W , totaling 24 W for the circuit. Closing the switch makes the switch and the wires
connected to it a zero-resistance branch. All of the current through A and B will go through the
switch and (b) lamp C goes out, with zero voltage across it. With less total resistance, the (c) current
12 V
in the battery
= 3 A becomes larger than before and (a) lamps A and B get brighter. (d) The
4Ω
voltage across each of A and B is 3 A 2 Ω = 6 V , larger than before. Each converts power
3 A 6 V = 18 W , totaling 36 W, which is (e) an increase.
a fa f
a fa f
a fa f
Q28.30
a fa f
The starter motor draws a significant amount of current from the battery while it is starting the car.
This, coupled with the internal resistance of the battery, decreases the output voltage of the battery
below its the nominal 12 V emf. Then the current in the headlights decreases.
Chapter 28
Q28.31
. Three runs in parallel:
Two runs in series:
two runs:
. Junction of one lift and
.
Gustav Robert Kirchhoff, Professor of Physics at Heidelberg and Berlin, was master of the
obvious. A junction rule: The number of skiers coming into any junction must be equal to the
number of skiers leaving. A loop rule: the total change in altitude must be zero for any skier
completing a closed path.
SOLUTIONS TO PROBLEMS
Section 28.1
P28.1
(a)
(b)
Electromotive Force
P=
a∆V f
2
R
a11.6 V f
becomes
20.0 W =
so
R = 6.73 Ω .
2
R
FIG. P28.1
∆V = IR
a
f
so
11.6 V = I 6.73 Ω
and
I = 1.72 A
ε = IR + Ir
so
15.0 V = 11.6 V + 1.72 A r
a
f
r = 1.97 Ω .
P28.2
(a)
(b)
P28.3
∆Vterm = IR
a
becomes
10.0 V = I 5.60 Ω
so
I = 1.79 A .
∆Vterm = ε − Ir
a
f
fa
becomes
10.0 V = ε − 1.79 A 0.200 Ω
so
ε = 10.4 V .
The total resistance is R =
f
3.00 V
= 5.00 Ω .
0.600 A
(a)
Rlamp = R − rbatteries = 5.00 Ω − 0.408 Ω = 4.59 Ω
(b)
0.408 Ω I 2
Pbatteries
=
= 0.081 6 = 8.16%
Ptotal
5.00 Ω I 2
a
a
f
f
133
FIG. P28.3
134
P28.4
Direct Current Circuits
f
12.6 V
(b)
Let I 1 and I 2 be the currents flowing through the battery and the
headlights, respectively.
Then,
I 1 = I 2 + 35.0 A , and ε − I 1 r − I 2 r = 0
b5.00 Ω + 0.080 0 Ωg = 2.48 A .
Then, ∆V = IR = a 2.48 A fa5.00 Ωf = 12.4 V .
b
R+r
=
g a
gb
ε = I 2 + 35.0 A 0.080 0 Ω + I 2 5.00 Ω = 12.6 V
giving
I 2 = 1.93 A .
Thus,
∆V2 = 1.93 A 5.00 Ω = 9.65 V .
a
fa
FIG. P28.4
f
so
f
Resistors in Series and Parallel
a
f
g a fb
g
Therefore, a 2.00 A fR = a1.60 A fb R + 3.00 Ωg or R = 12.0 Ω .
b
∆V = I 1 R1 = 2.00 A R1 and ∆V = I 2 R1 + R 2 = 1.60 A R1 + 3.00 Ω
1
P28.6
ε
Here ε = I R + r , so I =
Section 28.2
P28.5
a
(a)
(a)
Rp =
1
1
1
b1 7.00 Ωg + b1 10.0 Ωg = 4.12 Ω
Rs = R1 + R 2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω
(b)
∆V = IR
34.0 V = I 17.1 Ω
a
f
FIG. P28.6
I = 1.99 A for 4.00 Ω, 9.00 Ω resistors.
a1.99 Afa4.12 Ωf = 8.18 V
8.18 V = I a7.00 Ωf
Applying ∆V = IR ,
so
I = 1.17 A for 7.00 Ω resistor
a
8.18 V = I 10.0 Ω
so
P28.7
f
I = 0.818 A for 10.0 Ω resistor.
For the bulb in use as intended,
P
75.0 W
=
= 0.625 A
∆V
120 V
∆V
120 V
=
= 192 Ω .
R=
0.625 A
I
I=
and
FIG. P28.7
Now, presuming the bulb resistance is unchanged,
I=
Across the bulb is
so its power is
120 V
= 0.620 A .
193.6 Ω
a
f
P = I∆V = 0.620 Aa119 V f =
∆V = IR = 192 Ω 0.620 A = 119 V
73.8 W .
Chapter 28
P28.8
120 V = IReq = I
∆V2 =
P28.9
Iρ
=
A2 A2
FG ρ
HA
e
+
1
1
A1
IJ
K
ρ
ρ
ρ
+
+
, or Iρ =
A 2 A3 A 4
a120 V f
+
1
A2
+
1
A3
+
1
A4
j
e
a120 Vf
1
A1
+
1
A2
+
1
A3
+
1
A4
135
j
= 29.5 V
If we turn the given diagram on its side, we find that it is the same as figure
(a). The 20.0 Ω and 5.00 Ω resistors are in series, so the first reduction is
shown in (b). In addition, since the 10.0 Ω, 5.00 Ω, and 25.0 Ω resistors are
then in parallel, we can solve for their equivalent resistance as:
1
Req =
= 2.94 Ω .
1
1
1
10.0 Ω + 5.00 Ω + 25 .0 Ω
c
h
This is shown in figure (c), which in turn reduces to the circuit shown in
figure (d).
∆V
Next, we work backwards through the diagrams applying I =
and
R
∆V = IR alternately to every resistor, real and equivalent. The 12.94 Ω
resistor is connected across 25.0 V, so the current through the battery in
every diagram is
∆V
25.0 V
=
= 1.93 A .
I=
12.94 Ω
R
In figure (c), this 1.93 A goes through the 2.94 Ω equivalent resistor to give a
potential difference of:
∆V = IR = 1.93 A 2.94 Ω = 5.68 V .
a
fa
f
From figure (b), we see that this potential difference is the same across ∆Vab ,
the 10 Ω resistor, and the 5.00 Ω resistor.
*P28.10
(b)
Therefore, ∆Vab = 5.68 V .
(a)
Since the current through the 20.0 Ω resistor is also the current
through the 25.0 Ω line ab,
∆Vab 5.68 V
=
= 0.227 A = 227 mA .
I=
25.0 Ω
R ab
FIG. P28.9
We assume that the metal wand makes low-resistance contact with the person’s hand and that the
resistance through the person’s body is negligible compared to the resistance Rshoes of the shoe soles.
The equivalent resistance seen by the power supply is 1.00 MΩ + Rshoes . The current through both
50.0 V
resistors is
. The voltmeter displays
1.00 MΩ + Rshoes
50.0 V 1.00 MΩ
∆V = I 1.00 MΩ =
= ∆V .
1.00 MΩ + Rshoes
a
f
a
f
a
a
f
f
b
50.0 V 1.00 MΩ = ∆V 1.00 MΩ + ∆V Rshoes
1.00 MΩ 50.0 − ∆V
Rshoes =
.
∆V
(a)
We solve to obtain
(b)
With Rshoes → 0 , the current through the person’s body is
50.0 V
= 50.0 µA
The current will never exceed 50 µA .
1.00 MΩ
a
f
g
136
P28.11
Direct Current Circuits
(a)
Since all the current in the circuit must pass through the series
100 Ω resistor, P = I 2 R
2
Pmax = RI max
P
25.0 W
=
= 0.500 A
100 Ω
R
I max =
so
R eq = 100 Ω +
FG 1 + 1 IJ
H 100 100 K
−1
FIG. P28.11
Ω = 150 Ω
∆Vmax = R eq I max = 75.0 V
a
fa
f
P = I∆V = 0.500 A 75.0 V = 37.5 W total power
(b)
P1 = 25.0 W
a
fa
P2 = P3 = RI 2 100 Ω 0.250 A
P28.12
2
= 6.25 W
Using 2.00-Ω, 3.00-Ω, 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations:
Series
2.00 Ω
3.00 Ω
4.00 Ω
5.00 Ω
P28.13
f
Parallel
0.923 Ω
1.20 Ω
1.33 Ω
1.71 Ω
6.00 Ω
7.00 Ω
9.00 Ω
Mixed
1.56 Ω
2.00 Ω
2.22 Ω
3.71 Ω
4.33 Ω
5.20 Ω
The resistors may be arranged in patterns:
The potential difference is the same across either combination.
∆V = IR = 3 I
1+
c
1
1
R
+
1
500
R
=3
500
h
FG 1 + 1 IJ = 3
H R 500 K
so
R
and
R = 1 000 Ω = 1.00 kΩ .
FIG. P28.13
*P28.14
When S is open, R1 , R 2 , R3 are in series with the battery. Thus:
6V
(1)
R1 + R 2 + R3 = −3 = 6 kΩ .
10 A
When S is closed in position 1, the parallel combination of the two R 2 ’s is in series with R1 , R3 , and the
battery. Thus:
1
6V
= 5 kΩ.
(2)
R1 + R 2 + R3 =
2
1.2 × 10 −3 A
When S is closed in position 2, R1 and R 2 are in series with the battery. R3 is shorted. Thus:
6V
= 3 kΩ.
(3)
R1 + R 2 =
2 × 10 −3 A
From (1) and (3): R3 = 3 kΩ.
Subtract (2) from (1): R 2 = 2 kΩ .
From (3): R1 = 1 kΩ.
Answers: R1 = 1.00 kΩ , R 2 = 2.00 kΩ , R3 = 3.00 kΩ .
Chapter 28
P28.15
FG 1 + 1 IJ = 0.750 Ω
H 3.00 1.00 K
= a 2.00 + 0.750 + 4.00f Ω = 6.75 Ω
−1
Rp =
Rs
137
∆V 18.0 V
=
= 2.67 A
Rs 6.75 Ω
I battery =
a
P = I 2R:
P2 = 2.67 A
f a2.00 Ωf
2
P2 = 14.2 W in 2.00 Ω
a
f a4.00 Af = 28.4 W
= a 2.67 A fa 2.00 Ωf = 5.33 V,
= a 2.67 A fa 4.00 Ωf = 10.67 V
P4 = 2.67 A
∆V2
∆V4
2
in 4.00 Ω
b
∆Vp = 18.0 V − ∆V2 − ∆V4 = 2.00 V = ∆V3 = ∆V1
P28.16
P3 =
b∆V g = a2.00 V f
P1 =
b∆V g = a2.00 Vf
3
2
2
3.00 Ω
R3
1
= 1.33 W in 3.00 Ω
FIG. P28.15
2
= 4.00 W in 1.00 Ω
1.00 Ω
R1
g
Denoting the two resistors as x and y,
x + y = 690, and
1
1 1
= +
150 x y
a
f
690 − x + x
1
1
1
= +
=
150 x 690 − x
x 690 − x
a
f
2
x − 690 x + 103 500 = 0
x=
690 ±
a690f
− 414 000
2
x = 470 Ω
*P28.17
2
y = 220 Ω
a f
A certain quantity of energy ∆Eint = P time is required to raise the temperature of the water to
a∆V f where a∆V f is a constant.
R
∆V f ∆t a ∆V f 2 ∆t
a
=
. Then R = 2 R .
Thus comparing coils 1 and 2, we have for the energy
2
100°C . For the power delivered to the heaters we have P = I∆V =
2
2
R1
(a)
2
R2
When connected in parallel, the coils present equivalent resistance
Rp =
a f
a f
2
2
∆V ∆t p
∆V ∆t
2 R1
1
1
=
=
=
. Now
1 R1 + 1 R 2 1 R1 + 1 2 R1
3
2 R1 3
R1
∆t p =
a ∆V f ∆t = a ∆ V f ∆ t
2
(b)
1
For the series connection, Rs = R1 + R 2 = R1 + 2 R1 = 3 R1 and
∆t s = 3 ∆t .
R1
2
3 R1
s
2 ∆t
.
3
138
P28.18
Direct Current Circuits
(a)
a
f
a
∆V = IR :
33.0 V = I1 11.0 Ω
33.0 V = I 2 22.0 Ω
I 3 = 3.00 A
I 2 = 1.50 A
P = I 2R:
P1 = 3.00 A
P1 = 99.0 W
a
f a11.0 Ωf
2
a
f
fa
2
P2 = 1.50 A 22.0 Ω
P2 = 49.5 W
f
The 11.0- Ω resistor uses more power.
FIG. P28.18(a)
a f a fa f
P = I ∆V = 4.50 33.0 = 148 W
(b)
P1 + P2 = 148 W
(c)
Rs = R1 + R 2 = 11.0 Ω + 22.0 Ω = 33.0 Ω
a
f
∆V = IR :
33.0 V = I 33.0 Ω , so I = 1.00 A
P = I 2R:
P1 = 1.00 A
P1 = 11.0 W
a
f a11.0 Ωf
2
a
f a22.0 Ωf
P2 = 1.00 A
P2 = 22.0 W
2
FIG. P28.18(c)
The 22.0- Ω resistor uses more power.
(d)
(e)
*P28.19
g a f a33.0 Ωf =
P = I a ∆V f = a1.00 A fa33.0 V f = 33.0 W
b
P1 + P2 = I 2 R1 + R2 = 1.00 A
2
33.0 W
The parallel configuration uses more power.
(a)
The resistors 2, 3, and 4 can be combined to a single 2R resistor. This is in series with
resistor 1, with resistance R, so the equivalent resistance of the whole circuit is 3R. In series,
1
potential difference is shared in proportion to the resistance, so resistor 1 gets of the
3
2
battery voltage and the 2-3-4 parallel combination get of the battery voltage. This is the
3
1
potential difference across resistor 4, but resistors 2 and 3 must share this voltage. goes to
3
2
2 and to 3. The ranking by potential difference is ∆V4 > ∆V3 > ∆V1 > ∆V2 .
3
(b)
Based on the reasoning above the potential differences are
ε
2ε
4ε
2ε
.
∆V1 = , ∆V2 =
, ∆V3 =
, ∆V4 =
3
9
9
3
(c)
All the current goes through resistor 1, so it gets the most. The current then splits at the
parallel combination. Resistor 4 gets more than half, because the resistance in that branch is
less than in the other branch. Resistors 2 and 3 have equal currents because they are in
series. The ranking by current is I 1 > I 4 > I 2 = I 3 .
(d)
Resistor 1 has a current of I. Because the resistance of 2 and 3 in series is twice that of
resistor 4, twice as much current goes through 4 as through 2 and 3. The current through
I
2I
the resistors are I 1 = I , I 2 = I 3 = , I 4 =
.
3
3
continued on next page
Chapter 28
(e)
Increasing resistor 3 increases the equivalent resistance of the entire circuit. The current in
the circuit, which is the current through resistor 1, decreases. This decreases the potential
difference across resistor 1, increasing the potential difference across the parallel
combination. With a larger potential difference the current through resistor 4 is increased.
With more current through 4, and less in the circuit to start with, the current through
resistors 2 and 3 must decrease. To summarize, I 4 increases and I 1 , I 2 , and I 3 decrease .
(f)
If resistor 3 has an infinite resistance it blocks any current from passing through that branch,
and the circuit effectively is just resistor 1 and resistor 4 in series with the battery. The circuit
3
now has an equivalent resistance of 4R. The current in the circuit drops to of the original
4
4
current because the resistance has increased by . All this current passes through resistors 1
3
3I
3I
, I2 = I3 = 0, I4 =
.
and 4, and none passes through 2 or 3. Therefore I 1 =
4
4
Section 28.3
P28.20
Kirchhoff’s Rules
a f a fa f
+15.0 − 7.00 I1 − 2.00 5.00 = 0
5.00 = 7.00 I 1
so
I 1 = 0.714 A
so
I 2 = 1.29 A
I 3 = I 1 + I 2 = 2.00 A
0.714 + I 2 = 2.00
a f
a f
+ε − 2.00 1.29 − 5.00 2.00 = 0
P28.21
139
FIG. P28.20
ε = 12.6 V
We name currents I 1 , I 2 , and I 3 as shown.
From Kirchhoff’s current rule, I 3 = I 1 + I 2 .
Applying Kirchhoff’s voltage rule to the loop containing I 2 and I 3 ,
a f a f
8.00 = a 4.00fI + a6.00fI
12.0 V − 4.00 I 3 − 6.00 I 2 − 4.00 V = 0
3
2
Applying Kirchhoff’s voltage rule to the loop containing I 1 and I 2 ,
a f
a f
− 6.00 I 2 − 4.00 V + 8.00 I1 = 0
a8.00fI
1
a f
= 4.00 + 6.00 I 2 .
FIG. P28.21
Solving the above linear system, we proceed to the pair of simultaneous equations:
RS8 = 4I + 4I
T8I = 4 + 6 I
1
1
2
+ 6I2
or
2
RS8 = 4I + 10I
TI = 1.33I − 0.667
1
2
2
1
and to the single equation 8 = 4I 1 + 13.3 I 1 − 6.67
I1 =
and
14.7 V
= 0.846 A .
17.3 Ω
I 3 = I1 + I 2
Then
give
a
f
I 2 = 1.33 0.846 A − 0.667
I 1 = 846 mA, I 2 = 462 mA, I 3 = 1.31 A .
All currents are in the directions indicated by the arrows in the circuit diagram.
140
P28.22
Direct Current Circuits
The solution figure is shown to the right.
FIG. P28.22
P28.23
We use the results of Problem 28.21.
(a)
a f a fa
f
a12.0 Vfa1.31 Af120 s = 1.88 kJ .
∆U = ∆V I∆t = 4.00 V −0.462 A 120 s = −222 J .
By the 4.00-V battery:
By the 12.0-V battery:
(b)
a
f a8.00 Ωf120 s = 687 J
a0.462 Af a5.00 Ωf120 s = 128 J .
a0.462 Af a1.00 Ωf120 s = 25.6 J .
a1.31 Af a3.00 Ωf120 s = 616 J .
a1.31 Af a1.00 Ωf120 s = 205 J .
I 2 R∆t = 0.846 A
By the 8.00-Ω resistor:
.
2
By the 5.00-Ω resistor:
2
By the 1.00-Ω resistor:
2
By the 3.00-Ω resistor:
2
By the 1.00-Ω resistor:
(c)
2
−222 J + 1.88 kJ = 1.66 kJ from chemical to electrical.
687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from electrical to internal.
P28.24
We name the currents I 1 , I 2 , and I 3 as shown.
[2]
f a
f
80.0 − I a 4.00 kΩf − 60.0 − I a3.00 kΩf = 0
[3]
I 2 = I1 + I 3
[1]
(a)
a
70.0 − 60.0 − I 2 3.00 kΩ − I 1 2.00 kΩ = 0
3
2
Substituting for I 2 and solving the resulting simultaneous
equations yields
bthrough R g
I = 2.69 mA bthrough R g
I = 3.08 mA bthrough R g
∆V = −60.0 V − a3.08 mA fa3.00 kΩf =
I 1 = 0.385 mA
(b)
1
3
3
2
2
cf
Point c is at higher potential.
−69.2 V
FIG. P28.24
Chapter 28
P28.25
Label the currents in the branches as shown in the first figure.
Reduce the circuit by combining the two parallel resistors as shown
in the second figure.
Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain:
and
a2.71RfI + a1.71RfI
a1.71RfI + a3.71RfI
1
2
= 250
1
2
= 500 .
(a)
With R = 1 000 Ω, simultaneous solution of these equations yields:
I 1 = 10.0 mA
and
I 2 = 130.0 mA .
From Figure (b),
Vc − Va = I 1 + I 2 1.71R = 240 V .
Thus, from Figure (a),
I4 =
b
ga
f
Vc − Va
240 V
=
= 60.0 mA .
4R
4 000 Ω
(b)
Finally, applying Kirchhoff’s point rule at point a in Figure (a)
gives:
FIG. P28.25
I = I 4 − I 1 = 60.0 mA − 10.0 mA = +50.0 mA ,
or
P28.26
I = 50.0 mA from point a to point e .
Name the currents as shown in the figure to the right. Then w + x + z = y . Loop
equations are
−200 w − 40.0 + 80.0 x = 0
FIG. P28.26
−80.0 x + 40.0 + 360 − 20.0 y = 0
+360 − 20.0 y − 70.0 z + 80.0 = 0
Eliminate y by substitution.
Eliminate x.
Eliminate z = 17.5 − 13.5 w to obtain
R|x = 2.50w + 0.500
S|400 − 100 x − 20.0 w − 20.0 z = 0
T440 − 20.0 w − 20.0 x − 90.0 z = 0
RS350 − 270 w − 20.0 z = 0
T430 − 70.0w − 90.0z = 0
430 − 70.0 w − 1 575 + 1 215 w = 0
w=
Now
70.0
= 1.00 A upward in 200 Ω .
70.0
z = 4.00 A upward in 70.0 Ω
x = 3.00 A upward in 80.0 Ω
y = 8.00 A downward in 20.0 Ω
and for the 200 Ω,
a
fa
f
∆V = IR = 1.00 A 200 Ω = 200 V .
141
142
P28.27
Direct Current Circuits
Using Kirchhoff’s rules,
b g b g
10.0 + a1.00fI − a0.060 0fI = 0
12.0 − 0.010 0 I 1 − 0.060 0 I 3 = 0
2
and
3
I1 = I 2 + I 3
a
f a
f
10.0 + a1.00fI − b0.060 0 gI = 0
12.0 − 0.010 0 I 2 − 0.070 0 I 3 = 0
2
FIG. P28.27
3
Solving simultaneously,
I 2 = 0.283 A downward in the dead battery
and
I 3 = 171 A downward in the starter.
The currents are forward in the live battery and in the starter, relative to normal starting operation.
The current is backward in the dead battery, tending to charge it up.
P28.28
a f a fb g
= a1.00fI + a1.00fI + a5.00fb I − I + I g
= a3.00fb I − I g + a5.00fb I − I + I g
∆Vab = 1.00 I 1 + 1.00 I 1 − I 2
∆Vab
∆Vab
1
2
1
1
1
2
2
Let I = 1.00 A , I 1 = x , and I 2 = y .
Then, the three equations become:
FIG. P28.28
∆Vab = 2.00 x − y , or y = 2.00 x − ∆Vab
∆Vab = −4.00 x + 6.00 y + 5.00
and ∆Vab = 8.00 − 8.00 x + 5.00 y .
Substituting the first into the last two gives:
7.00 ∆Vab = 8.00 x + 5.00 and 6.00 ∆Vab = 2.00 x + 8.00 .
Solving these simultaneously yields ∆Vab =
Then, R ab =
P28.29
27
V
∆Vab
= 17
1.00 A
I
27
V.
17
R ab =
or
27
Ω .
17
We name the currents I 1 , I 2 , and I 3 as shown.
(a)
I1 = I 2 + I 3
Counterclockwise around the top loop,
a
f a
f
12.0 V − 2.00 Ω I 3 − 4.00 Ω I 1 = 0 .
Traversing the bottom loop,
a
f a
f
8.00 V − 6.00 Ω I 2 + 2.00 Ω I 3 = 0
4 1
1
I 1 = 3.00 − I 3 , I 2 = + I 3 , and I 3 = 909 mA .
3 3
2
(b)
a
fa
f
Va − 0.909 A 2.00 Ω = Vb
Vb − Va = −1.82 V
FIG. P28.29
Chapter 28
P28.30
143
We apply Kirchhoff’s rules to the second diagram.
50.0 − 2.00 I 1 − 2.00 I 2 = 0
(1)
20.0 − 2.00 I 3 + 2.00 I 2 = 0
(2)
I1 = I 2 + I 3
(3)
Substitute (3) into (1), and solve for I 1 , I 2 , and I 3
I 1 = 20.0 A ; I 2 = 5.00 A ; I 3 = 15.0 A .
Then apply P = I 2 R to each resistor:
a
f a f a2.00 Ωf =
F 5.00 AIJ a4.00 Ωf = 25.0 W
P =G
H 2 K
a2.00 Ωf :
a4.00 Ωf :
P = I 12 2.00 Ω = 20.0 A
1
P28.32
FIG. P28.30
(Half of I 2 goes through each)
a
f a
P = I 32 2.00 Ω = 15.0 A
3
P28.31
800 W
2
a2.00 Ωf :
Section 28.4
2
f a2.00 Ωf =
2
450 W .
RC Circuits
e
je
j
(a)
RC = 1.00 × 10 6 Ω 5.00 × 10 −6 F = 5.00 s
(b)
Q = Cε = 5.00 × 10 −6 C 30.0 V = 150 µC
(c)
It =
(a)
I t = − I 0 e − t RC
5.10 × 10 −6 C
Q
=
= 1.96 A
I0 =
RC
1 300 Ω 2.00 × 10 −9 F
ja
e
af
FG
H
f
IJ
K
LM
MN e
ε − t RC
−10.0
30.0
=
exp
e
6
R
1.00 × 10
1.00 × 10 6 5.00 × 10 −6
je
OP
=
j PQ
4.06 µA
FIG. P28.31
af
b
ge
j
L −9.00 × 10 s OP
I at f = −a1.96 A f exp M
MN b1 300 Ωge2.00 × 10 Fj PQ = −61.6 mA
L −8.00 × 10 s OP
= b5.10 µC g exp M
qat f = Qe
MN a1 300 Ωfe2.00 × 10 Fj PQ = 0.235 µC
−6
−9
(b)
(c)
P28.33
U=
−6
− t RC
−9
The magnitude of the maximum current is I 0 = 1.96 A .
a f
1
C ∆V
2
2
and ∆V =
Q
.
C
Q2
and when the charge decreases to half its original value, the stored energy is one2C
1
quarter its original value: U f = U 0 .
4
Therefore, U =
144
P28.34
Direct Current Circuits
so
qt
= 1 − e − t RC
Q
0.600 = 1 − e −0 .900 RC
or
e −0.900 RC = 1 − 0.600 = 0.400
thus
RC =
af
a
−0.900
= ln 0.400
RC
*P28.35
f
−0.900
= 0.982 s .
ln 0.400
a
f
We are to calculate
z
∞
e −2 t RC dt = −
0
P28.36
af
q t = Q 1 − e − t RC
z
∞
IJ
K
FG
H
RC −2 t RC 2dt
RC −2 t RC
−
=−
e
e
RC
2 0
2
e
je
∞
0
=−
j
(a)
τ = RC = 1.50 × 10 5 Ω 10.0 × 10 −6 F = 1.50 s
(b)
τ = 1.00 × 10 5 Ω 10.0 × 10 −6 F = 1.00 s
(c)
The battery carries current
10.0 V
= 200 µA .
50.0 × 10 3 Ω
The 100 kΩ carries current of magnitude
I = I 0 e − t RC =
e
je
j
FG 10.0 V IJ e
H 100 × 10 Ω K
200 µA + b100 µA ge
.
(a)
− t 1.00 s
3
.
− t 1.00 s
So the switch carries downward current
P28.37
RC −∞
RC
RC
e − e0 = −
0−1 = +
.
2
2
2
Call the potential at the left junction VL and at the right VR . After a
“long” time, the capacitor is fully charged.
VL = 8.00 V because of voltage divider:
10.0 V
= 2.00 A
5.00 Ω
VL = 10.0 V − 2.00 A 1.00 Ω = 8.00 V
IL =
a
fa f
F 2.00 Ω IJ a10.0 V f = 2.00 V
=G
H 2.00 Ω + 8.00 Ω K
Likewise,
VR
or
IR =
10.0 V
= 1.00 A
10.0 Ω
a
f a
fa
FIG. P28.37(a)
f
VR = 10.0 V − 8.00 Ω 1.00 A = 2.00 V .
(b)
Therefore,
∆V = VL − VR = 8.00 − 2.00 = 6.00 V .
Redraw the circuit
R=
and
so
1
= 3.60 Ω
1 9.00 Ω + 1 6.00 Ω
b
g b
g
RC = 3.60 × 10 −6 s
1
e − t RC =
10
t = RC ln 10 = 8. 29 µs .
FIG. P28.37(b)
Chapter 28
*P28.38
(a)
af
af
t = RC ln
FG
H
∆V0
t
= ln
RC
∆V
∆V0
= e + t RC
∆V
∆V
= e − t RC
∆V0
P28.39
3 000 V
We model the person’s body and street shoes
as shown. For the discharge to reach 100 V,
q t = Qe − t RC = C∆V t = C∆V0 e − t RC
FG ∆V IJ = 5 000 × 10 Ωe230 × 10
H ∆V K
6
0
e
−12
150 pF
IJ
K
000 I
JK =
j FGH 3100
F ln
t = 1 × 10 6 V A 230 × 10 −12 C V ln 30 = 782 µs
(a)
τ = RC = 4.00 × 10 6 Ω 3.00 × 10 −6 F = 12.0 s
(b)
I=
je
3.91 s
ε − t RC
12.0
=
e
e − t 12.0 s
6
R
4.00 × 10
a f
q = 36.0 µC 1 − e − t 12 .0
∆V0 =
FIG. P28.38(a)
j
q = Cε 1 − e − t RC = 3.00 × 10 −6 12.0 1 − e − t 12.0
P28.40
5 000 M Ω
80 pF
j
(b)
e
I = 3.00 µAe − t 12 .0
Q
C
FIG. P28.39
b g
Then, if q(t ) = Q e − t RC
∆V ( t ) = ∆V0 e − t RC
and
b∆V g = e
∆V ( t )
− t RC .
0
b g
When ∆V ( t ) = 1 ∆V0 , then
2
e − t RC =
−
Section 28.5
P28.41
FG IJ
H K
t
1
= ln
= − ln 2 .
RC
2
R=
Thus,
1
2
t
C ln 2
a f
.
Electrical Meters
e
j
∆V = I g rg = I − I g R p , or R p =
I g rg
eI − I
g
a f
j eI − I j
=
I g 60.0 Ω
g
Therefore, to have I = 0.100 A = 100 mA when I g = 0.500 mA :
Rp =
a0.500 mAfa60.0 Ωf =
99.5 mA
145
0.302 Ω .
FIG. P28.41
146
P28.42
Direct Current Circuits
a
f e
j
Applying Kirchhoff’s loop rule, − I g 75.0 Ω + I − I g R p = 0 .
Therefore, if I = 1.00 A when I g = 1.50 mA ,
Rp =
a f = e1.50 × 10 Aja75.0 Ωf =
eI − I j 1.00 A − 1.50 × 10 A
−3
I g 75.0 Ω
−3
g
P28.43
Series Resistor → Voltmeter
0.113 Ω .
b
25.0 = 1.50 × 10 −3 Rs + 75.0
∆V = IR :
FIG. P28.42
g
Rs = 16.6 kΩ .
Solving,
FIG. P28.43
P28.44
(a)
In Figure (a), the
emf sees an
equivalent
resistance of
200.00 Ω .
6.000 0 V
200.00 Ω
= 0.030 000 A
I=
6.0000 V
A
V
A
V
180.00 Ω
180.00 Ω
180.00 Ω
(a)
(b)
(c)
FIG. P28.44
(b)
20.000 Ω
20.000 Ω
20.000 Ω
b
f
ga
The terminal potential difference is
∆V = IR = 0.030 000 A 180.00 Ω = 5.400 0 V .
In Figure (b),
R eq =
F 1 + 1 I
GH 180.00 Ω 20 000 Ω JK
−1
= 178.39 Ω.
The equivalent resistance across the emf is 178.39 Ω + 0.500 00 Ω + 20.000 Ω = 198.89 Ω .
The ammeter reads
and the voltmeter reads
I=
ε
R
=
6.000 0 V
= 0.030 167 A
198.89 Ω
b
ga
F 1 + 1 I
GH 180.50 Ω 20 000 Ω JK
f
∆V = IR = 0.030 167 A 178.39 Ω = 5.381 6 V .
−1
(c)
In Figure (c),
= 178.89 Ω .
Therefore, the emf sends current through
Rtot = 178.89 Ω + 20.000 Ω = 198.89 Ω.
The current through the battery is
I=
but not all of this goes through the ammeter.
6.000 0 V
= 0.030 168 A
198.89 Ω
b
ga
f
The voltmeter reads
∆V = IR = 0.030 168 A 178.89 Ω = 5.396 6 V .
The ammeter measures current
I=
∆V 5.396 6 V
=
= 0.029 898 A .
R
180.50 Ω
The connection shown in Figure (c) is better than that shown in Figure (b) for accurate
readings.
Chapter 28
P28.45
Consider the circuit diagram shown, realizing that
I g = 1.00 mA . For the 25.0 mA scale:
a24.0 mAfbR
1
g a
fa
+ R 2 + R3 = 1.00 mA 25.0 Ω
f
FG
H
FIG. P28.45
IJ
K
25.0
R1 + R 2 + R3 =
Ω.
24.0
or
(1)
(2)
For the 100 mA scale:
a49.0 mAfbR + R g = a1.00 mAfb25.0 Ω + R g
49.0b R + R g = 25.0 Ω + R .
a99.0 mAfR = a1.00 mAfb25.0 Ω + R + R g
or
99.0 R1 = 25.0 Ω + R 2 + R3 .
(3)
For the 50.0 mA scale:
1
or
1
2
3
2
3
1
2
3
Solving (1), (2), and (3) simultaneously yields
R1 = 0.260 Ω , R 2 = 0.261 Ω , R3 = 0.521 Ω .
P28.46
∆V = IR
(a)
jb
e
20.0 V = 1.00 × 10 −3 A R1 + 60.0 Ω
g
R1 = 1.994 × 10 4 Ω = 19.94 kΩ
P28.47
FIG. P28.46
e
jb
g
R 2 = 30.0 kΩ
e
jb
g
R3 = 50.0 kΩ
(b)
50.0 V = 1.00 × 10 −3 A R 2 + R1 + 60.0 Ω
(c)
100 V = 1.00 × 10 −3 A R3 + R1 + 60.0 Ω
e
ja
Ammeter:
I g r = 0.500 A − I g 0.220 Ω
or
I g r + 0.220 Ω = 0.110 V
Voltmeter:
2.00 V = I g r + 2 500 Ω
a
f
b
f
(1)
g
(2)
Solve (1) and (2) simultaneously to find:
I g = 0.756 mA and r = 145 Ω .
FIG. P28.47
Section 28.6
Household Wiring and Electrical Safety
FG ρ IJ = a1.00 Af e1.70 × 10 Ω ⋅ mja16.0 ftfb0.304 8 m ftg =
H AK
π e0.512 × 10 mj
2
P28.48
147
2
2
(a)
P =I R=I
(b)
P = I 2 R = 100 0.101 Ω = 10.1 W
−8
−3
a
f
2
0.101 W
148
P28.49
Direct Current Circuits
(a)
P = I∆V :
1 500 W
P
=
= 12.5 A .
∆V
120 V
750 W
I=
= 6.25 A .
120 V
I=
So for the Heater,
For the Toaster,
I=
And for the Grill,
(b)
1 000 W
= 8.33 A .
120 V
12.5 + 6.25 + 8.33 = 27.1 A
The current draw is greater than 25.0 amps, so this circuit breaker would not be sufficient.
P28.50
2
2
I Al
R Al = I Cu
RCu
P28.51
(a)
RCu
I Cu =
R Al
I Al =
so
ρ Cu
1.70
20.0 = 0.776 20.0 = 15.5 A
I Cu =
2.82
ρ Al
a f
a f
Suppose that the insulation between either of your fingers and the conductor adjacent is a
chunk of rubber with contact area 4 mm 2 and thickness 1 mm. Its resistance is
e
je
j
10 13 Ω ⋅ m 10 −3 m
ρ
≈
≈ 2 × 10 15 Ω .
A
4 × 10 −6 m 2
R=
The current will be driven by 120 V through total resistance (series)
2 × 10 15 Ω + 10 4 Ω + 2 × 10 15 Ω ≈ 5 × 10 15 Ω .
It is: I =
(b)
∆V
120 V
~
~ 10 −14 A .
15
R
5 × 10 Ω
Vh
, where Vh
2
is the potential of the “hot” wire. The potential difference between your finger and thumb is
∆V = IR ~ 10 −14 A 10 4 Ω ~ 10 −10 V . So the points where the rubber meets your fingers are
The resistors form a voltage divider, with the center of your hand at potential
e
je
j
at potentials of
~
Vh
+ 10 −10 V
2
and
~
Vh
− 10 −10 V .
2
Additional Problems
P28.52
The set of four batteries boosts the electric potential of each bit of charge that goes through them by
4 × 1.50 V = 6.00 V . The chemical energy they store is
a
fb
g
∆U = q∆V = 240 C 6.00 J C = 1 440 J .
∆V 6.00 V
=
= 0.030 0 A .
200 Ω
R
The radio draws current
I=
So, its power is
P = ∆V I = 6.00 V 0.030 0 A = 0.180 W = 0.180 J s.
a f a
Then for the time the energy lasts, we have P =
We could also compute this from I =
Q
:
∆t
fb
E
:
∆t
∆t =
∆t =
g
E
P
=
1 440 J
= 8.00 × 10 3 s .
0.180 J s
240 C
Q
=
= 8.00 × 10 3 s = 2. 22 h .
I 0.030 0 A
149
Chapter 28
P28.53
aR + r f = FGH εP IJK R .
R + a 2 r − x fR − r = 0 .
R + a 2.40 − xfR − 1.44 = 0 ,
−a 2.40 − x f ± a 2.40 − x f
R=
ε
ε 2R
, so P = I 2 R =
or
2
R+r
R+r
ε2
2
, then R + r = xR or
Let x ≡
P
With r = 1. 20 Ω, this becomes
I=
a
f
a
2
f
2
With
(b)
For
2
− 5.76
2
ε = 9.20 V and
R=
2
2
which has solutions of
(a)
2
+4.21 ±
P = 12.8 W , x = 6.61 :
a4.21f
2
− 5.76
2
= 3.84 Ω or
ε = 9.20 V and
+1.59 ±
.
0.375 Ω .
P = 21.2 W , x ≡
a1.59f
2
ε2
= 3.99
P
− 5.76
1.59 ± −3.22
=
.
2
2
The equation for the load resistance yields a complex number, so there is no resistance
R=
that will extract 21.2 W from this battery. The maximum power output occurs when
R = r = 1.20 Ω, and that maximum is: Pmax =
P28.54
ε2
= 17.6 W .
4r
Using Kirchhoff’s loop rule for the closed loop, +12.0 − 2.00 I − 4.00 I = 0 , so I = 2.00 A
a
fa
f a fa
f
Vb − Va = +4.00 V − 2.00 A 4.00 Ω − 0 10.0 Ω = −4.00 V .
Thus, ∆Vab = 4.00 V and point a is at the higher potential .
P28.55
*P28.56
I=
ε
3R
Pseries = ε I =
I=
3ε
R
Pparallel = ε I =
(a)
Req = 3 R
(b)
Req =
(c)
Nine times more power is converted in the parallel connection.
(a)
We model the generator as a constant-voltage power supply.
Connect two light bulbs across it in series. Each bulb is designed to
P
100 W
=
= 0.833 A . Each has resistance
carry current I =
∆V 120 V
∆V
120 V
=
= 144 Ω . In the 240-V circuit the equivalent
R=
0.833 A
I
resistance is 144 Ω + 144 Ω = 288 Ω . The current is
∆V 240 V
=
= 0.833 A and the generator delivers power
I=
288 Ω
R
P = I∆V = 0.833 A 240 V = 200 W .
1
R
=
3
1R + 1R + 1R
b g b g b g
a
continued on next page
f
ε2
3R
3ε 2
R
FIG. P28.56(a)
150
Direct Current Circuits
(b)
The hot pot is designed to carry
current
I=
P
∆V
=
28.8 Ω
500 W
= 4.17 A .
120 V
144 Ω
240 V
It has resistance
R=
FIG. P28.56(b)
∆V 120 V
=
= 28.8 Ω .
4.17 A
I
4.17 A
= 5 , we can place five light bulbs in parallel and the hot
0.833 A
pot in series with their combination. The current in the generator is then 4.17 A and it
In terms of current, since
a
f
delivers power P = I∆V = 4.17 A 240 V = 1 000 W .
P28.57
The current in the simple loop circuit will be I =
(a)
∆Vter = ε − Ir =
(b)
I=
(c)
P = I2R = ε 2
εR
R+r
ε
R+r
Then 2R = R + r
P28.58
.
and
∆Vter → ε as R → ∞ .
and
I→
ε
as R → 0 .
r
ε2
− 2ε 2 R
dP
=
+
3
dR
R+r
R+r
R
aR + r f
ε
R+r
a
2
and
f a
f
2
FIG. P28.57
=0
R=r .
af
e
j
The potential difference across the capacitor
∆V t = ∆Vmax 1 − e − t RC .
Using 1 Farad = 1 s Ω ,
4.00 V = 10.0 V 1 − e
Therefore,
0.400 = 1.00 − e
Or
e
Taking the natural logarithm of both sides,
−
and
R=−
fLMN
a
e
j
− 3.00 ×10 5 Ω R
a
− 3.00 s
e
j
− 3 .00 × 10 5 Ω R
f Re10.0 ×10
−6
sΩ
j O.
PQ
.
= 0.600 .
3.00 × 10 5 Ω
= ln 0.600
R
a
f
3.00 × 10 5 Ω
= +5.87 × 10 5 Ω = 587 kΩ .
ln 0.600
a
f
Chapter 28
P28.59
Let the two resistances be x and y.
Ps
225 W
=
= 9.00 Ω
2
2
I
5.00 A
Then,
Rs = x + y =
and
Rp =
so
x 9.00 Ω − x
= 2.00 Ω
x + 9.00 Ω − x
a
a
y = 9.00 Ω − x
f
x
Pp
xy
50.0 W
= 2 =
= 2.00 Ω
2
x+y I
5.00 A
f
a
a
y
x
f
y
x 2 − 9.00 x + 18.0 = 0 .
f
Factoring the second equation,
ax − 6.00fax − 3.00f = 0
so
x = 6.00 Ω or x = 3.00 Ω .
Then, y = 9.00 Ω − x gives
y = 3.00 Ω or y = 6.00 Ω .
FIG. P28.59
The two resistances are found to be 6.00 Ω and 3.00 Ω .
P28.60
Let the two resistances be x and y.
Then, Rs = x + y =
Pp
xy
Ps
=
=
and
R
.
p
x + y I2
I2
From the first equation, y =
becomes
e
x Ps I 2 − x
e
2
j
x + Ps I − x
j
=
Pp
I
2
Using the quadratic formula, x =
Then, y =
(a)
(b)
FG P IJ x + P P
HI K I
s p
4
s
2
Ps ± Ps 2 − 4Ps Pp
2I 2
Ps + Ps 2 − 4Ps Pp
2I 2
and
= 0.
FIG. P28.60
.
Ps − Ps 2 − 4Ps Pp
2I 2
.
c ∑ R h − bε + ε g = 0
40.0 V − a 4.00 A f a 2.00 + 0.300 + 0.300 + RfΩ − a6.00 + 6.00 f V = 0 ;
ε−I
1
2
g a
fa
2
f
2
2
2
For the limiting resistor,
b
so
R = 4.40 Ω
a f a2.00 Ωf = 32.0 W .
P = I R = a 4.00 A f a0.600 Ωf = 9.60 W .
P = a 4.00 A f a 4.40 Ωf = 70.4 W .
P = I 2 R = 4.00 A
Inside the supply,
Inside both batteries together,
(c)
y
Ps ∓ Ps 2 − 4Ps Pp
Ps
.
−
=
x
gives
y
2I 2
I2
The two resistances are
P28.61
x
Ps
− x , and the second
I2
or x 2 −
y
x
P = I ε 1 + ε 2 = 4.00 A 6.00 + 6.00 V = 48.0 W
151
152
*P28.62
Direct Current Circuits
(a)
∆V1 = ∆V2
I 1 R1 = I 2 R 2
I R
R + R1
I = I 1 + I 2 = I1 + 1 1 = I1 2
R2
R2
I1 =
I2 =
(b)
R2
I2
FIG. P28.62(a)
I 1 R1
IR1
=
= I2
R2
R1 + R 2
b
The power delivered to the pair is P = I12 R1 + I 22 R2 = I 12 R1 + I − I1
dP
we want to find I 1 such that
= 0.
dI 1
ga f
dP
= 2 I 1 R1 + 2 I − I 1 −1 R 2 = 0
dI 1
I1 =
I1
I
IR 2
R1 + R 2
b
R1
g R . For minimum power
2
2
I 1 R1 − IR 2 + I 1 R 2 = 0
IR 2
R1 + R 2
This is the same condition as that found in part (a).
P28.63
Let Rm = measured value, R = actual value,
(a)
I R = current through the resistor R
I = current measured by the ammeter.
(a)
b
g
When using circuit (a), I R R = ∆V = 20 000 I − I R or R = 20 000
But since I =
∆V
∆V
and I R =
, we have
Rm
R
I
R
=
I R Rm
FIG. P28.63
bR − R g .
m
R = 20 000
When R > Rm , we require
bR − R g ≤ 0.050 0 .
b
When using circuit (b),
But since I R =
∆V
,
Rm
When Rm > R , we require
From (2) we find
(b)
R
and
Rm
(1)
m
R
g
Therefore, Rm ≥ R 1 − 0.050 0 and from (1) we find
(b)
LM I − 1OP .
NI Q
R ≤ 1 050 Ω .
a
f
I R R = ∆V − I R 0.5 Ω .
a
f
Rm = 0.500 + R .
bR
m
−R
R
g ≤ 0.050 0 .
R ≥ 10.0 Ω .
(2)
Chapter 28
P28.64
FG
H
dE
ε −1 RC
e
.
= P = εI = ε
dt
R
Then the total energy put out by the battery is
z z
z
dE =
FG
H
fz
∞
ε2
t
− RC exp −
R
RC
0
a
dE =
IJ FG − dt IJ = −ε C expFG − t IJ
K H RC K
H RC K
IJ
K
FG
H
∞
ε2
t
exp −
dt
R
RC
t=0
∞
2
= −ε 2 C 0 − 1 = ε 2C .
0
FG
H
IJ
K
The power delivered to the resistor is
ε2
dE
2t
.
= P = ∆VR I = I 2 R = R 2 exp −
dt
RC
R
So the total internal energy appearing in the resistor is
z z
IJ z expFG − 2t IJ FG − 2dt IJ = − ε C expFG − 2t IJ
z
K H RC K H RC K 2 H RC K
1
The energy finally stored in the capacitor is U = C a ∆V f
2
dE =
FG
H
∞
ε2
RC
−
R
2
battery.
(a)
2
0
conserved ε 2 C =
P28.65
IJ
K
The battery supplies energy at a changing rate
153
e
I=
0
2
=
1 2
Cε . Thus, energy of the circuit is
2
j
L
F ja10.0 V fM1 − e
N
−10.0
e 2.00 ×10 je1.00 ×10 j O =
−6
6
PQ
9.93 µC
FG IJ
H K
FG 10.0 V IJ e = 3.37 × 10 A = 33.7 nA
H 2.00 × 10 Ω K
dU d F 1 q I F q I dq F q I
=G J
= G JI
=
dt dt GH 2 C JK H C K dt H C K
dU F 9.93 × 10 C I
=
e3.37 × 10 Aj = 3.34 × 10 W = 334 nW
dt GH 1.00 × 10 C V JK
P
= Iε = e3.37 × 10 A ja10.0 V f = 3.37 × 10 W = 337 nW
−5.00
6
−8
2
−6
−8
−6
battery
IJ
K
ε 2C
ε 2C
0−1 =
.
2
2
dq
∆V − t RC
=
e
dt
R
I=
(d)
=−
FG
H
ε2
2t
exp −
dt
R
RC
0
q = C∆V 1 − e − t RC
e
(c)
∞
∞
1 2
1
ε C + ε 2C and resistor and capacitor share equally in the energy from the
2
2
q = 1.00 × 10 −6
(b)
dE =
−8
−7
−7
154
P28.66
Direct Current Circuits
Start at the point when the voltage has just reached
2
∆V
3
2
∆V and is
3
decaying towards 0 V with a time constant R 2 C
and the switch has just closed. The voltage is
∆V
Voltage
controlled
switch
a f LMN 32 ∆V OPQe .
1
We want to know when ∆V atf will reach ∆V .
3
1
L2 O
Therefore,
∆V = M ∆V P e
3
N3 Q
R2
− t R 2C
∆VC t =
+
R1
C
V ∆Vc
C
− t R2C
1
2
or
e − t R 2C =
or
t1 = R 2 C ln 2 .
FIG. P28.66
1
After the switch opens, the voltage is ∆V , increasing toward ∆V with time constant R1 + R 2 C :
3
b
LM 2 ∆V OPe
N3 Q
af
∆VC t = ∆V −
.
af
2
∆V
3
2
2
∆V = ∆V − ∆Ve − t b R1 + R2 gC
3
3
b
g
t 2 = R1 + R 2 C ln 2
So
(a)
g
∆VC t =
When
P28.67
b
− t R1 + R 2 C
g
or
e − t b R1 + R2 gC =
and
T = t1 + t 2 =
First determine the resistance of each light bulb: P =
a∆V f = a120 Vf
2
R=
bR
1
g
+ 2 R 2 C ln 2 .
2
R
2
60.0 W
P
a∆V f
1
.
2
= 240 Ω .
We obtain the equivalent resistance Req of the network of light
FIG. P28.67
bulbs by identifying series and parallel equivalent resistances:
Req = R1 +
1
= 240 Ω + 120 Ω = 360 Ω.
+ 1 R3
b1 R g b g
2
a∆V f = a120 Vf
2
The total power dissipated in the 360 Ω is
(b)
P=
The current through the network is given by P = I 2 Req : I =
The potential difference across R1 is
Req
360 Ω
2
= 40.0 W .
P
40.0 W 1
=
= A.
360 Ω
3
Req
∆V1 = IR1 =
FG 1 AIJ a240 Ωf =
H3 K
The potential difference ∆V23 across the parallel combination of R 2 and R3 is
∆V23 = IR 23 =
I=
1
FG 1 AIJ FG
H 3 K H b1 240 Ωg + b1 240 Ωg JK
40.0 V .
80.0 V .
Chapter 28
*P28.68
(a)
155
With the switch closed, current exists in a simple series
circuit as shown. The capacitors carry no current. For R 2
we have
P
2.40 V ⋅ A
=
= 18.5 mA .
P = I 2 R2
I=
R2
7 000 V A
The potential difference across R1 and C 1 is
jb
e
g
∆V = IR1 = 1.85 × 10 −2 A 4 000 V A = 74.1 V .
The charge on C 1
ja
e
FIG. P28.68(a)
f
Q = C1 ∆V = 3.00 × 10 −6 C V 74.1 V = 222 µC .
The potential difference across R 2 and C 2 is
jb
e
g
∆V = IR 2 = 1.85 × 10 −2 A 7 000 Ω = 130 V .
The charge on C 2
ja
e
f
Q = C 2 ∆V = 6.00 × 10 −6 C V 130 V = 778 µC .
The battery emf is
b
g
b
g
IR eq = I R1 + R 2 = 1.85 × 10 −2 A 4 000 + 7 000 V A = 204 V .
(b)
In equilibrium after the switch has been opened, no current
exists. The potential difference across each resistor is zero. The
full 204 V appears across both capacitors. The new charge C 2
e
ja
f
Q = C 2 ∆V = 6.00 × 10 −6 C V 204 V = 1 222 µC
for a change of 1 222 µC − 778 µC = 444 µC .
*P28.69
FIG. P28.68(b)
The battery current is
a150 + 45 + 14 + 4f mA = 213 mA .
(a)
The resistor with highest resistance is that
carrying 4 mA. Doubling its resistance will
reduce the current it carries to 2 mA. Then
the total current is
FIG. P28.69
211
=
a150 + 45 + 14 + 2f mA = 211 mA , nearly the same as before. The ratio is 213
(b)
The resistor with least resistance carries 150 mA. Doubling its resistance changes this current
to 75 mA and changes the total to
138
75 + 45 + 14 + 4 mA = 138 mA . The ratio is
= 0.648 , representing a much larger
213
reduction (35.2% instead of 0.9%).
a
(c)
0.991 .
f
This problem is precisely analogous. As a battery maintained a potential difference in parts
(a) and (b), a furnace maintains a temperature difference here. Energy flow by heat is
analogous to current and takes place through thermal resistances in parallel. Each resistance
can have its “R-value” increased by adding insulation. Doubling the thermal resistance of
the attic door will produce only a negligible (0.9%) saving in fuel. Doubling the thermal
resistance of the ceiling will produce a much larger saving. The ceiling originally has the
smallest thermal resistance.
156
*P28.70
Direct Current Circuits
From the hint, the equivalent resistance of
RT +
That is,
RT +
.
1
= R eq
1 RL + 1 R eq
RL R eq
= R eq
RL + R eq
2
RT RL + RT R eq + RL R eq = RL R eq + R eq
2
− RT R eq − RT RL = 0
R eq
R eq =
a fb
2a1f
RT ± RT2 − 4 1 − RT RL
g
Only the + sign is physical:
R eq =
P28.71
1
2
FH
IK
4RT RL + RT2 + RT .
For example, if
RT = 1 Ω.
And
RL = 20 Ω , R eq = 5 Ω.
(a)
After steady-state conditions have been reached, there is no DC current through the
capacitor.
b
g
I R3 = 0 steady-state .
Thus, for R3 :
For the other two resistors, the steady-state current is simply determined by the 9.00-V emf
across the 12-kΩ and 15-kΩ resistors in series:
For R1 and R 2 :
(b)
ε
IbR
1 + R2
g = R1 + R 2
=
9.00 V
a12.0 kΩ + 15.0 kΩf =
b
After the transient currents have ceased, the potential
difference across C is the same as the potential
difference across R 2 = IR 2 because there is no voltage
drop across R3 . Therefore, the charge Q on C is
b
a f
Q = C ∆V
R2
= 50.0 µC .
continued on next page
b g b
g
gb
ga
= C IR 2 = 10.0 µF 333 µA 15.0 kΩ
g
333 µA steady-state .
f
FIG. P28.71(b)
Chapter 28
(c)
When the switch is opened, the branch containing R1
is no longer part of the circuit. The capacitor discharges
through R 2 + R3 with a time constant of
b
g
g a
b
fb
g
R 2 + R3 C = 15.0 kΩ + 3.00 kΩ 10.0 µF = 0.180 s . The
initial current I i in this discharge circuit is determined
by the initial potential difference across the capacitor
applied to R 2 + R3 in series:
Ii =
a f
∆V
bR
2
b
C
+ R3
g
=
g b
b
a
g
f
f
ga
FIG. P28.71(c)
333 µA 15.0 kΩ
IR 2
=
= 278 µA .
R 2 + R3
15.0 kΩ + 3.00 kΩ
Thus, when the switch is opened, the current through R 2 changes instantaneously from
333 µA (downward) to 278 µA (downward) as shown in the graph. Thereafter, it decays
according to
I R 2 = I i e − t b R2 + R3 gC =
(d)
b278 µAge
a
− t 0 .180 s
f afor t > 0f
The charge q on the capacitor decays from Qi to
q = Qi e −
b
g
.
Qi
according to
5
t R 2 + R3 C
Qi
− t 0.180 s g
= Qi e b
5
5 = e t 0.180 s
t
180 ms
t = 0.180 s ln 5 = 290 ms
ln 5 =
a
*P28.72
(a)
First let us flatten the circuit on a 2-D plane
as shown; then reorganize it to a format
easier to read. Notice that the five resistors
on the top are in the same connection as
those in Example 28.5; the same argument
tells us that the middle resistor can be
removed without affecting the circuit. The
remaining resistors over the three parallel
branches have equivalent resistance
R eq =
(b)
fa f
FG 1 + 1 + 1 IJ
H 20 20 10 K
−1
= 5.00 Ω .
So the current through the battery is
∆V 12.0 V
=
= 2.40 A .
R eq 5.00 Ω
FIG. P28.72(a)
157
158
P28.73
Direct Current Circuits
∆V = ε e − t RC
FG ε IJ = FG 1 IJ t .
H ∆V K H RC K
F ε IJ versus t should
A plot of lnG
H ∆V K
so
ln
be a straight line with slope equal
1
to
.
RC
Using the given data values:
FIG. P28.73
(a)
A least-square fit to this data yields the graph above.
∑ xi = 282 ,
∑ xi yi = 244,
∑ = 1.86 × 10
∑ yi = 4.03 ,
x i2
4
af
,
4.87
N=8
11.1
19.4
c∑ x y h − c∑ x hc∑ y h = 0.011 8
30.8
N e∑ x j − c∑ x h
46.6
67.3
∑ x jc∑ y h − c∑ x hc∑ x y h
e
= 0.088 2
Intercept =
102.2
N e∑ x j − c∑ x h
F ε IJ = b0.011 8gt + 0.088 2
The equation of the best fit line is:
lnG
H ∆V K
Slope =
N
i i
i
i
i
i
2
i
(b)
i
2
2
i
2
i
i i
2
i
Thus, the time constant is
and the capacitance is
P28.74
(a)
1
R1 .
2
(b)
FG
H
5.55
0.109
4.93
0.228
4.34
0.355
3.72
0.509
3.09
0.695
2.47
0.919
1.83
1.219
.
R1
a
IJ
K
If R1 = 13.0 Ω and R 2 = 6.00 Ω, then R x = 2.75 Ω .
Ry
Rx
(1)
1 1
1
R1 + R x , or R x = R 2 − R1 .
2 2
4
b
c
Ry
For the second measurement, the equivalent circuit is
shown in Figure 2.
1
(2)
R ac = R 2 = R y + R x .
Thus,
2
Substitute (1) into (2) to obtain:
R2 =
0
1
1
=
= 84.7 s
slope 0.011 8
τ
84.7 s
= 8.47 µF .
C= =
R 10.0 × 10 6 Ω
R ab = R1 = R y + R y = 2 R y
Ry =
b
ln ε ∆V
∆V V
6.19
τ = RC =
For the first measurement, the equivalent circuit is as
shown in Figure 1.
so
af
ts
0
Figure 1
a
Ry
R2
Ry
Figure 2
FIG. P28.74
The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω .
c
Rx
g
Chapter 28
P28.75
The total resistance between points b and c is:
2.00 kΩ 3.00 kΩ
R=
= 1. 20 kΩ .
2.00 kΩ + 3.00 kΩ
The total capacitance between points d and e is:
C = 2.00 µF + 3.00 µF = 5.00 µF .
a
fa
2.00 kΩ
f
b
3.00 kΩ
The potential difference between point d and e in this series RC
circuit at any time is:
a
f
∆V = ε 1 − e − t RC = 120.0 V 1 − e −1 000 t
6
S
a
and q 2
f
b
ga
= C a ∆V f = b3.00 µFga120.0 V f 1 − e
*P28.76
(a)
2
C1 = 2.00 µF
d
e
C2 = 3.00 µF
120 V
+
-
f
.
FIG. P28.75
Therefore, the charge on each capacitor between points d and e is:
q1 = C1 ∆V = 2.00 µF 120.0 V 1 − e −1 000 t
c
159
6
=
b240 µCg 1 − e
= b360 µC g 1 − e
−1 000 t 6
−1 000 t 6
−1 000 t 6
.
Let i represent the current in the battery and i c the current charging the capacitor. Then i − i c is
q
the current in the voltmeter. The loop rule applied to the inner loop is +ε − iR − = 0 . The loop
C
dq
dq
rule for the outer perimeter is ε − iR − i − i c r = 0 . With i c = , this becomes ε − iR − ir + r = 0 .
dt
dt
q
ε
by substitution to obtain
Between the two loop equations we eliminate i = −
R RC
b g
a
ε − R+r
fFGH Rε − RCq IJK + dqdt r = 0
dq
R+r
R+r
ε+
q+ r =0
R
RC
dt
q
r
Rr dq
−
ε+ +
=0
R+r
C R + r dt
ε−
This is the differential equation required.
(b)
To solve we follow the same steps as on page 875.
FG
H
ε rC
dq ε R + r
R+r
= −
q=−
q−
dt R RrC
RrC
R+r
z
q
0
z
t
dq
R+r
=−
dt
q − ε rC R + r
RrC 0
a
f
IJ
K
F ε rc IJ
lnG q −
H R+rK
q
=−
0
F q − ε rc aR + r f I = − R + r t q − ε rc = − ε rc e
lnG
R+r
R+r
H − ε rc aR + r f JK RrC
r
Rr
where R =
q=
Cε e1 − e
j
r+R
R+r
− t Req C
a f
− R + r RrC t
eq
The voltage across the capacitor is VC =
(c)
t
R+r
t
RrC 0
q
r
−t R C
=
ε 1 − e eq .
C r+R
e
j
r
rε
ε 1−0 =
. If the switch is then opened,
r+R
r+R
the capacitor discharges through the voltmeter. Its voltage decays exponentially according
rε − t rC
.
to
e
r+R
As t → ∞ the capacitor voltage approaches
a f
160
Direct Current Circuits
ANSWERS TO EVEN PROBLEMS
P28.2
(a) 1.79 A ; (b) 10.4 V
P28.42
0.113 Ω
P28.4
(a) 12.4 V ; (b) 9.65 V
P28.44
P28.6
(a) 17.1 Ω ; (b) 1.99 A in 4 Ω and 9 Ω;
1.17 A in 7 Ω; 0.818 A in 10 Ω
(a) 30.000 mA , 5.400 0 V ;
(b) 30.167 mA , 5.381 6 V ;
(c) 29.898 mA ; 5.396 6 V
P28.8
29.5 V
P28.46
see the solution
P28.10
(a) see the solution; (b) no
P28.48
(a) 0.101 W; (b) 10.1 W
P28.12
see the solution
P28.50
15.5 A
P28.14
R1 = 1.00 kΩ; R 2 = 2.00 kΩ ; R3 = 3.00 kΩ
P28.52
2.22 h
P28.16
470 Ω and 220 Ω
P28.54
a is 4.00 V higher
P28.18
(a) 11.0 Ω ; (b) and (d) see the solution;
(c) 220 Ω; (e) Parallel
P28.56
(a) see the solution; 833 mA; 200 W;
(b) see the solution; 4.17 A; 1.00 kW
P28.20
I 1 = 714 mA ; I 2 = 1.29 A ; ε = 12.6 V
P28.58
587 kΩ
P28.22
see the solution
P28.60
P28.24
(a) 0.385 mA in R1 ; 2.69 mA in R3 ;
3.08 mA in R 2 ; (b) c higher by 69.2 V
P28.62
(a) I 1 =
1.00 A up in 200 Ω ; 4.00 A up in 70 Ω ;
3.00 A up in 80 Ω ; 8.00 A down in 20 Ω;
200 V
P28.64
see the solution
P28.28
see the solution
P28.66
bR
P28.30
800 W to the left-hand resistor; 25.0 W to
each 4 Ω; 450 W to the right-hand resistor
P28.68
(a) 222 µC ; (b) increase by 444 µC
P28.32
(a) −61.6 mA ; (b) 0.235 µC ; (c) 1.96 A
P28.70
see the solution
P28.34
0.982 s
P28.72
(a) 5.00 Ω; (b) 2.40 A
P28.36
(a) 1.50 s; (b) 1.00 s;
(c) 200 µA + 100 µA e − t 1.00 s
P28.74
(a) R x = R 2 −
P28.76
(a) and (b) see the solution; (c)
P28.26
b
g
P28.38
(a) 3.91 s; (b) 0.782 ms
P28.40
t
C ln 2
Ps + Ps 2 − 4Ps Pp
and
2I 2
Ps − Ps 2 − 4Ps Pp
2I 2
IR 2
IR1
;
; I2 =
R
+
R
1
2
1 + R2
(b) see the solution
1
bR
g
g
+ 2 R 2 C ln 2
R1
; (b) no; R x = 2.75 Ω
4
rε − t rC
e
r+R
29
Magnetic Fields
CHAPTER OUTLINE
29.1
29.2
29.3
29.4
29.5
29.6
Magnetic Fields and Forces
Magnetic Force Acting on a
Current-Carrying Conductor
Torque on a Current Loop in
a Uniform Magnetic Field
Motion of a Charged Particle
in a Uniform Magnetic Field
Applications Involving
Charged Particles Moving in
a Magnetic Field
The Hall Effect
ANSWERS TO QUESTIONS
Q29.1
The force is in the +y direction. No, the proton will not
continue with constant velocity, but will move in a circular
path in the x-y plane. The magnetic force will always be
perpendicular to the magnetic field and also to the velocity of
the proton. As the velocity changes direction, the magnetic
force on the proton does too.
Q29.2
If they are projected in the same direction into the same
magnetic field, the charges are of opposite sign.
Q29.3
Not necessarily. If the magnetic field is parallel or antiparallel
to the velocity of the charged particle, then the particle will
experience no magnetic force.
Q29.4
One particle veers in a circular path clockwise in the page,
while the other veers in a counterclockwise circular path. If the
magnetic field is into the page, the electron goes clockwise and
the proton counterclockwise.
Q29.5
Send the particle through the uniform field and look at its path. If the path of the particle is
parabolic, then the field must be electric, as the electric field exerts a constant force on a charged
particle. If you shoot a proton through an electric field, it will feel a constant force in the same
direction as the electric field—it’s similar to throwing a ball through a gravitational field. If the path
of the particle is helical or circular, then the field is magnetic—see Question 29.1. If the path of the
particle is straight, then observe the speed of the particle. If the particle accelerates, then the field is
electric, as a constant force on a proton with or against its motion will make its speed change. If the
speed remains constant, then the field is magnetic—see Question 29.3.
Q29.6
Similarities: Both can alter the velocity of a charged particle moving through the field. Both exert
forces directly proportional to the charge of the particle feeling the force. Positive and negative
charges feel forces in opposite directions. Differences: The direction of the electric force is parallel or
antiparallel to the direction of the electric field, but the direction of the magnetic force is
perpendicular to the magnetic field and to the velocity of the charged particle. Electric forces can
accelerate a charged particle from rest or stop a moving particle, but magnetic forces cannot.
161
162
Q29.7
Magnetic Fields
a
f
Since FB = q v × B , then the acceleration produced by a magnetic field on a particle of mass m is
q
aB =
v × B . For the acceleration to change the speed, a component of the acceleration must be in
m
the direction of the velocity. The cross product tells us that the acceleration must be perpendicular to
the velocity, and thus can only change the direction of the velocity.
a
f
Q29.8
The magnetic field in a cyclotron essentially keeps the charged particle in the electric field for a
longer period of time, and thus experiencing a larger change in speed from the electric field, by
forcing it in a spiral path. Without the magnetic field, the particle would have to move in a straight
line through an electric field over a distance that is very large compared to the size of the cyclotron.
Q29.9
(a)
The qv × B force on each electron is down. Since electrons are negative, v × B must be up.
With v to the right, B must be into the page, away from you.
(b)
Reversing the current in the coils would reverse the direction of B, making it toward you.
Then v × B is in the direction right × toward you = down, and qv × B will make the electron
beam curve up.
Q29.10
If the current is in a direction parallel or antiparallel to the magnetic field, then there is no force.
Q29.11
Yes. If the magnetic field is perpendicular to the plane of the loop, then it exerts no torque on the loop.
Q29.12
If you can hook a spring balance to the particle and measure the force on it in a known electric field,
F
then q = will tell you its charge. You cannot hook a spring balance to an electron. Measuring the
E
acceleration of small particles by observing their deflection in known electric and magnetic fields can
tell you the charge-to-mass ratio, but not separately the charge or mass. Both an acceleration
produced by an electric field and an acceleration caused by a magnetic field depend on the
q
properties of the particle only by being proportional to the ratio .
m
Q29.13
If the current loop feels a torque, it must be caused by a magnetic field. If the current loop feels no
torque, try a different orientation—the torque is zero if the field is along the axis of the loop.
Q29.14
The Earth’s magnetic field exerts force on a charged incoming cosmic ray,
tending to make it spiral around a magnetic field line. If the particle energy is
low enough, the spiral will be tight enough that the particle will first hit some
matter as it follows a field line down into the atmosphere or to the surface at a
high geographic latitude.
FIG. Q29.14
Q29.15
The net force is zero, but not the net torque.
Q29.16
Only a non-uniform field can exert a non-zero force on a magnetic dipole. If the dipole is aligned
with the field, the direction of the resultant force is in the direction of increasing field strength.
Chapter 29
163
Q29.17
The proton will veer upward when it enters the field and move in a counter-clockwise semicircular
arc. An electron would turn downward and move in a clockwise semicircular arc of smaller radius
than that of the proton, due to its smaller mass.
Q29.18
Particles of higher speeds will travel in semicircular paths of proportionately larger radius. They will
take just the same time to travel farther with their higher speeds. As shown in Equation 29.15, the
time it takes to follow the path is independent of particle’s speed.
Q29.19
The spiral tracks are left by charged particles gradually losing kinetic energy. A straight path might
be left by an uncharged particle that managed to leave a trail of bubbles, or it might be the
imperceptibly curving track of a very fast charged particle.
Q29.20
No. Changing the velocity of a particle requires an accelerating force. The magnetic force is proportional
to the speed of the particle. If the particle is not moving, there can be no magnetic force on it.
Q29.21
Increase the current in the probe. If the material is a semiconductor, raising its temperature may
increase the density of mobile charge carriers in it.
SOLUTIONS TO PROBLEMS
Section 29.1
P29.1
Magnetic Fields and Forces
(a)
up
(b)
out of the page, since the
charge is negative.
(c)
no deflection
(d)
into the page
FIG. P29.1
P29.2
At the equator, the Earth’s magnetic field is
horizontally north. Because an electron has
negative charge, F = qv × B is opposite in direction
to v × B . Figures are drawn looking down.
(a)
Down × North = East, so the force is
directed West .
(b)
North × North = sin 0° = 0 : Zero deflection .
(c)
West × North = Down, so the force is directed Up .
(d)
Southeast × North = Up, so the force is Down .
(a)
(c)
FIG. P29.2
(d)
164
P29.3
Magnetic Fields
e j
Therefore, B = B e− k j which indicates the
FB = qv × B ; FB − j = − e v i × B
negative z direction .
FIG. P29.3
P29.4
e
je
je
j
FB = qvB sin θ = 1.60 × 10 −19 C 3.00 × 10 6 m s 3.00 × 10 −1 T sin 37.0°
(a)
FB = 8.67 × 10 −14 N
a=
(b)
P29.5
F 8.67 × 10 −14 N
=
= 5.19 × 10 13 m s 2
m 1.67 × 10 −27 kg
e
je
j
F = ma = 1.67 × 10 −27 kg 2.00 × 10 13 m s 2 = 3.34 × 10 −14 N = qvB sin 90°
B=
F
3.34 × 10 −14 N
=
= 2.09 × 10 −2 T
−19
7
qv
1.60 × 10
C 1.00 × 10 m s
e
je
j
The right-hand rule shows that B must be in the −y direction to yield a force in
the +x direction when v is in the z direction.
P29.6
First find the speed of the electron.
∆K =
P29.7
P29.8
1
mv 2 = e∆V = ∆U :
2
v=
e
2 e∆ V
=
m
e
jb
2 1.60 × 10 −19 C 2 400 J C
e9.11 × 10
ja
je
−31
kg
j
FIG. P29.5
g = 2.90 × 10
7
ms
f
(a)
FB, max = qvB = 1.60 × 10 −19 C 2.90 × 10 7 m s 1.70 T = 7.90 × 10 −12 N
(b)
FB, min = 0 occurs when v is either parallel to or anti-parallel to B.
e
je
ja
f
FB = qvB sin θ
so
8.20 × 10 −13 N = 1.60 × 10 −19 C 4.00 × 10 6 m s 1.70 T sin θ
sin θ = 0.754
and
θ = sin −1 0.754 = 48.9° or 131° .
Gravitational force:
Electric force:
Magnetic force:
a
f
e
je
j
F = qE = e −1.60 × 10
C jb100 N C downg = 1.60 × 10
N up .
.
F = qv × B = e −1.60 × 10
C je6.00 × 10 m s E j × e50.0 × 10 N ⋅ s C ⋅ m N
j
Fg = mg = 9.11 × 10 −31 kg 9.80 m s 2 = 8.93 × 10 −30 N down .
e
B
−19
−17
−19
FB = −4.80 × 10 −17 N up = 4.80 × 10 −17 N down .
6
−6
Chapter 29
P29.9
FB = qv × B
i
j
k
a
f a f a f
v × B = +2 −4 +1 = 12 − 2 i + 1 + 6 j + 4 + 4 k = 10 i + 7 j + 8k
+1 +2 −3
v × B = 10 2 + 7 2 + 8 2 = 14.6 T ⋅ m s
jb
e
g
FB = q v × B = 1.60 × 10 −19 C 14.6 T ⋅ m s = 2.34 × 10 −18 N
P29.10
jb
e
g e
j
qE = −1.60 × 10 −19 C 20.0 N C k = −3.20 × 10 −18 N k
∑ F = qE + qv × B = ma
e−3.20 × 10
−e3.20 × 10
e1.92 × 10
j
e
j e
N jk − e1.92 × 10
C ⋅ m sji × B = e1.82 × 10
N jk
C ⋅ m sji × B = −e5.02 × 10
N jk
−18
−18
−15
je
j
N k − 1.60 × 10 −19 C 1.20 × 10 4 m s i × B = 9.11 × 10 −31 2.00 × 10 12 m s 2 k
−15
−18
−18
The magnetic field may have any x -component . Bz = 0 and By = −2.62 mT .
Section 29.2
P29.11
Magnetic Force Acting on a Current-Carrying Conductor
FB = ILB sin θ
with FB = Fg = mg
mg = ILB sin θ so
m
g = IB sin θ
L
I = 2.00 A
100 cm m
m
= 0.500 g cm
= 5.00 × 10 −2 kg m .
L
1 000 g kg
b
and
e5.00 × 10
Thus
−2
I
gFGH
JK
ja9.80f = a2.00fB sin 90.0°
B = 0.245 Tesla with the direction given by right-hand rule: eastward .
a
fa
f a
f
fa
fa
e−2.88jj N
P29.12
FB = IA × B = 2.40 A 0.750 m i × 1.60 T k =
P29.13
(a)
FB = ILB sin θ = 5.00 A 2.80 m 0.390 T sin 60.0° = 4.73 N
(b)
FB = 5.00 A 2.80 m 0.390 T sin 90.0° = 5.46 N
(c)
FB = 5.00 A 2.80 m 0.390 T sin 120° = 4.73 N
a
a
fa
fa
f
a
fa
fa
f
f
FIG. P29.11
165
166
P29.14
Magnetic Fields
FB
A
I=
=
mg I A × B
=
A
A
b
F
ge
j
0.040 0 kg m 9.80 m s 2
mg
=
= 0.109 A
BA
3.60 T
Bin
The direction of I in the bar is to the right .
FIG. P29.14
P29.15
a
f e j e j ej
+ K g + ∆E = b K
The work-energy theorem is bK
trasn
0 + 0 + Fs cos θ =
IdBL cos 0° =
v=
P29.16
trans
+ K rot
g
d
f
I
1
1
mv 2 + Iω 2
2
2
FG
H
IJ FG v IJ and IdBL = 3 mv
KH RK
4
4a 48.0 A fa0.120 mfa0.240 Tfa0.450 mf
=
3b0.720 kg g
2
1
1 1
mv 2 +
mR 2
2
2 2
4IdBL
=
3m
rot i
L
y
2
x
1.07 m s .
f e j e j ej
+ K g + ∆E = bK
The work-energy theorem is bK
z
FIG. P29.15
a
The rod feels force FB = I d × B = Id k × B − j = IdB i .
trans
0 + 0 + Fs cos θ =
IdBL cos 0° =
P29.17
B
The rod feels force FB = I d × B = Id k × B − j = IdB i .
rot i
trans
+ K rot
g
f
1
1
mv 2 + Iω 2
2
2
FG
H
1
1 1
mv 2 +
mR 2
2
2 2
IJ FG v IJ
KH RK
2
and v =
4IdBL
.
3m
The magnetic force on each bit of ring is
Ids × B = IdsB radially inward and upward, at
angle θ above the radial line. The radially
inward components tend to squeeze the ring
but all cancel out as forces. The upward
components IdsB sin θ all add to
I 2π rB sin θ up .
FIG. P29.17
Chapter 29
P29.18
For each segment, I = 5.00 A and B = 0.020 0 N A ⋅ m j .
Segment
P29.19
a f
FB = I A × B
A
ab
−0.400 m j
bc
0.400 m k
cd
−0.400 m i + 0.400 m j
da
0.400 m i − 0.400 m k
0
a40.0 mNfe− ij
a40.0 mNfe−k j
a40.0 mNfek + ij
FIG. P29.18
Take the x-axis east, the y-axis up, and the z-axis south. The field is
b
g
e j b
g
e j
B = 52.0 µT cos 60.0° − k + 52.0 µT sin 60.0° − j .
e j
ej
The current then has equivalent length: L ′ = 1.40 m − k + 0.850 m j
b
ge
j e
j
FB = IL ′ × B = 0.035 0 A 0.850 j − 1.40k m × −45.0 j − 26.0 k 10 −6 T
e
j
e j
FB = 3.50 × 10 −8 N −22.1i − 63.0 i = 2.98 × 10 −6 N − i = 2.98 µN west
Section 29.3
P29.20
(a)
Torque on a Current Loop in a Uniform Magnetic Field
2π r = 2.00 m
so
r = 0.318 m
j a
f
ja
f
e
µ = IA = 17.0 × 10 −3 A π 0.318
(b)
m 2 = 5.41 mA ⋅ m 2
τ = µ ×B
so
P29.21
2
e
τ = 5.41 × 10 −3 A ⋅ m 2 0.800 T = 4.33 mN ⋅ m
a
f
τ = µB sin θ so 4.60 × 10 −3 N ⋅ m = µ 0.250 sin 90.0°
µ = 1.84 × 10 −2 A ⋅ m 2 = 18.4 mA ⋅ m 2
FIG. P29.19
.
167
168
P29.22
Magnetic Fields
Let θ represent the unknown angle; L, the total length of the wire; and d, the length of one
side of the square coil. Then, using the definition of magnetic moment and the right-hand
rule in Figure 29.15, we find
(a)
FG L IJ d I at angle θ with the horizontal.
H 4d K
∑ τ = bµ × Bg − br × mgg = 0
FG ILBd IJ sina90.0°−θ f − FG mgd IJ sinθ = 0
H 4 K
H 2K
FG mgd IJ sinθ = FG ILBd IJ cos θ
H 4 K
H 2K
F ILB I = tan FG a3.40 Afa4.00 mfb0.010 0 Tg IJ =
θ = tan G
GH 2b0.100 kg ge9.80 m s j JK
H 2mg JK
µ = NAI :
2
µ=
At equilibrium,
and
−1
−1
2
τm =
(b)
P29.23
FG ILBd IJ cosθ = 1 a3.40 Afa4.00 mfb0.010 0 Tga0.100 mf cos 3.97° =
H 4 K
4
τ = NBAI sin φ
a
ja
fe
3.39 mN ⋅ m
f
τ = 100 0.800 T 0.400 × 0.300 m 2 1.20 A sin 60°
τ = 9.98 N ⋅ m
Note that φ is the angle between the magnetic
moment and the B field. The loop will rotate so as
to align the magnetic moment with the B field.
Looking down along the y-axis, the loop will rotate
in a clockwise direction.
P29.24
FIG. P29.23
From τ = µ × B = IA × B , the magnitude of the torque is IAB sin 90.0°.
(a)
Each side of the triangle is
40.0 cm
.
3
Its altitude is 13.3 2 − 6.67 2 cm = 11.5 cm and its area is
1
A = 11.5 cm 13.3 cm = 7.70 × 10 −3 m 2 .
2
Then τ = 20.0 A 7.70 × 10 −3 m 2 0.520 N ⋅ s C ⋅ m = 80.1 mN ⋅ m .
a
fa
a
(b)
f
fe
jb
Each side of the square is 10.0 cm and its area is 100 cm 2 = 10 −2 m 2 .
a
fe
ja
f
τ = 20.0 A 10 −2 m 2 0.520 T = 0.104 N ⋅ m
(c)
0.400 m
= 0.063 7 m
2π
A = π r 2 = 1.27 × 10 −2 m 2
r=
a
fe
ja
f
τ = 20.0 A 1.27 × 10 −2 m 2 0.520 = 0.132 N ⋅ m
(d)
g
The circular loop experiences the largest torque.
3.97° .
Chapter 29
P29.25
Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque of magnitude
µB sin θ on the dipole, tending to turn the dipole moment in the direction of decreasing θ. According
to Equations 8.16 and 10.22, the potential energy of the dipole-field system is given by
U −0 =
z
θ
a
µB sin θ dθ = µB − cos θ
f
90 .0 °
P29.26
169
(a)
θ
90 .0 °
= − µB cos θ + 0
U = −µ ⋅ B .
or
The field exerts torque on the needle tending to align it with the field, so the minimum
energy orientation of the needle is:
pointing north at 48.0° below the horizontal
e
je
j
where its energy is U min = − µB cos 0° = − 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = −5.34 × 10 −7 J .
It has maximum energy when pointing in the opposite direction,
south at 48.0° above the horizontal
e
je
j
where its energy is U max = − µB cos 180° = + 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = +5.34 × 10 −7 J .
P29.27
e
j
(b)
U min + W = U max : W = U max − U min = +5.34 × 10 −7 J − −5.34 × 10 −7 J = 1.07 µJ
(a)
τ = µ × B,
τ = µ × B = µB sin θ = NIAB sin θ
so
a
f b
τ max = NIAB sin 90.0° = 1 5.00 A π 0.050 0 m
(b)
g e3.00 × 10 Tj =
2
−3
118 µN ⋅ m
U = − µ ⋅ B, so − µB ≤ U ≤ + µB
a f
a
f b
Since µB = NIA B = 1 5.00 A π 0.050 0 m
g e3.00 × 10 Tj = 118 µJ ,
2
−3
the range of the potential energy is: −118 µJ ≤ U ≤ +118 µJ .
*P29.28
(a)
τ = µ × B = NIAB sin θ
e
ja
fb
g
F 2π rad IJ FG 1 min IJ =
N ⋅ mb3 600 rev mingG
H 1 rev K H 60 s K
τ max = 80 10 −2 A 0.025 m ⋅ 0.04 m 0.8 N A ⋅ m sin 90° = 6.40 × 10 −4 N ⋅ m
(b)
Pmax = τ maxω = 6.40 × 10 −4
(c)
In one half revolution the work is
b
g
= 2 NIAB = 2e6.40 × 10 N ⋅ mj = 1.28 × 10 J
In one full revolution, W = 2e1.28 × 10 Jj = 2.56 × 10
W = U max − U min = − µB cos 180°− − µB cos 0° = 2 µB
−4
−3
−3
(d)
Pavg =
W 2.56 × 10 −3 J
=
= 0.154 W
∆t
1 60 s
b g
The peak power in (b) is greater by the factor
π
.
2
−3
J .
0.241 W
170
Magnetic Fields
Section 29.4
P29.29
Motion of a Charged Particle in a Uniform Magnetic Field
B = 50.0 × 10 −6 T; v = 6.20 × 10 6 m s
(a)
Direction is given by the right-hand-rule: southward
FB = qvB sin θ
e
je
je
j
FB = 1.60 × 10 −19 C 6.20 × 10 6 m s 50.0 × 10 −6 T sin 90.0°
FIG. P29.29
= 4.96 × 10 −17 N
e
(b)
P29.30
je
1.67 × 10 −27 kg 6.20 × 10 6 m s
mv 2
mv 2
=
F=
so r =
r
F
4.96 × 10 −17 N
a f
1
mv 2 = q ∆V
2
P29.31
2
= 1.29 km
ja
1
3.20 × 10 −26 kg v 2 = 1.60 × 10 −19 C 833 V
2
e
j
e
The magnetic force provides the centripetal force: qvB sin θ =
r=
j
e
e
je
jb
f
v = 91.3 km s
mv 2
r
j
g
3.20 × 10 −26 kg 9.13 × 10 4 m s
mv
=
= 1.98 cm .
qB sin 90.0° 1.60 × 10 −19 C 0.920 N ⋅ s C ⋅ m
For each electron, q vB sin 90.0° =
mv 2
eBr
.
and v =
r
m
The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic:
K=
1
1
1
mv12i + 0 = mv12 f + mv 22 f
2
2
2
K=
e 2 B 2 R 22
e 2 B 2 R12
e2B2 2
1
1
R1 + R 22
m
+
=
m
2
2
2m
m2
m2
K=
P29.32
F
GH
e
I
JK
F
GH
jb
I
JK
e
g b0.010 0 mg + b0.024 0 mg
e 1.60 × 10 −19 C 0.044 0 N ⋅ s C ⋅ m
e
2 9.11 × 10
We begin with qvB =
−31
kg
j
2
2
2
qRB
mv 2
, so v =
.
R
m
The time to complete one revolution is T =
Solving for B, B =
j
2π m
= 6.56 × 10 −2 T .
qT
2π R
2π R
2π m
=
=
.
v
qRB m
qB
= 115 keV
Chapter 29
P29.33
a f
q ∆V =
1
mv 2
2
Also, qvB =
mv 2
r
or
v=
so
r=
a f
2 q ∆V
.
m
rp2 =
Therefore,
rα2
(a)
(b)
a f
2 m p ∆V
eB 2
2 m d ∆V
p
2
d
p
α
2
α
or
qRB = mv .
But
L = mvR = qR 2 B .
Therefore,
R=
Thus,
v=
ja
e
e1.60 × 10
2
p
−19
je
C 1.00 × 10 −3 T
j
= 0.050 0 m = 5.00 cm .
L
4.00 × 10 −25 J ⋅ s
=
= 8.78 × 10 6 m s .
−31
mR
9.11 × 10
kg 0.050 0 m
jb
e
g
f
P29.36
1
mv 2 = q ∆V
2
so
v=
so
r=
a f
2 q ∆V
m
a f
m 2 q ∆V m
qB
a f
r2 =
m 2 ∆V
⋅
q
B2
and
ar ′ f
m=
qB 2 r 2
2 ∆V
and
am ′f = bq′2gBa∆Varf′f
a f
2
4.00 × 10 −25 J ⋅ s
L
=
qB
P29.35
a f
2
p
mv 2
R
qvB =
mv
qB
2
p
2
−19
C 5.20 T
qB 1.60 × 10
ω=
=
= 4.98 × 10 8 rad s
m
1.67 × 10 −27 kg
r=
p
2
We begin with
a f
qB 2
rα = rd = 2 rp .
The conclusion is:
P29.34
a f.
2 m ∆V
a f = 2e2m ja∆V f = 2FG 2m a∆V f IJ = 2r
q B
eB
H eB K
F 2 m a ∆V f I = 2 r .
2m a ∆V f 2e 4m ja ∆V f
=
=
= 2G
q B
a2efB
H eB JK
rd2 =
and
a f
2 q ∆V
=
m
mv m
=
qB qB
2
=
m ′ 2 ∆V
⋅
q′
B2
2
2
a f FG IJ FG 2R IJ
H KH R K
2
so
m ′ q′ r ′
2e
= ⋅ 2 =
m
q r
e
2
= 8
171
172
P29.37
Magnetic Fields
E=
and
1
mv 2 = e∆V
2
evB sin 90° =
B=
mv m
=
eR eR
mv 2
R
2 e∆ V 1
=
m
R
1
B=
5.80 × 10 10 m
*P29.38
(a)
e
2m∆V
e
je
2 1.67 × 10 −27 kg 10.0 × 10 6 V
1.60 × 10
−19
C
j=
7.88 × 10 −12 T
v
At the moment shown in Figure 29.21, the particle must be
moving upward in order for the magnetic force on it to be
+
into the page, toward the center of this turn of its
spiral path. Throughout its motion it circulates clockwise.
(b)
B
FIG. P29.38(a)
After the particle has passed the middle of the bottle and
moves into the region of increasing magnetic field, the
magnetic force on it has a component to the left (as well as
a radially inward component) as shown. This force in the
–x direction slows and reverses the particle’s motion along
the axis.
v
B
F
FIG. P29.38(b)
(c)
The magnetic force is perpendicular to the velocity and does no work on the particle. The
particle keeps constant kinetic energy. As its axial velocity component decreases, its
tangential velocity component increases.
(d)
The orbiting particle constitutes a loop of current in the yz
q
plane and therefore a magnetic dipole moment I A = A
T
in the –x direction. It is like a little bar magnet with its N
pole on the left.
(e)
Problem 17 showed that a nonuniform magnetic field
exerts a net force on a magnetic dipole. When the dipole is
aligned opposite to the external field, the force pushes it
out of the region of stronger field. Here it is to the left, a
force of repulsion of one magnetic south pole on another
south pole.
+
N
FIG. P29.38(d)
B
N
S
S
FIG. P29.38(e)
S
Chapter 29
P29.39
e
ja
je
7.94 × 10 −3 m 1.60 × 10 −19 C 1.80 T
rqB
=
m=
v
4.60 × 10 5 m s
mv
so
r=
qB
m = 4.97 × 10 −27 kg
F 1u
GH 1.66 × 10
−27
I=
J
kg K
The particle is singly ionized: either a tritium ion,
Section 29.5
P29.40
2.99 u
+
3
1H
, or a helium ion,
+
3
2 He
.
Applications Involving Charged Particles Moving in a Magnetic Field
FB = Fe
so
qvB = qE
where
v=
2K
and K is kinetic energy of the electron.
m
E = vB =
P29.41
f
K=
2K
B=
m
a f
so
mv 2
r
r=
1
mv 2 = q ∆V
2
FB = qv × B =
9.11 × 10
j
(b)
r235 = 8. 23 cm
1.60 × 10
m 238
=
m 235
−19
j b0.015 0g =
244 kV m
a f
a f
2 q ∆V m 1
=
B
B
mv m
=
qB q
e
r238 =
−31
2 q ∆V
m
v=
2 238 × 1.66 × 10 −27 2 000
(a)
r238
=
r235
a fe
2 750 1.60 × 10 −19
a f
2m ∆V
q
FG 1 IJ = 8.28 × 10
H 1.20 K
−2
m = 8.28 cm
238.05
= 1.006 4
235.04
The ratios of the orbit radius for different ions are independent of ∆V and B.
P29.42
E 2 500 V m
=
= 7.14 × 10 4 m s .
B
0.035 0 T
In the velocity selector:
v=
In the deflection chamber:
2.18 × 10 −26 kg 7.14 × 10 4 m s
mv
r=
=
= 0.278 m .
qB
1.60 × 10 −19 C 0.035 0 T
e
e
je
jb
g
j
173
174
P29.43
Magnetic Fields
(a)
FB = qvB =
mv 2
R
ja
e
f
−19
C 0.450 T
v qBR qB 1.60 × 10
ω= =
=
=
= 4.31 × 10 7 rad s
−27
R mR
m
kg
1.67 × 10
(b)
P29.44
K=
v=
ja
e
fa
f
−19
C 0.450 T 1.20 m
qBR 1.60 × 10
=
= 5.17 × 10 7 m s
m
1.67 × 10 −27 kg
1
mv 2 :
2
j 12 e1.67 × 10 kg jv
kg je8.07 × 10 m sj
mv e1.67 × 10
r=
=
= 0.162 m
qB
e1.60 × 10 Cja5.20 Tf
e34.0 × 10
je
eV 1.60 × 10 −19 J eV =
6
−27
7
v = 8.07 × 10 m s
*P29.45
−27
2
7
−19
Note that the “cyclotron frequency” is an angular speed. The motion of the proton is described by
∑ F = ma :
q vB sin 90° =
qB=m
(a)
(b)
(c)
(d)
mv 2
r
v
= mω
r
e1.60 × 10 Cjb0.8 N ⋅ s C ⋅ mg FG kg ⋅ m IJ = 7.66 × 10 rad s
ω=
=
H N ⋅s K
m
e1.67 × 10 kgj
F 1 IJ = 2.68 × 10 m s
v = ω r = e7.66 × 10 rad sja0.350 mfG
H 1 rad K
F 1 eV IJ = 3.76 × 10
1
1
K = mv = e1.67 × 10
kg je 2.68 × 10 m sj G
2
2
H 1.6 × 10 J K
−19
qB
7
7
−27
2
7
2
−27
7
2
−19
a
P29.46
θ =ωt
FB = qvB =
B=
eV
The proton gains 600 eV twice during each revolution, so the number of revolutions is
3.76 × 10 6 eV
= 3.13 × 10 3 revolutions .
2 600 eV
(e)
6
t=
FG
H
mv 2
r
jb
IJ
K
θ
3.13 × 10 3 rev 2π rad
= 2.57 × 10 −4 s
=
ω 7.66 × 10 7 rad s 1 rev
4.80 × 10 −16 kg ⋅ m s
mv
=
= 3.00 T
qr
1.60 × 10 −19 C 1 000 m
e
f
g
Chapter 29
P29.47
θ = tan −1
FG 25.0 IJ = 68.2°
H 10.0 K
R=
and
1.00 cm
= 1.08 cm .
sin 68.2°
Ignoring relativistic correction, the kinetic energy of the electrons is
1
mv 2 = q∆V
2
From Newton’s second law
e
2 q ∆V
= 1.33 × 10 8 m s .
m
v=
so
mv 2
= qvB , we find the magnetic field
R
je
je
FIG. P29.47
j
9.11 × 10 −31 kg 1.33 × 10 8 m s
mv
B=
=
= 70.1 mT .
qR
1.60 × 10 −19 C 1.08 × 10 −2 m
Section 29.6
P29.48
e
The Hall Effect
1
nq
(a)
RH ≡
(b)
∆VH =
B=
P29.49
j
so
n=
1
1
=
= 7.44 × 10 28 m −3
−
19
qRH
C 0.840 × 10 −10 m3 C
1.60 × 10
e
je
j
IB
nqt
b
nqt ∆VH
I
g = e7.44 × 10
28
je
je
je
m −3 1.60 × 10 −19 C 0.200 × 10 −3 m 15.0 × 10 −6 V
20.0 A
j=
1.79 T
IB
, and given that I = 50.0 A , B = 1.30 T , and t = 0.330 mm, the number of charge
nqt
carriers per unit volume is
Since ∆VH =
n=
IB
= 1.28 × 10 29 m −3
e ∆VH t
b
g
The number density of atoms we compute from the density:
n0 =
F
GH
8.92 g 1 mole
cm 3 63.5 g
I F 6.02 × 10 atoms I F 10 cm I = 8.46 × 10
JK GH mole JK GH 1 m JK
23
3
6
3
So the number of conduction electrons per atom is
n 1.28 × 10 29
=
= 1.52
n 0 8.46 × 10 28
28
atom m3
175
176
P29.50
Magnetic Fields
∆VH =
(a)
IB
nqt
nqt
0.080 0 T
B
=
=
= 1.14 × 10 5 T V .
∆VH 0.700 × 10 −6 V
I
so
B=
Then, the unknown field is
e
je
FG nqt IJ b∆V g
HIK
H
j
B = 1.14 × 10 5 T V 0.330 × 10 −6 V = 0.037 7 T = 37.7 mT .
nqt
= 1.14 × 10 5 T V
I
(b)
e
n = 1.14 × 10 5 T V
P29.51
B=
b
nqt ∆VH
I
g = e8.49 × 10
e
n = 1.14 × 10 5 T V
so
j e1.60 × 10
0.120 A
−19
je
C 2.00 × 10
je
−3
j
m
j qtI
= 4.29 × 10 25 m −3 .
je
je
m −3 1.60 × 10 −19 C 5.00 × 10 −3 m 5.10 × 10 −12 V
28
j
8.00 A
B = 4.33 × 10 −5 T = 43.3 µT
Additional Problems
P29.52
(a)
The boundary between a region of strong magnetic field and a
region of zero field cannot be perfectly sharp, but we ignore the
thickness of the transition zone. In the field the electron moves on
an arc of a circle:
∑ F = ma :
q vB sin 90° =
mv 2
r
e
je
j
−19
C 10 −3 N ⋅ s C ⋅ m
q B 1.60 × 10
v
=ω =
=
= 1.76 × 10 8 rad s
r
m
9.11 × 10 −31 kg
e
j
FIG. P29.52(a)
The time for one half revolution is,
from
∆ θ = ω∆ t
∆t =
(b)
∆θ
ω
=
π rad
= 1.79 × 10 −8 s .
1.76 × 10 8 rad s
The maximum depth of penetration is the radius of the path.
Then
ja
e
f
v = ω r = 1.76 × 10 8 s −1 0.02 m = 3.51 × 10 6 m s
and
K=
1
1
mv 2 = 9.11 × 10 −31 kg 3.51 × 10 6 m s
2
2
e
= 35.1 eV .
je
j
2
= 5.62 × 10 −18 J =
5.62 × 10 −18 J ⋅ e
1.60 × 10 −19 C
Chapter 29
P29.53
(a)
Define vector h to have the downward direction of the current,
and vector L to be along the pipe into the page as shown. The
electric current experiences a magnetic force .
a
f
I h × B in the direction of L.
(b)
The sodium, consisting of ions and electrons, flows along the
pipe transporting no net charge. But inside the section of
length L, electrons drift upward to constitute downward
electric current J × area = J Lw .
FIG. P29.53
a f
The current then feels a magnetic force I h × B = JLwhB sin 90° .
This force along the pipe axis will make the fluid move, exerting pressure
F
JLwhB
=
= JLB .
area
hw
P29.54
∑ Fy = 0 :
+n − mg = 0
∑ Fx = 0 :
− µ k n + IB sin 90.0° = 0
B=
P29.55
b
a
ge
j
2
µ k mg 0.100 0.200 kg 9.80 m s
=
= 39.2 mT
10.0 A 0.500 m
Id
fa
f
The magnetic force on each proton, FB = qv × B = qvB sin 90° downward
perpendicular to velocity, causes centripetal acceleration, guiding it into a
circular path of radius r, with
mv 2
r
mv
.
r=
qB
qvB =
and
We compute this radius by first finding the proton’s speed:
Now,
(b)
K=
1
mv 2
2
v=
2K
=
m
FIG. P29.55
e
je
2 5.00 × 10 6 eV 1.60 × 10 −19 J eV
1.67 × 10
e
−27
je
kg
j
j = 3.10 × 10
7
m s.
1.67 × 10 −27 kg 3.10 × 10 7 m s
mv
=
= 6.46 m .
r=
qB
1.60 × 10 −19 C 0.050 0 N ⋅ s C ⋅ m
e
jb
g
From the figure, observe that
1.00 m
1m
=
r
6.46 m
α = 8.90°
sin α =
(a)
The magnitude of the proton momentum stays constant, and its final y component is
e
je
j
− 1.67 × 10 −27 kg 3.10 × 10 7 m s sin 8.90° = −8.00 × 10 −21 kg ⋅ m s .
177
178
P29.56
Magnetic Fields
e j e
j
If B = Bx i + By j + Bz k , FB = qv × B = e vi i × Bx i + By j + Bz k = 0 + evi B y k − evi Bz j .
(a)
Since the force actually experienced is FB = Fi j , observe that
B x could have any value , B y = 0 , and B z = −
Fi
.
evi
IJ
e j FGH
K
F
F I
= qv × B = − ee v i j × G B i + 0 j −
kJ =
ev K
H
(b)
If v = −vi i , then
F
FB = qv × B = e − vi i × Bx i + 0 j − i k = − Fi j .
evi
(c)
If q = − e and v = vi i , then
FB
i
x
i
i
− Fi j .
Reversing either the velocity or the sign of the charge reverses the force.
P29.57
(a)
The net force is the Lorentz force given by
a
f
j e4i − 1j − 2k j + e2 i + 3 j − 1k j × e2 i + 4j + 1k j N
F = qE + qv × B = q E + v × B
e
F = 3.20 × 10 −19
Carrying out the indicated operations, we find:
e3.52i − 1.60jj × 10 N .
3.52
FG F IJ = cos FG
H FK
GH a3.52f + a1.60f
−18
F=
θ = cos
(b)
P29.58
−1
−1
x
2
2
I
JJ =
K
24.4°
A key to solving this problem is that reducing the normal force will reduce
F
the friction force: FB = BIL or B = B .
IL
When the wire is just able to move,
∑ Fy = n + FB cos θ − mg = 0
so
n = mg − FB cos θ
and
f = µ mg − FB cos θ .
Also,
∑ Fx = FB sin θ − f = 0
so FB sin θ = f :
FB sin θ = µ mg − FB cos θ and FB =
We minimize B by minimizing FB :
dFB
cos θ − µ sin θ
= µmg
= 0 ⇒ µ sin θ = cos θ .
2
dθ
sin θ + µ cos θ
b
b
b gb
g
FIG. P29.58
g
g
FG 1 IJ = tan a5.00f = 78.7° for the smallest field, and
H µK
F
F µg I bm Lg
=G J
B=
IL H I K sin θ + µ cos θ
L a0.200fe9.80 m s j OP
0.100 kg m
=M
B
MN 1.50 A PQ sin 78.7°+a0.200f cos 78.7° = 0.128 T
Thus, θ = tan −1
−1
B
2
min
Bmin = 0.128 T pointing north at an angle of 78.7° below the horizontal
µmg
.
sin θ + µ cos θ
Chapter 29
*P29.59
179
The electrons are all fired from the electron gun with the same speed v in
Ui = K f
qV =
a− efa−∆V f = 12 m v
1
mv 2
2
e
2
v=
2 e∆V
me
For φ small, cos φ is nearly equal to 1. The time T of passage of each electron in the chamber is given
by
d = vT
T=d
FG m IJ
H 2 e∆ V K
e
12
Each electron moves in a different helix, around a different axis. If each completes just one
revolution within the chamber, it will be in the right place to pass through the exit port. Its
transverse velocity component v ⊥ = v sin φ swings around according to F⊥ = ma ⊥
qv ⊥ B sin 90° =
Then
*P29.60
mv ⊥2
r
FG IJ
H K
2π m e
B
e
eB =
12
=
me v⊥
2π
= m eω = m e
r
T
d
a2 ∆V f
12
B=
FG
H
2π 2m e ∆V
d
e
T=
IJ
K
FG
H
m e 2π
me
=d
2 e∆ V
eB
IJ
K
12
12
.
Let vi represent the original speed of the alpha particle. Let vα and v p represent the particles’
speeds after the collision. We have conservation of momentum 4m p vi = 4m p vα + m p v p and the
relative velocity equation vi − 0 = v p − vα . Eliminating vi ,
4 v p − 4 v α = 4 vα + v p
3 v p = 8 vα
vα =
3
vp .
8
For the proton’s motion in the magnetic field,
∑ F = ma
ev p B sin 90° =
m p v p2
eBR
= vp .
mp
R
For the alpha particle,
2 evα B sin 90° =
P29.61
4m p vα2
rα =
rα
2 m p vα
rα =
eB
2m p 3
2m p 3 eBR
3
vp =
R .
=
eB 8
4
eB 8 m p
Let ∆x 1 be the elongation due to the weight of the wire and let ∆x 2
be the additional elongation of the springs when the magnetic field
is turned on. Then Fmagnetic = 2 k∆x 2 where k is the force constant of
mg
. (The factor 2 is
2 ∆x 1
included in the two previous equations since there are 2 springs in
parallel.) Combining these two equations, we find
the spring and can be determined from k =
Fmagnetic = 2
FG mg IJ ∆x
H 2 ∆x K
1
2
=
mg∆x 2
; but FB = I L × B = ILB .
∆x 1
a
fa fe
a fb ge
FIG. P29.61
j
0.100 9.80 3.00 × 10 −3
mg∆x 2
24.0 V
Therefore, where I =
= 2.00 A , B =
=
= 0.588 T .
IL∆x1
12.0 Ω
2.00 0.050 0 5.00 × 10 −3
j
180
P29.62
Magnetic Fields
Suppose the input power is
a
f
I ~ 1 A = 10 0 A .
120 W = 120 V I :
FG 1 min IJ FG 2π rad IJ ~ 200 rad s
H 60 s K H 1 rev K
20 W = τω = τ b 200 rad sg τ ~ 10 N ⋅ m .
A ~ 10 m .
a3 cmf × a4 cmf, or
ω = 2 000 rev min
Suppose
−1
and the output power is
−3
Suppose the area is about
2
B ~ 10 −1 T .
Suppose that the field is
Then, the number of turns in the coil may be found from τ ≅ NIAB :
b
ge
je
j
0.1 N ⋅ m ~ N 1 C s 10 −3 m 2 10 −1 N ⋅ s C ⋅ m
N ~ 10
giving
*P29.63
3
.
The sphere is in translational equilibrium, thus
fs − Mg sin θ = 0 .
G
µ
(1)
θ
G
B
The sphere is in rotational equilibrium. If torques are taken about the
center of the sphere, the magnetic field produces a clockwise torque of
magnitude µB sin θ , and the frictional force a counterclockwise torque
of magnitude fs R , where R is the radius of the sphere. Thus:
fs R − µB sin θ = 0 .
(2)
From (1): fs = Mg sin θ . Substituting this in (2) and canceling out sin θ ,
one obtains
µB = MgR .
b
ge
j
f
(3)
fs
I
θ
Mg
FIG. P29.63
0.08 kg 9.80 m s 2
Mg
=
= 0.713 A . The current must be
π NBR π 5 0.350 T 0.2 m
counterclockwise as seen from above.
Now µ = NIπ R 2 . Thus (3) gives I =
P29.64
fa
Call the length of the rod L and the tension in each wire alone
∑ Fx = T sin θ − ILB sin 90.0° = 0
∑ Fy = T cos θ − mg = 0 ,
tan θ =
P29.65
a fa
ILB
IB
=
mg
mL g
b g
∑ F = ma or qvB sin 90.0° =
or
T sin θ = ILB
or
T cos θ = mg
or
B=
mv 2
r
∴ the angular frequency for each ion is
∆f = f12 − f14 =
FG
H
IJ e
K e
bm Lgg tanθ =
I
qB
v
=ω =
= 2π f and
r
m
ja
λg
tan θ
I
fF 1 − 1 I
G
J
j H 12.0 u 14.0 u K
1.60 × 10 −19 C 2.40 T
qB 1
1
=
−
2π m12 m14
2π 1.66 × 10 −27 kg u
∆f = f12 − f14 = 4.38 × 10 5 s −1 = 438 kHz
T
. Then, at equilibrium:
2
Chapter 29
P29.66
Let v x and v ⊥ be the components of the velocity of the positron parallel
to and perpendicular to the direction of the magnetic field.
(a)
The pitch of trajectory is the distance moved along x by the
positron during each period, T (see Equation 29.15)
fFGH 2πBqm IJK
e5.00 × 10 jacos 85.0°fa2π fe9.11 × 10 j =
p=
0.150e1.60 × 10 j
a
p = v x T = v cos 85.0°
−19
(b)
FIG. P29.66
−31
6
From Equation 29.13,
1.04 × 10 −4 m
r=
mv ⊥ mv sin 85.0°
=
Bq
Bq
r=
e9.11 × 10 je5.00 × 10 jasin 85.0°f =
a0.150fe1.60 × 10 j
−31
P29.67
6
−19
τ = IAB where the effective current due to the orbiting electrons is
I=
∆q q
=
∆t T
and the period of the motion is
T=
2π R
.
v
The electron’s speed in its orbit is found by requiring
keq2
R
2
=
mv 2
or
R
Substituting this expression for v into the equation for T, we find
e9.11 × 10 je5.29 × 10 j = 1.52 × 10 s .
T = 2π
e1.60 × 10 j e8.99 × 10 j
F q I 1.60 × 10 π e5.29 × 10 j a0.400f =
Therefore, τ = G J AB =
H T K 1.52 × 10
mR 3
q2 ke
9
−19
−11 2
−16
Use the equation for cyclotron frequency ω =
e1.60 × 10 Cje5.00 × 10 Tj =
a2π fe5.00 rev 1.50 × 10 sj
−19
T = 2π
ke
.
mR
−16
−19 2
m=
v=q
−11 3
−31
P29.68
1.89 × 10 −4 m
qB
qB
qB
=
or m =
ω 2π f
m
−2
−3
3.70 × 10 −24 N ⋅ m .
3.82 × 10 −25 kg .
181
182
P29.69
Magnetic Fields
(a)
K=
1
mv 2 = 6.00 MeV = 6.00 × 10 6 eV 1.60 × 10 −19 J eV
2
e
je
j
K = 9.60 × 10 −13 J
e
2 9.60 × 10
v=
1.67 × 10
FB = qvB =
R=
mv
=
qB
−13
−27
J
j = 3.39 × 10
kg
θ' x x x x x
B in = 1.00 T
x x x x x
45° x x x x x
45° x x x x x
R
x x x x x
x x x x x
45.0° x x x x x
x
7
ms
v
2
mv
so
R
1.67 × 10 −27 kg 3.39 × 10 7 m s
e
e1.60 × 10
je
−19
ja
C 1.00 T
f
FIG. P29.69
j = 0.354 m
a
f
Then, from the diagram, x = 2 R sin 45.0° = 2 0.354 m sin 45.0° = 0.501 m
P29.70
(b)
From the diagram, observe that θ ′ = 45.0° .
(a)
See graph to the right. The
Hall voltage is directly
proportional to the magnetic
field. A least-square fit to the
data gives the equation of the
best fitting line as:
120
100
80
∆V H (µV)
60
40
20
e
∆VH = 1.00 × 10
−4
j
V TB .
0
0
0.2
0.4
0.6
0.8
1.0
1.2
B (T)
(b)
Comparing the equation of
the line which fits the data
best to
∆VH =
FIG. P29.70
F 1 IB
GH nqt JK
observe that:
I
I
= 1.00 × 10 −4 V T, or t =
.
nqt
nq 1.00 × 10 −4 V T
e
j
Then, if I = 0.200 A , q = 1.60 × 10 −19 C , and n = 1.00 × 10 26 m −3 , the thickness of the sample is
t=
e1.00 × 10
0.200 A
26
m
−3
je1.60 × 10
−19
je
C 1.00 × 10 −4 V T
j
= 1.25 × 10 −4 m = 0.125 mm .
Chapter 29
P29.71
(a)
183
The magnetic force acting on ions in the blood stream will
deflect positive charges toward point A and negative
charges toward point B. This separation of charges
produces an electric field directed from A toward B. At
equilibrium, the electric force caused by this field must
balance the magnetic force, so
FG ∆V IJ
HdK
e160 × 10 Vj
∆V
=
v=
Bd b0.040 0 Tge3.00 × 10
qvB = qE = q
FIG. P29.71
−6
or
−3
j
m
= 1.33 m s .
No . Negative ions moving in the direction of v would be deflected toward point B, giving
(b)
A a higher potential than B. Positive ions moving in the direction of v would be deflected
toward A, again giving A a higher potential than B. Therefore, the sign of the potential
difference does not depend on whether the ions in the blood are positively or negatively
charged.
P29.72
When in the field, the particles follow a circular path
mv 2
mv
, so the radius of the path is: r =
according to qvB =
r
qB
(a)
qBh
mv
, that is, when v =
, the
qB
m
particle will cross the band of field. It will move in
a full semicircle of radius h, leaving the field at
2 h, 0 , 0 with velocity v = − vj .
When r = h =
b
(b)
g
f
FIG. P29.72
qBh
mv
, the particle will move in a smaller semicircle of radius r =
< h . It will
m
qB
leave the field at 2r, 0 , 0 with velocity v f = − vj .
When v <
b
(c)
v i = vj
g
qBh
mv
, the particle moves in a circular arc of radius r =
> h , centered at
qB
m
h
r, 0 , 0 . The arc subtends an angle given by θ = sin −1
. It will leave the field at the point
r
with coordinates r 1 − cos θ , h, 0 with velocity v = v sin θ i + v cos θ j .
When v >
b
FG IJ
HK
g
a
f
f
ANSWERS TO EVEN PROBLEMS
P29.2
(a) west; (b) no deflection; (c) up;
(d) down
P29.4
(a) 86.7 fN ; (b) 51.9 Tm s 2
P29.6
(a) 7.90 pN ; (b) 0
P29.8
Gravitational force: 8.93 × 10 −30 N down;
Electric force: 16.0 aN up ;
Magnetic force: 48.0 aN down
P29.10
By = −2.62 mT ; Bz = 0; Bx may have any
value
184
Magnetic Fields
P29.12
e−2.88jj N
P29.14
109 mA to the right
P29.16
FG 4IdBL IJ
H 3m K
P29.18
12
e j
= 40.0 mN e − k j ; F = a 40.0 mN fe i + k j
P29.50
(a) 37.7 mT ; (b) 4.29 × 10 25 m3
P29.52
(a) 17.9 ns; (b) 35.1 eV
P29.54
39.2 mT
P29.56
(a) Bx is indeterminate. By = 0 ; Bz =
Fab = 0; Fbc = 40.0 mN − i ;
Fcd
da
(b) −Fi j ; (c) −Fi j
P29.58
128 mT north at an angle of 78.7° below
the horizontal
P29.20
(a) 5.41 mA ⋅ m 2 ; (b) 4.33 mN ⋅ m
P29.22
(a)3.97° ; (b) 3.39 mN ⋅ m
P29.24
(a) 80.1 mN ⋅ m ; (b) 104 mN ⋅ m ;
(c) 132 mN ⋅ m ;
(d) The torque on the circle.
P29.62
P29.26
(a) minimum: pointing north at 48.0°
below the horizontal; maximum: pointing
south at 48.0° above the horizontal;
(b) 1.07 µJ
P29.64
λ g tan θ
I
P29.66
(a) 0.104 mm; (b) 0.189 mm
P29.68
3.82 × 10 −25 kg
P29.70
(a) see the solution;
empirically, ∆VH = 100 µ V T B ;
(b) 0.125 mm
P29.28
(a) 640 µN ⋅ m ; (b) 241 mW; (c) 2.56 mJ;
(d) 154 mW
P29.30
1.98 cm
P29.32
65.6 mT
P29.34
(a) 5.00 cm ; (b) 8.78 Mm s
P29.36
m′
=8
m
P29.38
see the solution
P29.40
244 kV m
P29.42
278 mm
P29.44
162 mm
P29.46
3.00 T
P29.48
(a) 7.44 × 10 28 m3 ; (b) 1.79 T
P29.60
P29.72
− Fi
;
evi
3R
4
B ~ 10 −1 T; τ ~ 10 −1 N ⋅ m; I ~ 1 A ;
A ~ 10 −3 m 2 ; N ~ 10 3
b
g
qBh
; The particle moves in a
m
semicircle of radius h and leaves the field
with velocity −vj;
(b) The particle moves in a smaller
mv
semicircle of radius
, attaining final
qB
velocity −vj;
(a) v =
(c) The particle moves in a circular arc of
mv
, leaving the field with
radius r =
qB
velocity v sin θ i + v cos θ j where
θ = sin −1
FG h IJ
HrK
30
Sources of the Magnetic Field
CHAPTER OUTLINE
30.1
30.2
30.3
30.4
30.5
30.6
30.7
30.8
30.9
The Biot-Savart Law
The Magnetic Force
Between Two Parallel
Conductors
Ampère’s Law
The Magnetic Field of a
Solenoid
Magnetic Flux
Gauss’s Law in Magnetism
Displacement Current and
the General Form of
Ampère’s Law
Magnetism in Matter
The Magnetic Field of the
Earth
ANSWERS TO QUESTIONS
Q30.1
It is not. The magnetic field created by a single loop of current
resembles that of a bar magnet—strongest inside the loop, and
decreasing in strength as you move away from the loop.
Neither is it in a uniform direction—the magnetic field lines
loop though the loop!
Q30.2
No magnetic field is created by a stationary charge, as the rate
of flow is zero. A moving charge creates a magnetic field.
Q30.3
The magnetic field created by wire 1 at the position of wire 2 is
into the paper. Hence, the magnetic force on wire 2 is in
direction down × into the paper = to the right, away from wire
1. Now wire 2 creates a magnetic field into the page at the
location of wire 1, so wire 1 feels force up ×into the paper = left,
away from wire 2.
FIG. Q30.3
185
186
Q30.4
Sources of the Magnetic Field
No total force, but a torque. Let wire one carry current in the y
direction, toward the top of the page. Let wire two be a millimeter
above the plane of the paper and carry current to the right, in the x
direction. On the left-hand side of wire one, wire one creates magnetic
field in the z direction, which exerts force in the i × k = − j direction on
wire two. On the right-hand side, wire one produces magnetic field in
2
e j
the − k direction and makes a i × − k = + j force of equal magnitude act
1
on wire two. If wire two is free to move, its center section will twist
counterclockwise and then be attracted to wire one.
FIG. Q30.4
Q30.5
Ampère’s law is valid for all closed paths surrounding a conductor, but not always convenient.
There are many paths along which the integral is cumbersome to calculate, although not impossible.
Consider a circular path around but not coaxial with a long, straight current-carrying wire.
Q30.6
The Biot-Savart law considers the contribution of each element of current in a conductor to
determine the magnetic field, while for Ampère’s law, one need only know the current passing
through a given surface. Given situations of high degrees of symmetry, Ampère’s law is more
convenient to use, even though both laws are equally valid in all situations.
Q30.7
If the radius of the toroid is very large compared to its cross-sectional area, then the field is nearly
uniform. If not, as in many transformers, it is not.
Q30.8
Both laws use the concept of flux—the “flow” of field lines through a surface to determine the field
strength. They also both relate the integral of the field over a closed geometrical figure to a
fundamental constant multiplied by the source of the appropriate field. The geometrical figure is a
surface for Gauss’s law and a line for Ampère’s.
Q30.9
Apply Ampère’s law to the circular path labeled 1 in the picture. Since there is no
current inside this path, the magnetic field inside the tube must be zero. On the
other hand, the current through path 2 is the current carried by the conductor.
Therefore the magnetic field outside the tube is nonzero.
FIG. Q30.9
Q30.10
Q30.11
The magnetic field inside a long solenoid is given by B =
(a)
If the length
(b)
If N is doubled, the magnetic field is doubled.
µ 0 NI
.
is doubled, the field is cut in half.
The magnetic flux is Φ B = BA cos θ . Therefore the flux is maximum when B is perpendicular to the
loop of wire. The flux is zero when there is no component of magnetic field perpendicular to the
loop—that is, when the plane of the loop contains the x axis.
Chapter 30
187
Q30.12
Maxwell included a term in Ampère’s law to account for the contributions to the magnetic field by
changing electric fields, by treating those changing electric fields as “displacement currents.”
Q30.13
M measures the intrinsic magnetic field in the nail. Unless the nail was previously magnetized, then
M starts out from zero. H is due to the current in the coil of wire around the nail. B is related to the
sum of M and H. If the nail is aluminum or copper, H makes the dominant contribution to B, but M
can add a little in the same or in the opposite direction. If the nail is iron, as it becomes magnetized
M can become the dominant contributor to B.
Q30.14
Magnetic domain alignment creates a stronger external magnetic field. The field of one piece of iron
in turn can align domains in another iron sample. A nonuniform magnetic field exerts a net force of
attraction on magnetic dipoles aligned with the field.
Q30.15
The shock misaligns the domains. Heating will also decrease magnetism.
Q30.16
Magnetic levitation is illustrated in Figure Q30.31. The Earth’s magnetic field is so weak that the
floor of his tomb should be magnetized as well as his coffin. Alternatively, the floor of his tomb could
be made of superconducting material, which exerts a force of repulsion on any magnet.
Q30.17
There is no magnetic material in a vacuum, so M must be zero. Therefore B = µ 0 H in a vacuum.
Q30.18
Atoms that do not have a permanent magnetic dipole moment have electrons with spin and orbital
magnetic moments that add to zero as vectors. Atoms with a permanent dipole moment have
electrons with orbital and spin magnetic moments that show some net alignment.
Q30.19
The magnetic dipole moment of an atom is the sum of the dipole moments due to the electrons’
orbital motions and the dipole moments due to the spin of the electrons.
Q30.20
M and H are in opposite directions. Section 30.8 argues that all atoms should be thought of as
weakly diamagnetic due to the effect of an external magnetic field on the motions of atomic
electrons. Paramagnetic and ferromagnetic effects dominate whenever they exist.
Q30.21
The effects of diamagnetism are significantly smaller than those of paramagnetism.
Q30.22
When the substance is above the Curie temperature, random thermal motion of the molecules
prevents the formation of domains. It cannot be ferromagnetic, but only paramagnetic.
Q30.23
A ferromagnetic substance is one in which the magnetic moments of the atoms are aligned within
domains, and can be aligned macroscopically. A paramagnetic substance is one in which the
magnetic moments are not naturally aligned, but when placed in an external magnetic field, the
molecules line their magnetic moments up with the external field. A diamagnetic material is one in
which the magnetic moments are also not naturally aligned, but when placed in an external
magnetic field, the molecules line up to oppose the external magnetic field.
Q30.24
(a)
B increases slightly
(b)
B decreases slightly
(c)
B increases significantly
Equations 30.33 and 30.34 indicate that, when each metal is in the solenoid, the total field is
B = µ 0 1 + χ H . Table 30.2 indicates that B is slightly greater than µ 0 H for aluminum and slightly less
for copper. For iron, the field can be made thousands of times stronger, as seen in Example 30.10.
b
g
188
Sources of the Magnetic Field
Q30.25
A “hard” ferromagnetic material requires much more energy per molecule than a “soft”
ferromagnetic material to change the orientation of the magnetic dipole moments. This way, a hard
ferromagnetic material is more likely to retain its magnetization than a soft ferromagnetic material.
Q30.26
The medium for any magnetic recording should be a hard ferromagnetic substance, so that thermal
vibrations and stray magnetic fields will not rapidly erase the information.
Q30.27
If a soft ferromagnetic substance were used, then the magnet would not be “permanent.” Any
significant shock, a heating/cooling cycle, or just rotating the magnet in the Earth’s magnetic field
would decrease the overall magnetization by randomly aligning some of the magnetic dipole
moments.
Q30.28
You can expect a magnetic tape to be weakly attracted to a magnet. Before you erase the information
on the tape, the net magnetization of a macroscopic section of the tape would be nearly zero, as the
different domains on the tape would have opposite magnetization, and be more or less equal in
number and size. Once your external magnet aligns the magnetic moments on the tape, there would
be a weak attraction, but not like that of picking up a paper clip with a magnet. A majority of the
mass of the tape is non-magnetic, and so the gravitational force acting on the tape will likely be
larger than the magnetic attraction.
Q30.29
To magnetize the screwdriver, stroke one pole of the magnet along the blade of the screwdriver
several or many times. To demagnetize the screwdriver, drop it on a hard surface a few times, or
heat it to some high temperature.
Q30.30
The north magnetic pole is near the south geographic pole. Straight up.
Q30.31
(a)
The magnets repel each other with a force equal to the weight of one of them.
(b)
The pencil prevents motion to the side and prevents the magnets from rotating under their
mutual torques. Its constraint changes unstable equilibrium into stable.
(c)
Most likely, the disks are magnetized perpendicular to their flat faces, making one face a
north pole and the other a south pole. One disk has its north pole on the top side and the
other has its north pole on the bottom side.
(d)
Then if either were inverted they would attract each other and stick firmly together.
SOLUTIONS TO PROBLEMS
Section 30.1
The Biot-Savart Law
b
g
µ 0 I µ 0 q v 2π R
=
= 12.5 T
2R
2R
P30.1
B=
P30.2
4π × 10 −7 T ⋅ m A 1.00 × 10 4 A
µ0I
=
= 2.00 × 10 −5 T = 20.0 µT
B=
2π R
2π 100 m
e
a
je
f
j
Chapter 30
P30.3
B=
(a)
FG
H
4µ 0 I
π
3π
cos − cos
4π a
4
4
189
IJ where a =
K
2
is the distance from any side to the center.
B=
4.00 × 10 −6
0.200
F
GH
I
JK
2
2
= 2 2 × 10 −5 T = 28.3 µT into the paper
+
2
2
FIG. P30.3
For a single circular turn with 4 = 2π R ,
(b)
ja f
a f
e
4π 2 × 10 −7 10.0
µ 0 I µ 0π I
=
=
= 24.7 µT into the paper
B=
2R
4
4 0. 400
e
ja
f
4π × 10 −7 1.00 A
µ0I
=
= 2.00 × 10 −7 T
2π r
2π 1.00 m
P30.4
B=
P30.5
For leg 1, ds × r = 0 , so there is no contribution to the field from this
segment. For leg 2, the wire is only semi-infinite; thus,
a
B=
F
GH
f
I
JK
µ0I
1 µ0I
=
into the paper .
2 2π x
4π x
FIG. P30.5
P30.6
We can think of the total magnetic field as the superposition of the field due to the long straight wire
µ I
(having magnitude 0 and directed into the page) and the field due to the circular loop (having
2π R
µ0I
magnitude
and directed into the page). The resultant magnetic field is:
2R
1 µ0I
B= 1+
directed into the page .
π 2R
FG
H
P30.7
IJ
K
b
g
For the straight sections ds × r = 0 . The quarter circle makes one-fourth the field of a full loop:
B=
1 µ0I µ0I
=
into the paper
4 2R
8R
B=
ja
8b0.030 0 mg
e4π × 10
−7
T ⋅ m A 5.00 A
f=
26.2 µT into the paper
190
P30.8
Sources of the Magnetic Field
Along the axis of a circular loop of radius R,
B=
µ 0 IR 2
e
B L
1
=M
B
MN bx Rg
or
2 x2 + R2
0
2
B0 ≡
where
j
32
OP
+ 1 PQ
3 2
µ0I
.
2R
FIG. P30.8
x R
0.00
1.00
2.00
3.00
4.00
5.00
*P30.9
B B0
1.00
0.354
0.0894
0.0316
0.0143
0.00754
Wire 1 creates at the origin magnetic field
B1 =
(a)
µ0I
µ I
right hand rule = 0 1
2π r
2π a
=
If the total field at the origin is
according to B 2 =
µ 0 I1
j
2π a
µ I
2µ 0 I1
j = 0 1 j + B 2 then the second wire must create field
2π a
2π a
µ 0 I1
µ I
j= 0 2
2π a
2π 2 a
.
a f
Then I 2 = 2 I 1 out of the paper = 2 I 1 k .
(b)
The other possibility is B1 + B 2 =
B2 =
*P30.10
µ I
2 µ 0 I1
− j = 0 1 j + B 2 . Then
2π a
2π a
e j
µ I
3µ 0 I1
−j = 0 2
2π a
2π 2 a
e j
a f
e j
I 2 = 6 I 1 into the paper = 6 I 1 − k
Every element of current creates magnetic field in the same direction, into the page, at the center of
the arc. The upper straight portion creates one-half of the field that an infinitely long straight wire
would create. The curved portion creates one quarter of the field that a circular loop produces at its
1 µ0I
center. The lower straight segment also creates field
.
2 2π r
The total field is
F 1 µ I + 1 µ I + 1 µ I I into the page =
GH 2 2π r 4 2r 2 2π r JK
F 0.284 15µ I IJ into the page.
=G
H r K
B=
0
0
0
0
FG
H
IJ
K
µ0I 1 1
+
into the plane of the paper
2r π 4
Chapter 30
*P30.11
(a)
191
Above the pair of wires, the field out of the page of the 50 A
e j
current will be stronger than the − k field of the 30 A
current, so they cannot add to zero. Between the wires,
both produce fields into the page. They can only add to
zero below the wires, at coordinate y = − y . Here the total
field is
B=
µ0I
2π r
+
µ0I
2π r
:
FIG. P30.11
OP
LM 50 A
30 A
−
+
k
k
MN d y + 0.28 mi e j y e jPQ
50 y = 30d y + 0.28 mi
50b− y g = 30b0.28 m − y g
−20 y = 30a0. 28 mf at y = −0.420 m
0=
µ0
2π
At y = 0.1 m the total field is B =
(b)
4π × 10 −7 T ⋅ m A
2π
B=
µ0I
2π r
+
µ0I
2π r
:
F 50 A e−kj + 30 A e−kjI = 1.16 × 10 Te−kj .
JK
GH a0.28 − 0.10f m
0.10 m
−4
The force on the particle is
e
je
je j e
je j
e j
F = qv × B = −2 × 10 −6 C 150 × 10 6 m s i × 1.16 × 10 −4 N ⋅ s C ⋅ m − k = 3.47 × 10 −2 N − j .
(c)
P30.12
dB =
B=
B=
e j
j
Fe = 3.47 × 10 −2 N + j = qE = −2 × 10 −6 C E .
So
E = −1.73 × 10 4 j N C .
µ0I d × r
4π r 2
µ0I
4π
e
We require
F
GH
1
6
FG
H
2π a
a
2
−
IJ
K
1
6
2π b
b
2
I
JK
µ0I 1 1
directed out of the paper
−
12 a b
192
*P30.13
Sources of the Magnetic Field
(a)
We use equation 30.4. For the distance a from the wire to
a
, a = 0. 288 7L . One
the field point we have tan 30° =
L 2
wire contributes to the field at P
µ0I
µ0I
cos θ 1 − cos θ 2 =
cos 30°− cos 150°
4π a
π
4 0. 288 7L
ga
b
g b
µ I a1.732 f
1.50 µ I
=
=
.
πL
π
L
4 b0.288 7 g
B=
0
f
a
P
I
F
GH
a
a f
b g
a f
b
g
f IJ
K
a
FIG. P30.13(a)
I
JK
As we showed in part (a), one whole side of the triangle
µ I 1.732
creates field at the center 0
. Now one-half of one
4π a
nearby side of the triangle will be half as far away from
point Pb and have a geometrically similar situation. Then it
µ I 1.732 2 µ 0 I 1.732
=
. The two halfcreates at Pb field 0
4π a
4π a 2
sides shown crosshatched in the picture create at Pb field
2 µ 0 I 1.732
4µ 0 I 1.732
6µ 0 I
=
=
. The rest of the
2
4π a
πL
4π 0.288 7L
triangle will contribute somewhat more field in the same
direction, so we already have a proof that the field at Pb is
F
GH
L
2 θ1
0
Each side contributes the same amount of field in the same
direction, which is perpendicularly into the paper in the
1.50 µ 0 I
4.50 µ 0 I
=
.
picture. So the total field is 3
πL
πL
(b)
θ2
f
a
a
f
FIG. P30.13(b)
stronger .
P30.14
Apply Equation 30.4 three times:
F
I toward you + µ I F
GH
JK
G
4π d H
I
µ I F −d
+
− cos 180°J toward you
G
4π a H d + a
K
µ I FH a + d − d a + d IK
B=
µ0I
d
cos 0 −
2
4π a
d + a2
0
0
2
B=
0
2
2
2
2
2π ad a 2 + d 2
2
away from you
Pb
a
d2 + a2
+
a
d 2 + a2
I away from you
JK
193
Chapter 30
P30.15
Take the x-direction to the right and the y-direction up in the plane of the
paper. Current 1 creates at P a field
e
−7
ja
2.00 × 10 T ⋅ m 3.00 A
µ I
B1 = 0 =
2π a
A 0.050 0 m
b
g
5.00 cm
I1
P
f
B1
B2
13.0 cm
12.0 cm
B1 = 12.0 µT downward and leftward, at angle 67.4° below the –x axis.
Current 2 contributes
I2
e2.00 × 10 T ⋅ mja3.00 Af clockwise perpendicular to 12.0 cm
Aa0.120 mf
−7
B2 =
FIG. P30.15
B 2 = 5.00 µT to the right and down, at angle –22.6°
Then,
Section 30.2
P30.16
b
ge
j b
B = b−11.1 µTg j − b1.92 µTg j = b −13.0 µTg j
ge
B = B1 + B 2 = 12.0 µT − i cos 67.4°− j sin 67.4° + 5.00 µT i cos 22.6°− j sin 22.6°
The Magnetic Force Between Two Parallel Conductors
Let both wires carry current in the x direction, the first at y = 0 and the
second at y = 10.0 cm .
(a)
e
ja
f
I2 = 8.00 A
f
4π × 10 −7 T ⋅ m A 5.00 A
µ0I
B=
k=
k
2π r
2π 0.100 m
a
y
I1 = 5.00 A
a
fa
f e
j e
je j
FB = I 2 × B = 8.00 A 1.00 m i × 1.00 × 10 −5 T k = 8.00 × 10 −5 N − j
B=
e j e
ja
f
f
4π × 10 −7 T ⋅ m A 8.00 A
µ0I
−k =
− k = 1.60 × 10 −5 T − k
2π r
2π 0.100 m
a
e j e
je j
B = 1.60 × 10 −5 T into the page
(d)
a
fa
x
FIG. P30.16(a)
FB = 8.00 × 10 −5 N toward the first wire
(c)
y = 10.0 cm
z
B = 1.00 × 10 −5 T out of the page
(b)
j
f e
je j e
je j
FB = I 1 × B = 5.00 A 1.00 m i × 1.60 × 10 −5 T − k = 8.00 × 10 −5 N + j
FB = 8.00 × 10 −5 N towards the second wire
194
P30.17
Sources of the Magnetic Field
By symmetry, we note that the magnetic forces on the top and bottom
segments of the rectangle cancel. The net force on the vertical segments of
the rectangle is (using Equation 30.11)
FG 1 − 1 IJ i = µ I I FG − a IJ i
H c + a c K 2π H cac + af K
e4π × 10 N A ja5.00 Afa10.0 Afa0.450 mf F −0.150 m I i
F=
GH a0.100 mfa0.250 mf JK
2π
F = e −2.70 × 10 i j N
µ 0 I1 I 2
2π
F = F1 + F2 =
−7
0 1 2
2
−5
or
*P30.18
F = 2.70 × 10 −5 N toward the left .
FIG. P30.17
To attract, both currents must be to the right. The attraction is described by
F = I 2 B sin 90° = I 2
So
µ0I
2π r
F
I
GjG 4π × 10 2πNa0⋅.s5 Cm⋅fm a20 Af JJ = 40.0 A
j K
He
F 2π r
= 320 × 10 −6 N m
µ 0 I1
e
I2 =
FIG. P30.18
−7
Let y represent the distance of the zero-field point below the upper wire.
Then
B=
µ0I
2π r
+
b
µ0I
2π r
g
20 0.5 m − y = 40 y
0=
F
GH
µ 0 20 A
40 A
away +
toward
y
2π
0.5 m − y
a
b
ga
g b
fIJK
f
20 0.5 m = 60 y
y = 0.167 m below the upper wire
*P30.19
Carrying oppositely directed currents, wires 1 and 2 repel
each other. If wire 3 were between them, it would have to
repel either 1 or 2, so the force on that wire could not be
zero. If wire 3 were to the right of wire 2, it would feel a
larger force exerted by 2 than that exerted by 1, so the total
force on 3 could not be zero. Therefore wire 3 must be to
the left of both other wires as shown. It must carry
downward current so that it can attract wire 2.
(a)
For the equilibrium of wire 3 we have
F1 on 3 = F2 on 3
a
f
1.5 20 cm + d = 4d
(b)
a
f
Wire 3
Wire 1
Wire 2
1.50 A
I3
4.00 A
20 cm
d
FIG. P30.19
a f
µ 0 1.50 A I 3
µ 0 4 A I3
=
2π d
2π 20 cm + d
30 cm
d=
= 12.0 cm to the left of wire 1
2.5
a
f
For the equilibrium of wire 1,
a
f
f
a fa f
a f
µ 0 I 3 1.5 A
µ 4 A 1.5 A
= 0
2π 12 cm
2π 20 cm
a
I3 =
12
4 A = 2.40 A down
20
We know that wire 2 must be in equilibrium because the forces on it are equal in magnitude
to the forces that it exerts on wires 1 and 3, which are equal because they both balance the
equal-magnitude forces that 1 exerts on 3 and that 3 exerts on 1.
Chapter 30
P30.20
The separation between the wires is
a
f
a = 2 6.00 cm sin 8.00° = 1.67 cm.
(a)
Because the wires repel, the currents are in
opposite directions .
(b)
Because the magnetic force acts horizontally,
µ I2
FB
= 0
= tan 8.00°
Fg 2π amg
mg 2π a
I2 =
Section 30.3
P30.21
µ0
FIG. P30.20
tan 8.00° so I = 67.8 A .
Ampère’s Law
Each wire is distant from P by
a0.200 mf cos 45.0° = 0.141 m.
Each wire produces a field at P of equal magnitude:
e
ja
f
2.00 × 10 −7 T ⋅ m A 5.00 A
µ0I
=
= 7.07 µT .
BA =
2π a
0.141 m
a
f
Carrying currents into the page, A produces at P a field of
7.07 µT to the left and down at –135°, while B creates a field
to the right and down at – 45°. Carrying currents toward
you, C produces a field downward and to the right at – 45°,
while D’s contribution is downward and to the left. The
total field is then
b
FIG. P30.21
g
4 7.07 µT sin 45.0° = 20.0 µT toward the bottom of the page
P30.22
µ0I
at the proton’s location. And we have a
2π d
balance between the weight of the proton and the magnetic force
Let the current I be to the right. It creates a field B =
e j e j
mg − j + qv − i ×
e
µ0I
k = 0 at a distance d from the wire
2π d
ej
je
je
je
j
1.60 × 10 −19 C 2.30 × 10 4 m s 4π × 10 −7 T ⋅ m A 1.20 × 10 −6 A
qvµ 0 I
d=
=
= 5.40 cm
2π mg
2π 1.67 × 10 −27 kg 9.80 m s 2
e
je
j
195
196
P30.23
Sources of the Magnetic Field
µ 0Ia
, where I a is the net current
2π ra
through the area of the circle of radius ra . In this case, I a = 1.00 A out of the page (the current in the
inner conductor), so
4π × 10 −7 T ⋅ m A 1.00 A
= 200 µT toward top of page .
Ba =
2π 1.00 × 10 −3 m
µ I
Similarly at point b : Bb = 0 b , where I b is the net current through the area of the circle having
2π rb
radius rb .
From Ampere’s law, the magnetic field at point a is given by Ba =
e
ja
e
f
j
Taking out of the page as positive, I b = 1.00 A − 3.00 A = −2.00 A , or I b = 2.00 A into the page.
Therefore,
4π × 10 −7 T ⋅ m A 2.00 A
= 133 µT toward bottom of page .
Bb =
2π 3.00 × 10 −3 m
e
P30.24
ja
e
j
f
µ0I
, the field will be one-tenth as large at a ten-times larger distance: 400 cm
2π r
(a)
In B =
(b)
B=
(c)
Call r the distance from cord center to field point and 2d = 3.00 mm the distance between
conductors.
µ I 1
µ I 2d
1
B= 0
−
= 0 2
2π r − d r + d
2π r − d 2
3.00 × 10 −3 m
−10
−7
so r = 1.26 m
7.50 × 10
T = 2.00 × 10 T ⋅ m A 2.00 A 2
r − 2.25 × 10 −6 m 2
The field of the two-conductor cord is weak to start with and falls off rapidly with distance.
a
4π × 10 −7 T ⋅ m 2.00 A
µ0I
µ I
k + 0 − k so B =
2π r1
2π r2
2π A
e j
FG
H
IJ
K
ja
e
(d)
f FG 1 − 1 IJ =
H 0.398 5 m 0.401 5 m K
f e
7.50 nT
j
The cable creates zero field at exterior points, since a loop in Ampère’s law encloses zero
total current. Shall we sell coaxial-cable power cords to people who worry about biological
damage from weak magnetic fields?
P30.25
(a)
One wire feels force due to the field of the other ninety-nine.
B=
ja fb
e
ge
j
j
−7
−2
µ 0 I 0 r 4π × 10 T ⋅ m A 99 2.00 A 0.200 × 10 m
=
= 3.17 × 10 −3 T
2
−2
2π R 2
2π 0.500 × 10 m
e
This field points tangent to a circle of radius 0.200 cm and exerts force
F = I × B toward the center of the bundle, on the single hundredth wire:
F
FB
(b)
a
fe
j
= IB sin θ = 2.00 A 3.17 × 10 −3 T sin 90° = 6.34 mN m
= 6.34 × 10 −3 N m inward
B ∝ r , so B is greatest at the outside of the bundle. Since each wire
carries the same current, F is greatest at the outer surface .
FIG. P30.25
Chapter 30
P30.26
(a)
(b)
*P30.27
e
ja fe
j
a
f
T ⋅ m A ja900fe14.0 × 10 A j
=
Binner
4π × 10 −7 T ⋅ m A 900 14.0 × 10 3 A
µ 0 NI
=
=
= 3.60 T
2π r
2π 0.700 m
Bouter
2 × 10 −7
µ 0 NI
=
=
2π r
e
197
3
1.30 m
1.94 T
We assume the current is vertically upward.
(a)
Consider a circle of radius r slightly less than R. It encloses no current so from
z
b g
B ⋅ ds = µ 0 I inside
B 2π r = 0
we conclude that the magnetic field is zero .
b
g
Now let the r be barely larger than R. Ampere’s law becomes B 2π R = µ 0 I ,
(b)
B=
so
µ0I
.
2π R
FIG. P30.27(a)
tangent to the wall of the cylinder in a counterclockwise sense .
The field’s direction is
(c)
Consider a strip of the wall of width dx and length . Its width is so small
compared to 2π R that the field at its location would be essentially
unchanged if the current in the strip were turned off.
Idx
The current it carries is I s =
up.
2π R
The force on it is
µ0I
µ I 2 dx
Idx
F = Is × B =
up × into page = 0 2 2 radially inward .
2π R 2π R
4π R
F
GH
I
JK
FIG. P30.27(c)
The pressure on the strip and everywhere on the cylinder is
P=
µ I 2 dx
F
= 02 2
=
A 4π R dx
µ0I 2
b2π Rg
2
inward .
The pinch effect makes an effective demonstration when an aluminum can crushes itself as
it carries a large current along its length.
z
2π rB
ja
e
2π 1.00 × 10 −3 0.100
f=
P30.28
From B⋅ d = µ 0 I ,
P30.29
Use Ampère’s law, B ⋅ ds = µ 0 I . For current density J, this becomes
z
z
B ⋅ ds = µ 0 J ⋅ dA .
(a)
I=
µ0
=
4π × 10
−7
500 A .
z
For r1 < R , this gives
B 2π r1 = µ 0
za
r1
fb
g
br 2π rdr and
0
FIG. P30.29
µ 0 br12
B=
for r1 < R or inside the cylinder .
3
b
(b)
g
When r2 > R , Ampère’s law yields
b2π r gB = µ z abr fb2π rdr g = 2πµ3bR
R
2
0
0
or B =
µ 0 bR 3
for r2 > R or outside the cylinder .
3r2
b
g
0
3
,
198
P30.30
Sources of the Magnetic Field
(a)
See Figure (a) to the right.
(b)
At a point on the z axis, the contribution from each wire has
µ0I
magnitude B =
and is perpendicular to the line
2π a 2 + z 2
from this point to the wire as shown in Figure (b). Combining (Currents are into the paper)
fields, the vertical components cancel while the horizontal
Figure (a)
components add, yielding
By = 2
F
GH 2π
µ0I
2
2
a +z
I
JK
sin θ =
µ0I
2
π a +z
2
F
GH
z
I = µ Iz
JK π ea + z j
0
2
a +z
2
2
2
The condition for a maximum is:
dBy
dz
=
a f + µ I = 0 , or µ I ea
π ea
π ea + z j
π ea + z j
− µ 0 Iz 2 z
2 2
2
2
2
j =0
+z j
2
− z2
2
2
0
0
Figure (b)
Thus, along the z axis, the field is a maximum at d = a .
Section 30.4
P30.31
*P30.32
FIG. P30.30
The Magnetic Field of a Solenoid
B = µ0
N
I so I =
e
j
1.00 × 10 −4 T 0.400 m
B
=
= 31.8 mA
µ 0 n 4π × 10 −7 T ⋅ m A 1 000
e
j
Let the axis of the solenoid lie along the y–axis from y = 0 to y = . We will determine the field at
y = a . This point will be inside the solenoid if 0 < a < and outside if a < 0 or a > . We think of
solenoid as formed of rings, each of thickness dy. Now I is the symbol for the current in each turn of
N
N
wire and the number of turns per length is
dy and
. So the number of turns in the ring is
FG IJ
H K
the current in the ring is I ring = I
FG IJ
H K
FG N IJ dy . Now we use the result of Example 30.3 for the field created
H K
by one ring:
Bring =
µ 0 I ring R 2
e
2 x2 + R2
j
32
where x is the name of the distance from the center of the ring, at location y, to the field point
x = a − y . Each ring creates field in the same direction, along our y–axis, so the whole field of the
solenoid is
µ 0 I ring R 2
µ 0 I N dyR 2
dy
µ 0 INR 2
=
=
B = ∑ Bring = ∑
.
32
3 2
3 2
2
2
2
2
2
all rings
0 2 a−y
0 2 a−y
2 x +R
+ R2
+ R2
e
j
z eb
b g
g j
To perform the integral we change variables to u = a − y .
B=
µ 0 INR 2
2
ze
a−
a
− du
u2 + R2
j
32
and then use the table of integrals in the appendix:
continued on next page
z eb
g
j
Chapter 30
a−
µ INR 2
−u
B= 0
2
2
R u2 + R2
(a)
(b)
If
µ 0 IN
2
=
a
LM
MMN
a
a2 + R2
a−
−
aa − f
2
+ R2
199
OP
PPQ
is much larger than R and a = 0,
we have B ≅
LM
MN
OP
PQ
µ 0 IN
µ IN
−
= 0 .
0−
2
2
2
This is just half the magnitude of the field deep within the solenoid. We would get the same
result by substituting a = to describe the other end.
P30.33
The field produced by the solenoid in its interior is given by
e j e
B = µ 0 nI − i = 4π × 10 −7 T ⋅ m A
e
jFGH 1030.0m IJK a15.0 Afe− ij
−2
j
B = − 5.65 × 10 −2 T i
The force exerted on side AB of the square current loop is
a
bF g
= IL × B = 0. 200 A
bF g
= 2.26 × 10 −4 N k
B AB
B AB
e
f e2.00 × 10
−2
j e
je j
m j × 5.65 × 10 −2 T − i
j
Similarly, each side of the square loop experiences a force, lying
in the plane of the loop, of
226 µN directed away from the center . From the above
result, it is seen that the net torque exerted on the square loop
by the field of the solenoid should be zero. More formally, the
magnetic dipole moment of the square loop is given by
a
fe
FIG. P30.33
j e− ij = −80.0 µA ⋅ m i
The torque exerted on the loop is then τ = µ × B = e −80.0 µA ⋅ m i j × e −5.65 × 10
µ = IA = 0.200 A 2.00 × 10 −2 m
2
2
2
Section 30.5
P30.34
a
f
(a)
bΦ g
(b)
The net flux out of the closed surface is zero: Φ B
B flat
= B ⋅ A = Bπ R 2 cos 180 − θ = − Bπ R 2 cos θ
b g
B curved
(a)
flat
b g
+ ΦB
curved
= 0.
= Bπ R 2 cos θ
z
e
j e
j
2
Φ B = B ⋅ dA = B ⋅ A = 5 i + 4 j + 3k T ⋅ 2.50 × 10 −2 m i
Φ B = 3.12 × 10 −3 T ⋅ m 2 = 3.12 × 10 −3 Wb = 3.12 mWb
(b)
bΦ g
B total
z
j
Ti = 0
Magnetic Flux
bΦ g
P30.35
−2
= B ⋅ dA = 0 for any closed surface (Gauss’s law for magnetism)
200
P30.36
Sources of the Magnetic Field
(a)
Φ B = B ⋅ A = BA where A is the cross-sectional area of the solenoid.
FG µ NI IJ eπ r j = 7.40 µWb
H K
F µ NI IJ π er − r j
= B ⋅ A = BA = G
H K
L e4π × 10 T ⋅ m Aja300fa12.0 Af OP
=M
PQπ a8.00f − a4.00f e10
MN
a0.300 mf
(b)
ΦB
2
0
ΦB =
2
2
0
2
1
−7
ΦB
Section 30.6
2
2
−3
m
j
2
= 2.27 µWb
Gauss’s Law in Magnetism
No problems in this section
Section 30.7
P30.37
P30.38
Displacement Current and the General Form of Ampère’s Law
a
0.100 A
dΦ E dQ dt
I
=
=
=
= 11.3 × 10 9 V ⋅ m s
∈0
∈0 8.85 × 10 −12 C 2 N ⋅ m 2
dt
(b)
I d =∈0
dΦ E
= I = 0.100 A
dt
dQ dt
dΦ E d
I
=
=
EA =
∈0
∈0
dt
dt
a f
(a)
dE
I
=
= 7.19 × 10 11 V m ⋅ s
dt ∈0 A
(b)
z
B ⋅ ds =∈0 µ 0
B=
Section 30.8
P30.39
f
(a)
(a)
LM
N
dΦ E
d
Q
⋅π r 2
so 2π rB =∈0 µ 0
dt
dt ∈0 A
a
fe
a f
j
OP
Q
−2
µ 0 Ir µ 0 0.200 5.00 × 10
=
= 2.00 × 10 −7 T
2
2A
2π 0.100
Magnetism in Matter
I=
ev
2π r
µ = IA =
F ev I π r
GH 2π r JK
2
= 9.27 × 10 −24 A ⋅ m 2
The Bohr model predicts the correct magnetic moment. However, the
“planetary model” is seriously deficient in other regards.
(b)
Because the electron is (–), its [conventional] current is clockwise, as seen
from above, and µ points downward .
FIG. P30.39
Chapter 30
F N I I so I = b2π r gB = 2π a0.100 mfa1.30 Tf =
GH 2π r JK
µN
5 000e 4π × 10 Wb A ⋅ mja 470f
P30.40
B = µnI = µ
P30.41
Assuming a uniform B inside the toroid is equivalent to assuming
NI
r << R ; then B0 ≈ µ 0
as for a tightly wound solenoid.
2π R
630 3.00
B0 = µ 0
= 0.001 89 T
2π 0.200
−7
a fa f
a f
b
277 mA
a fb
g
g B = 0.191 T
atoms m ja5.00fe9.27 × 10
J T j
=
With the steel, B = κ m B0 = 1 + χ B0 = 101 0.001 89 T
a
fa
fe
4.00 K 10.0% 8.00 × 10 27
TM
=
5.00 T
B
−24
3
FIG. P30.41
2
2.97 × 10 4
K⋅J
P30.42
C=
P30.43
B = µ 0 H + M so H =
P30.44
In B = µ 0 H + M we have 2.00 T = µ 0 M . But also M = xnµ B . Then B = µ 0 µ B xn where n is the
number of atoms per volume and x is the number of electrons per atom contributing.
2.00 T
B
=
= 2.02 .
x=
Then
µ 0 µ Bn 8.50 × 10 28 m −3 9.27 × 10 −24 N ⋅ m T 4π × 10 −7 T ⋅ m A
a
f
a
B
P30.47
T ⋅ m3
f
je
je
j
(a)
Comparing Equations 30.29 and 30.30, we see that the applied field is described by
B
C
B0 = µ 0 H . Then Eq. 30.35 becomes M = C 0 = µ 0 H , and the definition of susceptibility
T T
M C
(Eq. 30.32) is χ =
= µ0 .
H T
(b)
C=
Section 30.9
P30.46
2
− M = 2.62 × 10 6 A m
µ0
e
P30.45
201
ja
e
f
2.70 × 10 −4 300 K
χT
K ⋅A
=
= 6. 45 × 10 4
−7
µ0
T⋅m
4π × 10 T ⋅ m A
The Magnetic Field of the Earth
ja fa
e
Bh = Bcoil =
(b)
Bh = B sin φ → B =
(a)
Number of unpaired electrons =
Bh
12.6 µT
=
= 56.0 µT
sin φ sin 13.0°
FIG. P30.46
8.00 × 10 22 A ⋅ m 2
45
8
.
63
10
=
×
.
9.27 × 10 −24 A ⋅ m 2
Each iron atom has two unpaired electrons, so the number of iron atoms required is
1
8.63 × 10 45 .
2
e
j
e4.31 × 10 atomsje7 900 kg m j =
e8.50 × 10 atoms m j
45
(b)
f
4π × 10 −7 5.00 0.600
µ 0 NI
=
= 12.6 µT
2R
0.300
(a)
Mass =
3
28
3
4.01 × 10 20 kg
202
Sources of the Magnetic Field
Additional Problems
P30.48
B=
µ 0 IR 2
e
2 R2 + R
µ0
=
e
je
2 5 2 7.00 × 10 −5 T 6.37 × 10 6 m
e4π × 10
−7
T⋅m A
j
j
Consider a longitudinal filament of the strip of width dr as
shown in the sketch. The contribution to the field at point P
due to the current dI in the element dr is
dB =
µ 0 dI
2π r
dI = I
where
z z
B = dB =
b+w
b
P30.50
I=
2 5 2 BR
I = 2.01 × 10 9 A toward the west
so
P30.49
j
2 3 2
µ I
= 5 02
2 R
FG dr IJ
H wK
FG
H
IJ
K
FIG. P30.49
µ 0 Idr
µ0I
w
k=
k .
ln 1 +
b
2π wr
2π w
200 W
= 17 A
12 V
current going through the switch 60 cm from the compass. Suppose the dashboard contains little
iron, so µ ≈ µ 0 . Model the current as straight. Then,
Suppose you have two 100-W headlights running from a 12-V battery, with the whole
B=
e
j
a f
4π × 10 −7 17
µ0I
=
~ 10 −5 T .
2π r
2π 0.6
If the local geomagnetic field is 5 × 10 −5 T , this is ~ 10 −1 times as large, enough to affect the
compass noticeably.
P30.51
We find the total number of turns:
B=
µ 0 NI
b0.030 0 Tga0.100 mf A = 2.39 × 10
µ I e 4π × 10 T ⋅ mja1.00 A f
F 10.0 cm I = 200 closely wound turns
Each layer contains G
H 0.050 0 cm JK
F 2.39 × 10 I = 12 layers .
so she needs
GH 200 JK
N=
B
=
0
3
−7
3
The inner diameter of the innermost layer is 10.0 mm. The outer diameter of the outermost layer is
10.0 mm + 2 × 12 × 0.500 mm = 22.0 mm . The average diameter is 16.0 mm, so the total length of
wire is
e2.39 × 10 jπ e16.0 × 10
3
−3
j
m = 120 m .
Chapter 30
*P30.52
At a point at distance x from the left end of the bar, current
µ 0I2
to the left and
I 2 creates magnetic field B =
2π h 2 + x 2
x
above the horizontal at angle θ where tan θ = . This field
h
exerts force on an element of the rod of length dx
µ 0 I 2 dx
dF = I 1 × B = I 1
=
dF =
2π h 2 + x 2
µ 0 I 1 I 2 dx
x
2π h 2 + x 2
µ 0 I 1 I 2 xdx
e
2π h 2 + x 2
h2 + x 2
θ
x
h
B
203
I1
θ
I2
sin θ
FIG. P30.52
right hand rule
into the page
−k
je j
The whole force is the sum of the forces on all of the elements of the bar:
µ I I e −k j 2 xdx
µ I I e−k j
−
k
=
=
e
j
z 2π eh + x j
z h + x 4π lneh + x j
4π
µ I I e− k j
10 N a100 A fa 200 A fe − k j L a0.5 cmf + a10 cmf
=
lne h + j − ln h =
ln M
4π
A
MN a0.5 cmf
= 2 × 10 Ne − k j ln 401 = 1.20 × 10 Ne − k j
µ 0 I 1 I 2 xdx
F=
2
x =0
0 1 2
0 1 2
2
0
2
2
2
0
−7
0 1 2
2
2
2
2
2
−3
P30.53
2
2
2
OP
PQ
−2
On the axis of a current loop, the magnetic field is given by
where in this case I =
B=
q
µ 0 IR 2
e
2 x2 + R2
j
32
b2π ω g . The magnetic field is directed away from the center, with a magnitude
of
B=
P30.54
µ 0 ωR 2 q
e
2
4π x + R
j
2 32
=
a fa f e10.0 × 10 j =
4π b0.050 0g + a0.100f
µ 0 20.0 0.100
2
2
−6
2
32
1.43 × 10 −10 T .
On the axis of a current loop, the magnetic field is given by
where in this case I =
when x =
R
.
2
q
b 2π ω g
.
B=
Therefore,
B=
then
B=
µ 0 IR 2
e
2 x2 + R2
j
32
µ 0 ωR 2 q
e
4π x 2 + R 2
µ 0 ωR 2 q
4π
eRj
5
4
j
2 32
3 2
=
µ 0 qω
.
2.5 5πR
204
P30.55
Sources of the Magnetic Field
(a)
Bx =
Use equation 30.7 twice:
e
µ 0 IR 2
2 x2 + R2
j
3 2
If each coil has N turns, the field is just N times larger.
B = Bx1 + Bx 2
B=
(b)
Nµ 0 IR
2
LM
MM ex
N
2
2
2
+ R2
j
32
+
a R − xf
1
2
+ R2
32
OP
1
1
+
+R j
e2R + x − 2xRj PPQ
2 3 2
2
LM a fe
N
OP
PP
Q
FIG. P30.55
32
2
j
LMe
N
a
f e2 R
2
2
e
+ x 2 − 2 xR
j a
e
j
Again substituting x =
j
e
fOPQ
j
−5 2
j OPQ
−7 2
d 2B
R
=0 .
and canceling terms,
2
dx 2
“Helmholtz pair” → separation distance = radius
B=
2 µ 0 IR 2
b g
2 R2
2
+ R2
32
=
µ 0 IR 2
1
4
+1
For N turns in each coil, B =
*P30.57
1
−5 2
−5 2
dB Nµ 0 IR 2
3
3
=
− 2x x 2 + R 2
− 2 R 2 + x 2 − 2 xR
2x − 2R
dx
2
2
2
R
dB
Substituting x = and canceling terms,
=0 .
dx
2
−5 2
−7 2
d 2 B −3 Nµ 0 IR 2
=
− 5x 2 x 2 + R2
+ 2 R 2 + x 2 − 2 xR
x2 + R2
2
2
dx
−5 x − R
P30.56
LM
MM x
Ne
Nµ 0 IR 2
=
2
3 2
R
3
=
µ0I
for 1 turn.
1.40 R
e
j a f
a f
4π × 10 −7 100 10.0
µ 0 NI
=
= 1.80 × 10 −3 T .
1.40 R
1.40 0.500
Consider first a solid cylindrical rod of radius R carrying
current toward you, uniformly distributed over its crosssectional area. To find the field at distance r from its center we
consider a circular loop of radius r:
z
B ⋅ ds = µ 0 I inside
µ 0 Jr
µ J
B= 0 k×r
2
2
Now the total field at P inside the saddle coils is the field due to
a solid rod carrying current toward you, centered at the head of
vector a, plus the field of a solid rod centered at the tail of
vector a carrying current away from you.
µ J
µ J
B1 + B 2 = 0 k × r1 + 0 − k × r2
2
2
Now note a + r1 = r2
B 2π r = µ 0 π r 2 J
B=
e j
B1 + B 2 =
k
B1
P
r2
g
r1
a
k
FIG. P30.57
µ0 J
µ J
µ J
µ 0 Ja
k × r1 − 0 k × a + r1 = 0 a × k =
down in the diagram .
2
2
2
2
b
B2
205
Chapter 30
*
P30.58
From example 30.6, the upper sheet creates field
µ J
µ J
B = 0 s k above it and 0 s − k below it. Consider a
2
2
patch of the sheet of width w parallel to the z axis and
length d parallel to the x axis. The charge on it σ wd passes z
q σ wdv
d
a point in time , so the current it constitutes is =
v
t
d
σ wv
= σ v . Then the
and the linear current density is J s =
w
magnitude of the magnetic field created by the upper sheet
1
is µ 0 σ v . Similarly, the lower sheet in its motion toward
2
the right constitutes current toward the left. It creates
1
1
magnetic field µ 0σ v − k above it and µ 0 σ vk below it.
2
2
y
e j
w
+
+
x
–
–
+
–
+
–
d
FIG. P30.58
e j
(b)
Above both sheets and below both, their equal-magnitude fields add to zero .
(a)
Between the plates, their fields add to µ 0σ v − k = µ 0σ v away from you horizontally.
(c)
The upper plate exerts no force on itself. The field of the lower plate,
e j
a force on the current in the w- by d-section, given by
I × B = σ wvd i ×
e j
1
1
µ 0σ v − k = µ 0σ 2 v 2 wd j .
2
2
The force per area is
(d)
1
µ 0σ v −k will exert
2
e j
1 µ 0σ 2 v 2 wd
1
µ 0σ 2 v 2 up .
j=
2
2
wd
σ
−j =
2 ∈0
wσ 2
σ2
1
down =
down. To have µ 0σ 2 v 2 =
The electrical force per area is
2
2 ∈0 w
2 ∈0
require
e j
The electrical force on our section of the upper plate is qE lower = σ w
v=
1
µ 0 ∈0
1
=
4π × 10
−7
bTm AgbN TAmg8.85 × 10 eC
−12
2
Nm
2
jbAs Cg
This is the speed of light, not a possible speed for a metal plate.
2
wσ 2
−j .
2 ∈0
e j
σ2
we
2 ∈0
= 3.00 × 10 8 m s .
206
P30.59
Sources of the Magnetic Field
Model the two wires as straight parallel wires (!)
(a)
FB =
µ0I 2
(Equation 30.12)
2π a
FB =
e4π × 10 ja140f a2π fa0.100f
2π e1.00 × 10 j
2
−7
−3
= 2.46 N upward
(b)
2.46 N − m loop g
a loop =
m loop
= 107 m s 2 upward
FIG. P30.59
P30.60
(a)
µ0
Ids × r , the moving charge constitutes a bit of current as in I = nqvA . For a
4π r 2
µ0
nqA ds v × r . Next,
positive charge the direction of ds is the direction of v, so dB =
4π r 2
A ds is the volume occupied by the moving charge, and nA ds = 1 for just one charge.
Then,
In dB =
a f
a f
B=
a f
µ0
4π r 2
qv × r .
e4π × 10
je
−7
je
T ⋅ m A 1.60 × 10 −19 C 2.00 × 10 7 m s
j sin 90.0° =
3.20 × 10 −13 T
(b)
B=
(c)
FB = q v × B = 1.60 × 10 −19 C 2.00 × 10 7 m s 3.20 × 10 −13 T sin 90.0°
e
4π 1.00 × 10
e
j
−3 2
je
je
j
FB = 1.02 × 10 −24 N directed away from the first proton
(d)
Fe = qE =
k e q1 q 2
r2
e8.99 × 10
=
9
je
N ⋅ m 2 C 2 1.60 × 10 −19 C
j
2
e1.00 × 10 j
−3 2
Fe = 2.30 × 10 −22 N directed away from the first proton
Both forces act together. The electrical force is stronger by two orders of magnitude. It is
productive to think about how it would look to an observer in a reference frame moving
along with one proton or the other.
Chapter 30
P30.61
e
ja
g
f
(a)
4π × 10 −7 T ⋅ m A 24.0 A
µ0I
=
= 2.74 × 10 −4 T
B=
2π r
2π 0.017 5 m
(b)
At point C, conductor AB produces a field
b
1
2.74 × 10 −4 T − j ,
2
e
produces a field of
207
je j
1
2.74 × 10 −4 T − j ,
2
e
je j
conductor DE
BD produces no field, and AE produces
e j
negligible field. The total field at C is 2.74 × 10 −4 T − j .
*P30.62
a
fe
j e
je j e1.15 × 10
FB = I × B = 24.0 A 0.035 0 mk × 5 2.74 × 10 −4 T − j =
(d)
a=
(e)
The bar is already so far from AE that it moves through nearly constant magnetic field. The
force acting on the bar is constant, and therefore the bar’s acceleration is constant .
(f)
v 2f = vi2 + 2 ax = 0 + 2 0.384 m s 2 1.30 m , so v f =
∑ F = e1.15 × 10
m
3.0 × 10
−3
−3
2
ja
Each turn creates field at the center
IJ
K
N i
j = e0.384 m s ji
kg
N i
e
FG
H
−3
j
(c)
f
b0.999 m sgi
µ0I
. Together they create field
2R
FG
H
IJ
K
µ0I 1
1
1
4π × 10 −7 TmI 1
1
1
1
+
+…+
=
+
+…+
2 R1 R 2
R50
2A
5.05 5.15
9.95 10 −2 m
b
g
B
I
FIG. P30.62
= µ 0 I 50 m 6.93 = 347 µ 0 I m
P30.63
At equilibrium,
IB =
P30.64
(a)
b
FB
=
b g
2π a m g
µ 0 I A I B mg
=
or I B =
µ0IA
2π a
gb
ge
2π 0.025 0 m 0.010 0 kg m 9.80 m s 2
e4π × 10
−7
ja
T ⋅ m A 150 A
f
j=
The magnetic field due to an infinite sheet of current (or the
magnetic field at points near a large sheet of current) is given by
µ J
I
B = 0 s . The current density J s = and in this case the
2
equivalent current of the moving charged belt is
I=
dq d
dx
σ x = σ v; v = .
=
dt
dt dt
b g
µ σv
Therefore, J s = σ v and B = 0
.
2
(b)
81.7 A
FIG. P30.64
If the sheet is positively charged and moving in the direction shown, the magnetic field is
out of the page, parallel to the roller axes .
208
P30.65
Sources of the Magnetic Field
µ 0 I1
counterclockwise. The curved portions of the loop feels no
2π R
force since × B = 0 there. The straight portions both feel I × B forces to the right, amounting to
The central wire creates field B =
FB = I 2 2L
I=
2π rB
µ 0 I1
µ 0 I1 I 2 L
=
to the right .
πR
2π R
=
e
je
2π 9.00 × 10 3 1.50 × 10 −8
j=
P30.66
675 A
µ0
4π × 10
Flow of positive current is downward or negative charge flows upward .
P30.67
By symmetry of the arrangement, the magnitude of the net magnetic field
at point P is B = 8B0 x where B0 is the contribution to the field due to
L
current in an edge length equal to . In order to calculate B0 , we use the
2
Biot-Savart law and consider the plane of the square to be the yz-plane with
point P on the x-axis. The contribution to the magnetic field at point P due
to a current element of length dz and located a distance z along the axis is
given by Equation 30.3.
µ I d ×r
B0 = 0
.
4π
r2
From the figure we see that
−7
z
e L 4j + x
e L 4j + x + z
2
e j
r = x 2 + L2 4 + z 2 and d × r = dz sin θ = dz
2
FIG. P30.67
2
2
2
.
By symmetry all components of the field B at P cancel except the components along x
(perpendicular to the plane of the square); and
L2
B0 x = B0 cos φ where cos φ =
.
L2 4 + x 2
e j
Therefore, B0 x =
µ0I
4π
z
L2
0
sin θ cos φdz
and B = 8B0 x .
r2
Using the expressions given above for sin θ cos φ , and r, we find
B=
P30.68
(a)
µ 0 IL2
e e jj
2π x 2 + L2 4
e j
x 2 + L2 2
.
From Equation 30.9, the magnetic field produced by one loop at the center of the second
2
µ IR 2 µ 0 I π R
µ µ
= 0 3 where the magnetic moment of either loop
loop is given by B = 0 3 =
3
2x
2π x
2π x
e
e
j
j
is µ = I π R 2 . Therefore,
FG
H
µ µ
dB
=µ 0
Fx = µ
dx
2π
(b)
IJ FG 3 IJ = 3 µ eπ R Ij
K H x K 2π x
0
4
2
4
2
=
ja
e
3π µ 0 I 2 R 4
.
2
x4
fe
j
2
j
−7
−3
3π µ 0 I 2 R 4 3π 4π × 10 T ⋅ m A 10.0 A 5.00 × 10 m
=
Fx =
4
2
2
x4
5.00 × 10 −2 m
e
4
= 5.92 × 10 −8 N
Chapter 30
P30.69
There is no contribution from the straight portion of the
wire since ds × r = 0 . For the field of the spiral,
dB =
b
µ 0 I ds × r
4π
r2
a f
g
z
z
ze
B=
µ 0 I 2π ds sin θ r µ 0 I 2π
=
4π θ = 0
4π θ = 0
r2
B=
µ 0 I 2 π −2
µ I
r dr = − 0 r −1
4π θ = 0
4π
e j
(a)
2
2θ
θ =0
µ I
Substitute r = eθ : B = − 0 e −θ
4π
P30.70
jLMN FGH 34π IJK OPQ r1
2 dr sin
2π
0
µ I
µ0I
1 − e −2 π
= − 0 e −2 π − e 0 =
4π
4π
e
FIG. P30.69
j
out of the page.
B = B0 + µ 0 M
M=
B − B0
µ0
and M =
B − B0
µ0
Assuming that B and B0 are parallel, this
B − B0
becomes M =
.
µ0
The magnetization curve gives a plot of M
versus B0 .
(b)
The second graph is a plot of the relative
B
permeability
as a function of the applied
B0
field B0 .
FG IJ
H K
Relative Permeability
8 000
B/B0
4 000
0
0.0
1.0
2.0
B0 (mT)
FIG. P30.70
3.0
209
210
P30.71
Sources of the Magnetic Field
Consider the sphere as being built up of little rings of radius r,
centered on the rotation axis. The contribution to the field from
each ring is
dB =
e
µ 0 r 2 dI
2
2 x +r
b
dQ ωdQ
=
t
2π
where dI =
j
2 32
ω
dr
r
dx
x
R
ga f
dQ = ρdV = ρ 2π rdr dx
dB =
B=
µ 0 ρω r 3 drdx
e
2 x2 + r
j
2 3 2
z z
+R
R2 − x2
x =− R
r =0
Q
where ρ =
b4 3gπ R
3
µ 0 ρω r 3 drdx
32
2
x2 + r 2
e
FIG. P30.71
j
Let v = r 2 + x 2 , dv = 2rdr , and r 2 = v − x 2
LM
MN z
OP
PQ
µ ρω
LM2 v + e2x jv OPdx = µ ρω z LM2cR − x h + 2x FG 1 − 1 IJ OPdx
B=
z
4
4
H R x K PQ
N
Q
MN
O 2 µ ρω LM2 x − 4x + 2 ROPdx
µ ρω L x
− 4 x + 2 R Pdx =
B=
2
M
z
zN R
4
4
N R
Q
Q
I µ ρωR
2 µ ρω F 2 R
4R
−
+ 2R J =
B=
G
4 H 3R
2
3
K
B=
e
z z
x =− R v = x 2
R
0
2
12 R
2
R
z
R
v
dv − x
z
R2
2
v −3 2 dv dx
v= x2
R
0
x2
2
−1 2
x =− R v = x 2
2
x =− R
R
0
−R
2
0
3
0
P30.72
R2
2
−1 2 R
x2
x =− R
0
j
2
µ 0 ρω v − x dv
µ ρω
dx = 0
3 2
2
4
2v
R2
+R
2
2
2
0
Consider the sphere as being built up of little rings of radius r,
centered on the rotation axis. The current associated with each
rotating ring of charge is
ω
dr
r
dQ ω
ρ 2π rdr dx .
dI =
=
t
2π
ga f
b
dx
x
R
The magnetic moment contributed by this ring is
a f
dµ = A dI = π r 2
µ = πωρ
LM
MN
z z
ze
+R
x =− R
ω
2π
R2 − x 2
3
ga f
b
ρ 2π rdr dx = πωρr 3 drdx
OP
PQ
r dr dx = πωρ
r =0
z
+R
FH
x =− R
2
R −x
2
IK
FIG. P30.72
4
4
LM a f
MN
dx = πωρ
z
+R
eR
2
x =− R
F
GH
I
JK
µ=
πωρ + R 4
πωρ 4
2R 3
2R5
+
R − 2 R 2 x 2 + x 4 dx =
R 2R − 2R 2
4 x =− R
4
3
5
µ=
πωρ 5
πωρR 5 16
4 2
4πωρR 5
=
=
R 2− +
4
3 5
4
15
15
FG
H
j
IJ
K
FG IJ
H K
up
−x
4
OP
PQ
j
2 2
dx
211
Chapter 30
P30.73
Note that the current I exists in the conductor with a
I
current density J = , where
A
π a2
a2 a2
−
=
.
A = π a2 −
4
4
2
LM
N
P1
Bs
OP
Q
-B2
r
2I
.
Therefore, J =
π a2
To find the field at either point P1 or P2 , find Bs which
would exist if the conductor were solid, using Ampère’s
law. Next, find B1 and B2 that would be due to the
a
conductors of radius that could occupy the void where
2
the holes exist. Then use the superposition principle and
subtract the field that would be due to the part of the
conductor where the holes exist from the field of the
solid conductor.
(a)
e j, B
-B1
Bs
r + (a 2)
2
2
a/2
r
a/2
θ
P2
− B1′
θ θ
FIG. P30.73
b g .
=
At point P , B =
2π r
2π dr + b a 2gi
OP
µ Jπ a L 1
1
1
M
−
−
B=B −B −B =
2π M r 4dr − b a 2gi 4dr + b a 2 gi P
Q
N
L
O µ I L 2r − a O
µ a 2 I f M 4r − a − 2 r P
=
B=
M
P directed to the left
π r N 4r − a Q
2π M 4r er − e a 4jj P
MN
PQ
µ 0 J π a2
1
s
1
b g , and B
=
2π dr − b a 2gi
µ 0 Jπ a 2
2
µ 0 Jπ a 2
2
2
2
s
1
2
2
0
2
2
(b)
At point P2 , Bs =
2
0
2
2
2
2
2
e j and B′ = B′ =
µ 0 J π a2
1
2π r
2
b g
+ b a 2g
µ0 Jπ a 2
2π r
2
2
2
.
The horizontal components of B1′ and B2′ cancel while their vertical components add.
e j − 2FG µ J π a 4 IJ
GG 2π r + ea 4j JJ
2π r
H
K
OP µ a2I f L 2r O
µ J eπ a j LM
r
=
B=
1−
P
M1 −
2π r M 2er + e a 4jj P
2π r N 4r + a Q
MN
PQ
µ I L 2r + a O
=
M
P directed toward the top of the page
π r N 4r + a Q
B = Bs − B1′ cos θ − B2′ cos θ =
2
0
2
2
0
2
2
2
2
µ0 J π a2
2
0
2
2
r
2
e j
r + a2 4
2
0
2
2
2
ANSWERS TO EVEN PROBLEMS
P30.2
20.0 µT
P30.4
200 nT
P30.6
FG 1 + 1 IJ µ I into the page
H π K 2R
0
P30.8
see the solution
P30.10
FG 1 + 1 IJ µ I into the page
H π 4 K 2r
0
− B2′
212
P30.12
P30.14
P30.16
P30.18
Sources of the Magnetic Field
FG IJ
H K
µ I FH a + d − d
µ0I 1 1
out of the page
−
12 a b
2
0
2
a2 + d2
2
2π ad a + d
2
IK
into the page
(a) 10.0 µT ; (b) 80.0 µN toward wire 1;
(c) 16.0 µT ; (d) 80.0 µN toward wire 2
Parallel to the wires and
0.167 m below the upper wire
(a) opposite; (b) 67.8 A
P30.22
5.40 cm
P30.24
(a) 400 cm ; (b) 7.50 nT ; (c) 1.26 m; (d) zero
P30.26
(a) 3.60 T ; (b) 1.94 T
P30.28
500 A
P30.30
(a) see the solution; (b) d = a
P30.32
(a)
LM
MMN
a
a2 + R2
−
a−
aa − f
2
2.02
P30.46
(a) 12.6 µT ; (b) 56.0 µT
P30.48
2.01 GA west
P30.50
~ 10 −5 T , enough to affect the compass
noticeably
P30.52
12.0 mN −k
P30.54
P30.20
µ 0 IN
2
P30.44
+ R2
(b) see the solution
OP
PPQ ;
e j
µ 0 qω
2.5 5πR
P30.56
1.80 mT
P30.58
(a) µ 0σ v horizontally away from you;
1
(b) 0; (c) µ 0σ 2 v 2 up; (d) 3.00 × 10 8 m s
2
P30.60
(a) see the solution; (b) 3.20 × 10 −13 T;
(c) 1.02 × 10 −24 N away from the first
proton;
(d) 2.30 × 10 −22 N away from the first
proton
P30.62
347 µ 0 I m perpendicular to the coil
P30.64
(a)
1
µ 0σ v ; (b) out of the page,
2
parallel to the roller axes
P30.34
(a) −Bπ R 2 cos θ ; (b) Bπ R 2 cos θ
P30.36
(a) 7.40 µWb ; (b) 2.27 µWb
P30.66
675 A downward
P30.38
(a) 7.19 × 10 11 V m ⋅ s ; (b) 200 nT
P30.68
(a) see the solution; (b) 59.2 nN
P30.40
277 mA
P30.70
see the solution
P30.42
2.97 × 10 4
P30.72
4
πωρR 5 upward
15
K⋅J
T ⋅ m3
2
31
Faraday’s Law
CHAPTER OUTLINE
31.1
31.2
31.3
31.4
31.5
31.6
31.7
Faraday’s Law of Induction
Motional emf
Lenz’s Law
Induced emf and Electric
Fields
Generators and Motors
Eddy Currents
Maxwell’s Equations
ANSWERS TO QUESTIONS
Q31.1
Magnetic flux measures the “flow” of the magnetic field
through a given area of a loop—even though the field does not
actually flow. By changing the size of the loop, or the
orientation of the loop and the field, one can change the
magnetic flux through the loop, but the magnetic field will not
change.
Q31.2
The magnetic flux is Φ B = BA cos θ . Therefore the flux is
maximum when B is perpendicular to the loop of wire and zero
when there is no component of magnetic field perpendicular to
the loop. The flux is zero when the loop is turned so that the
field lies in the plane of its area.
Q31.3
The force on positive charges in the bar is F = q v × B . If the bar
is moving to the left, positive charge will move downward and
accumulate at the bottom end of the bar, so that an electric field
will be established upward.
a
f
Q31.4
No. The magnetic force acts within the bar, but has no influence on the forward motion of the bar.
Q31.5
By the magnetic force law F = q v × B : the positive charges in the moving bar will flow downward
and therefore clockwise in the circuit. If the bar is moving to the left, the positive charge in the bar
will flow upward and therefore counterclockwise in the circuit.
Q31.6
We ignore mechanical friction between the bar and the rails. Moving the conducting bar through the
magnetic field will force charges to move around the circuit to constitute clockwise current. The
downward current in the bar feels a magnetic force to the left. Then a counterbalancing applied
force to the right is required to maintain the motion.
Q31.7
A current could be set up in the bracelet by moving the bracelet through the magnetic field, or if the
field rapidly changed.
Q31.8
Moving a magnet inside the hole of the doughnut-shaped toroid will not change the magnetic flux
through any turn of wire in the toroid, and thus not induce any current.
a
f
213
214
Faraday’s Law
Q31.9
As water falls, it gains speed and kinetic energy. It then pushes against turbine blades, transferring
its energy to the rotor coils of a large AC generator. The rotor of the generator turns within a strong
magnetic field. Because the rotor is spinning, the magnetic flux through its turns changes in time as
−NdΦ B
. This induced emf is the
Φ B = BA cos ω t . Generated in the rotor is an induced emf of ε =
dt
voltage driving the current in our electric power lines.
Q31.10
Yes. Eddy currents will be induced around the circumference of the copper tube so as to fight the
changing magnetic flux by the falling magnet. If a bar magnet is dropped with its north pole
downwards, a ring of counterclockwise current will surround its approaching bottom end and a ring
of clockwise current will surround the receding south pole at its top end. The magnetic fields created
by these loops of current will exert forces on the magnet to slow the fall of the magnet quite
significantly. Some of the original gravitational energy of the magnet will appear as internal energy
in the walls of the tube.
Q31.11
Yes. The induced eddy currents on the surface of the aluminum will slow the descent of the
aluminum. It may fall very slowly.
Q31.12
The maximum induced emf will increase, increasing the terminal voltage of the generator.
Q31.13
The increasing counterclockwise current in
the solenoid coil produces an upward
magnetic field that increases rapidly. The
increasing upward flux of this field
through the ring induces an emf to
produce clockwise current in the ring. The
magnetic field of the solenoid has a
radially outward component at each point
on the ring. This field component exerts
upward force on the current in the ring
there. The whole ring feels a total upward
force larger than its weight.
FIG. Q31.13
Q31.14
Oscillating current in the solenoid produces an always-changing magnetic field. Vertical flux
through the ring, alternately increasing and decreasing, produces current in it with a direction that
is alternately clockwise and counterclockwise. The current through the ring’s resistance produces
internal energy at the rate I 2 R .
Q31.15
(a)
The south pole of the magnet produces an upward magnetic field that increases as the
magnet approaches. The loop opposes change by making its own downward magnetic field;
it carries current clockwise, which goes to the left through the resistor.
(b)
The north pole of the magnet produces an upward magnetic field. The loop sees decreasing
upward flux as the magnet falls away, and tries to make an upward magnetic field of its
own by carrying current counterclockwise, to the right in the resistor.
Chapter 31
Q31.16
(a)
The battery makes counterclockwise current
I 1 in the primary coil, so its magnetic field
B1 is to the right and increasing just after the
switch is closed. The secondary coil will
oppose the change with a leftward field B 2 ,
which comes from an induced clockwise
current I 2 that goes to the right in the
resistor.
(b)
At steady state the primary magnetic field is
unchanging, so no emf is induced in the
secondary.
(c)
The primary’s field is to the right and
decreasing as the switch is opened. The
secondary coil opposes this decrease by
making its own field to the right, carrying
counterclockwise current to the left in the
resistor.
215
FIG. Q31.16
Q31.17
The motional emf between the wingtips cannot be used to run a light bulb. To connect the light, an
extra insulated wire would have to be run out along each wing, making contact with the wing tip.
The wings with the extra wires and the bulb constitute a single-loop circuit. As the plane flies
through a uniform magnetic field, the magnetic flux through this loop is constant and zero emf is
generated. On the other hand, if the magnetic field is not uniform, a large loop towed through it will
generate pulses of positive and negative emf. This phenomenon has been demonstrated with a cable
unreeled from the Space Shuttle.
Q31.18
No, they do not. Specifically, Gauss’s law in magnetism prohibits magnetic monopoles. If magnetic
monopoles existed, then the magnetic field lines would not have to be closed loops, but could begin
or terminate on a magnetic monopole, as they can in Gauss’s law in electrostatics.
Q31.19
(a)
A current is induced by the changing magnetic flux through the a ring of the tube, produced
by the high frequency alternating current in the coil.
(b)
The higher frequency implies a greater rate of change in the magnetic field, for a larger
induced voltage.
(c)
The resistance of one cubic centimeter in the bulk sheet metal is low, so the I 2 R rate of
production of internal energy is low. At the seam, the current starts out crowded into a small
area with high resistance, so the temperature rises rapidly, and the edges melt together.
(d)
The edges must be in contact to allow the induced emf to create an electric current around
the circumference of the tube. Additionally, (duh) the two edges must be in contact to be
welded at all, just as you can’t glue two pieces of paper together without putting them in
contact with each other.
216
Faraday’s Law
SOLUTIONS TO PROBLEMS
Section 31.1
Faraday’s Law of Induction
Section 31.3
Lenz’s Law
a
f
∆Φ B ∆ NBA
=
= 500 mV
∆t
∆t
P31.1
ε=
P31.2
2.50 T − 0.500 T 8.00 × 10 −4 m 2
∆Φ B ∆ B ⋅ A
ε =
=
=
1.00 s
∆t
∆t
a f a
ε = 1.60 mV and I loop =
P31.3
ε = −N
fe
j FG 1 N ⋅ s IJ FG 1 V ⋅ C IJ
H 1 T ⋅C ⋅ mK H 1 N ⋅ mK
ε 1.60 mV
=
= 0.800 mA
R
2.00 Ω
FG
H
IJ
K
°− cos 0° I
f FGH cos 180
JK
0.200 s
cos θ f − cos θ i
∆BA cos θ
= − NBπ r 2
= −25.0 50.0 × 10 −6 T π 0.500 m
∆t
∆t
j a
e
2
ε = +9.82 mV
P31.4
dΦ B
ABmax − t τ
dB
e
= −A
=
dt
dt
τ
(a)
ε=−
(b)
e0.160 m ja0.350 Tf e
ε=
(c)
At t = 0 ε = 28.0 mV
2
P31.5
−4.00 2.00
2.00 s
= 3.79 mV
Noting unit conversions from F = qv × B and U = qV , the induced voltage is
a f
FG
H
IJ
K
a
fe
j
FG
H
+200 1.60 T 0.200 m 2 cos 0° 1 N ⋅ s
d B⋅A
0 − Bi A cos θ
ε = −N
= −N
=
dt
∆t
1 T ⋅C ⋅ m
20.0 × 10 −3 s
ε 3 200 V
= 160 A
I= =
R 20.0 Ω
P31.6
ε = −N
∆t =
a
N BA − 0
dΦ B
=−
dt
∆t
NBA
ε
=
f
e j = 500a0.200fπ e5.00 × 10 j
−2 2
NB π r 2
ε
10.0 × 10 3
= 7.85 × 10 −5 s
IJ FG 1 V ⋅ C IJ = 3 200 V
KH N ⋅m K
Chapter 31
P31.7
ε =
217
a f
d BA
dI
= 0.500 µ 0 nA = 0.480 × 10 −3 V
dt
dt
(a)
I ring =
ε 4.80 × 10 −4
=
= 1.60 A
R 3.00 × 10 −4
(b)
Bring =
µ0I
= 20.1 µT
2rring
(c)
Coil’s field points downward, and is increasing, so
Bring points upward .
FIG. P31.7
P31.8
ε =
a f
d BA
dI
∆I
= 0.500 µ 0 nA = 0.500 µ 0 nπ r22
dt
dt
∆t
µ 0nπ r22 ∆I
ε
=
R
2 R ∆t
(a)
I ring =
(b)
B=
(c)
The coil’s field points downward, and is increasing, so
Bring points upward .
µ 02nπ r22 ∆I
µ0I
=
2r1
4r1 R ∆t
FIG. P31.8
P31.9
(a)
dΦ B = B ⋅ dA =
(b)
ε=−
ε=−
LM
N
IJ OP = − LM µ L lnFG h + w IJ OP dI
K Q N 2π H h K Q dt
T ⋅ m A ja1.00 mf F 1.00 + 10.0 I
lnG
H 1.00 JK b10.0 A sg =
2π
FG
H
dΦ B
d µ 0 IL
h+w
=−
ln
dt
dt 2π
h
e4π × 10
−7
FG
H
z
h+ w
µ0I
µ 0 IL dx
µ 0 IL
h+ w
Ldx : Φ B =
=
ln
2π x
h
2π x
2π
h
IJ
K
0
−4.80 µV
The long wire produces magnetic flux into the page through the rectangle,
shown by the first hand in the figure to the right.
As the magnetic flux increases, the rectangle produces its own magnetic
field out of the page, which it does by carrying counterclockwise current
(second hand in the figure).
FIG. P31.9
218
P31.10
Faraday’s Law
b
g
Φ B = µ 0 nI Asolenoid
ε = −N
dΦ B
dI
2
= − Nµ 0 n π rsolenoid
dt
dt
e
e
j
g b600 A sg cosa120tf
jb
je
ε = −15.0 4π × 10 −7 T ⋅ m A 1.00 × 10 3 m −1 π 0.020 0 m
a f
2
ε = −14.2 cos 120 t mV
P31.11
For a counterclockwise trip around the left-hand loop, with
B = At
a f
d
At 2 a 2 cos 0° − I 1 5 R − I PQ R = 0
dt
e j
and for the right-hand loop,
a f
d
Ata 2 + I PQ R − I 2 3 R = 0
dt
where I PQ = I 1 − I 2 is the upward current in QP.
i
Aa + I R = I a3 Rf
5
2 Aa − 6 RI − e Aa
3
2
and
PQ
2
I PQ
ε =
2
2
PQ
I PQ =
P31.12
j
+ I PQ R = 0
Aa 2
upward, and since R = 0.100 Ω m 0.650 m = 0.065 0 Ω
23 R
b
e1.00 × 10
=
−3
b
ja
23 0.065 0 Ω
FG IJ
H K
f
T s 0.650 m
g
f
ga
2
= 283 µA upward .
∆Φ B
dB
A = N 0.010 0 + 0.080 0t A
=N
dt
∆t
b
g
b
g b
fe
j
At t = 5.00 s , ε = 30.0 0.410 T s π 0.040 0 m
P31.13
FIG. P31.11
d
2 Aa 2 − 5 R I PQ + I 2 − I PQ R = 0
Thus,
a
B = µ 0 nI = µ 0 n 30.0 A 1 − e −1.60 t
a fe
jz dA
Φ = µ na30.0 A fe1 − e
jπ R
dΦ
ε = −N
= − Nµ na30.0 A fπ R a1.60fe
dt
g
2
= 61.8 mV
z
Φ B = BdA = µ 0 n 30.0 A 1 − e −1.60 t
B
−1.60 t
0
B
a fe
a68.2 mVfe
ε = − 250 4π × 10
ε=
2
0
−7
−1.60 t
2
N A
2
−1.60 t
je400 m ja30.0 Af π b0.060 0 mg
−1
counterclockwise
FIG. P31.13
2
−1 −1.60 t
1.60 s e
219
Chapter 31
*P31.14
(a)
Each coil has a pulse of voltage tending to produce
counterclockwise current as the projectile
approaches, and then a pulse of clockwise voltage
as the projectile recedes.
v=
(b)
d
1.50 m
=
= 625 m s
t 2.40 × 10 −3 s
∆V
t
0
V1
V2
FIG. P31.14
P31.15
ε=
d
NA 2 ∆B cos θ
NBA 2 cos θ =
dt
∆t
e
j
e80.0 × 10 Vja0.400 sf
a50fe600 × 10 T − 200 × 10 Tj cosa30.0°f = 1.36 m
Length = 4AN = 4a1.36 mfa50f = 272 m
−3
ε ∆t
A=
=
N∆B cos θ
*P31.16
(a)
−6
−6
Suppose, first, that the central wire is long and straight. The enclosed current of unknown
amplitude creates a circular magnetic field around it, with the magnitude of the field given
by Ampere’s Law.
z
B ⋅ ds = µ 0 I : B =
µ 0 I max sin ω t
2π R
at the location of the Rogowski coil, which we assume is centered on the wire. This field
passes perpendicularly through each turn of the toroid, producing flux
B⋅A =
µ 0 I max A sin ω t
.
2π R
The toroid has 2π Rn turns. As the magnetic field varies, the emf induced in it is
ε = −N
µ I A d
d
sin ω t = − µ 0 I max nAω cos ω t .
B ⋅ A = −2π Rn 0 max
dt
2π R dt
This is an alternating voltage with amplitude ε max = µ 0 nAω I max . Measuring the amplitude
determines the size I max of the central current. Our assumptions that the central wire is long
and straight and passes perpendicularly through the center of the Rogowski coil are all
unnecessary.
(b)
If the wire is not centered, the coil will respond to stronger magnetic fields on one side, but
to correspondingly weaker fields on the opposite side. The emf induced in the coil is
proportional to the line integral of the magnetic field around the circular axis of the toroid.
Ampere’s Law says that this line integral depends only on the amount of current the coil
encloses. It does not depend on the shape or location of the current within the coil, or on
any currents outside the coil.
220
P31.17
Faraday’s Law
In a toroid, all the flux is confined to the inside of the
toroid.
B=
µ 0 NI 500 µ 0 I
=
2π r
2π r
z
Φ B = BdA =
ΦB =
=
*P31.18
FG
H
IJ
K
500 µ 0 I max
b+R
a sin ω t ln
R
2π
ε = N′
ε=
z
500 µ 0 I max
adr
sin ω t
r
2π
FG
H
IJ
K
FG
H
IJ
K
dΦ B
500 µ 0 I max
b+R
= 20
cos ω t
ω a ln
R
dt
2π
FIG. P31.17
g FGH a
f IJ
K
3.00 + 4.00 cm
10 4
4π × 10 −7 N A 2 50.0 A 377 rad s 0.020 0 m ln
cos ω t
2π
4.00 cm
ja
e
fb
gb
a0.422 Vf cos ω t
a
f
The upper loop has area π 0.05 m
ε = −N
2
= 7.85 × 10 −3 m 2 . The induced emf in it is
b
g
d
dB
BA cos θ = −1 A cos 0°
= −7.85 × 10 −3 m 2 2 T s = −1.57 × 10 −2 V .
dt
dt
The minus sign indicates that it tends to produce counterclockwise current, to make its own
magnetic field out of the page. Similarly, the induced emf in the lower loop is
ε = − NA cos θ
a
f
dB
2
= −π 0.09 m 2 T s = −5.09 × 10 −2 V = +5.09 × 10 −2 V to produce
dt
counterclockwise current in the lower loop, which becomes clockwise current in the upper loop .
The net emf for current in this sense around the figure 8 is
5.09 × 10 −2 V − 1.57 × 10 −2 V = 3.52 × 10 −2 V .
a
f
a
f
It pushes current in this sense through series resistance 2π 0.05 m + 2π 0.09 m 3 Ω m = 2.64 Ω.
ε 3.52 × 10 −2 V
= 13.3 mA .
The current is I = =
R
2.64 Ω
Section 31.2
Motional emf
Section 31.3
Lenz’s Law
P31.19
(a)
For maximum induced emf, with positive charge at the top of the antenna,
a
f
F+ = q + v × B , so the auto must move east .
(b)
e
ja
ε = BAv = 5.00 × 10 −5 T 1.20 m
fFGH 65.30 ×60010 s m IJK cos 65.0° =
3
4.58 × 10 −4 V
Chapter 31
P31.20
I=
ε
R
=
221
BAv
R
v = 1.00 m s
FIG. P31.20
P31.21
(a)
FB = I A × B = IAB
and
ε
R
ε = BAv
we get
FB =
I=
When
a f a1.20f a2.00f = 3.00 N .
2.50
BAv
B2A2 v
AB =
=
R
R
a f
2
2
6.00
The applied force is 3.00 N to the right .
(b)
P31.22
*P31.23
P = I2R =
FIG. P31.21
B2A2 v2
= 6.00 W or P = Fv = 6.00 W
R
FB = IAB and ε = BAv
IR
ε BAv
I= =
so B =
R
R
Av
I 2 AR
and I =
Av
FB v
= 0.500 A
R
(a)
FB =
(b)
I 2 R = 2.00 W
(c)
For constant force, P = F ⋅ v = 1.00 N 2.00 m s = 2.00 W .
a
fb
g
Model the magnetic flux inside the metallic tube as constant as it shrinks form radius R to radius r:
e j
F RI
= 2.50 TG J
HrK
2.50 T πR 2 = B f π r 2
Bf
2
a f
= 2.50 T 12
2
= 360 T
222
*P31.24
Faraday’s Law
Observe that the homopolar generator has no commutator and
produces a voltage constant in time: DC with no ripple. In time dt, the
disk turns by angle dθ = ωdt . The outer brush slides over distance rdθ .
The radial line to the outer brush sweeps over area
1
1
dA = rrdθ = r 2ωdt .
2
2
d
ε = −N B⋅ A
The emf generated is
dt
dA
1
ε = − 1 B cos 0°
= − B r 2ω
dt
2
(We could think of this as following from the result of Example 31.4.)
The magnitude of the emf is
FG
H
af
IJ
K
FIG. P31.24
FG 1 r ω IJ = b0.9 N ⋅ s C ⋅ mgLM 1 a0.4 mf b3 200 rev mingOPFG 2π rad rev IJ
H2 K
N2
QH 60 s min K
ε =B
2
2
ε = 24.1 V
A free positive charge q shown, turning with the disk, feels a magnetic force qv × B
radially
outward. Thus the outer contact is positive .
*P31.25
The speed of waves on the wire is
v=
T
µ
=
267 N ⋅ m
= 298 m s .
3 × 10 −3 kg
In the simplest standing-wave vibration state,
d NN = 0.64 m =
and f =
v
λ
=
λ
2
λ = 1.28 m
298 m s
= 233 Hz .
1.28 m
(a)
The changing flux of magnetic field through the circuit containing the wire will drive
current to the left in the wire as it moves up and to the right as it moves down. The emf will
have this same frequency of 233 Hz .
(b)
The vertical coordinate of the center of the wire is described by
a
f b
g
dx
= −a1.5 cmfb 2π 233 sg sinb 2π 233 t sg .
Its velocity is v =
dt
Its maximum speed is 1.5 cma 2π f 233 s = 22.0 m s .
x = A cos ω t = 1.5 cm cos 2π 233 t s .
The induced emf is ε = −BAv , with amplitude
a
f
ε max = BAv max = 4.50 × 10 −3 T 0.02 m 22 m s = 1.98 × 10 −3 V .
Chapter 31
P31.26
ε = −N
a
FG IJ
H K
d
∆A
BA cos θ = − NB cos θ
dt
∆t
f
ε = −1 0.100 T cos 0°
I=
223
a3.00 m × 3.00 m sin 60.0°f − a3.00 mf
2
0.100 s
1.21 V
= 0.121 A
10.0 Ω
= 1. 21 V
FIG. P31.26
The flux is into the page and decreasing. The loop makes its own magnetic field into the page by
carrying clockwise current.
P31.27
b
ε=
P31.28
gb
g
ω = 2.00 rev s 2π rad rev = 4.00π rad s
(a)
1
BωA 2 = 2.83 mV
2
B ext = B ext i and Bext decreases; therefore, the
induced field is B = B i (to the right) and the
0
0
current in the resistor is directed to the right .
(b)
e j
= B e+ ij is to the right, and the current in
B ext = B ext − i increases; therefore, the induced
field B0
0
the resistor is directed to the right .
(c)
e j
B ext = B ext − k into the paper and Bext decreases;
e j
therefore, the induced field is B0 = B0 − k into the
paper, and the current in the resistor is directed
to the right .
(d)
a
f
FIG. P31.28
By the magnetic force law, FB = q v × B . Therefore, a positive charge will move to the top of
the bar if B is into the paper .
224
P31.29
Faraday’s Law
(a)
The force on the side of the coil entering the field
(consisting of N wires) is
a f a f
F = N ILB = N IwB .
The induced emf in the coil is
ε =N
so the current is I =
a f
d Bwx
dΦ B
=N
= NBwv .
dt
dt
ε NBwv
=
counterclockwise.
R
R
The force on the leading side of the coil is then:
F=N
(b)
FG NBwv IJ wB =
H R K
N 2B2w 2 v
to the left .
R
Once the coil is entirely inside the field,
Φ B = NBA = constant ,
ε =0, I =0,
so
and
F= 0 .
FIG. P31.29
As the coil starts to leave the field, the flux decreases at the rate Bwv, so the magnitude of the
current is the same as in part (a), but now the current is clockwise. Thus, the force exerted on
the trailing side of the coil is:
(c)
F=
N 2B2 w 2 v
to the left again .
R
P31.30
Look in the direction of ba. The bar magnet creates a field into the page, and the field increases. The
loop will create a field out of the page by carrying a counterclockwise current. Therefore, current
must flow from b to a through the resistor. Hence, Va − Vb will be negative .
P31.31
Name the currents as shown in the diagram:
Left loop:
+ Bdv 2 − I 2 R 2 − I 1 R1 = 0
Right loop:
+ Bdv 3 − I 3 R3 + I 1 R1 = 0
At the junction:
I 2 = I1 + I 3
Then,
Bdv 2 − I 1 R 2 − I 3 R 2 − I 1 R1 = 0
I3 =
Bdv 3 I 1 R1
+
.
R3
R3
FIG. P31.31
b
Bdv 3 R 2 I 1 R1 R 2
−
=0
R3
R3
FG v R − v R IJ upward
HR R +R R +R R K
L b4.00 m sga15.0 Ωf − b2.00 m sga10.0 Ωf OP =
= b0.010 0 Tga0.100 mfM
MN a5.00 Ωfa10.0 Ωf + a5.00 Ωfa15.0 Ωf + a10.0 Ωfa15.0 Ωf PQ
I 1 = Bd
I1
g
Bdv 2 − I 1 R1 + R 2 −
So,
2
1
2
3
3
1
3
2
2
3
145 µA upward.
Chapter 31
Section 31.4
P31.32
(a)
Induced emf and Electric Fields
dB
= 6.00t 2 − 8.00t
dt
dΦ B
dt
ε =
At t = 2.00 s ,
E=
b
π R 2 dB dt
2π r2
g = 8.00π b0.025 0g
2π b0.050 0g
2
F = qE = 8.00 × 10 −21 N
FIG. P31.32
(b)
P31.33
2
When 6.00t − 8.00t = 0 ,
dB
= 0.060 0t
dt
ε =
t = 1.33 s
dΦ B
dB
= π r12
= 2π r1 E
dt
dt
At t = 3.00 s ,
E=
F π r I dB = 0.020 0 m e0.060 0 T s jb3.00 sgFG 1 N ⋅ s IJ
GH 2π r JK dt
H 1 T ⋅C ⋅ mK
2
2
1
2
1
E = 1.80 × 10 −3 N C perpendicular to r1 and counterclockwise
P31.34
(a)
z
E ⋅ dA =
dΦ B
dt
e j dBdt
2π rE = π r 2
(b)
Section 31.5
P31.35
FIG. P31.33
E=
so
b9.87 mV mg cosb100π tg
The E field is always opposite to increasing B. ∴ clockwise .
Generators and Motors
b
ga
fa
fa
f
(a)
ε max = NABω = 1 000 0.100 0.200 120π = 7.54 kV
(b)
ε t = NBAω sin ω t = NBAω sin θ
af
ε is maximal when sin θ = 1
or θ = ±
π
2
so the plane of coil is parallel to B .
FIG. P31.35
225
226
P31.36
Faraday’s Law
b
ε = −N
P31.37
rad I F 1 min I
gFGH 21πrev
JK GH 60 s JK = 314 rad s
T ⋅ m j cosb314t sg = +250e 2.50 × 10
For the alternator, ω = 3 000 rev min
dΦ B
d
= −250
2.50 × 10 −4
dt
dt
e
a
2
−4
g a f
jb
T ⋅ m 2 314 s sin 314t
f a f
(a)
ε = 19.6 V sin 314t
(b)
ε max = 19.6 V
e
ja
je
f
B = µ 0 nI = 4π × 10 −7 T ⋅ m A 200 m −1 15.0 A = 3.77 × 10 −3 T
e j
For the small coil, Φ B = NB ⋅ A = NBA cos ω t = NB π r 2 cos ω t .
ε=−
Thus,
dΦ B
= NBπ r 2ω sin ω t
dt
a fe
g e4.00π s j sinb4.00π tg = a28.6 mVf sinb4.00π tg
jb
ε = 30.0 3.77 × 10 −3 T π 0.080 0 m
P31.38
As the magnet rotates, the flux
through the coil varies
sinusoidally in time with Φ B = 0
at t = 0 . Choosing the flux as
positive when the field passes
from left to right through the
area of the coil, the flux at any
time may be written as
Φ B = − Φ max sin ω t so the
induced emf is given by
ε=−
2
−1
1
0.5
I/I max
0
0
0.5
1
1.5
–1
t/T = ( ω t/2 π )
FIG. P31.38
ε ω Φ max
=
cos ω t = I max cos ω t .
R
R
850 mA
M
120 V
2
–0.5
dΦ B
= ω Φ max cos ω t .
dt
The current in the coil is then I =
*P31.39
850 mA
To analyze the actual circuit, we model it as 120 V
a
f
(a)
The loop rule gives +120 V − 0.85 A 11.8 Ω − ε back = 0
(b)
The resistor is the device changing electrical work input into internal energy:
2
P = I 2 R = 0.85 A 11.8 Ω = 8.53 W .
(c)
With no motion, the motor does not function as a generator, and ε back = 0 . Then
a
fa
c
2
ε back = 110 V .
f
I = 10.2 A
a f
P = I R = a10.2 A f a11.8 Ωf = 1.22 kW
120 V − I c 11.8 Ω = 0
2
c
.
c
11.8 Ω
ε back
.
227
Chapter 31
P31.40
(a)
FG 1 π R IJω
H2 K
π
= a1.30 Tf a0. 250 mf b 4.00π rad sg
2
ε max
ε
2
ε max = BAω = B
t
2
ε max = 1.60 V
(b)
ε=
z
2π
0
P31.41
z
2π
ε
t
sin θdθ = 0
0
(c)
The maximum and average ε would remain
unchanged.
(d)
See Figure 1 at the right.
(e)
See Figure 2 at the right.
(a)
Φ B = BA cos θ = BA cos ω t = 0.800 T 0.010 0 m 2 cos 2π 60.0 t =
(b)
ε=−
(c)
I=
(d)
P = I2R =
(e)
P = Fv = τω so τ =
Section 31.6
P31.42
BAω
ε
dθ =
2π
2π
Figure 1
a
dΦ B
=
dt
ε
=
R
fe
Figure 2
FIG. P31.40
j
a f e8.00 mT ⋅ m j cosa377tf
2
a3.02 Vf sina377tf
a3.02 Af sina377tf
a9.10 Wf sin a377tf
2
P
=
ω
a24.1 mN ⋅ mf sin a377tf
2
Eddy Currents
The current in the magnet creates an
upward magnetic field, so the N and S poles on the
solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B increases as
the coil approaches, so a current is induced here to create a downward magnetic field. This is
clockwise current, so the S pole on the rail is shown correctly. On the rail behind the brake, the
upward magnetic flux is decreasing. The induced current in the rail will produce upward magnetic
field by being
counterclockwise as the picture correctly shows.
228
P31.43
Faraday’s Law
(a)
At terminal speed,
Mg = FB = IwB =
MgR
B 2ω 2
2
T
.
The emf is directly proportional to vT , but the
current is inversely proportional to R. A large R
means a small current at a given speed, so the loop
must travel faster to get FB = mg .
(b)
(c)
P31.44
2
T
vT =
or
Section 31.7
FG ε IJ wB = FG Bwv IJ wB = B w v
H RK H R K
R
FIG. P31.43
At a given speed, the current is directly proportional to the magnetic field. But the force is
proportional to the product of the current and the field. For a small B, the speed must
increase to compensate for both the small B and also the current, so vT ∝ B 2 .
Maxwell’s Equations
i
j
k
−e
E + v × B where v × B = 10.0 0
F = ma = qE + qv × B so a =
0 = −4.00 j
m
0
0 0.400
e−1.60 × 10 j 2.50 i + 5.00 j − 4.00 j = e−1.76 × 10 j 2.50 i + 1.00j
9.11 × 10
a = e −4.39 × 10 i − 1.76 × 10 jj m s
−19
a=
11
−31
11
P31.45
2
11
F = ma = qE + qv × B
j
i
k
e
a=
E + v × B where v × B = 200
0
0 = −200 0.400 j + 200 0.300 k
m
0.200 0.300 0.400
a
a
f
1.60 × 10 −19
50.0 j − 80.0 j + 60.0k = 9.58 × 10 7 −30.0 j + 60.0k
1.67 × 10 −27
a = 2.87 × 10 9 − j + 2k m s 2 = −2.87 × 10 9 j +5.75 × 10 9 k m s 2
a=
e
j
Additional Problems
P31.46
a
f e j FGH
a fLMN e
j OPQa f
ε = −a30.0fLMπ e 2.70 × 10 mj OPe3.20 × 10
N
Q
ε = −e7.22 × 10 V j cos 2π e523t s j
IJ
K
d
dB
BA cos θ = − N π r 2 cos 0°
dt
dt
2
d
ε = − 30.0 π 2.70 × 10 −3 m
1
50.0 mT + 3.20 mT sin 2π 523 t s −1
dt
ε = −N
−3
−3
2
−1
−3
a
j e
T 2π 523 s −1
f e
j
j cose2π 523t s j
−1
f
Chapter 31
P31.47
(a)
229
Doubling the number of turns.
Amplitude doubles: period unchanged
(b)
Doubling the angular velocity.
doubles the amplitude: cuts the period in half
(c)
Doubling the angular velocity while reducing the
number of turns to one half the original value.
FIG. P31.47
Amplitude unchanged: cuts the period in half
P31.48
P31.49
ε = −N
a
f
ja fFGH
IJ
K
∆
∆B
1.50 T − 5.00 T
BA cos θ = − N π r 2 cos 0°
= −1 0.005 00 m 2 1
= 0.875 V
∆t
∆t
20.0 × 10 −3 s
ε
e
0.875 V
= 43.8 A
0.020 0 Ω
(a)
I=
(b)
P = εI = 0.875 V 43.8 A = 38.3 W
R
=
e j
a
fa
f
In the loop on the left, the induced emf is
ε =
a
f b100 T sg = π V
dΦ B
dB
=A
= π 0.100 m
dt
dt
2
and it attempts to produce a counterclockwise
current in this loop.
In the loop on the right, the induced emf is
ε =
a
f b100 T sg = 2.25π V
dΦ B
= π 0.150 m
dt
2
FIG. P31.49
and it attempts to produce a clockwise current. Assume that I 1 flows down through the 6.00-Ω
resistor, I 2 flows down through the 5.00-Ω resistor, and that I 3 flows up through the 3.00-Ω
resistor.
From Kirchhoff’s junction rule:
I3 = I1 + I 2
(1)
Using the loop rule on the left loop:
6.00 I 1 + 3.00 I 3 = π
(2)
Using the loop rule on the right loop:
5.00 I 2 + 3.00 I 3 = 2.25π
(3)
Solving these three equations simultaneously,
I1 = 0.062 3 A , I 2 = 0.860 A , and I 3 = 0.923 A .
230
P31.50
Faraday’s Law
The emf induced between the ends of the moving bar is
a
fa
fb
g
ε = BAv = 2.50 T 0.350 m 8.00 m s = 7.00 V .
The left-hand loop contains decreasing flux away from you, so the induced current in it will be
clockwise, to produce its own field directed away from you. Let I 1 represent the current flowing
upward through the 2.00-Ω resistor. The right-hand loop will carry counterclockwise current. Let I 3
be the upward current in the 5.00-Ω resistor.
(a)
a f
+7.00 V − I a5.00 Ωf = 0
+7.00 V − I 1 2.00 Ω = 0
Kirchhoff’s loop rule then gives:
and
(b)
3
I1 = 3.50 A
I 3 = 1.40 A .
The total power dissipated in the resistors of the circuit is
g a
b
fa
f
P = εI1 + εI 3 = ε I 1 + I 3 = 7.00 V 3.50 A + 1.40 A = 34.3 W .
(c)
Method 1: The current in the sliding conductor is downward with value
I 2 = 3.50 A + 1.40 A = 4.90 A . The magnetic field exerts a force of
a
fa
fa
f
Fm = IAB = 4.90 A 0.350 m 2.50 T = 4.29 N directed
toward the right on this
conductor. An outside agent must then exert a force of 4.29 N
to the left to keep the bar
moving.
Method 2: The agent moving the bar must supply the power according to P = F ⋅ v = Fv cos 0° .
The force required is then:
F=
P31.51
P
34.3 W
=
= 4.29 N .
v 8.00 m s
Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm. Suppose we
use a bar magnet to produce field 10 −3 T through the coil in one direction along its axis. Suppose
we then flip the magnet to reverse the flux in 10 −1 s . The average induced emf is then
e jFGH
IJ
K
∆ BA cos θ
∆Φ B
cos 180°− cos 0°
= −N
= − NB π r 2
∆t
∆t
∆t
−2
2
ε = − 20 10 −3 T π 0.015 0 m
~ 10 −4 V
10 −1 s
ε = −N
a fe
jb
g FGH
IJ
K
Chapter 31
P31.52
I=
ε + ε induced
R
ε induced = −
and
a f
d
BA
dt
dv
= IBd
dt
dv IBd Bd
=
=
ε + ε induced
dt
m
mR
dv Bd
=
ε − Bvd
dt mR
u = ε − Bvd
du
dv
= − Bd
dt
dt
1 du Bd
−
=
u
Bd dt mR
F=m
b
a
To solve the differential equation, let
z
u
so
u0
f
za f
or
2
t
Bd
du
dt .
=−
u
mR
0
a f
Since v = 0 when t = 0 ,
u0 = ε
and
u = ε − Bvd
2
ε − Bvd = ε e − B
v=
Therefore,
ε
Bd
e1 − e
2 2
d t mR
.
2 2
− B d t mR
j
.
Φ B = BA = Bπ r 2 .
The enclosed flux is
The particle moves according to
∑ F = ma :
qvB sin 90° =
r=
−6
je
T ⋅ m 2 30 × 10 −9 C
Bπ m 2 v 2
q2B2
.
j a0.6 Tf =
2
2.54 × 10 5 m s
(a)
v=
(b)
Energy for the particle-electric field system is conserved in the firing process:
π m2
Ui = K f :
=
e15 × 10
mv 2
r
mv
.
qB
ΦB =
Then
ΦBq 2B
FIG. P31.52
Bd
u
t
=−
ln
u0
mR
2 2
u
= e − B d t mR .
u0
Integrating from t = 0 to t = t ,
*P31.53
g
e
π 2 × 10
q∆V =
−16
kg
j
2
1
mv 2
2
e
je
2 × 10 −16 kg 2.54 × 10 5 m s
mv 2
∆V =
=
2q
2 30 × 10 −9 C
e
j
j
2
= 215 V .
231
232
*P31.54
Faraday’s Law
(a)
Consider an annulus of radius r, width dr, height b, and resistivity ρ. Around its
circumference, a voltage is induced according to
ε = −N
b
g
d
d
B ⋅ A = −1 Bmax cos ω t π r 2 = + Bmaxπ r 2ω sin ω t .
dt
dt
b g
ρA ρ 2π r
=
.
Ax
bdr
The resistance around the loop is
The eddy current in the ring is dI =
b g
dPi = ε dI =
The instantaneous power is
b
g
Bmax π r 2ω sin ω t bdr Bmax rbωdr sin ω t
ε
.
=
=
2ρ
resistance
ρ 2π r
2
π r 3 bω 2 dr sin 2 ω t
Bmax
.
2ρ
sin 2 ω t =
The time average of the function
1 1
1
1
− cos 2ω t is − 0 =
2
2 2
2
so the time-averaged power delivered to the annulus is
dP =
2
π r 3 bω 2 dr
Bmax
.
4ρ
z z
P = dP =
The power delivered to the disk is
P=
P31.55
I
JK
2
Bmax
πbω 2 3
r dr
4ρ
0
2
2
Bmax
R 4 bω 2
π bω 2 R 4
π Bmax
−0 =
.
4ρ
4
16 ρ
2
Bmax
and P get 4 times larger.
(b)
When Bmax gets two times larger,
(c)
When f and ω = 2π f double, ω 2 and P get 4 times larger.
(d)
When R doubles,
I=
(a)
R 4 and P become 2 4 = 16 times larger.
ε B A
=
R R ∆t
so q = I∆t =
P31.56
F
GH
R
I=
b15.0 µTga0.200 mf
0.500 Ω
2
= 1.20 µC
dq ε
dΦ B
=
where ε = −N
so
dt
dt R
z
dq =
N
R
z
Φ2
dΦ B
Φ1
and the charge through the circuit will be Q =
(b)
IJ OP = BAN
KQ R
RQ a 200 Ωfe5.00 × 10 C j
=
=
so B =
NA a100fe 40.0 × 10 m j
Q=
LM
N
FG
H
N
π
BA cos 0 − BA cos
R
2
−4
−4
2
b
g
N
Φ 2 − Φ1 .
R
0.250 T .
Chapter 31
P31.57
ε
= 0.900 A
R
(a)
ε = BAv = 0.360 V
(b)
FB = IAB = 0.108 N
(c)
Since the magnetic flux B ⋅ A is in effect decreasing, the
induced current flow through R is from b to a. Point b is
I=
233
FIG. P31.57
at higher potential.
No . Magnetic flux will increase through a loop to the left of ab. Here counterclockwise
(d)
current will flow to produce upward magnetic field. The current in R is still from b to a.
P31.58
ε = BAv at a distance r from wire
ε =
F µ I I Av
GH 2π r JK
0
v
FIG. P31.58
P31.59
(a)
a
fe j
a
f
At time t, the flux through the loop is Φ B = BA cos θ = a + bt π r 2 cos 0° = π a + bt r 2 .
At t = 0 , Φ B = π ar 2 .
P31.60
a
ε=−
(c)
I=
(d)
P =εI= −
ε=−
π br 2
ε
= −
R
R
F
GH
a
f
d a + bt
dΦ B
= −π r 2
= −π br 2
dt
dt
(b)
f
π br 2
R
I e−π br j =
JK
2
π 2b 2r 4
R
FG IJ
H K
d
dB
NBA = −1
π a2 = π a2K
dt
dt
(a)
Q = Cε = Cπ a 2 K
(b)
B into the paper is decreasing; therefore, current will attempt to counteract this. Positive
charge will go to upper plate .
(c)
The changing magnetic field through the enclosed area induces an electric field ,
surrounding the B-field, and this pushes on charges in the wire.
234
P31.61
Faraday’s Law
The flux through the coil is Φ B = B ⋅ A = BA cos θ = BA cos ω t . The induced emf is
ε = −N
b
g
d cos ω t
dΦ B
= − NBA
= NBAω sin ω t .
dt
dt
a
fe
jb
ε max = NBAω = 60.0 1.00 T 0.100 × 0.200 m 2 30.0 rad s = 36.0 V
(b)
dΦ B ε
dΦ B
= , thus
dt
N
dt
=
max
ε max 36.0 V
=
= 0.600 V = 0.600 Wb s
N
60.0
(c)
At t = 0.050 0 s , ω t = 1.50 rad and
ε = ε max sin 1.50 rad = 36.0 V sin 1.50 rad = 35.9 V .
(d)
The torque on the coil at any time is
a
f a
f a
f
FG ε IJ FG ε IJ sin ω t .
H ω KH RK
a36.0 V f
ε
=
=
, sin ω t = 1.00 and τ =
ω R b30.0 rad sga10.0 Ωf
a
f
max
τ = µ × B = NIA × B = NAB I sin ω t =
When ε = ε max
P31.62
g
(a)
(a)
2
2
max
∆Φ B
, with N = 1 .
∆t
Taking a = 5.00 × 10 −3 m to be the radius of the washer, and h = 0.500 m ,
We use ε = −N
b
g
∆Φ B = B2 A − B1 A = A B2 − B1 = π a 2
F µ I − µ I I = a µ I FG 1 − 1 IJ = − µ ahI .
GH 2π ah + af 2π a JK 2 H h + a a K 2ah + af
0
2
0
0
The time for the washer to drop a distance h (from rest) is:
Therefore, ε =
µ 0 ahI
µ ahI
= 0
2 h + a ∆t 2 h + a
a f
−7
P31.63
g
µ 0 aI
0
∆t =
gh
−3
2
97.4 nV .
Since the magnetic flux going through the washer (into the plane of the paper) is decreasing
in time, a current will form in the washer so as to oppose that decrease. Therefore, the
current will flow in a clockwise direction .
Find an expression for the flux through a rectangular area “swept out”
I
by the bar in time t. The magnetic field at a distance x from wire is
B=
2h
.
g
a f 2 h = 2a h + a f 2
e4π × 10 T ⋅ m Aje5.00 × 10 mja10.0 Af e9.80 m s ja0.500 mf =
ε=
2
2b0.500 m + 0.005 00 mg
and
(b)
4.32 N ⋅ m .
z
µ0I
and Φ B = BdA . Therefore,
2π x
z
µ Ivt r + A dx
where vt is the distance the bar has moved in time t.
ΦB = 0
2π r x
Then, ε =
FG
H
dΦ B
µ 0 Iv
A
=
ln 1 +
dt
r
2π
IJ
K
.
r
v
A
FIG. P31.63
vt
Chapter 31
P31.64
The magnetic field at a distance x from a long wire is B =
through the loop.
dΦ B =
µ0I
Adx so
2π x
a f
Therefore,
P31.65
FG
H
z
ε=−
dΦ B µ 0 IAv w
µ 0 IAv w
ε
=
and I = =
dt
2π r r + w
R
2π Rr r + w
f
a
e
f
.
j
We are given
Φ B = 6.00t 3 − 18.0t 2 T ⋅ m 2
and
ε=−
dΦ B
= −18.0t 2 + 36.0t .
dt
which gives
dε
= −36.0t + 36.0 = 0
dt
t = 1.00 s .
Therefore, the maximum current (at t = 1.00 s ) is
I=
For the suspended mass, M:
Mg −
zb
v
0
ε BAv
=
R
R
Mg
dv
B2A2v
a=
=
−
dt m + M R M + m
f
a
z
t
dv
= dt where
α − βv 0
g
α=
v=
Therefore, the velocity varies with time as
(a)
ε = −N
f
I=
B2A2 v
= m + M a or
R
a
a
−18.0 + 36.0 V
ε
=
= 6.00 A .
R
3.00 Ω
∑ F = Mg − T = Ma .
∑ F = T − IAB = ma , where
For the sliding bar, m:
P31.67
IJ
K
µ 0 IA r + w dx µ 0 IA
w
=
ln 1 +
r
2π r x
2π
Maximum E occurs when
P31.66
µ0I
. Find an expression for the flux
2π x
ΦB =
a
f
Mg
B2A2
.
and β =
R M+m
M+m
a
f
2 2
MgR
α
1 − e−β t =
1 − e − B A t Ra M + m f
2 2
β
B A
e
j
dΦ B
dB
d
= − NA
= − NA
µ 0nI where A = area of coil
dt
dt
dt
b
g
N = number of turns in coil
(b)
and
n = number of turns per unit length in solenoid.
Therefore,
ε = Nµ 0 An
I=
∆V
R
235
ε =
a f b g
b g
g e
e
j b
ja f b
a1.19 Vf cosb120π tg
and
P = ∆VI =
d
4 sin 120π t = Nµ 0 An 480π cos 120π t
dt
2
ε = 40 4π × 10 −7 π 0.050 0 m 2.00 × 10 3 480π cos 120π t
cos 2 θ =
From
1
the average value of cos θ is , so
2
2
a
a
a1.19 V f
2
b
cos 2 120π t
8.00 Ω
1 1
+ cos 2θ
2 2
f
2
1 1.19 V
P =
= 88.5 mW .
2 8.00 Ω
f
g
g
.
236
*P31.68
Faraday’s Law
(a)
d θ a2
Ba 2 dθ
1
1
d
2
BA cos θ = −1 B
cos 0° = −
= − Ba 2 ω = − 0.5 T 0.5 m 2 rad s
dt
dt
2
2 dt
2
2
= −0.125 V = 0.125 V clockwise
a
ε = −N
fa
f
The – sign indicates that the induced emf produces clockwise current, to make its own
magnetic field into the page.
(b)
*P31.69
a
f
At this instant θ = ω t = 2 rad s 0.25 s = 0.5 rad . The arc PQ has length
rθ == 0.5 rad 0.5 m = 0.25 m . The length of the circuit is 0.5 m + 0.5 m + 0.25 m = 1.25 m its
0.125 V
resistance is 1.25 m 5 Ω m = 6.25 Ω . The current is
= 0.020 0 A clockwise .
6.25 Ω
a
fa
b
f
g
Suppose the field is vertically down. When an electron is moving away
from you the force on it is in the direction given by
b
g
qv × B c as − away × down = −
= − left = right .
Therefore, the electrons circulate clockwise.
FIG. P31.69
(a)
As the downward field increases, an emf is induced to produce some current that in turn
produces an upward field. This current is directed
counterclockwise, carried by
negative electrons moving clockwise. Therefore the original electron motion speeds up.
(b)
At the circumference, we have
∑ Fc = ma c :
q vBc sin 90° =
mv 2
r
mv = q rBc .
The increasing magnetic field Bav in the area enclosed by the orbit produces a tangential
electric field according to
dB
d
r dBav
E 2π r = π r 2 av
E=
E ⋅ ds = − B av ⋅ A
.
dt
dt
2 dt
dv
qE=m
.
An electron feels a tangential force according to ∑ Ft = mat :
dt
r dBav
r
dv
q
q B av = mv = q rBc
=m
Then
dt
2 dt
2
Bav = 2Bc .
and
z
P31.70
b g
The induced emf is ε = BAv where B =
y f = yi −
µ0I
, v f = vi + gt = 9.80 m s 2 t , and
2π y
e
j
1 2
gt = 0.800 m − 4.90 m s 2 t 2 .
2
e
j
f a0.300 mf 9.80 m s t = e1.18 × 10 jt
e
j 0.800 − 4.90t
2π 0.800 m − e 4.90 m s jt
e1.18 × 10 ja0.300f V = 98.3 µV .
At t = 0.300 s , ε =
0.800 − 4.90a0.300 f
ε=
e4π × 10
−7
ja
−4
T ⋅ m A 200 A
2
2
2
2
−4
2
V
Chapter 31
P31.71
237
The magnetic field produced by the current in the straight wire is
perpendicular to the plane of the coil at all points within the coil. The
µ I
magnitude of the field is B = 0 . Thus, the flux linkage is
2π r
NΦ B =
FG
H
z
IJ b
K
µ 0 NIL h+ w dr µ 0 NI max L
h+ w
=
ln
sin ω t + φ .
h
r
2π
2π
h
g
Finally, the induced emf is
FIG. P31.71
FG IJ b g
H K
e4π × 10 ja100fa50.0fa0.200 mfe200π s j lnFG 1 + 5.00 cm IJ cosbω t + φ g
ε=−
H 5.00 cm K
2π
ε = −a87.1 mV f cosb 200π t + φ g
The term sinbω t + φ g in the expression for the current in the straight wire does not change
ε=−
µ 0 NI max Lω
w
ln 1 +
cos ω t + φ
h
2π
−7
−1
appreciably when ω t changes by 0.10 rad or less. Thus, the current does not change appreciably
during a time interval
∆t <
0.10
e200π s j
−1
= 1.6 × 10 −4 s .
e
je
j
We define a critical length, c∆t = 3.00 × 10 8 m s 1.6 × 10 −4 s = 4.8 × 10 4 m equal to the distance to
which field changes could be propagated during an interval of 1.6 × 10 −4 s . This length is so much
larger than any dimension of the coil or its distance from the wire that, although we consider the
straight wire to be infinitely long, we can also safely ignore the field propagation effects in the
vicinity of the coil. Moreover, the phase angle can be considered to be constant along the wire in the
vicinity of the coil.
If the frequency w were much larger, say, 200π × 10 5 s −1 , the corresponding critical length would be
only 48 cm. In this situation propagation effects would be important and the above expression for ε
would require modification. As a “rule of thumb” we can consider field propagation effects for
circuits of laboratory size to be negligible for frequencies, f =
P31.72
Φ B = BA cos θ
ω
2π
, that are less than about 10 6 Hz.
dΦ B
= −ωBA sin θ ;
dt
I ∝ − sin θ
θ
τ ∝ IB sin θ ∝ − sin 2 θ
FIG. P31.72
238
Faraday’s Law
ANSWERS TO EVEN PROBLEMS
P31.2
0.800 mA
P31.40
P31.4
(a) see the solution; (b) 3.79 mV ;
(c) 28.0 mV
(a) 1.60 V ; (b) 0; (c) no change;
(d) and (e) see the solution
P31.42
both are correct; see the solution
78.5 µs
P31.44
P31.6
P31.8
µ 0 nπ r22
2R
µ 02nπ r22
(b)
4r1 R
(a)
∆I
counterclockwise;
∆t
∆I
; (c) upward
∆t
a f
P31.10
−14.2 mV cos 120 t
P31.12
61.8 mV
P31.14
(a) see the solution; (b) 625 m/s
P31.16
see the solution
P31.18
13.3 mA counterclockwise in the lower
loop and clockwise in the upper loop.
P31.20
1.00 m s
P31.22
(a) 500 mA; (b) 2.00 W ; (c) 2.00 W
P31.24
24.1 V with the outer contact positive
P31.26
121 mA clockwise
P31.28
(a) to the right; (b) to the right;
(c) to the right; (d) into the paper
P31.30
negative; see the solution
P31.32
(a) 8.00 × 10 −21 N downward
perpendicular to r1 ; (b) 1.33 s
P31.34
(a) 9.87 mV m cos 100π t ; (b) clockwise
P31.36
b
g b g
(a) a19.6 V f sina314tf ; (b) 19.6 V
P31.38
see the solution
P31.46
e−4.39i − 1.76jj10 m s
−a7.22 mV f cosb 2π 523 t sg
P31.48
(a) 43.8 A ; (b) 38.3 W
P31.50
(a) 3.50 A up in 2 Ω and 1.40 A up in 5 Ω;
(b) 34.3 W ; (c) 4.29 N
P31.52
see the solution
P31.54
(a)
P31.56
(a) see the solution; (b) 0.250 T
P31.58
see the solution
P31.60
(a) Cπ a 2 K ; (b) the upper plate;
(c) see the solution
P31.62
(a) 97.4 nV ; (b) clockwise
P31.64
µ 0 I Av w
2π Rr r + w
P31.66
P31.68
11
2
2
R 4 bω 2
πBmax
; (b) 4 times larger;
16 ρ
(c) 4 times larger; (d) 16 times larger
a
MgR
2 2
B A
f
1 − e−B
2 2
a
A t R M +m
f
(a) 0.125 V to produce clockwise current;
(b) 20.0 mA clockwise
P31.70
1.18 × 10 −4
; 98.3 µV
0.800 − 4.90t 2
P31.72
see the solution
32
Inductance
CHAPTER OUTLINE
32.1
32.2
32.3
32.4
32.5
32.6
Self-Inductance
RL Circuits
Energy in a Magnetic Field
Mutual Inductance
Oscillations in an LC Circuit
The RLC Circuit
ANSWERS TO QUESTIONS
Q32.1
The emf induced in an inductor is opposite to the direction of the
changing current. For example, in a simple RL circuit with current
flowing clockwise, if the current in the circuit increases, the
inductor will generate an emf to oppose the increasing current.
Q32.2
The coil has an inductance regardless of the nature of the current
in the circuit. Inductance depends only on the coil geometry and
its construction. Since the current is constant, the self-induced
emf in the coil is zero, and the coil does not affect the steady-state
current. (We assume the resistance of the coil is negligible.)
Q32.3
The inductance of a coil is determined by (a) the geometry of the
coil and (b) the “contents” of the coil. This is similar to the
parameters that determine the capacitance of a capacitor and the
resistance of a resistor. With an inductor, the most important
factor in the geometry is the number of turns of wire, or turns per
unit length. By the “contents” we refer to the material in which the
inductor establishes a magnetic field, notably the magnetic
properties of the core around which the wire is wrapped.
Q32.4
If the first set of turns is wrapped clockwise around a spool, wrap the second set counter-clockwise,
so that the coil produces negligible magnetic field. Then the inductance of each set of turns
effectively negates the inductive effects of the other set.
Q32.5
After the switch is closed, the back emf will not exceed that of the battery. If this were the case, then
the current in the circuit would change direction to counterclockwise. Just after the switch is opened,
the back emf can be much larger than the battery emf, to temporarily maintain the clockwise current
in a spark.
Q32.6
The current decreases not instantaneously but over some span of time. The faster the decrease in the
current, the larger will be the emf generated in the inductor. A spark can appear at the switch as it is
opened because the self-induced voltage is a maximum at this instant. The voltage can therefore
briefly cause dielectric breakdown of the air between the contacts.
Q32.7
When it is being opened. When the switch is initially standing open, there is no current in the
circuit. Just after the switch is then closed, the inductor tends to maintain the zero-current condition,
and there is very little chance of sparking. When the switch is standing closed, there is current in the
circuit. When the switch is then opened, the current rapidly decreases. The induced emf is created in
the inductor, and this emf tends to maintain the original current. Sparking occurs as the current
bridges the air gap between the contacts of the switch.
239
240
Inductance
Q32.8
A physicist’s list of constituents of the universe in 1829 might include matter, light, heat, the stuff of
stars, charge, momentum, and several other entries. Our list today might include the quarks,
electrons, muons, tauons, and neutrinos of matter; gravitons of gravitational fields; photons of
electric and magnetic fields; W and Z particles; gluons; energy; momentum; angular momentum;
charge; baryon number; three different lepton numbers; upness; downness; strangeness; charm;
topness; and bottomness. Alternatively, the relativistic interconvertability of mass and energy, and
of electric and magnetic fields, can be used to make the list look shorter. Some might think of the
conserved quantities energy, momentum, … bottomness as properties of matter, rather than as
things with their own existence. The idea of a field is not due to Henry, but rather to Faraday, to
whom Henry personally demonstrated self-induction. Still the thesis stated in the question has an
important germ of truth. Henry precipitated a basic change if he did not cause it. The biggest
difference between the two lists is that the 1829 list does not include fields and today’s list does.
Q32.9
The energy stored in the magnetic field of an inductor is proportional to the square of the current.
1
Doubling I makes U = LI 2 get four times larger.
2
Q32.10
The energy stored in a capacitor is proportional to the square of the electric field, and the energy
stored in an induction coil is proportional to the square of the magnetic field. The capacitor’s energy
is proportional to its capacitance, which depends on its geometry and the dielectric material inside.
The coil’s energy is proportional to its inductance, which depends on its geometry and the core
material. On the other hand, we can think of Henry’s discovery of self-inductance as fundamentally
new. Before a certain school vacation at the Albany Academy about 1830, one could visualize the
universe as consisting of only one thing, matter. All the forms of energy then known (kinetic,
gravitational, elastic, internal, electrical) belonged to chunks of matter. But the energy that
temporarily maintains a current in a coil after the battery is removed is not energy that belongs to
any bit of matter. This energy is vastly larger than the kinetic energy of the drifting electrons in the
wires. This energy belongs to the magnetic field around the coil. Beginning in 1830, Nature has
forced us to admit that the universe consists of matter and also of fields, massless and invisible,
known only by their effects.
Q32.11
The inductance of the series combination of inductor L1 and inductor L 2 is L1 + L 2 + M 12 , where
M 12 is the mutual inductance of the two coils. It can be defined as the emf induced in coil two when
the current in coil one changes at one ampere per second, due to the magnetic field of coil one
producing flux through coil two. The coils can be arranged to have large mutual inductance, as by
winding them onto the same core. The coils can be arranged to have negligible mutual inductance,
as separate toroids do.
Q32.12
The mutual inductance of two loops in free space—that is, ignoring the use of cores—is a maximum
if the loops are coaxial. In this way, the maximum flux of the primary loop will pass through the
secondary loop, generating the largest possible emf given the changing magnetic field due to the
first. The mutual inductance is a minimum if the magnetic field of the first coil lies in the plane of the
second coil, producing no flux through the area the second coil encloses.
Q32.13
The answer depends on the orientation of the solenoids. If they are coaxial, such as two solenoids
end-to-end, then there certainty will be mutual induction. If, however, they are oriented in such a
way that the magnetic field of one coil does not go through turns of the second coil, then there will
be no mutual induction. Consider the case of two solenoids physically arranged in a “T” formation,
but still connected electrically in series. The magnetic field lines of the first coil will not produce any
net flux in the second coil, and thus no mutual induction will be present.
Chapter 32
241
Q32.14
When the capacitor is fully discharged, the current in the circuit is a maximum. The inductance of
the coil is making the current continue to flow. At this time the magnetic field of the coil contains all
the energy that was originally stored in the charged capacitor. The current has just finished
discharging the capacitor and is proceeding to charge it up again with the opposite polarity.
Q32.15
The oscillations would eventually decrease, but perhaps with very small damping. The original
potential energy would be converted to internal energy within the wires. Such a situation
constitutes an RLC circuit. Remember that a real battery generally contains an internal resistance.
Q32.16
If R >
Q32.17
The condition for critical damping must be investigated to design a circuit for a particular purpose.
For example, in building a radio receiver, one would want to construct the receiving circuit so that it
is underdamped. Then it can oscillate in resonance and detect the desired signal. Conversely, when
designing a probe to measure a changing signal, such free oscillations are undesirable. An electrical
vibration in the probe would constitute “ringing” of the system, where the probe would measure an
additional signal—that of the probe itself! In this case, one would want to design a probe that is
critically damped or overdamped, so that the only signal measured is the one under study. Critical
damping represents the threshold between underdamping and overdamping. One must know the
condition for it to meet the design criteria for a project.
Q32.18
An object cannot exert a net force on itself. An object cannot create momentum out of nothing. A coil
can induce an emf in itself. When it does so, the actual forces acting on charges in different parts of
the loop add as vectors to zero. The term electromotive force does not refer to a force, but to a
voltage.
4L
4L
, then the oscillator is overdamped—it will not oscillate. If R <
, then the oscillator is
C
C
underdamped and can go through several cycles of oscillation before the radiated signal falls below
background noise.
SOLUTIONS TO PROBLEMS
Section 32.1
P32.1
P32.2
Self-Inductance
ε =L
jFGH
IJ
K
∆I
1.50 A − 0.200 A
= 3.00 × 10 −3 H
= 1.95 × 10 −2 V = 19.5 mV
∆t
0.200 s
e
Treating the telephone cord as a solenoid, we have:
ja f e
e
j
2
−7
−3
µ N 2 A 4π × 10 T ⋅ m A 70.0 π 6.50 × 10 m
=
L= 0
A
0.600 m
fFGH
a
0 − 0.500 A
∆I
= −2.00 H
∆t
0.010 0 s
P32.3
ε = −L
P32.4
L=
P32.5
ε back = −ε = L
I FG 1 V ⋅ s IJ =
JK H 1 H ⋅ A K
NΦ B
LI
→ ΦB =
= 240 nT ⋅ m 2
I
N
a
b
2
= 1.36 µH .
100 V
through each turn
g
ja
fa f
dI
d
=L
I max sin ω t = LωI max cos ω t = 10.0 × 10 −3 120π 5.00 cos ω t
dt
dt
f b
g a
e
f a f
ε back = 6.00π cos 120π t = 18.8 V cos 377 t
242
P32.6
Inductance
From ε = L
From L =
FG ∆I IJ , we have
H ∆t K
L=
NΦ B
, we have
I
ΦB
a fe
P32.8
N
ε =L
e
j e
500
f=
19.2 µT ⋅ m 2 .
j
(a)
At t = 1.00 s ,
ε = 360 mV
(b)
At t = 4.00 s ,
ε = 180 mV
(c)
ε = 90.0 × 10 −3 2t − 6 = 0
ja
e
f
t = 3.00 s .
FG 450 IJ b0.040 0 Ag =
H 0.120 K
(a)
B = µ 0 nI = µ 0
(b)
Φ B = BA = 3.33 × 10 −8 T ⋅ m 2
(c)
L=
188 µT
NΦ B
= 0.375 mH
I
B and Φ B are proportional to current; L is independent of current
a f e
2
*P32.11
ja
H 4.00 A
dI
d 2
= 90.0 × 10 −3
t − 6t V
dt
dt
(d)
P32.10
−3
−4
µ N 2 A µ 0 420 3.00 × 10
L= 0
=
= 4.16 × 10 −4 H
0.160
A
dI
dI −ε −175 × 10 −6 V
ε = −L → =
=
= −0.421 A s
dt
dt L
4.16 × 10 −4 H
when
P32.9
g
LI e 2. 40 × 10
=
=
j
2
P32.7
b
ε
24.0 × 10 −3 V
=
= 2.40 × 10 −3 H .
10.0 A s
∆I ∆t
(a)
−3
µ N 2 A µ 0 120 π 5.00 × 10
=
L= 0
0.090 0
A
(b)
Φ ′B =
j
2
= 15.8 µH
µm
µ N2A
ΦB → L = m
= 800 1.58 × 10 −5 H = 12.6 mH
µ0
A
e
j
We can directly find the self inductance of the solenoid:
ε = −L
dI
dt
+0.08 V = − L
0 − 1.8 A
0.12 s
L = 5.33 × 10 −3 Vs A =
e
j
µ
µ N π F 200 m I
=
=
G
J
A
H 2π N K
µ0N2A
.
A
Here A = π r 2 , 200 m = N 2π r , and A = N 10 −3 m . Eliminating extra unknowns step by step, we have
5.33 × 10
A=
−3
4 × 10
−3
µ N 2π r 2
Vs A = 0
A
WbmA
5.33 × 10 −3 AVs
= 0.750 m
0
2
2
0 40 000
4π A
m2
=
e
j
10 −7 40 000 m 2 Tm
A
A
Chapter 32
P32.12
L=
243
NΦ B NBA NA µ 0 NI
µ0N 2 A
=
≈
⋅
=
I
I
I 2π R
2π R
FIG. P32.12
P32.13
ε = ε 0 e − kt = − L
dI = −
ε 0 − kt
e dt
L
dI
dt
ε 0 − kt dq
e =
kL
dt
ε
Q = 20 .
k L
I=
If we require I → 0 as t → ∞ , the solution is
z z
Q = Idt =
Section 32.2
P32.14
I=
∞
ε 0 − kt
ε
e dt = − 20
kL
k
L
0
RL Circuits
ε
R
e1 − e j :
− Rt L
0.900
ε
R
=
ε
R
1 − e − R a3.00 s f
2.50 H
FG Ra3.00 sf IJ = 0.100
H 2.50 H K
exp −
R=
2.50 H
ln 10.0 = 1.92 Ω
3.00 s
a f ε e1 −Re j
I (A)
−t τ
P32.15
(a)
At time t,
It =
where
L
τ = = 0.200 s .
R
After a long time,
I max =
e
ε 1 − e −∞
R
1
0
R
a0.500f Rε = ε e1 − e R
so
0.500 = 1 − e − t 0. 200 s .
− t 0. 200 s
j a
e
Isolating the constants on the right, ln e − t 0. 200 s = ln 0.500
(b)
0.5
j=ε .
At I t = 0.500 I max
af
j
t (s)
0
f
t
= −0.693
0.200 s
and solving for t,
−
or
t = 0.139 s .
Similarly, to reach 90% of I max ,
0.900 = 1 − e − t τ
and
t = −τ ln 1 − 0.900 .
Thus,
t = − 0.200 s ln 0.100 = 0.461 s .
a
a
f a
I max
f
f
0.2
0.4
FIG. P32.15
0.6
244
P32.16
Inductance
Taking τ =
IR + L
L
,
R
dI
= 0 will be true if
dt
Because τ =
P32.17
P32.18
0
−t τ
L
= 2.00 × 10 −3 s = 2.00 ms
R
τ=
(b)
I = I max 1 − e − t τ =
(c)
I max =
I=
I 0 Re − t τ
L
, we have agreement with 0 = 0 .
R
(a)
.00 V I
J e1 − e
j FGH 64.00
ΩK
e
−0. 250 2.00
ε 6.00 V
=
= 1.50 A
R 4.00 Ω
0.800 = 1 − e
(d)
FG IJ
H K
F 1I
+ Le I e jG − J = 0 .
H τK
dI
1
= I0 e−t τ −
τ
dt
I = I 0 e −t τ :
− t 2.00 ms
a
f a
j=
0.176 A
FIG. P32.17
f
→ t = − 2.00 ms ln 0.200 = 3.22 ms
120
ε
1 − e −t τ =
1 − e −1.80 7.00 = 3.02 A
R
9.00
e
∆VR
j
e
j
= IR = a3.02fa9.00f = 27.2 V
∆VL = ε − ∆VR = 120 − 27.2 = 92.8 V
P32.19
Note: It may not be correct to call the voltage or emf across a coil a “potential
difference.” Electric potential can only be defined for a conservative electric
field, and not for the electric field around an inductor.
(a)
∆VR = IR = 8.00 Ω 2.00 A = 16.0 V
a
fa
f
and
∆VL = ε − ∆VR = 36.0 V − 16.0 V = 20.0 V .
Therefore,
∆VR 16.0 V
=
= 0.800 .
∆VL 20.0 V
a
fa
FIG. P32.19
f
∆VR = IR = 4.50 A 8.00 Ω = 36.0 V
(b)
∆VL = ε − ∆VR = 0
P32.20
a
f
After a long time, 12.0 V = 0. 200 A R . Thus, R = 60.0 Ω . Now, τ =
jb
e
g
L
gives
R
L = τ R = 5.00 × 10 −4 s 60.0 V A = 30.0 mH .
P32.21
e
e jFGH − τ1 IJK
dI
= − I max e − t τ
dt
j
I = I max 1 − e − t τ :
τ=
L 15.0 H
=
= 0.500 s :
R 30.0 Ω
ε
dI R
= I max e − t τ and I max =
dt L
R
dI R
ε 100 V
= I max e 0 = =
= 6.67 A s
dt L
L 15.0 H
(a)
t=0:
(b)
t = 1.50 s :
dI ε − t τ
= e
= 6.67 A s e −1.50 a0.500 f = 6.67 A s e −3.00 = 0.332 A s
dt L
b
g
b
g
Chapter 32
P32.22
e
j
I = I max 1 − e − t τ :
0.980 = 1 − e −3.00 ×10
−3
−3
τ
0.020 0 = e −3.00 ×10
τ =−
L
, so
R
τ=
P32.23
τ
3.00 × 10 −3
= 7.67 × 10 −4 s
ln 0.020 0
b
g
FIG. P32.22
ja f
e
L = τ R = 7.67 × 10 −4 10.0 = 7.67 mH
Name the currents as shown. By Kirchhoff’s laws:
I1 = I 2 + I 3
(1)
+10.0 V − 4.00 I 1 − 4.00 I 2 = 0
(2)
a f dIdt = 0
3
+10.0 V − 4.00 I 1 − 8.00 I 3 − 1.00
(3)
From (1) and (2),
+10.0 − 4.00 I 1 − 4.00 I 1 + 4.00 I 3 = 0
and
I 1 = 0.500 I 3 + 1.25 A .
Then (3) becomes
10.0 V − 4.00 0.500 I 3 + 1.25 A − 8.00 I 3 − 1.00
FIG. P32.23
a f dIdt = 0
b
g
a1.00 HfFGH dIdt IJK + a10.0 ΩfI = 5.00 V .
3
3
3
We solve the differential equation using Equations 32.6 and 32.7:
.00 V I
a f FGH 510.0
J 1− e a
ΩK
I 1 = 1.25 + 0.500 I 3 =
P32.24
P32.25
a0.500 Af 1 − e
1.50 A − a0.250 A fe
f
− 10.0 Ω t 1.00 H
I3 t =
=
−10 t s
L
L
=
, we get R =
R
C
3.00 H
= 1.00 × 10 3 Ω = 1.00 kΩ .
3.00 × 10 −6 F
(a)
Using τ = RC =
(b)
τ = RC = 1.00 × 10 3 Ω 3.00 × 10 −6 F = 3.00 × 10 −3 s = 3.00 ms
e
je
−10 t s
j
For t ≤ 0 , the current in the inductor is zero . At t = 0 , it starts to
grow from zero toward 10.0 A with time constant
τ=
a
f
10.0 mH
L
=
= 1.00 × 10 −4 s .
R
100 Ω
a
f
e
j a10.0 Afe1 − e
At t = 200 µs , I = a10.00 A fe1 − e
j = 8.65 A .
For 0 ≤ t ≤ 200 µs , I = I max 1 − e − t τ =
−10 000 t s
j
.
−2.00
Thereafter, it decays exponentially as I = I 0 e
a
f
a
f
I = 8.65 A e −10 000b t − 200 µsg s = 8.65 A e −10 000 t
− t′ τ
FIG. P32.25
, so for t ≥ 200 µs ,
s + 2.00
e
j
= 8.65 e 2.00 A e −10 000 t s =
a63.9 Afe
−10 000 t s
.
245
246
P32.26
Inductance
ε
I=
(b)
Initial current is 1.00 A: ∆V12 = 1.00 A 12.00 Ω = 12.0 V
R
=
12.0 V
= 1.00 A
12.0 Ω
(a)
a
fa
a
f
fb
g
∆V1 200 = 1.00 A 1 200 Ω = 1.20 kV
∆VL = 1.21 kV .
(c)
dI
R
= − I max e − Rt L
dt
L
dI
− L = ∆VL = I max R e − Rt L .
dt
I = I max e − Rt L :
and
b
12.0 V = 1 212 V e −1 212 t
so
9.90 × 10 −3 = e −606 t .
2.00
t = 7.62 ms .
L 0.140
=
= 28.6 ms
R 4.90
ε 6.00 V
= 1.22 A
I max = =
R 4.90 Ω
τ=
(a)
e
I = I max 1 − e − t τ
e
P32.28
g
Solving
Thus,
P32.27
FIG. P32.26
−t τ
j
so
a
f
j
t = −τ ln 0.820 = 5.66 ms
= 0.820 :
e
e
0. 220 = 1.22 1 − e − t τ
j a
fe
j
(b)
I = I max 1 − e −10.0 0.028 6 = 1.22 A 1 − e −350 = 1.22 A
(c)
I = I max e − t τ
(a)
and
0.160 = 1.22 e − t τ
so
t = −τ ln 0.131 = 58.1 ms .
a
(c)
f
For a series connection, both inductors carry equal currents at every instant, so
same for both. The voltage across the pair is
dI
dI
dI
= L1 + L 2
Leq
so
dt
dt
dt
(b)
FIG. P32.27
dI
dI
dI
= L1 1 = L 2 2 = ∆VL
dt
dt
dt
∆VL ∆VL ∆VL
=
+
Thus,
Leq
L1
L2
Leq
where
and
dI
is the
dt
Leq = L1 + L 2 .
I = I 1 + I 2 and
dI dI 1 dI 2
=
+
.
dt dt
dt
1
1
1
=
+
.
Leq L1 L 2
dI
dI
dI
+ Req I = L1 + IR1 + L 2
+ IR 2
dt
dt
dt
dI
are separate quantities under our control, so functional equality requires
Now I and
dt
both Leq = L1 + L 2
and
Req = R1 + R 2 .
Leq
continued on next page
Chapter 32
∆V = Leq
(d)
dI
dI
dI
dI dI 1 dI 2
=
+
+ Req I = L1 1 + R1 I 1 = L 2 2 + R 2 I 2 where I = I 1 + I 2 and
.
dt
dt
dt
dt dt
dt
We may choose to keep the currents constant in time. Then,
1
1
1
=
+
.
Req R1 R 2
We may choose to make the current swing through 0. Then,
1
1
1
=
+
.
Leq L1 L 2
This equivalent coil with resistance will be equivalent to the pair of real inductors for
all other currents as well.
Section 32.3
P32.29
L=
P32.30
(a)
Energy in a Magnetic Field
e
j
−4
NΦ B 200 3.70 × 10
1
1
=
= 42.3 mH so U = LI 2 = 0.423 H 1.75 A
I
1.75
2
2
a
f
e
The magnetic energy stored in the field equals u times the volume of the solenoid (the
volume in which B is non-zero).
j a0.260 mfπ b0.031 0 mg
a f LMN e
2
= 6.32 kJ
j OPQ
2
68.0 π 0.600 × 10 −2
N2A
L = µ0
= µ0
= 8.21 µH
A
0.080 0
1
1
2
U = LI 2 = 8. 21 × 10 −6 H 0.770 A = 2.44 µJ
2
2
ja
e
(a)
U=
(b)
I=
FG IJ
H K
ε
1 2 1
LI = L
2
2 2R
FG ε IJ 1 − e b
H RK
g
− RL t
R
t = ln 2
L
u =∈0
z
∞
*P32.34
= 0.064 8 J .
j
2
P32.33
2
2
e
P32.32
f
4.50 T
B2
=
= 8.06 × 10 6 J m3 .
2 µ 0 2 1.26 × 10 −6 T ⋅ m A
U = uV = 8.06 × 10 6 J m3
P32.31
fa
The magnetic energy density is given by
µ=
(b)
a
0
2
f
=
Lε 2
8R2
so
so
E2
= 44.2 nJ m3
2
e −2 Rt L dt = −
z
a0.800fa500f = 27.8 J
8a30.0f
ε FεI
= G J 1− e b g
2R H R K
2
=
2
− RL t
t=
u=
FG
H
IJ
K
→e
b g
− RL t
=
1
2
L
0.800
ln 2 =
ln 2 = 18.5 ms
30.0
R
B2
= 995 µJ m3
2µ 0
L ∞ −2 Rt L −2 Rdt
L −2 Rt L
e
e
=−
L
2R 0
2R
∞
0
=−
a f
L −∞
L
L
e − e0 =
0−1 =
2R
2R
2R
e
j
247
248
P32.35
Inductance
a
fa
1 2 1
LI = 4.00 H 0.500 A
2
2
(a)
U=
(b)
When the current is 1.00 A,
Kirchhoff’s loop rule reads
f
2
U = 0.500 J
a
fa
f
+22.0 V − 1.00 A 5.00 Ω − ∆VL = 0 .
Then
∆VL = 17.0 V .
The power being stored in the inductor is
I∆VL = 1.00 A 17.0 V = 17.0 W .
a
a
P32.36
fa
P = I∆V = 0.500 A 22.0 V
(c)
f
fa
P = 11.0 W
I=
From Equation 32.7,
ε
e1 − e j .
− Rt L
R
ε
= 2.00 A .
R
At that time, the inductor is fully energized and P = I ∆V = 2.00 A 10.0 V = 20.0 W .
(a)
The maximum current, after a long time t , is
I=
a f a
P32.37
a
f a5.00 Ωf =
2
(b)
Plost = I 2 R = 2.00 A
(c)
Pinductor = I ∆Vdrop = 0
(d)
U=
e j
a10.0 Hfa2.00 Af
=
LI 2
2
u =∈0
Therefore
∈0
E2
2
E2
B2
=
2
2µ 0
6.80 × 10 5 V m
3.00 × 10 8 m s
fa
f
20.0 W
2
2
We have
B = E ∈0 µ 0 =
P32.38
FIG. P32.35
f
= 20.0 J
B2
.
2µ 0
and
u=
so
B 2 =∈0 µ 0 E 2
= 2.27 × 10 −3 T .
The total magnetic energy is the volume integral of the energy density, u =
Because B changes with position, u is not constant. For B = B0
FG R IJ
HrK
2
,
u=
B2
.
2µ 0
FB
GH 2 µ
2
0
0
I FG R IJ
JK H r K
4
.
Next, we set up an expression for the magnetic energy in a spherical shell of radius r and thickness
dr. Such a shell has a volume 4π r 2 dr , so the energy stored in it is
j FGH 2π µB R IJK rdr .
e
2
0
dU = u 4π r 2 dr =
0
4
2
We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside the
sphere. This gives
U=
2π B02 R 3
µ0
=
e
j e6.00 × 10 mj
e1.26 × 10 T ⋅ m Aj
2π 5.00 × 10 −5 T
2
−6
6
3
= 2.70 × 10 18 J .
Chapter 32
Section 32.4
P32.39
249
Mutual Inductance
af
I1 t = I max e −αt sin ω t with I max = 5.00 A , α = 0.025 0 s −1 , and ω = 377 rad s
dI 1
= I max e −α t −α sin ω t + ω cos ω t .
dt
b
g
a fh
a fh
dI 1
= 5.00 A s e −0.020 0 − 0.025 0 sin 0.800 377 + 377 cos 0.800 377
dt
b
At t = 0.800 s ,
g
b
g c
c
dI 1
= 1.85 × 10 3 A s .
dt
Thus, ε 2 = −M
P32.40
ε 2 = −M
bε g
2 max
dI 1
:
dt
M=
−ε 2
+3.20 V
=
= 1.73 mH .
dI 1 dt 1.85 × 10 3 A s
dI 1
= − 1.00 × 10 −4 H 1.00 × 10 4 A s cos 1 000t
dt
e
j b
je
g
= 1.00 V
ε2
M=
P32.42
Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at
µ I
distance x from the wire is B = 0 , and this field passes perpendicularly through the plane of the
2π x
loop. The flux through the loop is
dI 1 dt
z
=
96.0 mV
= 80.0 mH
1.20 A s
P32.41
z
za
f
Φ B = B ⋅ dA = BdA = B Adx =
FG
H
z
IJ
K
µ 0 IA 1.70 mm dx µ 0 IA
1.70
=
ln
.
2π 0. 400 mm x
2π
0.400
The mutual inductance between the wire and the loop is then
M=
FG
H
a f e
IJ
K
M = 7.81 × 10 −10 H = 781 pH
P32.43
je
ja f
1 4π × 10 −7 T ⋅ m A 2.70 × 10 −3 m
N 2 Φ 12 N 2 µ 0 IA
N µ A
1.70
=
= 2 0 1.45 =
ln
1.45
I1
2π I
0. 400
2π
2π
e
j
(a)
−6
N B Φ BA 700 90.0 × 10
=
= 18.0 mH
M=
3.50
IA
(b)
−6
Φ A 400 300 × 10
=
= 34.3 mH
LA =
IA
3.50
(c)
ε B = −M
e
j
a
fb
dI A
= − 18.0 mH 0.500 A s = −9.00 mV
dt
g
250
*P32.44
Inductance
The large coil produces this field at the center of the small coil:
N 1 µ 0 I 1 R12
e
2 x 2 + R12
j
32
. The field is normal to
the area of the small coil and nearly uniform over this area, so it produces flux
N 1 µ 0 I 1 R12
Φ 12 =
e
2
2x +
j
3 2
R12
π R 22 through the face area of the small coil. When current I 1 varies, this is the
emf induced in the small coil:
ε 2 = −N2
P32.45
2
2
N 1 N 2 πµ 0 R12 R 22 dI 1
dI 1
N 1 N 2 πµ 0 R12 R 22
d N 1 µ 0 R1 π R 2
I
=
−
=
−
M
=
so
M
.
1
3 2
32
dt 2 x 2 + R 2 3 2
dt
dt
2 x 2 + R12
2 x 2 + R12
1
e
j
e
e
j
With I = I 1 + I 2 , the voltage across the pair is:
∆V = − L1
dI 1
dI
dI
dI
dI
− M 2 = − L 2 2 − M 1 = − Leq
.
dt
dt
dt
dt
dt
dI 1 ∆V M dI 2
=
+
dt
L1 L1 dt
−
So,
−L 2
and
a f
(a)
1 2
+ M2
j dIdt
2
b
FIG. P32.45
g
[1]
g
[2]
= ∆V L1 − M .
dI 2 ∆V M dI 1
=
+
dt
L 2 L 2 dt
By substitution,
−
leads to
e− L L
+ M2
j dIdt = ∆V bL
2
Adding [1] to [2],
e− L L
+ M2
j dIdt = ∆V bL
+ L2 − 2 M .
So,
Leq = −
L1 L 2 − M 2
∆V
=
.
dI dt
L1 + L 2 − 2 M
Section 32.5
1 2
1 2
1
1
−M .
g
Oscillations in an LC Circuit
b g
At different times, U C
I max =
a f
C
∆V
L
max
(b)
2
dI 2 M ∆V
M dI 2
+
+
= ∆V
dt
L1
L1 dt
e− L L
P32.46
j
=
max
b g
= UL
max
so
LM 1 Ca∆V f OP
N2
Q
2
=
max
1.00 × 10 −6 F
40.0 V = 0.400 A .
10.0 × 10 −3 H
a
f
FG 1 LI IJ
H2 K
2
max
Chapter 32
P32.47
P32.48
LM 1 Ca∆V f OP
N2
Q
2
=
max
FG 1 LI IJ
H2 K
2
max
b g
so ∆VC
max
=
251
L
20.0 × 10 −3 H
I max =
0.100 A = 20.0 V
C
0.500 × 10 −6 F
a
f
When the switch has been closed for a long time, battery, resistor, and
coil carry constant current I max =
ε
. When the switch is opened,
R
current in battery and resistor drops to zero, but the coil carries this
same current for a moment as oscillations begin in the LC loop.
We interpret the problem to mean that the voltage amplitude of these
1
1 2
2
.
oscillations is ∆V , in C ∆V = LI max
2
2
a f
e0.500 × 10 Fja150 Vf a250 Ωf
C a ∆V f
C a ∆V f R
=
=
Then, L =
ε
I
a50.0 Vf
2
2
2
2
max
P32.49
2
P32.51
P32.52
2
= 0.281 H .
This radio is a radiotelephone on a ship, according to frequency assignments made by international
treaties, laws, and decisions of the National Telecommunications and Information Administration.
1
The resonance frequency is
f0 =
.
2π LC
1
1
Thus,
C=
=
= 608 pF .
2
2
−6
6
2π f 0 L
2π 6.30 × 10 Hz 1.05 × 10 H
b
P32.50
2
−6
2
FIG. P32.50
f=
1
2π LC
L=
:
1
(a)
f=
(b)
Q = Qmax
(c)
I=
(a)
f=
(b)
(c)
1
b 2π f g C
2
=
e
j e
1
a f e8.00 × 10 j
2π 120
2
1
−6
−6
2π
j
= 0.220 H
= 135 Hz
b0.082 0 Hge17.0 × 10 Fj
cos ω t = b180 µCg cosb847 × 0.001 00g =
2π LC
a fa f a
119 µC
f
dQ
= −ω Qmax sin ω t = − 847 180 sin 0.847 = −114 mA
dt
1
=
1
a0.100 Hfe1.00 × 10 Fj
Q = Cε = e1.00 × 10 Fja12.0 V f = 12.0 µC
2π LC
−6
2π
= 503 Hz
−6
1 2 1 2
Cε = LI max
2
2
I max = ε
(d)
=
g
FIG. P32.52
C
1.00 × 10 −6 F
= 12 V
= 37.9 mA
L
0.100 H
At all times
U=
ja
1 2 1
Cε = 1.00 × 10 −6 F 12.0 V
2
2
e
f
2
= 72.0 µJ .
252
P32.53
Inductance
ω=
1
LC
1
=
a3.30 Hfe840 × 10
Q = Q max cos ω t , I =
2
e 105 × 10
=
e
−6
je
cos 1.899 × 10 4 rad s 2.00 × 10 −3 s
UC =
Q
2C
(b)
UL =
2
Qmax
sin 2 ω t
1 2 1
2
LI = Lω 2 Qmax
sin 2 ω t =
2
2
2C
(c)
Section 32.6
(a)
e105 × 10 Cj
=
j
2
e
2e840 × 10
jj
2
= 6.03 J
b g
je
sin 2 1.899 × 10 4 rad s 2.00 × 10 −3 s
−12
F
j
j=
0.529 J
U total = UC + U L = 6.56 J
The RLC Circuit
ωd =
(b)
Rc =
(a)
ω0 =
(b)
ωd =
(c)
e
2 840 × 10 −12
b g
FG IJ
H K
1
R
−
2L
LC
∆ω
ω0
2
fd =
Therefore,
P32.56
j
(a)
UL
P32.55
F
= 1.899 × 10 4 rad s
dQ
= −ω Qmax sin ω t
dt
−6
P32.54
−12
=
e
1
je
2.20 × 10 −3 1.80 × 10 −6
F 7.60
−G
j GH 2e2.20 × 10
−3
I
J
j JK
2
= 1.58 × 10 4 rad s
ωd
= 2.51 kHz .
2π
4L
= 69.9 Ω
C
1
LC
=
1
a0.500fe0.100 × 10 j
−6
FG IJ
H K
1
R
−
2L
LC
2
= 4.47 krad s
= 4.36 krad s
= 2.53% lower
Choose to call positive current clockwise in Figure 32.21. It drains charge from the capacitor
dQ
. A clockwise trip around the circuit then gives
according to I = −
dt
dI
Q
+ − IR − L = 0
C
dt
Q dQ
d dQ
R+L
+ +
= 0 , identical with Equation 32.28.
C dt
dt dt
Chapter 32
*P32.57
The period of damped oscillation is T =
capacitor is Q = Qmax e − RT
2L
= Qmax e −2π R
after one oscillation it is U = U 0 e
e 2π R
Lω d
=
1
0.99
b
Lω d =
2 Lω d
. The energy is proportional to the charge squared, so
= 0.99U 0 . Then
g
F
GH
R2
2π 2 Ω
1
= 1 250 Ω = L
− 2
LC 4L
0.001 005
1.563 × 10 6 Ω 2 =
2π
. After one oscillation the charge returning to the
ωd
− 2π R Lω d
2π 2 Ω
= ln 1.010 1 = 0.001 005
Lω d
a f
2Ω
L
−
C
4
I
JK
12
2
L
= 1.563 × 10 6 Ω 2
C
We are also given
1
ω = 2π × 10 3 s =
LC =
LC
1
e2π × 10 sj
3
2
= 2.533 × 10 −8 s 2
Solving simultaneously,
C = 2.533 × 10 −8 s 2 L
L2
= 1.563 × 10 6 Ω 2
L = 0.199 H
2.533 × 10 −8 s 2
2.533 × 10 −8 s 2
C=
= 127 nF = C
0.199 H
P32.58
(a)
Q = Qmax e − Rt 2 L cos ω d t
so
I max ∝ e − Rt 2 L
0.500 = e − Rt 2 L
and
Rt
= − ln 0.500
2L
so
Q = 0.500Qmax = 0.707Qmax
t=−
(b)
a
f
t=−
a
f
FG IJ
H K
2L
2L
ln 0.500 = 0.693
R
R
2
U 0 ∝ Qmax
and U = 0.500U 0
a
f
FG IJ
H K
2L
2L
ln 0.707 = 0.347
R
R
253
(half as long)
254
Inductance
Additional Problems
*P32.59
(a)
Let Q represent the magnitude of the opposite charges on the plates of a parallel plate
capacitor, the two plates having area A and separation d. The negative plate creates electric
Q
Q2
toward itself. It exerts on the positive plate force F =
field E =
toward the
2 ∈0 A
2 ∈0 A
Q
. The energy density is
negative plate. The total field between the plates is
∈0 A
uE =
Q2
Q2
1
1
∈0 E 2 = ∈0 2 2 =
. Modeling this as a negative or inward pressure, we
2
2
∈0 A
2 ∈0 A 2
have for the force on one plate F = PA =
(b)
Q2
, in agreement with our first analysis.
2 ∈0 A 2
The lower of the two current sheets shown creates
µ J
above it magnetic field B = 0 s − k . Let A and w
2
represent the length and width of each sheet. The
upper sheet carries current J s w and feels force
e j
Js
Js
y
x
µ J
µ wAJ s2 F = IA × B = J s wA 0 s i × − k = 0
j.
2
2
e j
The force per area is P =
(c)
z
µ 0 J s2
F
=
.
Aw
2
FIG. P32.59(b)
µ 0 Js µ 0 Js −k +
− k = µ 0 J s k , with
2
2
. Outside the space they enclose, the fields of the separate sheets are
Between the two sheets the total magnetic field is
magnitude B = µ 0 J s
e j
e j
in opposite directions and add to zero .
P32.60
1
B2 =
µ 02 J s2
µ 0 J s2
=
2µ 0
2
(d)
uB =
(e)
This energy density agrees with the magnetic pressure found in part (b).
2µ 0
With Q = Qmax at t = 0 , the charge on the capacitor at any time is Q = Q max cos ω t where ω =
The energy stored in the capacitor at time t is then
2
Q 2 Qmax
cos 2 ω t = U 0 cos 2 ω t .
=
2C
2C
1
1
cos ω t =
and
When U = U 0 ,
4
2
U=
Therefore,
The inductance is then:
t
LC
=
π
3
or
1
ω t = π rad .
3
t2 π 2
=
.
LC
9
L=
9t 2
.
π 2C
1
LC
.
Chapter 32
P32.61
a
ε L = −L
(b)
Q = Idt =
(c)
z za
t
t
0
0
f
20.0t dt = 10.0t 2
−Q
−10.0t 2
=
= − 10.0 MV s 2 t 2
C
1.00 × 10 −6 F
e
e
e
jt
j
j
2
−10.0t 2
Q2 1 2
1
2
≥ LI , or
When
≥ 1.00 × 10 −3 20.0t ,
−6
2C 2
2
2 1.00 × 10
e
then 100 t 4 ≥ 400 × 10 −9
P32.62
f
d 20.0t
dI
= − 1.00 mH
= −20.0 mV
dt
dt
(a)
∆VC =
fa
(a)
ε L = −L
(b)
I=
2
a f
dI
d
= −L
Kt = − LK
dt
dt
dQ
,
dt
z z
t
t
0
0
Q = Idt = Ktdt =
so
∆VC =
When
f
. The earliest time this is true is at t = 4.00 × 10 −9 s = 63.2 µs .
b g
1
C ∆VC
2
2
=
1 2
Kt
2
−Q
Kt 2
= −
C
2C
F
GH
I
JK
1 K 2t 4
1
= L K 2t2
C
2
2
2
4C
1 2
LI ,
2
e
j
t = 2 LC
Thus
P32.63
ja
j e
and
(c)
FG IJ
H K
1 Q2
1 Q
=
2 C
2C 2
2
+
1 2
LI
2
so
The flux through each turn of the coil is
I=
3Q 2
.
4CL
ΦB =
LI
Q
=
N
2N
3L
C
where N is the number of turns.
P32.64
B=
(a)
255
µ 0 NI
2π r
z
Φ B = BdA =
L=
z
b
a
FG IJ
HK
z
µ 0 NI
µ NIh b dr µ 0 NIh
b
=
hdr = 0
ln
a
2π r
2π a r
2π
FG IJ
HK
NΦ B
µ0N2h
b
ln
=
I
a
2π
a f b0.010 0g lnFG 12.0 IJ = 91.2 µH
H 10.0 K
2π
µ N F A I µ a500 f F 2.00 × 10
=
G J = 2π GH 0.110
2π H R K
µ 0 500
(b)
L=
(c)
Lappx
FIG. P32.64
0
2
2
0
2
−4
m2
I=
JK
90.9 µH , only 0.3% different.
256
P32.65
Inductance
(a)
B=
So the coil creates flux through itself
Φ B = BA cos θ =
When the current it carries changes,
ε L = −N
a
f
2π r = 3 0.3 m
e
2
2 R +0
L≈
so
(b)
Nµ 0 IR 2
At the center,
j
2 3 2
=
Nµ 0 I
.
2R
Nµ 0 I
π
π R 2 cos 0° = Nµ 0 IR .
2R
2
FG IJ
H K
dΦ B
π
dI
dI
≈ −N
Nµ 0 R = − L
2
dt
dt
dt
π 2
N µ0R .
2
r ≈ 0.14 m
π
L ≈ 1 2 4π × 10 −7 T ⋅ m A 0.14 m = 2.8 × 10 −7 H
2
L ~ 100 nH
so
ja
e je
(c)
P32.66
(a)
−7
L 2.8 × 10 V ⋅ s A
=
= 1.0 × 10 −9 s
R
270 V A
L
~ 1 ns
R
If unrolled, the wire forms the diagonal of a
0.100 m (10.0 cm) rectangle as shown. The length of
this rectangle is
L′ =
a9.80 mf − a0.100 mf
2
2
f
9.80 m
0.100 m
L′
.
FIG. P32.66(a)
The mean circumference of each turn is C = 2π r ′ , where r ′ =
radius of each turn. The number of turns is then:
a
a
f a
f
f
2
2
N=
9.80 m − 0.100 m
L′
=
= 127 .
C 2π 24.0 + 0.644 2 × 10 −3 m
(b)
R=
−8
ρA 1.70 × 10 Ω ⋅ m 10.0 m
=
= 0.522 Ω
2
A
π 0.322 × 10 −3 m
(c)
L=
µN 2 A 800 µ 0 L ′
=
A′
A′
C
L=
ja
e
e
e
800 4π × 10 −7
0.100 m
24.0 + 0.644
mm is the mean
2
f
j
FG IJ π ar ′f
H K
j FG a9.80 mf − a0.100 mf IJ
GH π a24.0 + 0.644f × 10 m JK
2
L = 7.68 × 10 −2 H = 76.8 mH
2
2
2
−3
2
π
LMFG 24.0 + 0.644 IJ × 10
NH 2 K
−3
m
OP
Q
2
Chapter 32
P32.67
257
From Ampere’s law, the magnetic field at distance r ≤ R is found as:
b g
e j
B 2π r = µ 0 J π r 2 = µ 0
F I
GH π R
2
I eπ r j , or B = µ Ir
JK
2π R
2
0
2
.
The magnetic energy per unit length within the wire is then
z
F I
GH JK
z
R
µ I2 R
µ I 2 R4
µ0I2
U
B2
=
=
.
2π rdr = 0 4 r 3 dr = 0 4
A 0 2µ 0
4
16π
4π R 0
4π R
b
g
This is independent of the radius of the wire.
P32.68
The primary circuit (containing the battery and solenoid) is an RL
circuit with R = 14.0 Ω , and
jb
e
ge
j
2
−7
12 500 1.00 × 10 −4
µ N 2 A 4π × 10
L= 0
=
= 0.280 H .
A
0.070 0
(a)
The time for the current to reach 63.2% of the maximum value
is the time constant of the circuit:
τ=
(b)
L 0.280 H
=
= 0.020 0 s = 20.0 ms .
R
14.0 Ω
The solenoid’s average back emf is
where
f
Thus,
(c)
FIG. P32.68
FG ∆I IJ = LFG I − 0 IJ
H ∆t K H ∆t K
F ∆V IJ = 0.632FG 60.0 V IJ = 2.71 A .
= 0.632G
I = 0.632 I
HRK
H 14.0 Ω K
F 2.71 A I = 37.9 V .
ε = a0.280 H fG
H 0.020 0 s JK
f
εL = L
max
L
The average rate of change of flux through each turn of the overwrapped concentric coil is
the same as that through a turn on the solenoid:
jb
a f e
ga
fe
4π × 10 −7 T ⋅ m A 12 500 0.070 0 m 2.71 A 1.00 × 10 −4 m 2
∆Φ B µ 0 n ∆I A
=
=
0.020 0 s
∆t
∆t
j
= 3.04 mV
(d)
The magnitude of the average induced emf in the coil is ε L = N
FG ∆Φ IJ and magnitude of
H ∆t K
the average induced current is
I=
FG
H
IJ
K
ε L N ∆Φ B
820
=
=
3.04 × 10 −3 V = 0.104 A = 104 mA .
R
R ∆t
24.0 Ω
e
j
B
258
P32.69
Inductance
b g
ε − b I + I gR
ε − I + I 2 R1 − I 2 R2 = 0 .
Left-hand loop:
Outside loop:
2
1
−L
dI
=0.
dt
dI
=0.
dt
This is of the same form as Equation 32.6, so its solution is of the same
form as Equation 32.7:
ε′
It =
1 − e − R ′t L .
R′
Eliminating I 2 gives
ε ′ − IR ′ − L
af
But R ′ =
b
b
af
It =
g
g
j.
ε
1 − e − R ′t L
R1
e
When switch is closed, steady current I 0 = 1.20 A . When the switch is
opened after being closed a long time, the current in the right loop is
I = I 0 e − R2 t
so
e Rt L =
Therefore,
P32.71
j
ε ′ ε R 2 R1 + R 2
ε
=
.
=
R ′ R1 R 2 R1 + R 2
R1
R1 R 2
R2ε
and ε ′ =
, so
R1 + R 2
R1 + R 2
Thus
P32.70
e
FIG. P32.69
(a)
I0
I
L
a
FG IJ
H K
I
Rt
= ln 0 .
L
I
and
fa
f
1.00 Ω 0.150 s
R2 t
=
= 0.095 6 H = 95.6 mH .
L=
ln I 0 I
ln 1.20 A 0.250 A
b g
b
FIG. P32.70
g
While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of the
circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer
loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:
a
f
e
j
+ ε 0 − 2.00 + 6.00 × 10 3 Ω 9.00 × 10 −3 A = 0
+ ε 0 = 72.0 V with end b at the higher potential
(b)
FIG. P32.71(b)
(c)
After the switch is opened, the current around the outer loop decays as
I = I max e − Rt L with I max = 9.00 mA , R = 8.00 kΩ , and L = 0.400 H .
Thus, when the current has reached a value I = 2.00 mA , the elapsed time is:
t=
FG L IJ lnFG I IJ = FG 0.400 H IJ lnFG 9.00 IJ = 7.52 × 10
H R K H I K H 8.00 × 10 Ω K H 2.00 K
max
3
−5
s = 75.2 µs .
Chapter 32
P32.72
(a)
IL = 0
∆V = ε 0
+ – L
The instant after the switch is closed, the situation
is as shown in the circuit diagram of Figure (a). The
requested quantities are:
I L = 0 , IC =
ε0
ε
, IR = 0
R
R
∆VR = ε 0
∆VC= 0
∆VL = ε 0 , ∆VC = 0 , ∆VR = ε 0
(b)
I R = ε 0 /R
Q=0
+ε –
0
I C = ε 0 /R
After the switch has been closed a long time, the
steady-state conditions shown in Figure (b) will
exist. The currents and voltages are:
Figure (a)
IL = 0
∆V = 0
+ – L
I L = 0 , IC = 0 , I R = 0
IR = 0
Q = Cε 0
∆VL = 0 , ∆VC = ε 0 , ∆VR = 0
∆VR = 0
∆VC= ε 0
+ε –
0
Figure (b)
FIG. P32.72
P32.73
When the switch is closed, as shown in
figure (a), the current in the inductor is I:
12.0 − 7.50 I − 10.0 = 0 → I = 0.267 A .
When the switch is opened, the initial
current in the inductor remains at 0.267 A.
a0.267 AfR ≤ 80.0 V
IR = ∆V :
R ≤ 300 Ω
(a)
(b)
FIG. P32.73
P32.74
jb
e
ge
j
2
(a)
−7
−4
2
µ N 2 A 4π × 10 T ⋅ m A 1 000 1.00 × 10 m
L1 = 0 1 =
= 2.51 × 10 −4 H = 251 µH
A1
0.500 m
(b)
M=
M=
(c)
b
e4π × 10
−7
jb
ga fe
T ⋅ m A 1 000 100 1.00 × 10 −4 m 2
0.500 m
ε 1 = −M
j = 2.51 × 10
dI 2
dI
dQ1
M dI 2
, or I 1 R1 = − M 2 and I 1 =
=−
dt
dt
dt
R1 dt
M
Q1 = −
R1
Q1 =
g
N 2 Φ 2 N 2 Φ 1 N 2 BA N 2 µ 0 N 1 A 1 I 1 A µ 0 N 1 N 2 A
=
=
=
=
I1
I1
I1
I1
A1
z
tf
dI 2 = −
0
e2.51 × 10
−5
MI 2 i
M
M
I 2 f − I 2i = −
0 − I 2i =
R1
R1
R1
d
ja
H 1.00 A
1 000Ω
i
f = 2.51 × 10
b
−8
g
C = 25.1 nC
−5
H = 25.1 µH
259
260
P32.75
Inductance
It has a magnetic field, and it stores energy, so L =
(b)
Every field line goes through the rectangle between the conductors.
(c)
Φ = LI so
L=
Φ 1
=
I I
z
I2
is non-zero.
w− a
BdA
y=a
F µ I + µ I I = 2 µ Ix dy = 2µ x ln y
GH 2π y 2π bw − yg JK I z 2π y 2π
µ x F w − aI
L=
lnG
H a JK .
π
L=
1
I
z
w−a
xdy
0
0
0
0
a
w−a
.
a
0
Thus
P32.76
2U
(a)
For an RL circuit,
af
I t = I max e
b g
− RL t
af
It
R
− RL t
= 1 − 10 −9 = e b g ≅ 1 − t
I max
L
:
e3.14 × 10 je10 j =
=
b2.50 yrge3.16 × 10 s yrj
−8
R
t = 10 −9
L
so
Rmax
−9
7
3.97 × 10 −25 Ω .
(If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm 2 , its resistance
would be at least 10 −6 Ω ).
P32.77
a
fe
1 2 1
LI = 50.0 H 50.0 × 10 3 A
2
2
j
2
= 6.25 × 10 10 J
(a)
UB =
(b)
Two adjacent turns are parallel wires carrying current in the same direction. Since the loops
have such large radius, a one-meter section can be regarded as straight.
B=
Then one wire creates a field of
µ0I
.
2π r
This causes a force on the next wire of F = IAB sin θ
F = IA
giving
Evaluating the force,
e
F = 4π × 10
µ0I
µ AI 2
sin 90° = 0
.
2π r
2π r
a1.00 mf 50.0 × 10 A
j 2π ea0.250 mf j
3
−7
N A
2
2
= 2 000 N .
Chapter 32
P32.78
P
1.00 × 10 9 W
=
= 5.00 × 10 3 A
∆V 200 × 10 3 V
P = I∆V
I=
From Ampere’s law,
B 2π r = µ 0 I enclosed or B =
(a)
261
b g
µ 0 I enclosed
.
2π r
I enclosed = 5.00 × 10 3 A
At r = a = 0.020 0 m ,
FIG. P32.73
(b)
(c)
−7
je
T ⋅ m A 5.00 × 10 3 A
B=
At r = b = 0.050 0 m ,
I enclosed = I = 5.00 × 10 3 A
and
e4π × 10
B=
z
U = udV =
z
b
a
e4π × 10
U=
(d)
e4π × 10
and
−7
b
−7
2
0
2µ 0
b
2π 0.050 0 m
2
4π
je
T ⋅ m A 5.00 × 10 3 A
4π
je
T ⋅ m A 5.00 × 10 3 A
a f b2π rAdr g = µ I A
Br
g
2π 0.020 0 m
g
j = 0.050 0 T =
50.0 mT .
j = 0.020 0 T =
20.0 mT .
FG IJ
HK
z
b
b
dr µ 0 I 2 A
=
ln
a
r
4π
a
j e1 000 × 10 mj lnFG 5.00 cmIJ = 2.29 × 10
H 2.00 cm K
2
3
6
J = 2.29 MJ
The magnetic field created by the inner conductor exerts a force of repulsion on the current
in the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small
rectangular section of the outer cylinder of length A and width w.
F w I
e5.00 × 10 AjGH 2π b0.050
0 mg JK
3
It carries a current of
and experiences an outward force
e5.00 × 10 Ajw Ae20.0 × 10 Tj sin 90.0° .
2π b0.050 0 mg
F
F e5.00 × 10 A je 20.0 × 10 Tj
=
=
The pressure on it is P = =
A wA
2π b0.050 0 mg
3
F = IAB sin θ =
−3
3
−3
318 Pa .
262
P32.79
Inductance
jb
e
f
ga
4π × 10 −7 T ⋅ m A 1 400 2.00 A
µ 0 NI
=
= 2.93 × 10 −3 T upward
1.20 m
A
(a)
B=
(b)
2.93 × 10 −3 T
B2
u=
=
= 3.42 J m3
2 µ 0 2 4π × 10 −7 T ⋅ m A
e
e
j
2
j e
b
jFGH 1 N1 J⋅ m IJK = 3.42 N m
2
g
= 3.42 Pa
(c)
To produce a downward magnetic field, the surface of the superconductor must carry a
clockwise current.
(d)
The vertical component of the field of the solenoid exerts an inward force on the
superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field
diverges and has a radially outward horizontal component. This component exerts upward
force on the clockwise superconductor current. The total force on the core is upward . You
can think of it as a force of repulsion between the solenoid with its north end pointing up,
and the core, with its north end pointing down.
(e)
a
fLMN e
j OPQ =
F = PA = 3.42 Pa π 1.10 × 10 −2 m
2
1.30 × 10 −3 N
Note that we have not proven that energy density is pressure. In fact, it is not in some cases;
Equation 21.2 shows that the pressure is two-thirds of the translational energy density in an
ideal gas.
ANSWERS TO EVEN PROBLEMS
P32.2
1.36 µH
P32.4
240 nWb
P32.6
19.2 µWb
P32.8
(a) 360 mV ; (b) 180 mV ; (c) t = 3.00 s
P32.10
(a) 15.8 µH ; (b) 12.6 mH
P32.12
P32.26
(a) 1.00 A ; (b) ∆V12 = 12.0 V ,
∆V1 200 = 1.20 kV , ∆VL = 1.21 kV ; (c)
7.62 ms
P32.28
(a), (b), and (c) see the solution;
(d) yes; see the solution
P32.30
(a) 8.06 MJ m3 ; (b) 6.32 kJ
P32.32
(a) 27.8 J ; (b) 18.5 ms
see the solution
P32.34
see the solution
P32.14
1.92 Ω
P32.36
(a) 20.0 W ; (b) 20.0 W ; (c) 0;(d) 20.0 J
P32.16
see the solution
P32.18
92.8 V
P32.20
30.0 mH
P32.22
7.67 mH
P32.24
(a) 1.00 kΩ; (b) 3.00 ms
P32.38
2π B02 R 3
µ0
P32.40
1.00 V
P32.42
781 pH
P32.44
M=
= 2.70 × 10 18 J
N1 N 2 πµ 0 R12 R 22
e
2 x 2 + R12
j
32
Chapter 32
P32.46
400 mA
P32.48
281 mH
P32.50
220 mH
P32.52
(a) 503 Hz ; (b) 12.0 µC ; (c) 37.9 mA ;
(d) 72.0 µJ
P32.54
(a) 2.51 kHz; (b) 69.9 Ω
P32.56
see the solution
P32.58
(a) 0.693
P32.60
P32.62
P32.64
(a) see the solution; (b) 91.2 µH ;
(c) 90.9 µH , 0.3% smaller
P32.66
(a) 127; (b) 0.522 Ω; (c) 76.8 mH
P32.68
(a) 20.0 ms; (b) 37.9 V; (c) 3.04 mV;
(d) 104 mA
P32.70
95.6 mH
P32.72
(a) I L = 0 , I C =
P32.74
(a) 251 µH ; (b) 25.1 µH ; (c) 25.1 nC
P32.76
3.97 × 10 −25 Ω
P32.78
(a) 50.0 mT; (b) 20.0 mT; (c) 2.29 MJ;
(d) 318 Pa
FG 2L IJ ; (b) 0.347FG 2L IJ
H RK
H RK
9t 2
ε0
ε
, IR = 0 ,
R
R
∆VL = ε 0 , ∆VC = 0 , ∆VR = ε 0 ;
(b) I L = 0 , I C = 0 , I R = 0 ,
∆VL = 0 , ∆VC = ε 0 , ∆VR = 0
2
π C
(a) ε L = − LK ; (b) ∆Vc =
(c) t = 2 LC
− Kt 2
;
2C
263
33
Alternating Current Circuits
CHAPTER OUTLINE
33.1
33.2
33.3
33.4
33.5
33.6
33.7
33.8
33.9
AC Sources
Resistors in an AC Circuit
Inductors in an AC Circuit
Capacitors in an AC Circuit
The RLC Series Circuit
Power in an AC Circuit
Resonance in a Series
RLC Circuit
The Transformer and Power
Transmission
Rectifiers and Filters
ANSWERS TO QUESTIONS
Q33.1
If the current is positive half the time and negative half the
time, the average current can be zero. The rms current is not
zero. By squaring all of the values of the current, they all
become positive. The average (mean) of these positive values is
also positive, as is the square root of the average.
Q33.2
∆Vavg =
Q33.3
AC ammeters and voltmeters read rms values. With an
oscilloscope you can read a maximum voltage, or test whether
the average is zero.
Q33.4
Suppose the voltage across an inductor varies sinusoidally.
Then the current in the inductor will have its instantaneous
1
peak positive value cycle after the voltage peaks. The voltage
4
1
is zero and going positive cycle (90°) before the current is
4
zero and going positive.
∆Vmax
∆Vmax
, ∆Vrms =
2
2
Q33.5
If it is run directly from the electric line, a fluorescent light tube can dim considerably twice in every
cycle of the AC current that drives it. Looking at one sinusoidal cycle, the voltage passes through
zero twice. We don’t notice the flickering due to a phenomenon called retinal imaging. We do not
notice that the lights turn on and off since our retinas continue to send information to our brains
after the light has turned off. For example, most TV screens refresh at between 60 to 75 times per
second, yet we do not see the evening news flickering. Home video cameras record information at
frequencies as low as 30 frames per second, yet we still see them as continuous action. A vivid
display of retinal imaging is that persistent purple spot you see after someone has taken a picture of
you with a flash camera.
Q33.6
The capacitive reactance is proportional to the inverse of the frequency. At higher and higher
frequencies, the capacitive reactance approaches zero, making a capacitor behave like a wire. As the
frequency goes to zero, the capacitive reactance approaches infinity—the resistance of an open
circuit.
Q33.7
The second letter in each word stands for the circuit element. For an inductor L, the emf ε leads the
current I—thus ELI. For a capacitor C, the current leads the voltage across the device. In a circuit in
which the capacitive reactance is larger than the inductive reactance, the current leads the source
emf—thus ICE.
265
266
Alternating Current Circuits
Q33.8
The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams
throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across different
circuit elements are not simultaneously at their maximum values. Do not forget that an inductor can
induce an emf in itself and that the voltage across it is 90° ahead of the current in the circuit in phase.
Q33.9
In an RLC series circuit, the phase angle depends on the source frequency. At very low frequency
the capacitor dominates the impedance and the phase angle is near –90°. The phase angle is zero at
the resonance frequency, where the inductive and capacitive reactances are equal. At very high
frequencies φ approaches +90° .
Q33.10
−90° ≤ φ ≤ 90° . The extremes are reached when there is no significant resistance in the circuit.
Q33.11
The resistance remains unchanged, the inductive resistance doubles, and the capacitive reactance is
reduced by one half.
Q33.12
The power factor, as seen in equation 33.29, is the cosine of the phase angle between the current and
applied voltage. Maximum power will be delivered if ∆V and I are in phase. If ∆V and I are 90° out
of phase, the source voltage drives a net current of zero in each cycle and the average power is zero.
Q33.13
The person is doing work at a rate of P = Fv cos θ . One can consider the emf as the “force” that
moves the charges through the circuit, and the current as the “speed” of the moving charges. The
cos θ factor measures the effectiveness of the cause in producing the effect. Theta is an angle in real
space for the vacuum cleaner and phi is the analogous angle of phase difference between the emf
and the current in the circuit.
Q33.14
As mentioned in Question 33.5, lights that are powered by alternating current flicker or get slightly
brighter and dimmer at twice the frequency of the AC power source. Even if you tried using two
banks of lights, one driven by AC 180° of phase from the other, you would not have a stable light
source, but one that exhibits a “ripple” in intensity.
Q33.15
In 1881, an assassin shot President James Garfield. The bullet was lost in his body. Alexander
Graham Bell invented the metal detector in an effort to save the President’s life. The coil is preserved
in the Smithsonian Institution. The detector was thrown off by metal springs in Garfield’s mattress, a
new invention itself. Surgeons went hunting for the bullet in the wrong place and Garfield died.
Q33.16
As seen in Example 33.8, it is far more economical to transmit at high voltage than at low voltage, as
the I 2 R loss on the transmission line is significantly lower. Transmitting power at high voltage
permits the use of step-down transformers to make “low” voltages and high currents available to the
end user.
Q33.17
Insulation and safety limit the voltage of a transmission line. For an underground cable, the
thickness and dielectric strength of the insulation between the conductors determines the maximum
voltage that can be applied, just as with a capacitor. For an overhead line on towers, the designer
must consider electrical breakdown of the surrounding air, possible accidents, sparking across the
insulating supports, ozone production, and inducing voltages in cars, fences, and the roof gutters of
nearby houses. Nuisance effects include noise, electrical noise, and a prankster lighting a hand-held
fluorescent tube under the line.
Q33.18
No. A voltage is only induced in the secondary coil if the flux through the core changes in time.
Chapter 33
Q33.19
267
This person needs to consider the difference between the power delivered by a power plant and
I 2 R losses in transmission lines. At lower voltages, transmission lines must carry higher currents to
transmit the same power, as seen in Example 33.8. The high transmitted current at low voltage
actually results in more internal energy production than a lower current at high voltage. In his
2
∆V
, the ∆V does not represent the line voltage but the potential difference between the
formula
R
ends of one conductor. This is very small when the current is small.
a f
Q33.20
The Q factor determines the selectivity of the radio receiver. For example, a receiver with a very low
Q factor will respond to a wide range of frequencies and might pick up several adjacent radio
stations at the same time. To discriminate between 102.5 MHz and 102.7 MHz requires a high-Q
circuit. Typically, lowering the resistance in the circuit is the way to get a higher quality resonance.
Q33.21
Both coils are wrapped around the same core so that nearly all of the magnetic flux created by the
primary passes through the secondary coil, and thus induces current in the secondary when the
current in the primary changes.
Q33.22
The frequency of a DC signal is zero, making the capacitive reactance at DC infinite. The capacitor
then acts as an open switch. An AC signal has a non-zero frequency, and thus the capacitive
reactance is finite, allowing a signal to pass from Circuit A to Circuit B.
SOLUTIONS TO PROBLEMS
Section 33.1
AC Sources
Section 33.2
Resistors in an AC Circuit
af
b g
b g
a f a283 V f sina628tf
P33.1
∆v t = ∆Vmax sin ω t = 2 ∆Vrms sin ω t = 200 2 sin 2π 100 t =
P33.2
∆Vrms =
P33.3
170 V
2
(a)
P=
(b)
R=
= 120 V
b∆V g
rms
2
R
a120 Vf
100 W
→R=
a120 Vf
2
75.0 W
= 193 Ω
2
= 144 Ω
Each meter reads the rms value.
∆Vrms =
I rms =
100 V
2
= 70.7 V
∆Vrms 70.7 V
=
= 2.95 A
R
24.0 Ω
FIG. P33.3
268
P33.4
Alternating Current Circuits
∆v R = ∆Vmax sin ω t
(a)
b
a
g
f
∆v R = 0.250 ∆Vmax , so sin ω t = 0. 250 , or ω t = sin −1 0.250 .
The smallest angle for which this is true is ω t = 0.253 rad Thus, if t = 0.010 0 s ,
ω=
0.253 rad
= 25.3 rad s .
0.010 0 s
b
a
t=
P33.5
2.89 rad
= 0.114 s .
25.3 rad s
i R = I max sin ω t
b
b0.007 00gω = sin a0.600f = 0.644
−1
b
f = 14.6 Hz .
so
g
P = I rms ∆Vrms and ∆Vrms = 120 V for each bulb (parallel circuit), so:
I1 = I 2 =
I3 =
P1
∆Vrms 120 V
150 W
=
= 1.25 A , and R1 =
=
= 96.0 Ω = R 2
∆Vrms 120 V
1.25 A
I1
P3
∆Vrms
100 W
120 V
=
= 0.833 A , and R3 =
=
= 144 Ω .
∆Vrms 120 V
0.833 A
I3
∆Vmax = 15.0 V and Rtotal = 8. 20 Ω + 10.4 Ω = 18.6 Ω
I max =
∆Vmax 15.0 V
=
= 0.806 A = 2 I rms
Rtotal
18.6 Ω
2
Pspeaker = I rms
Rspeaker =
Section 33.3
P33.8
g
0.600 = sin ω 0.007 00 .
becomes
and ω = 91.9 rad s = 2π f
P33.7
f
f
Thus,
P33.6
a
g
The second time when ∆v R = 0.250 ∆Vmax , ω t = sin −1 0.250 again. For this occurrence,
ω t = π − 0.253 rad = 2.89 rad (to understand why this is true, recall the identity
sin π − θ = sin θ from trigonometry). Thus,
(b)
FG 0.806 A IJ a10.4 Ωf =
H 2 K
2
3.38 W
Inductors in an AC Circuit
For I max = 80.0 mA , I rms =
bX g
L min
=
80.0 mA
2
= 56.6 mA
Vrms
50.0 V
=
= 884 Ω
I rms 0.056 6 A
X L = 2π fL → L =
XL
884 Ω
≥
≥ 7.03 H
2π f 2π 20.0
a f
Chapter 33
P33.9
XL =
(a)
L=
ω=
P33.10
ω
=
13.3
= 0.042 4 H = 42.4 mH
2π 50.0
a f
∆Vmax 100
=
= 40.0 Ω
2.50
I max
XL
40.0
=
= 942 rad s
L
42.4 × 10 −3
a
f
a
At 50.0 Hz, X L = 2π 50.0 Hz L = 2π 50.0 Hz
I max =
P33.11
XL
XL =
(b)
∆Vmax 100
=
= 13.3 Ω
I max
7.50
b
2 ∆Vrms
∆Vmax
=
XL
XL
g = 2 a100 V f =
45.0 Ω
fFGH 2πXa60.0 Hzf IJK = 5060..00 a54.0 Ωf = 45.0 Ω
3.14 A .
IJ a f a fb
K b
ge
i atf = a5.60 A f sina1.59 radf = 5.60 A
af
iL t =
FG
H
L 60.0 Hz
g
80.0 V sin 65.0π 0.015 5 − π 2
∆Vmax
π
=
sin ω t −
ωL
2
65.0π rad s 70.0 × 10 −3 H
j
L
P33.12
b
g
ω = 2π f = 2π 60.0 s = 377 rad s
b
gb
g
X L = ω L = 377 s 0.020 0 V ⋅ s A = 7.54 Ω
I rms =
∆Vrms 120 V
=
= 15.9 A
XL
7.54 Ω
a f
F 2π a60.0f ⋅ 1 s IJ = a22.5 Af sin 120° = 19.5 A
sin ω t = a 22.5 A f sinG
H s 180 K
1
= b0.020 0 V ⋅ s A ga19.5 A f = 3.80 J
2
I max = 2 I rms = 2 15.9 A = 22.5 A
af
i t = I max
P33.13
U=
1 2
Li
2
L=
NΦ B
X ∆VL , max
where Φ B is the flux through each turn. NΦ B , max = LI max = L
ω
I
XL
NΦ B , max =
2
d
d
2 ∆VL , rms
2π f
i=
FG T ⋅ C ⋅ m IJ FG N ⋅ m IJ FG J IJ =
2π a60.0f H N ⋅ s K H J K H V ⋅ C K
120 V ⋅ s
0.450 T ⋅ m 2 .
i
269
270
Alternating Current Circuits
Section 33.4
P33.14
Capacitors in an AC Circuit
1
1
:
< 175 Ω
2π f C 2π f 22.0 × 10 −6
XC =
(a)
e
1
e
j
f > 41.3 Hz
<f
ja f
2π 22.0 × 10 −6 175
P33.15
a f
a f
1
1
, so X 44 = X 22 : X C < 87.5 Ω
C
2
XC ∝
(b)
b
2 ∆Vrms
I max = 2 I rms =
XC
g = 2 b∆V g2π f C
rms
a f b ge
j
2 a 240 V f2π b50.0 sge 2.20 × 10 F j = 235 mA
(a)
I max = 2 120 V 2π 60.0 s 2.20 × 10 −6 C V = 141 mA
(b)
I max =
−6
b
g
b
g
b
Q max = C ∆Vmax = C
P33.17
I max = ∆Vmax ω C = 48.0 V 2π 90.0 s −1 3.70 × 10 −6 F = 100 mA
P33.18
XC =
b
af
2 ∆Vrms =
a
g
2C ∆Vrms
g
P33.16
fa fe
je
j
1
1
=
= 2.65 Ω
ω C 2π 60.0 s 1.00 × 10 −3 C V
b
ge
j
f LM
N
vC t = ∆Vmax sin ω t , to be zero at t = 0
2 120 V
∆Vmax
60 s −1
+ 90.0° = 64.0 A sin 120°+90.0° = −32.0 A
iC =
sin ω t + φ =
sin 2π
XC
2.65 Ω
180 s −1
b
Section 33.5
P33.19
a
g
f
a fe
j
1
1
X =
=
= 719 Ω
ω C 2π a50.0 fe 4.43 × 10 j
Z = R + b X − X g = 500 + a126 − 719 f = 776 Ω
∆V
= I Z = e 250 × 10 ja776f = 194 V
F X − X IJ = tan FG 126 − 719 IJ = −49.9° . Thus, the
φ = tan G
H 500 K
H R K
X L = ω L = 2π 50.0 400 × 10 −3 = 126 Ω
(a)
−6
2
max
L
ωL=
L
1
1
→ω =
=
ωC
LC
ω
2π
C
= 2.79 kHz
2
2
2
−3
max
−1
(b)
f=
f a
The RLC Series Circuit
C
P33.20
OP a
Q
C
−1
1
e57.0 × 10 je57.0 × 10 j
−6
−6
= 1.75 × 10 4 rad s
FIG. P33.19
Current leads the voltage.
Chapter 33
P33.21
P33.22
e
je
j
(a)
X L = ω L = 2π 50.0 s −1 250 × 10 −3 H = 78.5 Ω
(b)
XC =
(c)
Z = R 2 + XL − XC
(d)
I max =
(e)
φ = tan −1
(a)
Z = R 2 + X L − XC
1
= 2π 50.0 s −1 2.00 × 10 −6 F
ωC
e
je
b
g
2
271
j
−1
= 1.59 kΩ
= 1.52 kΩ
∆Vmax
210 V
=
= 138 mA
Z
1.52 × 10 3 Ω
LM X
N
b
L
OP
Q
a
f
− XC
= tan −1 −10.1 = −84.3°
R
a fa
g
2
a
= 68.0 2 + 16.0 − 101
f
f
2
= 109 Ω
X L = ω L = 100 0.160 = 16.0 Ω
XC =
(b)
(c)
1
1
=
= 101 Ω
ω C 100 99.0 × 10 −6
a fe
j
∆Vmax 40.0 V
=
= 0.367 A
109 Ω
Z
I max =
X L − X C 16.0 − 101
=
= −1.25 :
68.0
R
φ = −0.896 rad = −51.3°
tan φ =
ω = 100 rad s
I max = 0.367 A
P33.23
a fa
f
X L = 2π f L = 2π 60.0 0.460 = 173 Ω
1
1
=
= 126 Ω
XC =
2π f C 2π 60.0 21.0 × 10 −6
a fe
(a)
(b)
*P33.24
φ = −0.896 rad = −51.3°
j
X L − X C 173 Ω − 126 Ω
=
= 0.314
150 Ω
R
φ = 0.304 rad = 17.4°
tan φ =
Since X L > X C , φ is positive; so voltage leads the current .
a
f
For the source-capacitor circuit, the rms source voltage is ∆Vs = 25.1 mA X C . For the circuit with
a
resistor, ∆Vs = 15.7 mA
a
f
2
f
R +
X C2
inductor, ∆Vs = 68.2 mA X L − X C
b
∆Vs = I R 2 + X L − X C
a25.1 mAfX
C
I = 19.3 mA
=I
g
a
f
= a 25.1 mA f X
= 25.1 mA X C . This gives R = 1.247 X C . For the circuit with ideal
2
b1.247 X g + b0.368 X g
C
2
C
2
C.
So X L − X C = 0.368 0 X C . Now for the full circuit
272
P33.25
Alternating Current Circuits
XC =
1
1
=
= 1.33 × 10 8 Ω
2π f C 2π 60.0 Hz 20.0 × 10 −12 F
Z=
e50.0 × 10 Ωj + e1.33 × 10 Ωj
a
8
rms body
e
je
j
1
1
=
= 49.0 Ω
ω C 2π 50.0 65.0 × 10 −6
a fe
j
X = ω L = 2π a50.0fe185 × 10 j = 58.1 Ω
Z = R + b X − X g = a 40.0f + a58.1 − 49.0f
−3
2
I max =
L
C
2
2
2
= 41.0 Ω
∆Vmax 150
=
= 3.66 A
41.0
Z
FIG. P33.26
a fa f
(a)
∆VR = I max R = 3.66 40 = 146 V
(b)
∆VL = I max X L = 3.66 58.1 = 212.5 = 212 V
(c)
∆VC = I max X C = 3.66 49.0 = 179.1 V = 179 V
(d)
∆VL − ∆VC = 212.5 − 179.1 = 33.4 V
a fa f
a fa f
R = 300 Ω
FG 500 s IJ a0.200 Hf = 200 Ω
Hπ K
L F 500 s IJ e11.0 × 10 FjOP
1
= M 2π G
X =
K
ωC N H π
Q
Z = R + b X − X g = 319 Ω and
F X − X IJ = 20.0°
φ = tan G
H R K
−1
C
2
−1
XL = 200 Ω
−1
X L = ω L = 2π
*P33.28
≈ 1.33 × 10 8 Ω
= I rms R body = 3.77 × 10 −5 A 50.0 × 10 3 Ω = 1.88 V
L
P33.27
2
5 000 V
∆Vrms
=
= 3.77 × 10 −5 A
8
Z
1.33 × 10 Ω
b ∆V g
XC =
j
2
3
I rms =
P33.26
fe
L
L
C
−6
−1
= 90.9 Ω
XL - XC = 109 Ω
2
{
XC = 90.9 Ω
Z
φ
R = 300 Ω
C
FIG. P33.27
Let X c represent the initial capacitive reactance. Moving the plates to half their original separation
1
doubles the capacitance and cuts X C =
in half. For the current to double, the total impedance
ωC
must be cut in half: Zi = 2Z f ,
b
R 2 + R − XC
g
2
F
GH
FG
H
= 4 R2 + R −
b
R 2 + X L − XC
XC
2
IJ IJ
KK
2
2 R 2 − 2 RX C + X C2 = 8 R 2 − 4RX C + X C2
XC = 3 R
g
2
FG
H
= 2 R2 + XL −
XC
2
IJ
K
2
,
Chapter 33
P33.29
a
fa
f
X L = 2π 100 Hz 20.5 H = 1.29 × 10 4 Ω
∆Vrms 200 V
=
= 50.0 Ω
Z=
I rms
4.00 A
(a)
bX
− XC
L
g
2
b
= Z 2 − R 2 = 50.0 Ω
X L − X C = 1.29 × 10 4 Ω −
∆VL , rms = I rms X L
(b)
g − b35.0 Ωg
2
2
1
= ±35.7 Ω
2π 100 Hz C
b g
= a 4.00 A fe1.29 × 10 Ωj =
4
C = 123 nF or 124 nF
FIG. P33.29
51.5 kV
Notice that this is a very large voltage!
Section 33.6
P33.30
Power in an AC Circuit
b
gb
g
1
= b1 000 sge50.0 × 10 F j
X =
ωC
Z = R + bX − X g
Z = a 40.0f + a50.0 − 20.0f = 50.0 Ω
b∆V g = 100 V
I
=
(a)
X L = ω L = 1 000 s 0.050 0 H = 50.0 Ω
−6
C
2
L
C
−1
= 20.0 Ω
2
2
FIG. P33.30
2
rms
rms
50.0 Ω
Z
I rms = 2.00 A
φ = arctan
FG X
H
b
g
IJ
K
− XC
R
30.0 Ω
φ = arctan
= 36.9°
40.0 Ω
P33.31
L
a
f
(b)
P = ∆Vrms I rms cos φ = 100 V 2.00 A cos 36.9° = 160 W
(c)
2
PR = I rms
R = 2.00 A 40.0 Ω = 160 W
a
f
2
ω = 1 000 rad s ,
R = 400 Ω ,
∆Vmax = 100 V ,
ω L = 500 Ω ,
Z= R
I max =
2
F
1 I
+ Gω L −
H ω C JK
C = 5.00 × 10 −6 F ,
F 1 I = 200 Ω
GH ω C JK
2
= 400 2 + 300 2 = 500 Ω
∆Vmax 100
=
= 0.200 A
Z
500
2
The average power dissipated in the circuit is P = I rms
R=
a0.200 Af a400 Ωf =
2
2
P=
L = 0.500 H
8.00 W
F I IR.
GH 2 JK
2
max
273
274
P33.32
Alternating Current Circuits
b
Z = R 2 + X L − XC
bX
L
g
g
or X L − X C = Z 2 − R 2
g a75.0 Ωf − a45.0 Ωf = 60.0 Ω
FG X − X IJ = tan FG 60.0 Ω IJ = 53.1°
H 45.0 Ω K
H R K
2
− XC =
φ = tan −1
b
2
L
2
−1
C
∆Vrms 210 V
=
= 2.80 A
Z
75.0 Ω
P = ∆Vrms I rms cos φ = 210 V 2.80 A cos 53.1° = 353 W
I rms =
b
P33.33
g
b
g
a f
2
I rms
R
tan φ =
(b)
X L − XC
R
f a
f
a
f
b
a f
3
2
so
1.29 × 10 = 9.00 R
becomes
tan −37.0° =
a
gb
g
g = a20.0f + a9.42f
f
X L − XC
:
16
and
R = 16.0 Ω .
so
X L − X C = −12.0 Ω .
X L = ω L = 2π 60.0 s 0.025 0 H = 9.42 Ω
b
Z = R 2 + X L − XC
2
2
2
Ω = 22.1 Ω
∆Vrms 120 V
=
= 5.43 A
22.1 Ω
Z
(a)
I rms =
(b)
φ = tan −1
(c)
We require φ = 0 . Thus, X L = X C :
9. 42 Ω =
and
C = 281 µF .
FG 9.42 IJ = 25.2°
H 20.0 K
b
(d)
b
b ∆V g
rms b
=
power factor = cos φ = 0.905 .
so
rms b
b
e
1
rms b
2
rms d
R
=
b
a20.0 Ωfa120 Vfa5.43 Afa0.905f =
Consider a two-wire transmission line:
I rms =
P
FG P IJ b2R g = P
100
H ∆V K
ρd b ∆V g
=
=
P
100
2
Thus,
1
rms
R1
rms
A
or
R1 =
or
A=
2
200P
and the diameter is
109 V
R1
2
Rline =
and power loss = I rms
∆Vrms
j
2π 60.0 s −1 C
b ∆V g
=
g bI g cos φ
Rb ∆V g b I g cos φ
Pb = Pd or ∆Vrms
rms d
P33.35
fa
P = I rms ∆Vrms cos φ = 9.00 180 cos −37.0° = 1.29 × 10 3 W
(a)
P=
P33.34
a
2r =
.
∆Vrms
b∆V g
rms
2
R1
200P
a f
π 2r
4
RL
2
=
200 ρP d
b ∆V g
rms
800 ρP d
a f
π ∆V
2
.
2
FIG. P33.35
Chapter 33
P33.36
One-half the time, the left side of the generator is positive, the top
diode conducts, and the bottom diode switches off. The power
supply sees resistance
LM 1 + 1 OP
N 2R 2R Q
−1
= R and the power is
b∆V g
rms
R
LM 1 + 1 OP
N 3R R Q
−1
=
7R
4
.
R
P=
and
~
Section 33.7
P33.37
P33.38
rms
2
Req
rms
=
b
4 ∆Vrms
∆V
R + 4 ∆Vrms
FIG. P33.36
2
.
7R
b
2
g
g
2
7R
2
=
Resonance in a Series RLC Circuit
e
1
j
ω 0 = 2π 99.7 × 10 6 = 6.26 × 10 8 rad s =
C=
b ∆V g
b ∆V g
The overall time average power is:
1
ω 02 L
=
1
At resonance,
LC
= 1.82 pF
e6.26 × 10 j e1.40 × 10 j
8 2
−6
1
1
= 2π f L and
2π f C
2π f
b gL
2
=C .
The range of values for C is 46.5 pF to 419 pF .
*P33.39
(a)
(b)
1
f=
2π LC
1
1
A C
= 6.33 × 10 −13 F
=
2
2
−
2
10
12
As
4π f L 4π 10 s 400 × 10
Vs
C=
κ ∈0 A κ ∈0 A 2
=
d
d
A=
e
FG Cd IJ
Hκ ∈ K
0
(c)
FG IJ
H K
C=
2
12
=
j
F 6.33 × 10 F × 10
GH 1 × 8.85 × 10
−13
−3
−12
R
2
The other half of the time the right side of the generator is positive,
the upper diode is an open circuit, and the lower diode has zero
resistance. The equivalent resistance is then
Req = R +
2R
R
F
mm
I
JK
12
= 8.46 × 10 −3 m
X L = 2π f L = 2π × 10 10 s × 400 × 10 −12 Vs A = 25.1 Ω
b
11 ∆Vrms
14R
g
2
.
275
276
P33.40
P33.41
Alternating Current Circuits
L = 20.0 mH , C = 1.00 × 10 −7 , R = 20.0 Ω , ∆Vmax = 100 V
(a)
The resonant frequency for a series –RLC circuit is f =
(b)
At resonance,
I max =
(c)
From Equation 33.38,
Q=
(d)
∆VL , max = X L I max = ω 0 LI max = 2.24 kV
1
The resonance frequency is ω 0 =
XL = ω L =
FG
H
IJ L = 2
LC K
2
b
Z = R 2 + X L − XC
g
2
LC
1
= 3.56 kHz .
LC
∆Vmax
= 5.00 A .
R
ω 0L
= 22.4 .
R
. Thus, if ω = 2ω 0 ,
L
C
= R 2 + 2.25
1
2π
FG L IJ
H CK
1
LC 1 L
=
=
2C
2 C
ωC
and
XC =
so
I rms =
∆Vrms
=
Z
∆Vrms
b g
2
R + 2.25 L C
and the energy delivered in one period is E = P ∆t :
b∆V g R FG 2π IJ = b∆V g RC eπ
E=
R + 2.25bL C g H ω K R C + 2.25L
rms
2
rms
2
2
2
b
4π ∆Vrms
j
LC =
g RC
2
LC
.
2
4R C + 9.00L
With the values specified for this circuit, this gives:
E=
P33.42
a
f a10.0 Ωfe100 × 10 Fj e10.0 × 10 Hj
4a10.0 Ωf e100 × 10 F j + 9.00e10.0 × 10 H j
4π 50.0 V
2
32
−6
2
−6
FG
H
IJ L = 2
LC K
2
1
b
LC
.
L
C
Then Z = R 2 + X L − X C
g
2
12
−3
The resonance frequency is ω 0 =
XL = ω L =
−3
= R 2 + 2.25
FG L IJ
H CK
= 242 mJ .
Thus, if
ω = 2ω 0 ,
and
XC =
so
I rms =
LC 1 L
1
=
=
.
2C
2 C
ωC
∆Vrms
=
Z
and the energy delivered in one period is
b∆V g R FG 2π IJ = b∆V g RC eπ
E = P∆t =
R + 2.25bL C g H ω K R C + 2.25L
rms
2
2
rms
2
2
j
LC =
b
4π ∆Vrms
2
g RC
2
4R C + 9.00L
LC
.
∆Vrms
2
b g
R + 2.25 L C
Chapter 33
P33.43
1
For the circuit of Problem 22, ω 0 =
1
=
= 251 rad s
e160 × 10 Hje99.0 × 10 Fj
ω L b 251 rad sge160 × 10 H j
=
= 0.591
Q=
LC
−3
−6
−3
0
68.0 Ω
R
For the circuit of Problem 23, Q =
.
ω 0L
1 L
1
460 × 10 −3 H
L
=
=
=
= 0.987 .
R
R LC R C 150 Ω 21.0 × 10 −6 F
The circuit of Problem 23 has a sharper resonance.
Section 33.8
P33.44
The Transformer and Power Transmission
a
f
1
120 V = 9.23 V
13
(a)
∆V2 , rms =
(b)
∆V1 , rms I 1 , rms = ∆V2 , rms I 2 , rms
a120 Vfa0.350 Af = a9.23 V fI
42.0 W
= 4.55 A for a transformer with no energy loss.
9.23 V
I 2 , rms =
P = 42.0 W from part (b).
(c)
P33.45
b ∆V g
=
b∆V g
=
out max
out rms
P33.46
(a)
2 , rms
b
N2
∆Vin
N1
g
max
a971 Vf =
2 , rms
FG 2 000 IJ a170 Vf = 971 V
H 350 K
687 V
2
d ∆V
=
i = NN d∆V i
2
1
d
N2 =
1 , rms
i
d
i
(b)
I 1 , rms ∆V1 , rms = I 2 , rms ∆V2, rms
(c)
0.950 I 1 , rms ∆V1, rms = I 2 , rms ∆V2, rms
d
i
d
b2 200ga80f =
I 1, rms =
i
I 1 , rms =
110
1 600 windings
a1.50fb2 200g =
110
a1.20fb2 200g =
110a0.950f
30.0 A
25.3 A
277
278
P33.47
Alternating Current Circuits
The rms voltage across the transformer primary is
N1
∆V2, rms
N2
d
i
so the source voltage is ∆Vs , rms = I 1 , rms Rs +
d∆V
The secondary current is
2, rms
d
RL
i
N1
∆V2 , rms .
N2
d
i
FIG. P33.47
i , so the primary current is
N 2 ∆V2 , rms
= I 1 , rms .
N1
RL
Then ∆Vs , rms =
and Rs =
P33.48
i
+
N 1 RL
N 1 RL
d
N 2 ∆V2 , rms
F
∆V
i GGH
s , rms
N2
∆V1 , rms
N1
d
−
d
N 1 ∆V2 , rms
i
N2
d
N 1 ∆V2 , rms
N2
i
i IJ = 5a50.0 Ωf FG 80.0 V − 5a25.0 V f IJ =
JK 2a25.0 Vf H
2
K
∆V2 , rms =
(b)
I 2 , rms ∆V2 , rms = 0.900 I 1, rms ∆V1, rms
d
i
d
i
120 V I
J a120 Vf
e10.0 × 10 Vj = 0.900FGH 24.0
ΩK
3
∆V2 , rms
I 2, rms = 54.0 mA
10.0 × 10 3 V
= 185 kΩ
0.054 A
(c)
Z2 =
(a)
R = 4.50 × 10 −4 Ω M 6.44 × 10 5 m = 290 Ω and I rms =
I 2 , rms
e
2
Ploss = I rms
(b)
(c)
=
87.5 Ω .
N 2 ∆V2 , rms 10.0 × 10 3 V
=
=
= 83.3
120 V
N 1 ∆V1 , rms
(a)
I 2, rms
P33.49
d
N 2 ∆V2 , rms Rs
je
j
R = a10.0 A f a 290 Ωf = 29.0 kW
P
5.00 × 10 6 W
=
= 10.0 A
∆Vrms 5.00 × 10 5 V
2
Ploss 2.90 × 10 4
=
= 5.80 × 10 −3
P
5.00 × 10 6
It is impossible to transmit so much power at such low voltage. Maximum power transfer
occurs when load resistance equals the line resistance of 290 Ω , and is
e4.50 × 10 Vj
2 ⋅ 2a 290 Ωf
3
2
= 17.5 kW far below the required 5 000 kW.
Chapter 33
Section 33.9
*P33.50
(a)
279
Rectifiers and Filters
Input power = 8 W
a f
Useful output power = I∆V = 0.3 A 9 V = 2.7 W
useful output 2.7 W
=
= 0.34 = 34%
efficiency =
total input
8W
(b)
Total input power = Total output power
8 W = 2.7 W + wasted power
wasted power = 5.3 W
(c)
*P33.51
(a)
I=
a fa fFGH 861400d s IJK FGH 11WsJ IJK = 1.29 × 10 JFGH 3.6$0.135
J
× 10 J K
8
E = P ∆t = 8 W 6 31 d
2
The input voltage is ∆Vin = IZ = I R +
∆Vout = IR . The gain ratio is
(b)
(c)
(a)
b
∆Vout
1
→ ∞ and
→ 0
ωC
∆Vin
As ω → ∞ ,
∆Vout
1
R
→ 0 and
→ = 1
ωC
∆Vin
R
1
=
2
=I R
2
∆Vout
IR
=
∆Vin
I R2 + 1 ω C
As ω → 0 ,
R2 +
P33.52
X C2
F1I
+G
H ω C JK
g
$4.8
6
=
2
2
. The output voltage is
R
b
R2 + 1 ω C
g
.
2
R
b
R2 + 1 ω C
g
2
1
= 4R 2
ω 2C 2
ω 2C 2 =
1
3R2
ω = 2π f =
1
f=
3 RC
b
The input voltage is ∆Vin = IZ = I R 2 + X C2 = I R 2 + 1 ω C
∆Vout = IX C =
I ωC
∆Vout
I
=
. The gain ratio is
∆Vin
ωC
I R2 + 1 ω C
b
g
g
2
2
1
2π 3 RC
. The output voltage is
=
1 ωC
b
R2 + 1 ω C
g
2
.
1 ωC
∆Vout
1
→ ∞ and R becomes negligible in comparison. Then
→
= 1 . As
ωC
∆Vin
1 ωC
∆Vout
1
→ 0 and
ω → ∞,
→ 0 .
ωC
∆Vin
(b)
As ω → 0 ,
(c)
1
=
2
f=
1 ωC
2
b
R + 1 ωC
3
2π RC
g
2
R2 +
F1I
GH ω C JK
2
=
4
ω C2
2
R 2ω 2 C 2 = 3
ω = 2π f =
3
RC
280
P33.53
Alternating Current Circuits
∆Vout
=
∆Vin
For this RC high-pass filter,
(a)
When
then
R
2
R + X C2
.
∆Vout
= 0.500 ,
∆Vin
0.500 Ω
a0.500 Ωf
2
+ X C2
= 0.500 or X C = 0.866 Ω.
If this occurs at f = 300 Hz, the capacitance is
C=
(b)
1
1
=
= 6.13 × 10 −4 F = 613 µF .
2π f X C 2π 300 Hz 0.866 Ω
a
fa
f
With this capacitance and a frequency of 600 Hz,
XC =
1
a
fe
2π 600 Hz 6.13 × 10 −4 F
∆Vout
=
∆Vin
R
2
R +
X C2
=
j
= 0.433 Ω
0.500 Ω
a0.500 Ωf + a0.433 Ωf
2
2
= 0.756 .
FIG. P33.53
P33.54
For the filter circuit,
(a)
(b)
∆Vout
=
∆Vin
XC
2
R + X C2
.
1
1
=
= 3.32 × 10 4 Ω
2π f C 2π 600 Hz 8.00 × 10 −9 F
At f = 600 Hz,
XC =
and
∆Vout
=
∆Vin
a
fe
j
3.32 × 10 4 Ω
a90.0 Ωf + e3.32 × 10 Ωj
2
4
2
≈ 1.00 .
1
1
=
= 33.2 Ω
3
2π f C 2π 600 × 10 Hz 8.00 × 10 −9 F
At f = 600 kHz ,
XC =
and
∆Vout
=
∆Vin
e
je
33.2 Ω
a90.0 Ωf + a33.2 Ωf
2
2
= 0.346 .
j
281
Chapter 33
P33.55
∆Vout
=
∆Vin
(a)
2
R
b
R + XL − XC
g
2
1
=
4 8.00 Ω
At 200 Hz:
a
a8.00 Ωf
2
.
f + 400π L − 1 400π C
a8.00 Ωf + LM8 000π L − 8 0001 π C OP = 4a8.00 Ωf .
N
Q
2
2
2
At 4 000 Hz:
2
1
= −13.9 Ω .
400π C
1
= +13.9 Ω.
8 000π L −
8 000π C
400π L −
At the low frequency, X L − X C < 0 . This reduces to
For the high frequency half-voltage point,
When X L = X C ,
(c)
X L = X C requires
(d)
At 200 Hz,
FG
H
∆Vout
∆Vout
=
∆Vin
∆Vin
f0 =
1
2π LC
IJ
K
= 1.00 .
2π
e
5.80 × 10
∆Vout R 1
= = and X C > X L ,
∆Vin
Z 2
−4
je
H 5.46 × 10 −5 F
j
= 894 Hz .
R
φ
XL - XC
so the phasor diagram is as shown:
φ = − cos −1
FG R IJ = − cos FG 1 IJ so
H ZK
H 2K
or
Z
−1
φ
∆Vin
or
φ
φ
∆Vout
R
At f0 , X L = X C so
FIG. P33.55(d)
∆Vout and ∆Vin have a phase difference of 0° .
∆Vout R 1
= = and X L − X C > 0 .
∆Vin
Z 2
Thus, φ = cos −1
(e)
FG 1 IJ = 60.0°
H 2K
∆Vout lags ∆Vin by 60.0° .
or
At 200 Hz and at 4 kHz,
d ∆V
P=
i = db1 2g∆V
R
in, rms
d ∆V
At f , P =
R
(f)
i = d ∆V
R
a10.0 Vf = 1.56 W
8a8.00 Ωf
a10.0 V f = 6.25 W .
=
2a8.00 Ωf
Hj
= 0.408 .
2
in, max
R
2
out, rms
0
i = b1 2g b1 2g∆V
2
2
out, rms
i = b1 2g ∆V
2
in, rms
R
in, max
R
a
fe
−4
ω 0 L 2π f0 L 2π 894 Hz 5.80 × 10
=
=
We take: Q =
8.00 Ω
R
R
2
∆Vout
∆Vin
Z
XL - XC
∆Vout leads ∆Vin by 60.0° .
At 4 000 Hz,
[2]
max
1
=
[1]
C = 54.6 µF and L = 580 µH .
Solving Equations (1) and (2) simultaneously gives
(b)
FIG. P33.55(a)
2
2
=
2
.
282
Alternating Current Circuits
Additional Problems
P33.56
af
The equation for ∆v t during the first period (using y = mx + b ) is:
a f 2b∆VT gt − ∆V
max
∆v t =
a ∆v f
a ∆v f
z ∆vatf dt = b∆VT g z LMNT2 t − 1OPQ dt
b∆V g FG T IJ 2t T − 1 = b∆V g a+1f − a−1f = b∆V g
=
H 2K 3
6
3
T
a∆vf = b∆V g = ∆V
1
=
T
2
ave
2
ave
2
max
2 T
0
2
FIG. P33.56
0
3 t =T
2
max
2
3
3
max
2
t=0
2
max
=
LC
2
max
3
ave
1
ω0 =
T
max
∆Vrms =
P33.57
max
3
1
b0.050 0 Hge5.00 × 10 Fj
−6
= 2 000 s −1
so the operating angular frequency of the circuit is
ω=
ω0
= 1 000 s −1 .
2
b∆V g Rω
R ω + L eω − ω j
a400f a8.00fb1 000g
P=
a8.00f b1 000g + b0.050 0g a1.00 − 4.00f × 10
Using Equation 33.37, P =
rms
2
2
2
*P33.58
2
2
bQ ≈ 12.5g
2 2
0
FIG. P33.57
2
2
2
2
2
2
6 2
= 56.7 W .
The angular frequency is ω = 2π 60 s = 377 s . When S is open, R, L, and C are in series with the
source:
b
R 2 + X L − XC
20 V I
g = FGH ∆VI IJK = FGH 0.183
J
AK
2
s
2
2
= 1.194 × 10 4 Ω 2 .
When S is in position 1, a parallel combination of two R’s presents equivalent resistance
with L and C:
FG R IJ + b X
H 2K
2
L
− XC
20 V I
g = FGH 0.298
J
AK
2
2
(1)
R
, in series
2
= 4.504 × 10 3 Ω 2 .
(2)
When S is in position 2, the current by passes the inductor. R and C are in series with the source:
R 2 + X C2 =
FG 20 V IJ
H 0.137 A K
2
= 2.131 × 10 4 Ω 2 .
Take equation (1) minus equation (2):
3 2
R = 7. 440 × 10 3 Ω 2
4
continued on next page
R = 99.6 Ω
(3)
Chapter 33
283
(only the positive root is physical.) Now equation (3) gives
a f
X C = 2.131 × 10 4 − 99.6
b
C = ωX C
g
−1
b
2 12
Ω = 106.7 Ω =
g
−1
= 377 s 106.7 Ω
1
(only the positive root is physical.)
ωC
= 2.49 × 10 −5 F = C .
Now equation (1) gives
a f
X L − X C = ± 1.194 × 10 4 − 99.6
2 12
Ω = ±44.99 Ω
X L = 106.7 Ω + 44.99 Ω = 61.74 Ω or 151.7 Ω = ω L
L=
P33.59
XL
ω
= 0.164 H or 0.402 H = L
The resistance of the circuit is R =
∆V 12.0 V
=
= 19.0 Ω .
I
0.630 A
The impedance of the circuit is Z =
∆Vrms
24.0 V
=
= 42.1 Ω .
0.570 A
I rms
Z 2 = R 2 + ω 2 L2
L=
*P33.60
1
ω
Z2 − R2 =
1
377
a42.1f − a19.0f
2
2
= 99.6 mH
The lowest-frequency standing-wave state is NAN. The distance between the clamps we represent
λ
T
as L = d NN = . The speed of transverse waves on the string is v = fλ =
= f 2L . The magnetic
µ
2
force on the wire oscillates at 60 Hz, so the wire will oscillate in resonance at 60 Hz.
b g
T
2
= 60 s 4L2
0.019 kg m
e
j
T = 274 kg ms 2 L2
Any values of T and L related according to this expression will work, including
if L = 0.200 m T = 10.9 N . We did not need to use the value of the current and magnetic field. If
we assume the subsection of wire in the field is 2 cm wide, we can find the rms value of the
magnetic force:
a fa
fb
g
FB = IAB sin θ = 9 A 0.02 m 0.015 3T sin 90° = 2.75 mN .
So a small force can produce an oscillation of noticeable amplitude if internal friction is small.
P33.61
(a)
When ω L is very large, the bottom branch carries negligible current. Also,
negligible compared to 200 Ω and
top branch.
(b)
Now
1
will be
ωC
45.0 V
= 225 mA flows in the power supply and the
200 Ω
1
→ ∞ and ω L → 0 so the generator and bottom branch carry 450 mA .
ωC
284
P33.62
Alternating Current Circuits
(a)
With both switches closed, the current goes only through generator
and resistor.
af
it =
(b)
(c)
(d)
P=
∆Vmax
cos ω t
R
b
1 ∆Vmax
2
R
af
g
2
∆Vmax
it =
2
2 2
R +ω L
1
, so
ω 0C
C=
(e)
At this resonance frequency,
(f)
U=
b g
1
C ∆VC
2
U max =
2
=
1 2 2
CI X C
2
b
b
1 2
1 ∆Vmax
= LI max
=
L
2
2
R2
U max
(h)
Now ω = 2ω 0 =
So φ = arctan
(i)
Now ω L =
(a)
I R, rms =
2
LC
0
0
1
.
ω 02 L
g
g
2
b ∆V g L
1
=
ω 02 C 2
max
2
2
2R
2
.
F ω L − b1 ω Cg I = arctanF 2
GH
GH R JK
1 1
2 ωC
F ω L − b1 ω Cg I .
GH R JK
Z= R .
1 2
1 ∆Vmax
CI max X C2 = C
2
2
R2
(g)
(b)
FG ω L IJ OP
H R KQ
0 = φ = arctan
For
We require ω 0 L =
P33.63
LM
N
cos ω t + arctan
FIG. P33.62
1
ω=
2LC
=
b g
LC− 12
R
LC
I=
JK
F 3 LI
GH 2 R C JK
arctan
.
ω0
.
2
∆Vrms 100 V
=
= 1.25 A
80.0 Ω
R
IR
φ
The total current will lag the applied voltage as seen in the phasor IL
diagram at the right.
∆Vrms
100 V
I L , rms =
=
= 1.33 A
XL
2π 60.0 s −1 0.200 H
e
ja
Thus, the phase angle is: φ = tan −1
∆V
FI
GH I
f
L , rms
R , rms
I
FIG. P33.63
I = tan FG 1.33 A IJ =
JK
H 1.25 A K
−1
46.7° .
Chapter 33
P33.64
285
Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day, or
$36 per 30 days at 10¢ per kWh). Suppose the transmission line is at 20 kV. Then
20 000 500 W
P
I rms =
~ 10 3 A .
=
20 000 V
∆Vrms
b
f
ga
If the transmission line had been at 200 kV, the current would be only ~ 10 2 A .
P33.65
R = 200 Ω , L = 663 mH , C = 26.5 µF , ω = 377 s −1 , ∆Vmax = 50.0 V
ω L = 250 Ω ,
(a)
F 1 I = 100 Ω , Z =
GH ω C JK
g
2
= 250 Ω
∆Vmax 50.0 V
=
= 0. 200 A
250 Ω
Z
X − XC
φ = tan −1 L
= 36.8° ( ∆V leads I)
R
I max =
FG
H
P33.66
b
R 2 + X L − XC
IJ
K
(b)
∆VR, max = I max R = 40.0 V at φ = 0°
(c)
∆VC , max =
(d)
∆VL , max = I maxω L = 50.0 V at φ = +90.0° ( ∆V leads I)
I max
= 20.0 V
ωC
at φ = −90.0° (I leads ∆V )
af b
L = 2.00 H , C = 10.0 × 10 −6 F , R = 10.0 Ω , ∆v t = 100 sin ω t
(a)
g
The resonant frequency ω 0 produces the maximum current and thus the maximum power
delivery to the resistor.
1
1
ω0 =
=
= 224 rad s
LC
2.00 10.0 × 10 −6
a fe
j
(b)
b∆V g = a100f =
P=
2R
2a10.0f
(c)
I rms =
max
2
2
∆Vrms
=
Z
2
I rms
R=
1 2
I rms
2
e j
500 W
∆Vrms
b
d
R2 + ω L − 1 ω C
max
R
gi
2
and
or
bI g
rms max
b ∆V g
rms
2
Z
=
∆Vrms
R
2
R=
b
1 ∆Vrms
2
R2
g
2
R.
F
1 I
This occurs where Z = 2 R :
R + Gω L −
H ω C JK = 2R
ω L C − 2 Lω C − R ω C + 1 = 0
or
L C ω − e 2LC + R C jω + 1 = 0
LMa2.00f e10.0 × 10 j OPω − LM2a2.00fe10.0 × 10 j + a10.0f e10.0 × 10 j OPω + 1 = 0 .
N
Q N
Q
2
2
4 2
2
2
2
2
−6 2
2
2
2
2
4
2
−6
2
2
2
4
2
−6 2
2
2
2
Solving this quadratic equation, we find that
ω 2 = 51 130 , or 48 894
ω 1 = 48 894 = 221 rad s
ω 2 = 51 130 = 226 rad s .
and
286
P33.67
Alternating Current Circuits
(a)
From Equation 33.41,
N 1 ∆V1
=
.
N 2 ∆V2
Let input impedance
Z1 =
so that
N 1 Z1 I 1
=
.
N 2 Z2 I 2
∆V1
I1
Z2 =
But from Eq. 33.42,
I 1 ∆V2 N 2
=
=
.
I 2 ∆V1 N 1
N1
=
N2
So, combining with the previous result we have
P33.68
2
P = I rms
R=
FG ∆V IJ
H Z K
rms
Z1
Z2
a120 V f a40.0 Ωf : Z =
Z
2
2
R , so 250 W =
2
a120f a40.0f
a40.0f + 2π f a0.185f − 1 2π f e65.0 × 10 j
F
GH
R2 + ω L −
2
250 =
1=
2
2
−6
2 304 f 2
1 600 f 2 + 1.351 1 f 4 − 5 692.3 f 2 + 5 995 300
2
f =
6 396.3 ±
.
8 000
= 31.6
8.00
N1
Z1
=
=
N2
Z2
(b)
∆V2
I2
and the output impedance
1
ωC
I
JK
2
576 000 f 2
and
250 =
so
1.351 1 f 4 − 6 396.3 f 2 + 5 995 300 = 0
e
1 600 f 2 + 1.162 4 f 2 − 2 448.5
b6 396.3g − 4b1.351 1gb5 995 300g = 3 446.5 or 1 287.4
2b1.351 1g
2
f = 58.7 Hz or 35.9 Hz
P33.69
IR =
∆Vrms
;
R
IL =
∆Vrms
;
ωL
b
g
2
OPF 1 I
QGH ∆V R JK
(b)
tan φ =
IC − IL
1
1
= ∆Vrms
−
IR
XC X L
C
OP
Q
−1
LM
N
I rms =
1
−
XL
bω Cg
= ∆Vrms
(a)
L1
tan φ = R M
NX
∆Vrms
FG 1 IJ + FGω C − 1 IJ
H R K H ω LK
I R2
+ IC − IL
IC =
2
2
rms
FIG. P33.69
j
2
Chapter 33
P33.70
(a)
F
GH
1
1
+ ωC−
2
ω
L
R
I rms = ∆Vrms
b
∆Vrms → ∆Vrms
1
f=
(b)
2π LC
=
g
max
I
JK
2
when ω C =
1
ωL
1
2π 200 × 10
−3
287
e
H 0.150 × 10 −6 F
j
= 919 Hz
∆Vrms 120 V
=
= 1.50 A
80.0 Ω
R
∆Vrms
120 V
=
= 1.60 A
IL =
−
ωL
374 s 1 0.200 H
IR =
ja
e
b g a
FIG. P33.70
f
fe
je
j
I
= I + b I − I g = a1.50f + b0.006 73 − 1.60g =
L I − I OP = tan L 0.006 73 − 1.60 O = −46.7°
φ = tan M
MN 1.50 PQ
N I Q
I C = ∆Vrms ω C = 120 V 374 s −1 0.150 × 10 −6 F = 6.73 mA
(c)
(d)
2
R
rms
−1
C
C
2
L
2
2
2.19 A
−1
L
R
The current is lagging the voltage .
P33.71
(a)
X L = X C = 1 884 Ω when f = 2 000 Hz
1 884 Ω
XL
=
= 0.150 H and
2π f 4 000π rad s
1
1
=
= 42.2 nF
C=
2π f X C
4 000π rad s 1 884 Ω
L=
(b)
XL
b g b
= 2π f a0.150 H f
Z=
a40.0 Ωf + b X
2
gb
XC =
L
− XC
Impedence, Ω
FIG. P33.71(b)
g
2
g
1
b2π f ge4.22 × 10 Fj
−8
f (Hz) X L ( Ω)
283
300
X C ( Ω) Z ( Ω)
12 600 1 2300
600
565
6 280
5 720
800
754
4 710
3 960
1 000
942
3 770
2 830
1 500
1 410
2 510
1 100
2 000
1 880
1 880
40
3 000
2 830
1 260
1 570
4 000
3 770
942
2 830
6 000
5 650
628
5 020
10 000
9 420
377
9 040
288
P33.72
Alternating Current Circuits
1
ω0 =
ω
ωL Ω
ω0
999.0
0.9990
999.1
0.9991
999.3
0.9993
999.5
0.9995
999.7
0.9997
999.9
0.9999
1.0000 1000
1.0001 1000.1
1.0003 1000.3
1.0005 1000.5
1.0007 1000.7
1.0009 1000.9
1.0010 1001
a f
= 1.00 × 10 6 rad s
LC
For each angular frequency, we find
b
Z = R2 + ω L − 1 ω C
then I =
1.00 V
Z
2
a
g
2
f
P = I 1.00 Ω .
and
The full width at half maximum is:
b
g
1.000 5 − 0.999 5 ω 0
∆ω
∆f =
=
2π
2π
3 −1
1.00 × 10 s
∆f =
= 159 Hz
2π
while
R
2π L
=
e
1.00 Ω
2π 1.00 × 10 −3 H
j
a f Z aΩf
a f
1
Ω
ωC
1001.0
2.24
0.19984
1000.9
2.06
0.23569
1000.7
1.72
0.33768
1000.5
1.41
0.49987
1000.3
1.17
0.73524
1000.1
1.02
0.96153
1000.0
1.00
1.00000
999.9
1.02
0.96154
999.7
1.17
0.73535
999.5
1.41
0.50012
999.3
1.72
0.33799
999.1
2.06
0.23601
999.0
2.24
0.20016
P = I 2R W
= 159 Hz .
1.0
0.8
I 2R
(W)
0.6
0.4
0.2
0.0
0.996
0.998
1
ω/ω 0
1.002
1.004
FIG. P33.72
P33.73
∆Vout
=
∆Vin
(a)
R
2
b
R + 1 ωC
g
2
=
R
2
b
R + 1 2π f C
g
C
2
∆Vout 1
1
= when
=R 3 .
2
V
ωC
∆ in
Hence, f =
ω
1
=
= 1.84 kHz .
2π 2π RC 3
continued on next page
∆Vin
R
FIG. P33.73
∆Vout
Chapter 33
(b)
289
Log Gain versus Log Frequency
0
–1
Log ∆ V out/∆ V in
–2
–3
–4
0
1
2
3
4
5
6
Log f
FIG. P33.73(b)
ANSWERS TO EVEN PROBLEMS
P33.2
(a) 193 Ω ; (b) 144 Ω
P33.32
353 W
P33.4
(a) 25.3 rad/s; (b) 0.114 s
P33.34
(a) 5.43 A ; (b) 0.905 ; (c) 281 µF ; (d) 109 V
P33.6
1.25 A and 96.0 Ω for bulbs 1 and 2;
0.833 A and 144 Ω for bulb 3
P33.36
P33.8
7.03 H or more
P33.38
46.5 pF to 419 pF
P33.10
3.14 A
P33.40
P33.12
3.80 J
(a) 3.56 kHz; (b) 5.00 A ; (c) 22.4 ;
(d) 2.24 kV
P33.14
P33.16
(a) greater than 41.3 Hz ;
(b) less than 87.5 Ω
b
2C ∆Vrms
g
P33.18
–32.0 A
P33.20
2.79 kHz
P33.22
(a) 109 Ω ; (b) 0.367 A ; (c) I max = 0.367 A ,
ω = 100 rad s, φ = −0.896 rad
P33.24
19.3 mA
P33.26
(a) 146 V ; (b) 212 V ; (c) 179 V ; (d) 33.4 V
P33.28
XC = 3 R
P33.30
(a) 2.00 A ; (b) 160 W ; (c) see the solution
P33.42
b
11 ∆Vrms
g
2
14R
b
4π ∆Vrms
g RC
2
LC
2
4R C + 9L
P33.44
(a) 9.23 V ; (b) 4.55 A ; (c) 42.0 W
P33.46
(a) 1 600 turns ; (b) 30.0 A ; (c) 25.3 A
P33.48
(a) 83.3 ; (b) 54.0 mA ; (c) 185 kΩ
P33.50
(a) 0.34; (b) 5.3 W; (c) $4.8
P33.52
(a) see the solution; (b) 1; 0; (c)
P33.54
(a) 1.00; (b) 0.346
P33.56
see the solution
P33.58
R = 99.6 Ω, C = 24.9 µF , L = 164 mH or
402 mH
3
2π RC
290
P33.60
P33.62
Alternating Current Circuits
L = 0.200 m and T = 10.9 N , or any values
related by T = 274 kg ms 2 L2
e
j
a f ∆VR cos ω t ; (b) P = b∆V2R g ;
L
∆V
F ω L IJ OP ;
cos Mω t + tan G
(c) iat f =
H R KQ
N
R +ω L
b∆V g L ;
1
(d) C =
; (e) Z = R ; (f)
max
max
(a) i t =
(g)
(i)
b
ω 02 L
2
∆Vmax L
2
g
2R
1
2LC
~ 10 3 A
P33.66
(a) 224 rad s ; (b) 500 W ;
(c) 221 rad s and 226 rad s
P33.68
either 58.7 Hz or 35.9 Hz
P33.70
(a) 919 Hz ;
(b) I R = 1.50 A , I L = 1.60 A , I C = 6.73 mA ;
(c) 2.19 A ; (d) −46.7° ; current lagging
P33.72
see the solution
2
−1
max
2
P33.64
2 2
max
2
2
2R
; (h) tan −1
F 3 LI;
GH 2R C JK
34
Electromagnetic Waves
CHAPTER OUTLINE
34.1
34.2
34.3
34.4
34.5
34.6
Maxwell’s Equations and
Hertz’s Discoveries
Plane Electromagnetic
Waves
Energy Carried by
Electromagnetic Waves
Momentum and Radiation
Pressure
Production of
Electromagnetic Waves by
an Antenna
The Spectrum of
Electromagnetic Waves
ANSWERS TO QUESTIONS
Q34.1
Radio waves move at the speed of light. They can travel around
the curved surface of the Earth, bouncing between the ground
and the ionosphere, which has an altitude that is small when
compared to the radius of the Earth. The distance across the
lower forty-eight states is approximately 5 000 km, requiring a
5 × 10 6 m
~ 10 −2 s . To go halfway around the
transit time of
3 × 10 8 m s
Earth takes only 0.07 s. In other words, a speech can be heard
on the other side of the world before it is heard at the back of a
large room.
Q34.2
The Sun’s angular speed in our sky is our rate of rotation,
360°
= 15° h . In 8.3 minutes it moves west by
24 h
1h
θ = ω t = 15° h
8.3 min = 2.1° . This is about four
60 min
times the angular diameter of the Sun.
b
gFGH
IJ a
K
f
Q34.3
Energy moves. No matter moves. You could say that electric and magnetic fields move, but it is nicer
to say that the fields at one point stay at that point and oscillate. The fields vary in time, like sports
fans in the grandstand when the crowd does the wave. The fields constitute the medium for the
wave, and energy moves.
Q34.4
No. If a single wire carries DC current, it does not emit electromagnetic waves. In this case, there is a
constant magnetic field around the wire. Alternately, if the cable is a coaxial cable, it ideally does not
emit electromagnetic waves even while carrying AC current.
Q34.5
Acceleration of electric charge.
Q34.6
The changing magnetic field of the solenoid induces eddy currents in the conducting core. This is
accompanied by I 2 R conversion of electrically-transmitted energy into internal energy in the
conductor.
Q34.7
A wire connected to the terminals of a battery does not radiate electromagnetic waves. The battery
establishes an electric field, which produces current in the wire. The current in the wire creates a
magnetic field. Both fields are constant in time, so no electromagnetic induction or “magneto-electric
induction” happens. Neither field creates a new cycle of the other field. No wave propagation
occurs.
291
292
Q34.8
Q34.9
Electromagnetic Waves
No. Static electricity is just that: static. Without acceleration of the charge, there can be no
electromagnetic wave.
Sound
Light
The world of sound extends to the top of the
atmosphere and stops there; sound requires a
material medium. Sound propagates by a chain
reaction of density and pressure disturbances
recreating each other. Sound in air moves at
hundreds of meters per second. Audible sound
has frequencies over a range of three decades
(ten octaves) from 20 Hz to 20 kHz. Audible
sound has wavelengths of ordinary size (1.7 cm
to 17 m). Sound waves are longitudinal.
The universe of light fills the whole universe.
Light moves through materials, but faster in a
vacuum. Light propagates by a chain reaction of
electric and magnetic fields recreating each
other. Light in air moves at hundreds of millions
of meters per second. Visible light has
frequencies over a range of less than one octave,
from 430 to 750 Terahertz. Visible light has
wavelengths of very small size (400 nm to
700 nm). Light waves are transverse.
Sound and light can both be reflected, refracted, or absorbed to produce internal energy. Both have
amplitude and frequency set by the source, speed set by the medium, and wavelength set by both
source and medium. Sound and light both exhibit the Doppler effect, standing waves, beats,
interference, diffraction, and resonance. Both can be focused to make images. Both are described by
wave functions satisfying wave equations. Both carry energy. If the source is small, their intensities
both follow an inverse-square law. Both are waves.
Q34.10
The Poynting vector S describes the energy flow associated with an electromagnetic wave. The
direction of S is along the direction of propagation and the magnitude of S is the rate at which
electromagnetic energy crosses a unit surface area perpendicular to the direction of S.
Q34.11
Photons carry momentum. Recalling what we learned in Chapter 9, the impulse imparted to a
particle that bounces elastically is twice that imparted to an object that sticks to a massive wall.
Similarly, the impulse, and hence the pressure exerted by a photon reflecting from a surface must be
twice that exerted by a photon that is absorbed.
Q34.12
Different stations have transmitting antennas at different locations. For best reception align your
rabbit ears perpendicular to the straight-line path from your TV to the transmitting antenna. The
transmitted signals are also polarized. The polarization direction of the wave can be changed by
reflection from surfaces—including the atmosphere—and through Kerr rotation—a change in
polarization axis when passing through an organic substance. In your home, the plane of
polarization is determined by your surroundings, so antennas need to be adjusted to align with the
polarization of the wave.
Q34.13
You become part of the receiving antenna! You are a big sack of salt water. Your contribution usually
increases the gain of the antenna by a few tenths of a dB, enough to noticeably improve reception.
Q34.14
On the TV set, each side of the dipole antenna is a 1/4 of the wavelength of the VHF radio wave. The
electric field of the wave moves free charges in the antenna in electrical resonance, giving maximum
current in the center of the antenna, where the cable connects it to the receiver.
Q34.15
The loop antenna is essentially a solenoid. As the UHF radio wave varies the magnetic field inside
the loop, an AC emf is induced in the loop as described by Faraday’s and Lenz’s laws. This signal is
then carried down a cable to the UHF receiving circuit in the TV. An excellent reference for antennas
and all things radio is the ARRL Handbook.
Chapter 34
293
Q34.16
The voltage induced in the loop antenna is proportional to the rate of change of the magnetic field in
the wave. A wave of higher frequency induces a larger emf in direct proportion. The instantaneous
voltage between the ends of a dipole antenna is the distance between the ends multiplied by the
electric field of the wave. It does not depend on the frequency of the wave.
Q34.17
The radiation resistance of a broadcast antenna is the equivalent resistance that would take the same
power that the antenna radiates, and convert it into internal energy.
Q34.18
Consider a typical metal rod antenna for a car radio. The rod detects the electric field portion of the
carrier wave. Variations in the amplitude of the incoming radio wave cause the electrons in the rod
to vibrate with amplitudes emulating those of the carrier wave. Likewise, for frequency modulation,
the variations of the frequency of the carrier wave cause constant-amplitude vibrations of the
electrons in the rod but at frequencies that imitate those of the carrier.
Q34.19
The frequency of EM waves in a microwave oven, typically 2.45 GHz, is chosen to be in a band of
frequencies absorbed by water molecules. The plastic and the glass contain no water molecules.
Plastic and glass have very different absorption frequencies from water, so they may not absorb any
significant microwave energy and remain cool to the touch.
Q34.20
People of all the world’s races have skin the same color in the infrared. When you blush or exercise
or get excited, you stand out like a beacon in an infrared group picture. The brightest portions of
your face show where you radiate the most. Your nostrils and the openings of your ear canals are
bright; brighter still are just the pupils of your eyes.
Q34.21
Light bulbs and the toaster shine brightly in the infrared. Somewhat fainter are the back of the
refrigerator and the back of the television set, while the TV screen is dark. The pipes under the sink
show the same weak sheen as the walls until you turn on the faucets. Then the pipe on the right
turns very black while that on the left develops a rich glow that quickly runs up along its length. The
food on your plate shines; so does human skin, the same color for all races. Clothing is dark as a rule,
but your bottom glows like a monkey’s rump when you get up from a chair, and you leave behind a
patch of the same blush on the chair seat. Your face shows you are lit from within, like a jack-olantern: your nostrils and the openings of your ear canals are bright; brighter still are just the pupils
of your eyes.
Q34.22
Welding produces ultraviolet light, along with high intensity visible and infrared.
Q34.23
12.2 cm waves have a frequency of 2.46 GHz. If the Q value of the phone is low (namely if it is
cheap), and your microwave oven is not well shielded (namely, if it is also cheap), the phone can
likely pick up interference from the oven. If the phone is well constructed and has a high Q value,
then there should be no interference at all.
294
Electromagnetic Waves
SOLUTIONS TO PROBLEMS
Section 34.1
*P34.1
(a)
Maxwell’s Equations and Hertz’s Discoveries
The rod creates the same electric field that it would if
stationary. We apply Gauss’s law to a cylinder of radius
r = 20 cm and length :
q
E ⋅ dA = inside
∈0
λl
E 2π rl cos 0° =
∈0
z
b g
E=
(b)
e
FIG. P34.1
2
35 × 10 C m N ⋅ m
λ
j = 3.15 × 10 3 j N C .
radially outward =
2π ∈0 r
2π 8.85 × 10 −12 C 2 0.2 m
e
ja
f
e
je
j
The charge in motion constitutes a current of 35 × 10 −9 C m 15 × 10 6 m s = 0.525 A . This
current creates a magnetic field.
B=
(c)
j
−9
µ0I
2π r
=
e4π × 10
ja
2π a0. 2 mf
−7
T ⋅ m A 0.525 A
fk=
5.25 × 10 −7 k T
The Lorentz force on the electron is F = qE + qv × B
e
je
j e
e− jj N + 2.02 × 10 e+ jj N =
j FGH
je
F = −1.6 × 10 −19 C 3.15 × 10 3 j N C + −1.6 × 10 −19 C 240 × 10 6 i m s × 5.25 × 10 −7 k
F = 5.04 × 10 −16
Section 34.2
P34.2
(a)
−17
e j
4.83 × 10 −16 − j N
Plane Electromagnetic Waves
Since the light from this star travels at 3.00 × 10 8 m s
the last bit of light will hit the Earth in
6.44 × 10 18 m
8
3.00 × 10 m s
= 2.15 × 10 10 s = 680 years .
Therefore, it will disappear from the sky in the year 2 004 + 680 = 2.68 × 10 3 C.E. .
The star is 680 light-years away.
∆x 1. 496 × 10 11 m
=
= 499 s = 8.31 min
v
3 × 10 8 m s
(b)
∆t =
(c)
8
∆x 2 3.84 × 10 m
=
= 2.56 s
∆t =
v
3 × 10 8 m s
(d)
6
∆x 2π 6.37 × 10 m
∆t =
=
= 0.133 s
v
3 × 10 8 m s
(e)
∆t =
e
j
e
j
∆x 10 × 10 3 m
=
= 3.33 × 10 −5 s
v
3 × 10 8 m s
N⋅s
C ⋅m
IJ
K
Chapter 34
1
P34.3
v=
P34.4
E
=c
B
or
1
=
κµ 0 ∈0
1.78
c = 0.750 c = 2.25 × 10 8 m s
220
= 3.00 × 10 8
B
so B = 7.33 × 10 −7 T = 733 nT .
P34.5
fλ = c
(a)
a
f
f 50.0 m = 3.00 × 10 8 m s
or
f = 6.00 × 10 6 Hz = 6.00 MHz .
so
E
=c
B
(b)
or
22.0
= 3.00 × 10 8
Bmax
so
Bmax = −73.3k nT .
k=
(c)
2π
λ
=
2π
= 0.126 m −1
50.0
e
j
ω = 2π f = 2π 6.00 × 10 6 s −1 = 3.77 × 10 7 rad s
and
b
g
e
j
B = Bmax cos kx − ω t = −73.3 cos 0.126 x − 3.77 × 10 7 t k nT .
P34.6
ω = 2π f = 6.00π × 10 9 s −1 = 1.88 × 10 10 s −1
k=
2π
λ
=
ω 6.00π × 10 9 s −1
=
= 20.0π = 62.8 m −1
c 3.00 × 10 8 m s
b
g e
E = 300 V m cos 62.8 x − 1.88 × 10 10 t
P34.7
j
Bmax =
B=
100 V m
E
=
= 3.33 × 10 −7 T = 0.333 µT
c 3.00 × 10 8 m s
(b)
λ=
2π
2π
=
= 0.628 µm
k
1.00 × 10 7 m −1
(c)
f=
λ
=
3.00 × 10 8 m s
6. 28 × 10
−7
m
= 4.77 × 10 14 Hz
g e
B = 1.00 µT cos 62.8 x − 1.88 × 10 10 t
(a)
c
b
300 V m
E
=
= 1.00 µT
c 3.00 × 10 8 m s
j
295
296
P34.8
Electromagnetic Waves
b
E = Emax cos kx − ω t
g
ga f
ga f
ge j
ga f
∂E
= − Emax sin kx − ω t k
∂x
∂E
= − Emax sin kx − ω t −ω
∂t
∂2E
= − Emax cos kx − ω t k 2
∂x 2
∂2E
= − Emax cos kx − ω t −ω
∂t 2
b
b
b
b
∂E
We must show:
∂x 2
= µ 0 ∈0
∂2E
∂t 2
.
b
e j
2
a f
g
2
b
g
− k 2 Emax cos kx − ω t = − µ 0 ∈0 −ω Emax cos kx − ω t .
That is,
k2
But this is true, because
ω
2
F1I
=G J
H fλ K
2
=
1
c2
= µ 0 ∈0 .
The proof for the wave of magnetic field follows precisely the same steps.
P34.9
In the fundamental mode, there is a single loop in the standing wave between the plates. Therefore,
the distance between the plates is equal to half a wavelength.
a
f
λ = 2L = 2 2.00 m = 4.00 m
f=
Thus,
P34.10
c
λ
d A to A = 6 cm ± 5% =
λ = 12 cm ± 5%
a
=
3.00 × 10 8 m s
= 7.50 × 10 7 Hz = 75.0 MHz .
4.00 m
λ
2
fe
j
v = λ f = 0.12 m ± 5% 2.45 × 10 9 s −1 = 2.9 × 10 8 m s ± 5%
Section 34.3
Energy Carried by Electromagnetic Waves
P34.11
S=I=
P34.12
Sav =
1 000 W m 2
Energy
I
=u= =
= 3.33 µJ m3
c 3.00 × 10 8 m s
Unit Volume
U Uc
=
= uc
At V
P
4.00 × 10 3 W
=
2
4π r
4π 4.00 × 1 609 m
b
g
2
= 7.68 µW m 2
Emax = 2 µ 0 cSav = 0.076 1 V m
b
ga
f
b
g
∆Vmax = Emax L = 76.1 mV m 0.650 m = 49.5 mV amplitude
or 35.0 mV (rms)
Chapter 34
P34.13
a
3
P
250 × 10 W
=
2
4π r
4π 8.04 × 10 3 W
I=
e
100 W
a
P34.16
f
4π 1.00 m
2
j
2
= 307 µW m 2
= 7.96 W m 2
I
= 2.65 × 10 −8 J m3 = 26.5 nJ m 3
c
u=
P34.15
g
r = 5.00 mi 1 609 m mi = 8.04 × 10 3 m
S=
P34.14
fb
297
(a)
uE =
1
u = 13.3 nJ m3
2
(b)
uB =
1
u = 13.3 nJ m3
2
(c)
I = 7.96 W m 2
Power output = (power input)(efficiency).
Thus,
Power input =
and
A=
I=
Power out 1.00 × 10 6 W
=
= 3.33 × 10 6 W
eff
0.300
P
3.33 × 10 6 W
=
= 3.33 × 10 3 m 2 .
I 1.00 × 10 3 W m 2
2
Bmax
c
P
=
2µ 0
4π r 2
Bmax =
F P
GH 4π r
I FG 2 µ IJ = e10.0 × 10 ja2fe4π × 10 j =
JK H c K 4π 5.00 × 10 3.00 × 10
e
je
j
−7
3
2
0
3 2
8
5.16 × 10 −10 T
Since the magnetic field of the Earth is approximately 5 × 10 −5 T , the Earth’s field is some 100 000
times stronger.
P34.17
(a)
P = I 2 R = 150 W
jb
e
g
A = 2π rL = 2π 0.900 × 10 −3 m 0.080 0 m = 4.52 × 10 −4 m 2
(b)
S=
P
= 332 kW m 2
A
B=
µ 0 1.00
µ0I
=
= 222 µT
2π r 2π 0.900 × 10 −3
E=
∆V IR
150 V
=
=
= 1.88 kV m
∆x
0.080
0m
L
e
Note:
(points radially inward)
a f
S=
EB
µ0
j
= 332 kW m 2
298
*P34.18
Electromagnetic Waves
e
j
2
2
3 × 10 6 V m
Emax
I=
=
2 µ 0 c 2 4π × 10 −7 T ⋅ m A 3 × 10 8 m s
(a)
e
je
FG J IJ FG C IJ FG T ⋅ C ⋅ m IJ FG N ⋅ m IJ
j H V ⋅CK H A ⋅sKH N ⋅s KH J K
2
I = 1.19 × 10 10 W m 2
P34.19
e
10
F 5 × 10
W m jπ G
H 2
−3
2
m
I
JK
2
= 2.34 × 10 5 W
(b)
P = IA = 1.19 × 10
(a)
E ⋅ B = 80.0 i + 32.0 j − 64.0k N C ⋅ 0.200 i + 0.080 0 j + 0.290 k µT
e
jb g e
E ⋅ B = a16.0 + 2.56 − 18.56f N ⋅ s C ⋅ m =
2
S=
(b)
1
µ0
E×B=
2
j
0
e80.0 i + 32.0 j − 64.0kj N C × e0.200 i + 0.080 0 j + 0.290kj µT
4π × 10 −7 T ⋅ m A
e6.40k − 23.2 j − 6.40k + 9.28i − 12.8 j + 5.12 ij × 10
S=
−6
W m2
4π × 10 −7
e11.5i − 28.6 jj W m
S=
*P34.20
2
= 30.9 W m 2 at −68.2° from the +x axis.
The energy put into the water in each container by electromagnetic radiation can be written as
eP ∆t = eIA∆t where e is the percentage absorption efficiency. This energy has the same effect as heat
in raising the temperature of the water:
eIA∆t = mc∆T = ρVc∆T
∆T =
eI 2 ∆t eI∆t
=
ρ c
ρ 3c
where is the edge dimension of the container and c the specific heat of water. For the small
container,
0.7 25 × 10 3 W m 2 480 s
∆T =
= 33.4° C .
10 3 kg m3 0.06 m 4 186 J kg ⋅° C
e
e
ja
For the larger,
∆T =
P34.21
j
f
e
j
0.91 25 J s ⋅ m 2 480 s
e0.12 m j4 186 J ° C
2
= 21.7° C .
We call the current I rms and the intensity I. The power radiated at this frequency is
b
gb
g
P = 0.010 0 ∆Vrms I rms =
b
0.010 0 ∆Vrms
g
2
R
= 1.31 W .
If it is isotropic, the intensity one meter away is
I=
P
1.31 W
=
A 4π 1.00 m
Bmax =
a
2µ 0 I
=
c
f
2
= 0.104 W m 2 = Sav =
e
je
c
2
Bmax
2µ 0
2 4π × 10 −7 T ⋅ m A 0.104 W m 2
8
3.00 × 10 m s
j=
29.5 nT
Chapter 34
P34.22
(a)
efficiency =
(b)
Sav =
FG
H
299
IJ
K
useful power output
700 W
× 100% =
× 100% = 50.0%
total power input
1 400 W
P
700 W
=
= 2.69 × 10 5 W m 2
A
0.068 3 m 0.038 1 m
b
gb
g
Sav = 269 kW m 2 toward the oven chamber
(c)
Sav =
2
Emax
2µ 0 c
e
je
je
j
Emax = 2 4π × 10 −7 T ⋅ m A 3.00 × 10 8 m s 2.69 × 10 5 W m 2 = 1.42 × 10 4 V m
= 14.2 kV m
P34.23
Emax
:
c
7.00 × 10 5 N C
(a)
Bmax =
(b)
E2
I = max :
2µ 0 c
(c)
I=
(a)
e10.0 × 10 j W
I=
π e0.800 × 10 mj
(b)
uav =
(a)
E = cB = 3.00 × 10 8 m s 1.80 × 10 −6 T = 540 V m
Bmax =
3.00 × 10 8 m s
= 2.33 mT
e7.00 × 10 j
I=
= 650 MW m
2e 4π × 10 je3.00 × 10 j
Lπ
O
P = IA = e6.50 × 10 W m jM e1.00 × 10 mj P =
N4
Q
5 2
2
−7
P
:
A
8
8
2
−3
2
510 W
−3
P34.24
P34.25
−3
2
= 4.97 kW m 2
I 4.97 × 10 3 J m 2 ⋅ s
=
= 16.6 µJ m3
c
3.00 × 10 8 m s
e
B2
je
e1.80 × 10 j
=
j
−6 2
= 2.58 µJ m3
(b)
uav =
(c)
Sav = cuav = 3.00 × 10 8 2.58 × 10 −6 = 773 W m 2
(d)
This is 77.3% of the intensity in Example 34.5 . It may be cloudy, or the Sun may be setting.
µ0
4π × 10
e
−7
je
j
300
Electromagnetic Waves
Section 34.4
P34.26
P34.27
P34.28
Momentum and Radiation Pressure
2Sav
c
The pressure P upon the mirror is
P=
where A is the cross-sectional area of the beam and
Sav =
The force on the mirror is then
F = PA =
Therefore,
F=
For complete absorption, P =
(a)
P
.
A
e
FG IJ
H K
2 P
2P
A=
.
c A
c
2 100 × 10 −3
e
3 × 10
8
j
j=
6.67 × 10 −10 N .
25.0
S
=
= 83.3 nPa .
c 3.00 × 10 8
e
2 1 340 W m 2
The radiation pressure is
8
3.00 × 10 m s
j = 8.93 × 10
2
−6
N m2 .
Multiplying by the total area, A = 6.00 × 10 5 m 2 gives: F = 5.36 N .
(b)
The acceleration is:
a=
5.36 N
F
=
= 8.93 × 10 −4 m s 2 .
m 6 000 kg
(c)
It will arrive at time t where
d=
1 2
at
2
t=
or
P34.29
I=
(a)
(b)
(c)
2d
=
a
2
Emax
P
=
π r 2 2µ 0 c
Emax =
b
P 2µ 0 c
πr
2
15 × 10 −3 J s
3.00 × 10 8 m s
p=
g=
1.90 kN C
a1.00 mf =
50.0 pJ
U 5 × 10 −11
=
= 1.67 × 10 −19 kg ⋅ m s
c 3.00 × 10 8
e
j
2 3.84 × 10 8 m
e
8.93 × 10
−4
m s2
j
= 9.27 × 10 5 s = 10.7 days .
Chapter 34
P34.30
(a)
If PS is the total power radiated by the Sun, and rE and rM are the radii of the orbits of the
planets Earth and Mars, then the intensities of the solar radiation at these planets are:
IE =
IM =
and
PS
4π rE2
PS
.
2
4π rM
FG r IJ = e1 340 W m jF 1.496 × 10
GH 2.28 × 10
Hr K
2
IM = IE
Thus,
(b)
2
E
11
11
M
I
J
mK
m
Mars intercepts the power falling on its circular face:
e
jLNM e
j e
If Mars behaves as a perfect absorber, it feels pressure P =
and force
(d)
F = PA =
2
= 577 W m 2 .
j OQP =
2
PM = I M π R M
= 577 W m 2 π 3.37 × 10 6 m
(c)
301
2
2.06 × 10 16 W .
SM I M
=
c
c
P
IM
2.06 × 10 16 W
2
π RM
= M =
= 6.87 × 10 7 N .
8
c
c
3.00 × 10 m s
e
j
The attractive gravitational force exerted on Mars by the Sun is
Fg =
GM S M M
rM2
e6.67 × 10
=
−11
je
je
N ⋅ m 2 kg 2 1.991 × 10 30 kg 6.42 × 10 23 kg
e2.28 × 10 mj
11
2
j = 1.64 × 10
21
N
which is ~ 10 13 times stronger than the repulsive force of part (c).
P34.31
(a)
The total energy absorbed by the surface is
U=
(b)
FG 1 IIJ At = LM 1 e750 W m jOPe0.500 × 1.00 m ja60.0 sf =
H 2 K N2
Q
2
2
11.3 kJ .
The total energy incident on the surface in this time is 2U = 22.5 kJ , with U = 11.3 kJ being
absorbed and U = 11.3 kJ being reflected. The total momentum transferred to the surface is
b
g a
F U I F 2U IJ = 3U = 3e11.3 × 10 Jj =
p=G J +G
H c K H c K c 3.00 × 10 m s
p = momentum from absorption + momentum from reflection
3
8
1.13 × 10 −4 kg ⋅ m s
f
302
* P34.32
Electromagnetic Waves
The radiation pressure on the disk is P =
Then F =
af
af
H x 0 + H y 0 − mgr sin θ +
∑τ = 0
mg
π r 2 Ir
=0
c
2
π 0.4 m 10 7 W s 2 s
a
f
g a
b
Hy
PA
fe
r
F 1 kg m I
j GH 1 W s JK
π r 2I
= sin −1
mgc
0.024 kg m 2 9.8 m 3 × 10 8 m
−1
Hx
π r 2I
.
c
Take torques about the hinge:
θ = sin
S I F
F
= = =
.
c c A π r2
2
3
θ
= sin −1 0.071 2 = 4.09°
FIG. P34.32
Section 34.5
P34.33
P34.34
Production of Electromagnetic Waves by an Antenna
λ=
c
= 536 m
f
so
h=
λ
= 134 m
4
λ=
c
= 188 m
f
so
h=
λ
= 46.9 m
4
P=
a∆V f
R
2
a f
or P ∝ ∆V
2
af
∆y
∆V = − Ey ⋅ ∆y = E y ⋅ cos θ
θ
receiving
antenna
∆V ∝ cos θ so P ∝ cos 2 θ
a
f
a
f
a
f
(a)
θ = 15.0° : P = Pmax cos 2 15.0° = 0.933 Pmax = 93.3%
(b)
θ = 45.0° : P = Pmax cos 2 45.0° = 0.500Pmax = 50.0%
(c)
θ = 90.0° : P = Pmax cos 2 90.0° = 0
FIG. P34.34
Chapter 34
P34.35
(a)
303
Constructive interference occurs when
d cos θ = nλ for some integer n.
F I
GH JK
λ
λ
=n
= 2n
d
λ 2
n = 0 , ± 1, ± 2 , …
cos θ = n
∴ strong signal @ θ = cos −1 0 = 90° , 270°
(b)
Destructive interference occurs when
d cos θ =
FG 2n + 1 IJ λ :
H 2 K
cos θ = 2n + 1
a f
∴ weak signal @ θ = cos −1 ±1 = 0° , 180°
FIG. P34.35
P34.36
For the proton,
∑ F = ma
mv 2
.
R
2πR 2π m
=
T=
.
v
qB
qvB sin 90.0° =
yields
The period of the proton’s circular motion is therefore:
*P34.37
The frequency of the proton’s motion is
f=
1
.
T
The charge will radiate electromagnetic waves at this frequency, with
λ=
2π mc
c
= cT =
.
f
qB
(a)
b
g
1
µ 0 J max cos kx − ω t k applies for x > 0 , since it describes a wave
2
1
moving in the i direction. The electric field direction must satisfy S =
E × B as i = j × k so
The magnetic field B =
µ0
the direction of the electric field is j when the cosine is positive. For its magnitude we have
1
E = cB , so altogether we have E = µ 0 cJ max cos kx − ω t j .
2
b
(b)
S=
1
E×B=
b
g
g
1 1 2 2
µ 0 cJ max cos 2 kx − ω t i
µ0 4
µ0
1
2
S = µ 0 cJ max
cos 2 kx − ω t i
4
b
g
(c)
The intensity is the magnitude of the Poynting vector averaged over one or more cycles. The
1
1
2
average of the cosine-squared function is , so I = µ 0 cJ max
.
2
8
(d)
J max =
8I
=
µ0c
e
8 570 W m 2
4π × 10
−7
j
bTm Ag3 × 10
8
ms
= 3.48 A m
304
Electromagnetic Waves
Section 34.6
P34.38
P34.39
P34.40
The Spectrum of Electromagnetic Waves
From the electromagnetic spectrum chart and accompanying text discussion, the following
identifications are made:
Frequency, f
Wavelength, λ =
2 Hz = 2 × 10 0 Hz
2 kHz = 2 × 10 3 Hz
2 MHz = 2 × 10 6 Hz
2 GHz = 2 × 10 9 Hz
2 THz = 2 × 10 12 Hz
2 PHz = 2 × 10 15 Hz
2 EHz = 2 × 10 18 Hz
2 ZHz = 2 × 10 21 Hz
2 YHz = 2 × 10 24 Hz
150 Mm
150 km
150 m
15 cm
150 µm
150 nm
150 pm
150 fm
150 am
Wavelength, λ
Frequency, f =
2 km = 2 × 10 3 m
2 m = 2 × 10 0 m
2 mm = 2 × 10 −3 m
2 µm = 2 × 10 −6 m
2 nm = 2 × 10 −9 m
2 pm = 2 × 10 −12 m
2 fm = 2 × 10 −15 m
2 am = 2 × 10 −18 m
1.5 × 10 5 Hz
1.5 × 10 8 Hz
1.5 × 10 11 Hz
1.5 × 10 14 Hz
1.5 × 10 17 Hz
1.5 × 10 20 Hz
1.5 × 10 23 Hz
1.5 × 10 26 Hz
f=
(a)
(b)
c
λ
=
3.00 × 10 8 m s
5.50 × 10 −7 m
f=
λ
=
Classification
Radio
Radio
Radio
Microwave
Infrared
Ultraviolet
X-ray
Gamma ray
Gamma ray
c
λ
Classification
Radio
Radio
Microwave
Infrared
Ultraviolet/X-ray
X-ray/Gamma ray
Gamma ray
Gamma ray
= 5.45 × 10 14 Hz
3 × 10 8 m s
~ 10 8 Hz
1.7 m
radio wave
1 000 pages, 500 sheets, is about 3 cm thick so one sheet is about 6 × 10 −5 m thick.
f=
P34.41
c
c
f
3.00 × 10 8 m s
6 × 10 −5 m
~ 10 13 Hz
infrared
(a)
fλ = c
gives
e5.00 × 10
19
(b)
fλ = c
gives
e4.00 × 10
9
j
λ = 6.00 × 10 −12 m = 6.00 pm
j
λ = 0.075 m = 7.50 cm
Hz λ = 3.00 × 10 8 m s :
Hz λ = 3.00 × 10 8 m s :
Chapter 34
P34.42
P34.43
(a)
λ=
c 3.00 × 10 8 m s
=
= 261 m
f 1 150 × 10 3 s −1
so
180 m
= 0.690 wavelengths
261 m
(b)
λ=
8
c 3.00 × 10 m s
=
= 3.06 m
f
98.1 × 10 6 s −1
so
180 m
= 58.9 wavelengths
3.06 m
Time to reach object =
b
g e
305
1
1
total time of flight = 4.00 × 10 −4 s = 2.00 × 10 −4 s .
2
2
e
je
j
j
Thus, d = vt = 3.00 × 10 8 m s 2.00 × 10 −4 s = 6.00 × 10 4 m = 60.0 km .
P34.44
The time for the radio signal to travel 100 km is:
∆t r =
The sound wave travels 3.00 m across the room in:
∆t s =
100 × 10 3 m
8
3.00 × 10 m s
= 3.33 × 10 −4 s .
3.00 m
= 8.75 × 10 −3 s .
343 m s
Therefore, listeners 100 km away will receive the news before the people in the news room by a
total time difference of
∆t = 8.75 × 10 −3 s − 3.33 × 10 −4 s = 8.41 × 10 −3 s .
P34.45
8
c 3.00 × 10 m s
=
= 4.00 × 10 6 m .
f
75.0 Hz
The wavelength of an ELF wave of frequency 75.0 Hz is
λ=
The length of a quarter-wavelength antenna would be
L = 1.00 × 10 6 m = 1.00 × 10 3 km
or
L = 1 000 km
b
.621 mi I
gFGH 01.00
J=
km K
Thus, while the project may be theoretically possible, it is not very practical.
P34.46
(a)
For the AM band,
λ max =
λ min =
(b)
For the FM band,
λ max =
λ min =
c
f min
c
fmax
c
f min
c
fmax
=
=
=
=
3.00 × 10 8 m s
540 × 10 3 Hz
3.00 × 10 8 m s
1 600 × 10 3 Hz
3.00 × 10 8 m s
88.0 × 10 6 Hz
3.00 × 10 8 m s
108 × 10 6 Hz
= 556 m
= 187 m .
= 3.41 m
= 2.78 m .
621 mi .
306
Electromagnetic Waves
Additional Problems
P34.47
(a)
P = SA :
(b)
cB 2
S = max
2µ 0
S=
P34.48
jLMN e
e
P = 1 340 W m 2 4π 1.496 × 10 11 m
2
Emax
2µ 0 c
2 µ 0S
=
c
e
j OPQ =
2
3.77 × 10 26 W
je
2 4π × 10 −7 N A 2 1 340 W m 2
j=
3.35 µT
so
Bmax =
so
Emax = 2 µ 0 cS = 2 4π × 10 −7 3.00 × 10 8 1 340 = 1.01 kV m
8
3.00 × 10 m s
e
jb
je
g
Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sun
is 60°. Then the target area you fill in the Sun’s field of view is
a1.7 mfa0.3 mf cos 30° = 0.4 m .
U = IAt = e1 340 W m j a0.6 fa0.5fe0.4 m j b3 600 sg ~ 10
2
P34.49
Now I =
P U
=
A At
(a)
ε=−
af
2
a
dΦ B
d
=−
BA cos θ
dt
dt
f
2
ε = −A
af
b
6
J .
g
b
d
Bmax cos ω t cos θ = ABmaxω sin ω t cos θ
dt
g
ε t = 2π 2 r 2 fBmax cos θ sin 2π f t
ε t = 2π fBmax A sin 2π f t cos θ
ε max = 2π 2 r 2 f Bmax cos θ
Thus,
where θ is the angle between the magnetic field and the normal to the loop.
(b)
If E is vertical, B is horizontal, so the plane of the loop should be vertical
and the plane should contain the line of sight of the transmitter .
P34.50
(a)
Fgrav =
GM S m
2
=
FG GM
HR
2
S
3
R
where M S = mass of Sun, r = radius of particle and R = distance from Sun to particle.
Since
Frad =
Sπ r 2
,
c
FG
H
Frad
1
=
Fgrav
r
(b)
IJ LMρFG 4 π r IJ OP
KN H 3 KQ
IJ FG 3SR IJ ∝ 1 .
K H 4cGM ρ K r
2
S
From the result found in part (a), when Fgrav = Frad ,
we have r =
r=
3SR 2
4cGM S ρ
e
je
j
3 214 W m 2 3.75 × 10 11 m
e
je
je
2
je
4 6.67 × 10 −11 N ⋅ m 2 kg 2 1.991 × 10 30 kg 1 500 kg m3 3.00 × 10 8 m s
= 3.78 × 10 −7 m
j
Chapter 34
P34.51
Emax
= 6.67 × 10 −16 T
c
(a)
Bmax =
(b)
Sav =
(c)
P = Sav A = 1.67 × 10 −14 W
(d)
F = PA =
2
Emax
= 5.31 × 10 −17 W m 2
2µ 0 c
FG S IJ A =
HcK
av
5.56 × 10 −23 N (≅ the weight of
FIG. P34.51
3 000 H atoms!)
P34.52
(a)
307
j a
e
f
The power incident on the mirror is: PI = IA = 1 340 W m 2 π 100 m
2
e
= 4.21 × 10 7 W .
j
The power reflected through the atmosphere is PR = 0.746 4.21 × 10 7 W = 3.14 × 10 7 W .
PR
3.14 × 10 7 W
=
A π 4.00 × 10 3 m
= 0.625 W m 2
(b)
S=
(c)
Noon sunshine in Saint Petersburg produces this power-per-area on a horizontal surface:
e
j
2
PN
= 0.746 1 340 W m 2 sin 7.00° = 122 W m 2 .
A
e
j
The radiation intensity received from the mirror is
F 0.625 W m I 100% =
GH 122 W m JK
2
2
0.513% of that from the noon Sun in January.
1
2
∈0 Emax
2
2u
= 95.1 mV m
∈0
P34.53
u=
P34.54
The area over which we model the antenna as radiating is the lateral surface of a cylinder,
Emax =
ja
e
f
A = 2π r = 2π 4.00 × 10 −2 m 0.100 m = 2.51 × 10 −2 m 2 .
(a)
The intensity is then: S =
(b)
The standard is:
e
0.600 W
P
=
= 23.9 W m 2 .
A 2.51 × 10 −2 m 2
0.570 mW cm 2 = 0.570 mW cm 2
jFGH 1.001.00× 10mWW IJK FGH 1.001×.0010m cm IJK = 5.70 W m
−3
While it is on, the telephone is over the standard by
4
2
2
23.9 W m 2
5.70 W m 2
2
= 4.19 times .
.
308
P34.55
Electromagnetic Waves
(a)
Bmax =
k=
2π
λ
175 V m
Emax
=
= 5.83 × 10 −7 T
8
c
3.00 × 10 m s
=
2π
= 419 rad m
0.015 0 m
ω = kc = 1.26 × 10 11 rad s
Since S is along x, and E is along y, B must be in the z direction . (That is S ∝ E × B .)
P34.56
Emax Bmax
= 40.6 W m 2
2µ 0
(b)
S av =
(c)
Pr =
(d)
∑ F = PA = e2.71 × 10
a=
S av =
e40.6 W m ji
2
2S
= 2.71 × 10 −7 N m 2
c
m
−7
m
je
N m 2 0.750 m 2
0.500 kg
j = 4.06 × 10
−7
m s2
Of the intensity
S = 1 340 W m 2
the 38.0% that is reflected exerts a pressure
P1 =
2Sr 2 0.380 S
=
.
c
c
The absorbed light exerts pressure
P2 =
S a 0.620S
=
.
c
c
a
a=
e406 nm s ji
2
f
Altogether the pressure at the subsolar point on Earth is
P34.57
e
j
2
1.38S 1.38 1 340 W m
= P1 + P2 =
=
= 6.16 × 10 −6 Pa
c
3.00 × 10 8 m s
(a)
Ptotal
(b)
1.01 × 10 5 N m 2
Pa
=
= 1.64 × 10 10 times smaller than atmospheric pressure
Ptotal 6.16 × 10 −6 N m 2
(a)
P=
F I
=
A c
a = 3.03 × 10 −9 m s 2 and
100 J s
IA P
= =
= 3.33 × 10 −7 N = 110 kg a
c
c 3.00 × 10 8 m s
x=
1 2
at
2
b
2x
= 8.12 × 10 4 s = 22.6 h
a
t=
(b)
F=
b
g b
gb
g b
g
b
g
0 = 107 kg v − 3.00 kg 12.0 m s − v = 107 kg v − 36.0 kg ⋅ m s + 3.00 kg v
v=
36.0
= 0.327 m s
110
t = 30.6 s
g
Chapter 34
P34.58
The mirror intercepts power
j a
e
f
P = I 1 A1 = 1.00 × 10 3 W m 2 π 0.500 m
2
309
= 785 W .
In the image,
785 W
(a)
I2 =
P
:
A2
I2 =
(b)
I2 =
2
Emax
so
2µ 0 c
Emax = 2 µ 0 cI 2 = 2 4π × 10 −7 3.00 × 10 8 6.25 × 10 5 = 21.7 kN C
b
0.400P∆t = mc∆T
a
= 625 kW m 2
2
e
Bmax =
(c)
g
π 0.020 0 m
f b
je
je
Emax
= 72.4 µT
c
gb
ga
0. 400 785 W ∆t = 1.00 kg 4 186 J kg⋅° C 100° C − 20.0° C
∆t =
P34.59
f
5
3.35 × 10 J
= 1.07 × 10 3 s = 17.8 min
314 W
Think of light going up and being absorbed by the bead which presents a face area π rb2 .
The light pressure is P =
P34.60
j
S I
= .
c c
Iπ rb2
4
= mg = ρ π rb3 g
3
c
(a)
F =
(b)
P = IA = 8.32 × 10 7 W m 2 π 2.00 × 10 −3 m
e
and
je
I=
F
GH
4ρ gc 3m
3
4π ρ
j
2
I
JK
13
= 8.32 × 10 7 W m 2
= 1.05 kW
Think of light going up and being absorbed by the bead, which presents face area π rb2 .
S
I F
.
If we take the bead to be perfectly absorbing, the light pressure is P = av = =
c
c A
(a)
F = Fg
so
I=
F c Fg c mgc
=
=
.
A
A
π rb2
From the definition of density,
ρ=
m
m
=
V
4 3 π rb3
so
b g
1 F b 4 3gπ ρ I
=G
r
H m JK
mgc F 4π ρ I
I=
G J
π H 3m K
13
.
b
Substituting for rb ,
(b)
P = IA
23
F 4ρ I F m I
= gc G J G J
H 3 K HπK
FG IJ
H K
4π r 2 ρ gc 3m
P=
3
4πρ
23
13
13
F
GH
4ρgc 3m
=
3
4π ρ
I
JK
13
.
310
P34.61
Electromagnetic Waves
c 3.00 × 10 8 m s
=
= 1.50 cm
f
20.0 × 10 9 s −1
(a)
λ=
(b)
U = P ∆t = 25.0 × 10 3 J s 1.00 × 10 −9 s = 25.0 × 10 −6 J
a f e
je
j
= 25.0 µJ
(c)
uav =
FIG. P34.61
U
U
=
V
π r2
U
=
25.0 × 10 −6 J
=
e j eπ r jca∆tf π b0.060 0 mg e3.00 × 10
2
2
je
m s 1.00 × 10 −9 s
8
j
uav = 7.37 × 10 −3 J m3 = 7.37 mJ m 3
(d)
Bmax =
(e)
P34.62
(a)
(b)
2uav
=
∈0
Emax =
e
2 7.37 × 10 −3 J m3
8.85 × 10
−12
C N ⋅ m2
= 4.08 × 10 4 V m = 40.8 kV m
Emax 4.08 × 10 4 V m
=
= 1.36 × 10 −4 T = 136 µT
c
3.00 × 10 8 m s
F = PA =
FG S IJ A = u
H cK
av A
jb
e
= 7.37 × 10 −3 J m3 π 0.060 0 m
On the right side of the equation,
eC
2
N ⋅m
2
j
2
2
jbm sg
3
= 8.33 × 10 −5 N = 83.3 µN
N ⋅ m2 ⋅ C 2 ⋅ m2 ⋅ s3
=
2
4
C ⋅s ⋅m
jb
e
F = ma = qE or
F = ma c = m
e
C2 m s2
g
3
=
N ⋅m J
= = W.
s
s
g
−19
C 100 N C
qE 1.60 × 10
=
= 1.76 × 10 13 m s 2 .
a=
−31
m
9.11 × 10
kg
The radiated power is then:
(c)
j
2
F v I = qvB so
GH r JK
2
P=
v=
q2 a2
6π ∈0 c 3
e1.60 × 10 j e1.76 × 10 j
=
6π e8.85 × 10 je3.00 × 10 j
−19 2
13 2
−12
8 3
= 1.75 × 10 −27 W .
qBr
.
m
ja
e
2
fa
j
2
f
1.60 × 10 −19 0.350 0.500
v2 q 2B2r
=
=
= 5.62 × 10 14 m s 2 .
The proton accelerates at a =
2
2
−
27
r
m
1.67 × 10
e
The proton then radiates P =
P34.63
P=
q2 a2
6π ∈0 c 3
e1.60 × 10 j e5.62 × 10 j
=
6π e8.85 × 10 je3.00 × 10 j
−19 2
14 2
−12
8 3
= 1.80 × 10 −24 W .
P
S Power
60.0 W
=
=
=
= 6.37 × 10 −7 Pa
c
Ac
2π r c 2π 0.050 0 m 1.00 m 3.00 × 10 8 m s
b
ga
fe
j
Chapter 34
P34.64
b g
FG IJ
H K
e3.00 × 10 jb0.060 0g =
P
Therefore, θ =
=
2 cκ 2e3.00 × 10 je1.00 × 10 j
F = PA =
P AA P
P
SA
=
= ,τ =F
= , and τ = κθ .
c
c
c
2
2c
−3
−11
8
P34.65
P34.66
The light intensity is
I = Sav =
The light pressure is
P=
For the asteroid,
PA = ma
3.00 × 10 −2 deg .
E2
.
2µ 0 c
1
S
E2
=
= ∈0 E 2 .
c 2µ 0 c 2 2
and
a=
∈0 E 2 A
.
2m
f = 90.0 MHz , Emax = 2.00 × 10 −3 V m = 200 mV m
(a)
λ=
c
= 3.33 m
f
1
= 1.11 × 10 −8 s = 11.1 ns
f
E
Bmax = max = 6.67 × 10 −12 T = 6.67 pT
c
T=
(b)
b
g
E = 2.00 mV m cos 2π
e
FG x − t IJ j
H 3.33 m 11.1 ns K
j
2
b
(c)
2.00 × 10 −3
E2
= 5.31 × 10 −9 W m 2
I = max =
2 µ 0 c 2 4π × 10 −7 3.00 × 10 8
(d)
I = cuav
(e)
2 5.31 × 10 −9
2I
P=
=
= 3.54 × 10 −17 Pa
c
3.00 × 10 8
e
so
a fe
je
j
uav = 1.77 × 10 −17 J m 3
j
g
B = 6.67 pT k cos 2π
FG x − t IJ
H 3.33 m 11.1 ns K
311
312
*P34.67
Electromagnetic Waves
(a)
m = ρV = ρ
F 6m IJ
r =G
H ρ 4π K
14 3
πr
23
F 6b8.7 kg g I
=G
GH e990 kg m j4π JJK =
= 2π a0.161 mf = 0.163 m
13
13
0.161 m
3
1
4π r 2
2
2
2
(b)
A=
(c)
I = eσ T 4 = 0.970 5.67 × 10 −8 W m 2 ⋅ K 4 304 K
(d)
P = IA = 470 W m 2 0.163 m 2 = 76.8 W
(e)
I=
e
= 470 W m 2
g
12
e
je
je
= 8π × 10 −7 Tm A 3 × 10 8 m s 470 W m 2
j
12
= 595 N C
Emax = cBmax
Bmax =
(h)
4
2
Emax
2µ 0 c
b
(g)
f
j
Emax = 2 µ 0 cI
(f)
ja
e
595 N C
3 × 10 8 m s
= 1.98 µT
The sleeping cats are uncharged and nonmagnetic. They carry no macroscopic current. They
are a source of infrared radiation. They glow not by visible-light emission but by infrared
emission.
Each kitten has radius rk
b
g
2π 0.072 8 m
2
F 6a0.8f IJ
=G
H 990 × 4π K
13
= 0.072 8 m and radiating area
= 0.033 3 m 2 . Eliza has area 2π
e
j
FG 6a5.5f IJ
H 990 × 4π K
23
= 0.120 m 2 . The total glowing
area is 0.120 m 2 + 4 0.033 3 m 2 = 0.254 m 2 and has power output
e
2
j
2
P = IA = 470 W m 0.254 m = 119 W .
P34.68
(a)
(b)
At steady state, Pin = Pout and the power radiated out is Pout = eσAT 4 .
.
e
j
e
j
Thus,
0.900 1 000 W m 2 A = 0.700 5.67 × 10 −8 W m 2 ⋅ K 4 AT 4
or
L
900 W m
T=M
MN 0.700e5.67 × 10 W m
2
−8
2
⋅K4
OP
j PQ
14
= 388 K = 115° C .
The box of horizontal area A, presents projected area A sin 50.0° perpendicular to the
sunlight. Then by the same reasoning,
e
j
e
j
0.900 1 000 W m 2 A sin 50.0° = 0.700 5.67 × 10 −8 W m 2 ⋅ K 4 AT 4
L e900 W m j sin 50.0° OP
T=M
MN 0.700e5.67 × 10 W m ⋅ K j PQ
2
or
−8
2
4
14
= 363 K = 90.0° C .
Chapter 34
P34.69
313
We take R to be the planet’s distance from its star. The planet, of radius r, presents a
projected area π r 2 perpendicular to the starlight. It radiates over area 4π r 2 .
e j
e
F 6.00 × 10
GH 4π R
R=
23
W
2
e
j
eI in π r 2 = eσ 4π r 2 T 4
At steady-state, Pin = Pout :
I π r = eσ 4π r T
JK e j e j
2
2
4
so that 6.00 × 10 23 W = 16πσR 2 T 4
6.00 × 10 23 W
6.00 × 10 23 W
=
4
16πσ T
16π 5.67 × 10 −8 W m 2 ⋅ K 4 310 K
ja
e
f
4
= 4.77 × 10 9 m = 4.77 Gm .
ANSWERS TO EVEN PROBLEMS
P34.2
(a) 2.68 × 10 3 AD ; (b) 8.31 min ; (c) 2.56 s;
(d) 0.133 s; (e) 33.3 µs
P34.4
733 nT
P34.6
E = 300 V m cos 62.8 x − 1.88 × 10 10 t ;
b
g e
j
B = b1.00 µTg cose62.8 x − 1.88 × 10 t j
10
P34.8
see the solution
P34.10
2.9 × 10 8 m s ± 5%
P34.12
49.5 mV
P34.14
3
P34.30
(a) 577 W m 2 ; (b) 2.06 × 10 16 W ;
(c) 68.7 MN ; (d) The gravitational force is
~ 10 13 times stronger and in the opposite
direction.
P34.32
4.09°
P34.34
(a) 93.3% ; (b) 50.0% ; (c) 0
P34.36
3
(a) 13.3 nJ m ; (b) 13.3 nJ m ;
516 pT, ~ 10 5 times stronger than the
Earth’s field
P34.18
(a) 11.9 GW m 2 ; (b) 234 kW
P34.20
33.4°C for the smaller container and 21.7°C
for the larger
P34.22
(a) 50.0% ;
(b) 269 kW m 2 toward the oven chamber ;
(c) 14. 2 kV m
P34.24
(a) 4.97 kW m 2 ; (b) 16.6 µJ m 3
P34.26
667 pN
P34.28
(a) 5.36 N ; (b) 893 µm s 2 ; (c) 10.7 days
eB
P34.38
see the solution
P34.40
(a) ~ 10 8 Hz radio wave;
(b) ~ 10 13 Hz infrared light
P34.42
(a) 0.690 wavelengths ;
(b) 58.9 wavelengths
P34.44
The radio audience gets the news 8.41 ms
sooner.
P34.46
(a) 187 m to 556 m; (b) 2.78 m to 3.41 m
P34.48
~ 10 6 J
P34.50
(a) see the solution; (b) 378 nm
P34.52
(a) 31.4 MW; (b) 0.625 W m 2 ; (c) 0.513%
P34.54
(a) 23.9 W m 2 ; (b) 4.19 times the standard
P34.56
(a) 6.16 µPa ; (b) 1.64 × 10 10 times less than
atmospheric pressure
(c) 7.96 W m 2
P34.16
2π m p c
314
P34.58
P34.60
P34.62
Electromagnetic Waves
(a) 625 kW m 2 ;
(b) 21.7 kN C and 72.4 µT ; (c) 17.8 min
F 16mρ I
(a) G
H 9π JK
2
13
F 16π mρ I
gc ; (b) G
H 9 JK
(a) see the solution;
(b) 17.6 Tm s 2 , 1.75 × 10 −27 W ;
(c) 1.80 × 10 −24 W
P34.64
2
3.00 × 10 −2 deg
2
P34.66
(a) 3.33 m, 11.1 ns , 6.67 pT ;
b
g FGH 3.33x m − 11.1t ns IJK j ;
F x − t IJ ;
B = b6.67 pTgk cos 2π G
H 3.33 m 11.1 ns K
(b) E = 2.00 mV m cos 2π
13
r 2 gc
(c) 5.31 nW m 2 ; (d) 1.77 × 10 −17 J m3 ;
(e) 3.54 × 10 −17 Pa
P34.68
(a) 388 K; (b) 363 K
35
The Nature of Light and the
Laws of Geometric Optics
CHAPTER OUTLINE
35.1
35.2
35.3
35.4
35.5
35.6
35.7
35.8
35.9
The Nature of Light
Measurements of the Speed
of Light
The Ray Approximation in
Geometric Optics
Reflection
Refraction
Huygen’s Principle
Dispersion and Prisms
Total Internal Reflection
Fermat‘s Principle
Q35.3
ANSWERS TO QUESTIONS
Q35.1
The ray approximation, predicting sharp shadows, is valid for
λ << d . For λ ~ d diffraction effects become important, and the
light waves will spread out noticeably beyond the slit.
Q35.2
Light travels through a vacuum at a speed of 300 000 km per
second. Thus, an image we see from a distant star or galaxy
must have been generated some time ago. For example, the star
Altair is 16 light-years away; if we look at an image of Altair
today, we know only what was happening 16 years ago. This
may not initially seem significant, but astronomers who look at
other galaxies can gain an idea of what galaxies looked like
when they were significantly younger. Thus, it actually makes
sense to speak of “looking backward in time.”
Sun
Moon
partial eclipse
no eclipse
no eclipse
total eclipse
(full shadow)
Note: Figure not at all to scale
Earth’s surface
FIG. Q35.3
315
316
Q35.4
The Nature of Light and the Laws of Geometric Optics
With a vertical shop window, streetlights and his own
reflection can impede the window shopper’s clear view of
the display. The tilted shop window can put these
reflections out of the way. Windows of airport control
towers are also tilted like this, as are automobile
windshields.
FIG. Q35.4
Q35.5
We assume that you and the child are
always standing close together. For a flat
wall to make an echo of a sound that you
make, you must be standing along a normal
to the wall. You must be on the order of
100 m away, to make the transit time
sufficiently long that you can hear the echo
separately from the original sound. Your
sound must be loud enough so that you can
hear it even at this considerable range. In
the picture, the dashed rectangle represents
an area in which you can be standing. The
arrows represent rays of sound.
Now suppose two vertical
perpendicular walls form an inside corner
that you can see. Some of the sound you
radiate horizontally will be headed
generally toward the corner. It will reflect
from both walls with high efficiency to
reverse in direction and come back to you.
You can stand anywhere reasonably far
away to hear a retroreflected echo of sound
you produce.
If the two walls are not
perpendicular, the inside corner will not
produce retroreflection. You will generally
hear no echo of your shout or clap.
If two perpendicular walls have a
reasonably narrow gap between them at
the corner, you can still hear a clear echo. It
is not the corner line itself that retroreflects
the sound, but the perpendicular walls on
both sides of the corner. Diagram (b)
applies also in this case.
(a)
(b)
FIG. Q35.4
Chapter 35
317
Q35.6
The stealth fighter is designed so that adjacent panels are not joined at right angles, to prevent any
retroreflection of radar signals. This means that radar signals directed at the fighter will not be
channeled back toward the detector by reflection. Just as with sound, radar signals can be treated as
diverging rays, so that any ray that is by chance reflected back to the detector will be too weak in
intensity to distinguish from background noise. This author is still waiting for the automotive
industry to utilize this technology.
Q35.7
An echo is an example of the reflection of sound. Hearing the noise of a distant highway on a cold
morning, when you cannot hear it after the ground warms up, is an example of acoustical refraction.
You can use a rubber inner tube inflated with helium as an acoustical lens to concentrate sound in
the way a lens can focus light. At your next party, see if you can experimentally find the
approximate focal point!
Q35.8
No. If the incidence angle is zero, then the ray does not change direction. Also, if the ray travels from
a medium of relatively high index of refraction to one of lower index of refraction, it will bend away
from the normal.
Q35.9
Suppose the light moves into a medium of higher refractive index. Then its wavelength decreases.
The frequency remains constant. The speed diminishes by a factor equal to the index of refraction.
Q35.10
If a laser beam enters a sugar solution with a concentration gradient (density and index of refraction
increasing with depth) then the laser beam will be progressively bent downward (toward the
normal) as it passes into regions of greater index of refraction.
Q35.11
As measured from the diagram, the incidence angle is 60°, and the refraction angle is 35°. Using
sin θ 2 v 2
sin 35° v 2
=
=
and the speed of light in Lucite is 2.0 × 10 8 m s .
, then
equation 35.3,
sin 60°
sin θ 1 v1
c
The frequency of the light does not change upon refraction. Knowing the wavelength in a
vacuum, we can use the speed of light in a vacuum to determine the frequency: c = fλ , thus
e
j
3.00 × 10 8 = f 632.8 × 10 −9 , so the frequency is 474.1 THz. To find the wavelength of light in Lucite,
e
j
we use the same wave speed relation, v = fλ , so 2.0 × 10 8 = 4.741 × 10 14 λ , so λ Lucite = 420 nm .
Q35.12
Blue light would be refracted at a smaller angle from the normal, since the index of refraction for
blue light—a smaller wavelength than red light—is larger.
Q35.13
The index of refraction of water is 1.33, quite different from 1.00 for air. Babies learn that the
refraction of light going through the water indicates the water is there. On the other hand, the index
of refraction of liquid helium is close to that of air, so it gives little visible evidence of its presence.
Q35.14
The outgoing beam would be a rainbow, with the different colors of light traveling parallel to each
other. The white light would undergo dispersion upon refraction into the slab, with blue light
bending towards the normal more than the red light. Upon refraction out of the block, all rays of
light would exit the slab at the same angle at which they entered the slab, but offset from each other.
Q35.15
Diamond has higher index of refraction than glass and consequently a smaller critical angle for total
internal reflection. A brilliant-cut diamond is shaped to admit light from above, reflect it totally at
the converging facets on the underside of the jewel, and let the light escape only at the top. Glass
will have less light internally reflected.
318
The Nature of Light and the Laws of Geometric Optics
Q35.16
Light coming up from underwater is bent away from the normal. Therefore the part of the oar that is
submerged appears bent upward.
Q35.17
Highly silvered mirrors reflect about 98% of the incident light. With a 2-mirror periscope, that results
in approximately a 4% decrease in intensity of light as the light passes through the periscope. This
may not seem like much, but in low-light conditions, that lost light may mean the difference
between being able to distinguish an enemy armada or an iceberg from the sky beyond. Using
prisms results in total internal reflection, meaning that 100% of the incident light is reflected through
the periscope. That is the “total” in total internal reflection.
Q35.18
Sound travels faster in the warmer air, and thus the sound traveling through the warm air aloft will
refract much like the light refracting through the nonuniform sugar-water solution in Question
35.10. Sound that would normally travel up over the tree-tops can be refracted back towards the
ground.
Q35.19
The light with the greater change in speed will have the larger deviation. If the glass has a higher
index than the surrounding medium, X travels slower in the glass.
Q35.20
Immediately around the dark shadow of my head, I see a halo brighter than the rest of the dewy
grass. It is called the heiligenschein. Cellini believed that it was a miraculous sign of divine favor
pertaining to him alone. Apparently none of the people to whom he showed it told him that they
could see halos around their own shadows but not around Cellini’s. Thoreau knew that each person
had his own halo. He did not draw any ray diagrams but assumed that it was entirely natural.
Between Cellini’s time and Thoreau’s, the Enlightenment and Newton’s explanation of the rainbow
had happened. Today the effect is easy to see, whenever your shadow falls on a retroreflecting
traffic sign, license plate, or road stripe. When a bicyclist’s shadow falls on a paint stripe marking the
edge of the road, her halo races along with her. It is a shame that few people are sufficiently curious
observers of the natural world to have noticed the phenomenon.
Q35.21
Suppose the Sun is low in the sky and an observer faces away from the Sun toward a large uniform
rain shower. A ray of light passing overhead strikes a drop of water. The light is refracted first at the
front surface of the drop, with the violet light deviating the most and the red light the least. At the
back of the drop the light is reflected and it returns to the front surface where it again undergoes
refraction with additional dispersion as it moves from water into air. The rays leave the drop so that
the angle between the incident white light and the most intense returning violet light is 40°, and the
angle between the white light and the most intense returning red light is 42°. The observer can see a
ring of raindrops shining violet, a ring with angular radius 40° around her shadow. From the locus of
directions at 42° away from the antisolar direction, the observer receives red light. The other spectral
colors make up the rainbow in between. An observer of a rainbow sees violet light at 40° angular
separation from the direction opposite the Sun, then the other spectral colors, and then red light on
the outside the rainbow, with angular radius 42°.
Q35.22
At the altitude of the plane the surface of the Earth need not block off the lower half of the rainbow.
Thus, the full circle can be seen. You can see such a rainbow by climbing on a stepladder above a
garden sprinkler in the middle of a sunny day. Set the sprinkler for fine mist. Do not let the slippery
children fall from the ladder.
Q35.23
Total internal reflection occurs only when light moving originally in a medium of high index of
refraction falls on an interface with a medium of lower index of refraction. Thus, light moving from
air (n = 1) to water (n = 1.33 ) cannot undergo total internal reflection.
Chapter 35
Q35.24
319
A mirage occurs when light changes direction as it moves between batches of air having different
indices of refraction because they have different densities at different temperatures. When the sun
makes a blacktop road hot, an apparent wet spot is bright due to refraction of light from the bright
sky. The light, originally headed a little below the horizontal, always bends up as it first enters and
then leaves sequentially hotter, lower-density, lower-index layers of air closer to the road surface.
SOLUTIONS TO PROBLEMS
Section 35.1
The Nature of Light
Section 35.2
Measurements of the Speed of Light
P35.1
The Moon’s radius is 1.74 × 10 6 m and the Earth’s radius is 6.37 × 10 6 m . The total distance traveled
by the light is:
e
j
d = 2 3.84 × 10 8 m − 1.74 × 10 6 m − 6.37 × 10 6 m = 7.52 × 10 8 m .
7.52 × 10 8 m
= 2.995 × 10 8 m s = 299.5 Mm s .
2.51 s
This takes 2.51 s, so v =
e
jb
g
8
∆x 2 1.50 × 10 km 1 000 m km
c=
=
= 2. 27 × 10 8 m s = 227 Mm s
t
22.0 min 60.0 s min
P35.2
∆x = ct ;
P35.3
The experiment is most convincing if the wheel turns fast enough to pass outgoing light through
2
one notch and returning light through the next: t =
c
F2 I
θ = ω t = ωG J
HcK
a
fb
g
e
j a f
j
2.998 × 10 8 2π 720
cθ
ω=
=
= 114 rad s .
2
2 11. 45 × 10 3
so
e
The returning light would be blocked by a tooth at one-half the angular speed, giving another data
point.
P35.4
(a)
For the light beam to make it through both slots, the time for the light to travel the distance d
must equal the time for the disk to rotate through the angle θ, if c is the speed of light,
d θ
dω
= , so c =
.
c ω
θ
(b)
We are given that
d = 2.50 m , θ =
c=
dω
θ
FG
H
IJ
K
FG
H
IJ
K
2π rad
1.00° π rad
= 2.91 × 10 −4 rad , ω = 5 555 rev s
= 3.49 × 10 4 rad s
60.0 180°
1.00 rev
a2.50 mfe3.49 × 10
=
2.91 × 10
−4
4
rad s
rad
j = 3.00 × 10
8
m s = 300 Mm s
320
The Nature of Light and the Laws of Geometric Optics
Section 35.3
The Ray Approximation in Geometric Optics
Section 35.4
Reflection
Section 35.5
Refraction
*P35.5
(a)
(b)
Let AB be the originally horizontal ceiling, BC its originally vertical
normal, AD the new ceiling and DE its normal. Then angle
BAD = φ . By definition DE is perpendicular to AD and BC is
perpendicular to AB. Then the angle between DE extended and BC
is φ because angles are equal when their sides are perpendicular,
right side to right side and left side to left side.
Now CBE = φ is the angle of incidence of the vertical light beam. Its
angle of reflection is also φ. The angle between the vertical incident
beam and the reflected beam is 2φ .
A
D
E
B
C
FIG. P35.5(a)
A
D
E
P35.6
FIG. P35.5(b)
(c)
1.40 cm
= 0.001 94
tan 2φ =
720 cm
(a)
From geometry,
1.25 m = d sin 40.0°
so
d = 1.94 m .
(b)
φ = 0.055 7°
Mirror 2
Mirror 2
50.0° above the horizontal
P
1.25 m
Mirror 1
or parallel to the incident ray.
50°
Light
beam
40.0°
i2= 50°
40°
d
40° i1= 40°
50°
1.25 m
50°
Mirror 1
FIG. P35.6
*P35.7
(a)
Method One:
The incident ray makes angle α = 90°−θ 1
with the first mirror. In the picture, the law of reflection implies
that
θ 1 = θ 1′ .
FIG. P35.7
Then
β = 90°−θ 1′ = 90 − θ 1 = α .
In the triangle made by the mirrors and the ray passing between them,
Further,
and
β + 90°+γ = 180°
γ = 90°− β
δ = 90°−γ = β = α
∈= δ = α .
Thus the final ray makes the same angle with the first mirror as did the incident ray. Its
direction is opposite to the incident ray.
continued on next page
Chapter 35
321
Method Two:
The vector velocity of the incident light has a component v y perpendicular to the first mirror
and a component v x perpendicular to the second. The v y component is reversed upon the
first reflection, which leaves v x unchanged. The second reflection reverses v x and leaves v y
unchanged. The doubly reflected ray then has velocity opposite to the incident ray.
The incident ray has velocity v x i + v y j + v z k . Each reflection reverses one component and
(b)
leaves the other two unchanged. After all the reflections, the light has velocity
− v x i − v y j − v z k , opposite to the incident ray.
P35.8
Mirror
The incident light reaches the left-hand mirror at distance
Mirror
a1.00 mf tan 5.00° = 0.087 5 m
above its bottom edge. The reflected light first reaches the
right-hand mirror at height
b
g
1.00 m
reflected beam
2 0.087 5 m = 0.175 m .
It bounces between the mirrors with this distance between
points of contact with either.
FIG. P35.8
five times from the right-hand mirror and six times from the left .
the light reflects
Let d represent the perpendicular distance from the
person to the mirror. The distance between lamp and
peson measured parallel to the mirror can be written in
d tan φ
two ways: 2d tan θ + d tan θ = d tan φ . The condition on the
2d
2d
d
distance traveled by the light is
=
+
. We
cos φ cos θ cos θ
have the two equations 3 tan θ = tan φ and
2 cos θ = 3 cos φ . To eliminate φ we write
9 sin 2 θ sin 2 φ
=
cos 2 θ
cos 2 φ
e
j
F 4 I
4 cos θ sin θ = cos θ G 1 − cos θ J
H 9 K
4
4 sin θ = 1 − e1 − sin θ j
36 sin θ = 9 − 4 + 4 sin θ
9
2
2
2
2
2
2
5
32
θ = 23.3°
2
θ
φ
2d tan θ
θ
d tan θ
2d
d
d
FIG. P35.9
4 cos 2 θ = 9 cos 2 φ
9 cos 2 φ sin 2 θ = cos 2 θ 1 − cos 2 φ
sin 2 θ =
1.00 m
1.00 m
= 5.72
0.175 m
Since
*P35.9
5.00°
2
322
P35.10
The Nature of Light and the Laws of Geometric Optics
sin θ 2 =
Using Snell’s law,
n1
sin θ 1
n2
θ 2 = 25.5°
λ2 =
*P35.11
λ1
= 442 nm .
n1
FIG. P35.10
The law of refraction n1 sin θ 1 = n 2 sin θ 2 can be put into the more general form
c
c
sin θ 1 =
sin θ 2
v1
v2
sin θ 1 sin θ 2
=
v1
v2
In this form it applies to all kinds of waves that move through space.
sin θ 2
sin 3.5°
=
343 m s 1 493 m s
sin θ 2 = 0.266
θ 2 = 15.4°
The wave keeps constant frequency in
f=
v1 v 2
=
λ1 λ 2
λ2 =
P35.12
P35.13
a
f
v 2 λ 1 1 493 m s 0.589 m
=
= 2.56 m
343 m s
v1
c
= 4.74 × 10 14 Hz
f=
(b)
λ glass =
λ air 632.8 nm
=
= 422 nm
1.50
n
(c)
v glass =
c air 3.00 × 10 8 m s
=
= 2.00 × 10 8 m s = 200 Mm s
1.50
n
λ
=
3.00 × 10 8 m s
(a)
6.328 × 10
−7
m
n1 sin θ 1 = n 2 sin θ 2
sin θ 1 = 1.333 sin 45°
a fa
f
sin θ 1 = 1.33 0.707 = 0.943
θ 1 = 70.5°→ 19.5° above the horizon
FIG. P35.13
*P35.14
We find the angle of incidence:
n1 sin θ 1 = n 2 sin θ 2
1.333 sin θ 1 = 1.52 sin 19.6°
θ 1 = 22.5°
The angle of reflection of the beam in water is then also 22.5° .
Chapter 35
P35.15
(a)
323
n1 sin θ 1 = n 2 sin θ 2
1.00 sin 30.0° = n sin 19.24°
n = 1.52
P35.16
P35.17
= 4.74 × 10 14 Hz in air and in syrup.
(d)
v=
8
c 3.00 × 10 m s
=
= 1.98 × 10 8 m s = 198 Mm s
1.52
n
(b)
λ=
v 1.98 × 10 8 m s
=
= 417 nm
f
4.74 × 10 14 s
(a)
Flint Glass:
v=
8
c 3.00 × 10 m s
=
= 1.81 × 10 8 m s = 181 Mm s
1.66
n
(b)
Water:
v=
8
c 3.00 × 10 m s
=
= 2.25 × 10 8 m s = 225 Mm s
1.333
n
(c)
Cubic Zirconia:
v=
8
c 3.00 × 10 m s
=
= 1.36 × 10 8 m s = 136 Mm s
2.20
n
λ
=
3.00 × 10 8 m s
f=
6.328 × 10 −7 m
n1 sin θ 1 = n 2 sin θ 2 :
n 2 = 1.90 =
P35.18
c
(c)
c
:
v
1.333 sin 37.0° = n 2 sin 25.0°
v=
c
= 1.58 × 10 8 m s = 158 Mm s
1.90
sin θ 1 = n w sin θ 2
θ 1 = 62°
a
f
1
1
sin θ 1 =
sin 90.0°−28.0° = 0.662
1.333
1.333
θ 2 = sin −1 0.662 = 41.5°
sin θ 2 =
a
28°
f
3.00 m
d
=
= 3.39 m
h=
tan θ 2 tan41.5°
air
n = 1.00
θ2
water
n = 1.333
h
3.0 m
FIG. P35.18
324
P35.19
The Nature of Light and the Laws of Geometric Optics
θ 2 = sin −1
n1 sin θ 1 = n 2 sin θ 2 :
θ 2 = sin −1
FG n sinθ IJ
H n K
RS1.00 sin 30° UV =
T 1.50 W
1
1
2
19.5°
θ 2 and θ 3 are alternate interior angles formed by the ray cutting
parallel normals.
FIG. P35.19
θ 3 = θ 2 = 19.5°
So,
1.50 sin θ 3 = 1.00 sin θ 4
θ 4 = 30.0°
*P35.20
For
α + β = 90°
with
θ 1′ + α + β + θ 2 = 180°
we have
θ 1′ + θ 2 = 90° .
Also,
θ 1′ = θ 1
and
1 sin θ 1 = n sin θ 2 .
Then,
sin θ 1 = n sin 90 − θ 1 = n cos θ 1
b
sin θ 1
= n = tan θ 1
cos θ 1
P35.21
FIG. P35.20
g
At entry,
n1 sin θ 1 = n 2 sin θ 2
or
1.00 sin 30.0° = 1.50 sin θ 2
θ 1 = tan −1 n .
θ 2 = 19.5° .
The distance h the light travels in the medium is given by
cos θ 2 =
or
h=
2.00 cm
h
2.00 cm
= 2.12 cm .
cos19.5°
FIG. P35.21
α = θ 1 − θ 2 = 30.0°−19.5° = 10.5° .
The angle of deviation upon entry is
The offset distance comes from sin α =
P35.22
d
:
h
The distance, h, traveled by the light is h =
The speed of light in the material is
Therefore,
a
f
d = 2.21 cm sin 10.5° = 0.388 cm .
2.00 cm
= 2.12 cm
cos19.5°
..
c 3.00 × 10 8 m s
=
= 2.00 × 10 8 m s
1.50
h
..
h 2.12 × 10 −2 m
= 1.06 × 10 −10 s = 106 ps .
t= =
v 2.00 × 10 8 m s
v=
Chapter 35
P35.23
325
Applying Snell’s law at the air-oil interface,
n air sin θ = n oil sin 20.0°
yields
θ = 30.4° .
Applying Snell’s law at the oil-water interface
n w sin θ ′ = n oil sin 20.0°
yields
θ ′ = 22.3° .
FIG. P35.23
*P35.24
For sheets 1 and 2 as described,
n1 sin 26.5° = n 2 sin 31.7°
0.849n1 = n 2
For the trial with sheets 3 and 2,
n 3 sin 26.5° = n 2 sin 36.7°
0.747n 3 = n 2
Now
0.747n 3 = 0.849n1
n 3 = 1.14n1
For the third trial,
n1 sin 26.5° = n 3 sin θ 3 = 1.14n1 sin θ 3
θ 3 = 23.1°
P35.25
Consider glass with an index of refraction of 1.5, which is 3 mm thick. The speed of light in the glass
is
3 × 10 8 m s
= 2 × 10 8 m s .
1.5
The extra travel time is
3 × 10 −3 m
2 × 10 8 m s
−
3 × 10 −3 m
3 × 10 8 m s
For light of wavelength 600 nm in vacuum and wavelength
the extra optical path, in wavelengths, is
3 × 10 −3 m
4 × 10 −7 m
−
~ 10 −11 s .
600 nm
= 400 nm in glass,
1.5
3 × 10 −3 m
6 × 10 −7 m
~ 10 3 wavelengths .
326
P35.26
The Nature of Light and the Laws of Geometric Optics
Refraction proceeds according to
a1.00f sinθ = a1.66f sinθ
(a)
For the normal component of velocity to be constant,
v1 cos θ 1 = v 2 cos θ 2
or
acf cos θ = FGH 1.c66 IJK cos θ
We multiply Equations (1) and (2), obtaining:
sin θ 1 cos θ 1 = sin θ 2 cos θ 2
or
sin 2θ 1 = sin 2θ 2 .
1
1
2
.
2.
(1)
(2)
The solution θ 1 = θ 2 = 0 does not satisfy Equation (2) and must be rejected. The physical
solution is 2θ 1 = 180°−2θ 2 or θ 2 = 90.0°−θ 1 . Then Equation (1) becomes:
sin θ 1 = 1.66 cos θ 1 , or tan θ 1 = 1.66
θ 1 = 58.9° .
which yields
(b)
P35.27
Light entering the glass slows down and makes a smaller angle with the normal. Both effects
reduce the velocity component parallel to the surface of the glass, so that component cannot
remain constant, or will remain constant only in the trivial case θ 1 = θ 2 = 0 .
See the sketch showing the path of the
light ray. α and γ are angles of incidence at
mirrors 1 and 2.
For triangle abca,
2α + 2γ + β = 180°
or
b
g
β = 180°−2 α + γ .
(1)
Now for triangle bcdb,
a90.0°−α f + b90.0°−γ g + θ = 180°
or
θ =α +γ .
FIG. P35.27
(2)
Substituting Equation (2) into Equation (1) gives β = 180°−2θ .
Note: From Equation (2), γ = θ − α . Thus, the ray will follow a path like that shown only if α < 0 .
For α > 0 , γ is negative and multiple reflections from each mirror will occur before the incident and
reflected rays intersect.
327
Chapter 35
Section 35.6
*P35.28
Huygen’s Principle
(a)
For the diagrams of contour lines and wave fronts and rays, see Figures (a) and (b) below.
As the waves move to shallower water, the wave fronts bend to become more nearly parallel
to the contour lines.
(b)
For the diagrams of contour lines and wave fronts and rays, see Figures (c) and (d) below.
We suppose that the headlands are steep underwater, as they are above water. The rays are
everywhere perpendicular to the wave fronts of the incoming refracting waves. As shown,
the rays bend toward the headlands and deliver more energy per length at the headlands.
(a) Contour lines
(b) Wave fronts
and rays
(c) Contour lines
(d) Wave fronts
and rays
FIG. P35.28
Section 35.7
P35.29
Dispersion and Prisms
From Fig 35.21
n v = 1.470 at 400 nm
and
n r = 1.458 at 700 nm.
Then
1.00 sin θ = 1. 470 sin θ v
and
1.00 sin θ = 1. 458 sin θ r
FG sinθ IJ − sin FG sinθ IJ
H 1.458 K
H 1.470 K
FG sin 30.0° IJ − sin FG sin 30.0° IJ = 0.171°
H 1.458 K
H 1.470 K
δ r − δ v = θ r − θ v = sin −1
∆δ = sin −1
P35.30
a
−1
−1
f
n 700 nm = 1.458
(a)
(b)
a1.00f sin 75.0° = 1.458 sinθ
2
; θ 2 = 41.5°
θ1
Let
θ 3 + β = 90.0° , θ 2 + α = 90.0° then α + β + 60.0° = 180° .
So
60.0°−θ 2 − θ 3 = 0 ⇒ 60.0°−41.5° = θ 3 = 18.5° .
(c)
1.458 sin 18.5° = 1.00 sin θ 4
(d)
γ = θ 1 − θ 2 + β − 90.0°−θ 4
b
g
θ 4 = 27.6°
b
g
γ = 75.0°−41.5°+a90.0°−18.5°f − a90.0°−27.6°f =
42.6°
α
60.0°
β
θ2
θ3
FIG. P35.30
θ4
γ
328
P35.31
The Nature of Light and the Laws of Geometric Optics
Taking Φ to be the apex angle and δ min to be the angle of minimum deviation, from Equation 35.9,
the index of refraction of the prism material is
n=
Solving for δ min ,
P35.32
b
g
sin Φ + δ min 2
δ min
b g
F ΦI
= 2 sin G n sin J − Φ = 2 sin a 2.20f sina 25.0°f − 50.0° =
H 2K
sin Φ 2
−1
−1
b
g b
g
Note for use in every part:
Φ + 90.0°−θ 2 + 90.0°−θ 3 = 180°
so
θ3 = Φ −θ 2 .
α = θ1 −θ 2 .
At the first surface the deviation is
At exit, the deviation is
β = θ 4 −θ3 .
The total deviation is therefore
δ = α + β = θ1 +θ 4 −θ 2 −θ3 = θ1 +θ 4 − Φ .
(a)
86.8° .
At entry:
n1 sin θ 1 = n 2 sin θ 2
Thus,
θ 3 = 60.0°−30.0° = 30.0° .
At exit:
1.50 sin 30.0° = 1.00 sin θ 4
FIG. P35.32
FG sin 48.6° IJ = 30.0° .
H 1.50 K
or
θ 2 = sin −1
or
θ 4 = sin −1 1.50 sin 30.0° = 48.6°
a
f
so the path through the prism is symmetric when θ 1 = 48.6° .
(b)
δ = 48.6°+48.6°−60.0° = 37.2°
(c)
At entry:
(d)
At exit:
sin 45.6°
⇒ θ 2 = 28. 4°
1.50
sin θ 4 = 1.50 sin 31.6° ⇒ θ 4 = 51.7°
At entry:
sin θ 2 =
At exit:
P35.33
sin θ 2 =
a
f
sin 51.6°
⇒ θ 2 = 31.5°
1.50
sin θ 4 = 1.50 sin 28.5° ⇒ θ 4 = 45.7°
At the first refraction,
a
f
θ 3 = 60.0°−28.4° = 31.6° .
δ = 45.6°+51.7°−60.0° = 37.3° .
θ 3 = 60.0°−31.5° = 28.5° .
δ = 51.6°+45.7°−60.0° = 37.3° .
1.00 sin θ 1 = n sin θ 2 .
The critical angle at the second surface is given by n sin θ 3 = 1.00 :
FG 1.00 IJ = 41.8° .
H 1.50 K
or
θ 3 = sin −1
But,
θ 2 = 60.0°−θ 3 .
it is necessary that
θ 2 > 18.2° .
Since sin θ 1 = n sin θ 2 , this becomes
sin θ 1 > 1.50 sin 18.2° = 0.468
or
θ 1 > 27.9° .
FIG. P35.33
Thus, to avoid total internal reflection at the second surface (i.e., have θ 3 < 41.8° )
Chapter 35
P35.34
At the first refraction,
1.00 sin θ 1 = n sin θ 2 .
The critical angle at the second surface is given by
FG IJ
H K
g + b90.0°−θ g + Φ = 180°
n sin θ 3 = 1.00 , or
θ 3 = sin −1
But
b90.0°−θ
which gives
θ 2 = Φ −θ3 .
2
θ1
1.00
.
n
Φ
θ2
θ3
3
FIG. P35.34
FG 1.00 IJ and avoid total internal reflection at the second surface,
HnK
F 1.00 IJ .
θ > Φ − sin G
it is necessary that
HnK
L
F 1.00 IJ OP
Since sin θ = n sin θ , this requirement becomes
sin θ > n sin MΦ − sin G
H n KQ
N
F L
F 1.00 IJ OPIJ .
θ > sin G n sin MΦ − sin G
or
H n K QK
H N
θ > sin FH n − 1 sin Φ − cos ΦIK .
Through the application of trigonometric identities,
Thus, to have θ 3 < sin −1
−1
2
1
−1
1
2
−1
1
−1
1
P35.35
For the incoming ray,
sin θ 2 =
Using the figure to the right,
bθ g
2 violet
bθ g
2 red
For the outgoing ray,
and sin θ 4 = n sin θ 3 :
sin θ 1
.
n
= sin −1
bθ g
bθ g
4 violet
n sin θ = 1 . From Table 35.1,
(a)
θ = sin −1
(b)
θ = sin −1
(c)
θ = sin −1
FG 1 IJ = 24.4°
H 2.419 K
FG 1 IJ = 37.0°
H 1.66 K
FG 1 IJ = 49.8°
H 1.309 K
FIG. P35.35
= sin −1 1.66 sin 32.52° = 63.17°
= sin −1 1.62 sin 31.78° = 58.56° .
b g
P35.36
FG sin 50.0° IJ = 27.48°
H 1.66 K
FG sin 50.0° IJ = 28.22° .
H 1.62 K
= sin −1
The angular dispersion is the difference ∆θ 4 = θ 4
Total Internal Reflection
2
θ 3 = 60.0°−θ 2
4 red
Section 35.8
−1
violet
b g
− θ4
red
= 63.17°−58.56° = 4.61° .
329
330
P35.37
P35.38
The Nature of Light and the Laws of Geometric Optics
FG n IJ
Hn K
FG 1.333 IJ =
H 2.419 K
FG 1.333 IJ =
H 1.66 K
sin θ c =
n2
:
n1
θ c = sin −1
(a)
Diamond:
θ c = sin −1
(b)
Flint glass:
θ c = sin −1
(c)
Ice:
Since n 2 > n1 , there is no critical angle .
sin θ c =
n air
1.00
=
= 0.735
n pipe 1.36
2
1
33.4°
53.4°
θ c = 47.3°
Geometry shows that the angle of refraction at the end is
φ = 90.0°−θ c = 90.0°−47.3° = 42.7° .
1.00 sin θ = 1.36 sin 42.7°
Then, Snell’s law at the end,
FIG. P35.38
θ = 67.2° .
gives
The 2-µm diameter is unnecessary information.
P35.39
sin θ c =
n2
n1
b
gb
g
n 2 = n1 sin 88.8° = 1.000 3 0.999 8 = 1.000 08
*P35.40
(a)
A ray along the inner edge will escape if any ray escapes. Its angle of
R−d
and by n sin θ > 1 sin 90° . Then
incidence is described by sin θ =
R
n R−d
nd
.
>1
nR − nd > R
nR − R > nd
R>
n−1
R
a
(b)
(c)
P35.41
FIG. P35.39
f
As d → 0 , Rmin → 0 .
This is reasonable.
As n increases, Rmin decreases.
This is reasonable.
As n decreases toward 1, Rmin increases.
This is reasonable.
Rmin =
e
j=
1.40 100 × 10 −6 m
0.40
350 × 10 −6 m
From Snell’s law, n1 sin θ 1 = n 2 sin θ 2 .
At the extreme angle of viewing, θ 2 = 90.0°
a1.59fbsinθ g = a1.00f sin 90.0° .
1
So
θ 1 = 39.0° .
Therefore, the depth of the air bubble is
rp
rd
<d<
tan θ 1
tan θ 1
or
1.08 cm < d < 1.17 cm .
FIG. P35.41
FIG. P35.40
Chapter 35
P35.42
(a)
(b)
(c)
331
sin θ 2 v 2
=
sin θ 1 v1
and
θ 2 = 90.0° at the critical angle
sin 90.0° 1 850 m s
=
sin θ c
343 m s
so
θ c = sin −1 0.185 = 10.7° .
a
f
Sound can be totally reflected if it is traveling in the medium where it travels slower: air .
Sound in air falling on the wall from most directions is 100% reflected , so the wall is a
good mirror.
P35.43
For plastic with index of refraction n ≥ 1.42
reflection is given by
θ c = sin −1
surrounded by air, the critical angle for total internal
FG 1 IJ ≤ sin FG 1 IJ = 44.8° .
H nK
H 1.42 K
−1
In the gasoline gauge, skylight from above travels down the plastic. The rays close to the vertical are
totally reflected from the sides of the slab and from both facets at the lower end of the plastic, where
it is not immersed in gasoline. This light returns up inside the plastic and makes it look bright.
Where the plastic is immersed in gasoline, with index of refraction about 1.50, total internal
reflection should not happen. The light passes out of the lower end of the plastic with little reflected,
making this part of the gauge look dark. To frustrate total internal reflection in the gasoline, the
index of refraction of the plastic should be n < 2.12 .
since
Section 35.9
P35.44
θ c = sin −1
FG 1.50 IJ = 45.0° .
H 2.12 K
Fermat‘s Principle
Assume the lifeguard’s initial path makes angle θ 1 with the northsouth normal to the shoreline, and angle θ 2 with this normal in
the water. By Fermat’s principle, his path should follow the law of
refraction:
sin θ 1
sin θ 1 v1 7.00 m s
.
=
=
= 5.00 or θ 2 = sin −1
sin θ 2 v 2 1.40 m s
5
FG
H
IJ
K
FIG. P35.44
The lifeguard on land travels eastward a distance x = 16.0 m tan θ 1 . Then in the water, he travels
a f
L F sinθ
or 26.0 m = a16.0 mf tan θ + a 20.0 mf tan Msin G
N H 5
a
a
f
f
a
f
26.0 m − x = 20.0 m tan θ 2 further east. Thus, 26.0 m = 16.0 m tan θ 1 + 20.0 m tan θ 2
−1
1
We home in on the solution as follows:
θ 1 (deg)
right-hand side
1
IJ OP .
KQ
50.0
60.0
54.0
54.8
54.81
22.2 m
31.2 m
25.3 m
25.99 m
26.003 m
The lifeguard should start running at 54.8° east of north .
332
The Nature of Light and the Laws of Geometric Optics
Additional Problems
*P35.45
Scattered light leaves the center of the photograph (a) in all
horizontal directions between θ 1 = 0° and 90° from the
normal. When it immediately enters the water (b), it is
gathered into a fan between 0° and θ 2 max given by
(a)
θ 1 max
n1 sin θ 1 = n 2 sin θ 2
1.00 sin 90 = 1.333 sin θ 2 max
θ 2 max = 48.6°
θ 2 max
(b)
The light leaves the cylinder without deviation, so the
viewer only receives light from the center of the
photograph when he has turned by an angle less than
48.6°. When the paperweight is turned farther, light at the
back surface undergoes total internal reflection (c). The
viewer sees things outside the globe on the far side.
(c)
FIG. P35.45
P35.46
af
a
FG
H
af
(a)
(b)
IJ
K
The total time interval required to traverse the atmosphere is
∆t =
z z
h
af
dx h n x
=
dx :
v 0 c
0
z LMN
IJ OP
KQ
FG
H
∆t =
1h
n − 1.000
1.000 +
x dx
c0
h
∆t =
h n − 1.000
+
c
ch
a
f F h I = h FG n + 1.000 IJ
GH 2 JK c H 2 K
2
.
h
.
c
n + 1.000
times larger .
2
The travel time in the absence of an atmosphere would be
Thus, the time in the presence of an atmosphere is
P35.47
f
Let n x be the index of refraction at distance x below the top of the atmosphere and n x = h = n be
its value at the planet surface.
n − 1.000
Then,
n x = 1.000 +
x.
h
FG
H
IJ
K
Let the air and glass be medium 1 and 2, respectively. By Snell’s law,
n 2 sin θ 2 = n1 sin θ 1
or
1.56 sin θ 2 = sin θ 1 .
But the conditions of the problem are such that θ 1 = 2θ 2 .
1.56 sin θ 2 = sin 2θ 2 .
We now use the double-angle trig identity suggested.
1.56 sin θ 2 = 2 sin θ 2 cos θ 2
or
cos θ 2 =
Thus, θ 2 = 38.7° and θ 1 = 2θ 2 = 77.5° .
1.56
= 0.780 .
2
Chapter 35
P35.48
(a)
θ 1′ = θ 1 = 30.0°
333
n1 sin θ 1 = n 2 sin θ 2
1.00 sin 30.0° = 1.55 sin θ 2
θ 2 = 18.8°
(b)
θ 2 = sin −1
θ 1′ = θ 1 = 30.0°
= sin −1
FG n sinθ IJ
H n K
FG 1.55 sin 30.0° IJ =
H 1 K
1
FIG. P35.48
1
2
50.8°
(c), (d) The other entries are computed similarly, and are shown in the table below.
(c) air into glass, angles in degrees
incidence
0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
reflection
0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
refraction
0
6.43
12.7
18.8
24.5
29.6
34.0
37.3
39.4
40.2
(d) glass into air, angles in degrees
incidence
0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
reflection
0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
refraction
0
15.6
32.0
50.8
85.1
none*
none*
none*
none*
none*
*total internal reflection
P35.49
1
3
= .
43 4
For water,
sin θ c =
Thus
θ c = sin −1 0.750 = 48.6°
and
d = 2 1.00 m tan θ c
a
f
a
a
f
f
d = 2.00 m tan 48.6° = 2.27 m .
P35.50
FIG. P35.49
Call θ 1 the angle of incidence and of reflection on the
left face and θ 2 those angles on the right face. Let α
represent the complement of θ 1 and β be the
complement of θ 2 . Now α = γ and β = δ because
they are pairs of alternate interior angles. We have
A = γ +δ =α + β
and
B = α + A + β = α + β + A = 2A .
FIG. P35.50
334
P35.51
The Nature of Light and the Laws of Geometric Optics
(a)
We see the Sun moving from east to west across the sky. Its angular speed is
ω=
2π rad
∆θ
=
= 7.27 × 10 −5 rad s .
∆t 86 400 s
The direction of sunlight crossing the cell from the window changes at this rate, moving on
the opposite wall at speed
a
fe
j
v = rω = 2.37 m 7.27 × 10 −5 rad s = 1.72 × 10 −4 m s = 0.172 mm s .
(b)
The mirror folds into the cell the motion that would occur in a room twice as wide:
b
g
v = rω = 2 0.174 mm s = 0.345 mm s .
(c), (d) As the Sun moves southward and upward at 50.0°, we may regard the corner of the window
as fixed, and both patches of light move northward and downward at 50.0° .
* P35.52
(a)
45.0°
(b)
Yes
as shown in the first figure to the right.
If grazing angle is halved, the number of reflections from
the side faces is doubled.
P35.53
FIG. P35.52
Horizontal light rays from the setting Sun pass
above the hiker. The light rays are twice refracted
and once reflected, as in Figure (b). The most
intense light reaching the hiker, that which
represents the visible rainbow, is located between
angles of 40° and 42° from the hiker’s shadow.
The hiker sees a greater percentage of the violet
inner edge, so we consider the red outer edge. The
radius R of the circle of droplets is
a
Figure (a)
f
R = 8.00 km sin 42.0° = 5.35 km .
Then the angle φ, between the vertical and the
radius where the bow touches the ground, is given
by
cos φ =
2.00 km 2.00 km
=
= 0.374
5.35 km
R
φ = 68.1° .
or
The angle filled by the visible bow is
360°− 2 × 68.1° = 224°
a
f
224°
= 62.2% of a circle .
so the visible bow is
360°
Figure (b)
FIG. P35.53
Chapter 35
P35.54
Light passing the top of the pole makes an angle of incidence
φ 1 = 90.0°−θ . It falls on the water surface at distance from the pole
L−d
s1 =
tan θ
φ 2 from 1.00 sin φ 1 = n sin φ 2 .
and has an angle of refraction
s 2 = d tan φ 2
Then
and the whole shadow length is
FG FG IJ IJ
H H KK
L−d
F F cos θ IJ IJ
=
+ d tanG sin G
H H n KK
tan θ
2.00 m
F F cos 40.0° IJ IJ =
=
+ a 2.00 mf tanG sin G
H H 1.33 K K
tan 40.0°
s1 + s 2 =
s1 + s 2
sin φ 1
L−d
+ d tan sin −1
tan θ
n
−1
−1
P35.55
335
FIG. P35.54
3.79 m
As the beam enters the slab,
1.00 sin 50.0° = 1. 48 sin θ 2
θ 2 = 31.2° .
giving
FIG. P35.55
1.55 mm
The beam then strikes the top of the slab at x1 =
from the left end. Thereafter, the beam
tan 31.2°
strikes a face each time it has traveled a distance of 2 x 1 along the length of the slab. Since the slab is
420 mm long, the beam has an additional 420 mm − x1 to travel after the first reflection. The number
of additional reflections is
420 mm − x 1 420 mm − 1.55 mm tan 31.2°
=
= 81.5
2 x1
3.10 mm tan 31.2°
or 81 reflections
since the answer must be an integer. The total number of reflections made in the slab is then 82 .
P35.56
(a)
(b)
LM
N
S1′
n − n1
= 2
S1
n 2 + n1
OP = L 1.52 − 1.00 O
Q MN 1.52 + 1.00 PQ
2
2
= 0.042 6
LM
N
S1′
n − n1
= 2
S1
n 2 + n1
If medium 1 is glass and medium 2 is air,
OP = L 1.00 − 1.52 O
Q MN 1.00 + 1.52 PQ
2
There is no difference .
P35.57
(a)
n1 = 1
n2 = n
With
and
FG
H
S1′
n −1
=
S1
n +1
The remaining intensity must be transmitted:
the reflected fractional intensity is
FG
H
S2
n −1
=1−
S1
n +1
continued on next page
IJ = an + 1f − an − 1f
K
an + 1f
2
2
2
2
=
IJ
K
2
.
n 2 + 2n + 1 − n 2 + 2n − 1
a f
n +1
2
=
4n
a f
n+1
2
.
2
= 0.042 6 .
336
The Nature of Light and the Laws of Geometric Optics
(b)
FG
H
At entry,
S2
n −1
=1−
S1
n +1
At exit,
S3
= 0.828 .
S2
Overall,
S3
S
= 3
S1
S2
2
FG IJ FG S IJ = a0.828f
H KH S K
2
2
2
= 0.828 .
= 0.685
1
68.5% .
or
P35.58
IJ = 4a2.419f
K a2.419 + 1f
Define T =
4n
an + 1f
as the transmission coefficient for one
2
encounter with an interface. For diamond and air, it is 0.828, as
in Problem 57.
As shown in the figure, the total amount transmitted is
a
T2 +T2 1−T
a
f
+…+ T 2 1 − T
2
f
a
+T2 1−T
2n
f
4
a
+T2 1−T
f
6
+…
We have 1 − T = 1 − 0.828 = 0.172 so the total transmission is
a0.828f
f + a0.172f + a0.172f +… .
To sum this series, define F = 1 + a0.172 f + a0.172 f + a0.172 f + … .
Note that a0.172f F = a0.172 f + a0.172 f + a0.172 f +… , and
1 + a0.172 f F = 1 + a0.172f + a0.172 f + a0.172 f + … = F .
1
.
Then, 1 = F − a0.172f F or F =
1 − a0.172 f
a0.828f = 0.706 or 70.6% .
The overall transmission is then
1 − a0.172 f
2
a
1 + 0.172
2
4
6
2
2
2
2
2
4
4
6
FIG. P35.58
6
4
6
2
2
2
2
P35.59
Define n1 to be the index of refraction of the surrounding medium and n 2 to
be that for the prism material. We can use the critical angle of 42.0° to find the
n
ratio 2 :
n1
n 2 sin 42.0° = n1 sin 90.0° .
n2
1
=
= 1.49 .
n1 sin 42.0°
So,
Call the angle of refraction θ 2 at the surface 1. The ray inside the prism forms
a triangle with surfaces 1 and 2, so the sum of the interior angles of this
triangle must be 180°.
90.0°−θ 2 + 60.0°+ 90.0°−42.0° = 180° .
Thus,
b
Therefore,
Applying Snell’s law at surface 1,
sin θ 1 =
FG n IJ sinθ
Hn K
2
1
2
= 1.49 sin 18.0°
g
a
θ 2 = 18.0° .
n1 sin θ 1 = n 2 sin 18.0°
θ 1 = 27.5° .
f
FIG. P35.59
Chapter 35
*P35.60
(a)
As the mirror turns through angle θ, the angle of incidence
increases by θ and so does the angle of reflection. The incident ray
is stationary, so the reflected ray turns through angle 2θ . The
angular speed of the reflected ray is 2ω m . The speed of the dot of
light on the circular wall is 2ω m R .
(b)
The two angles marked θ in the figure to the right are equal
because their sides are perpendicular, right side to right side and
left side to left side.
We have
P35.61
(a)
cos θ =
d
2
x +d
2
=
ds
dx
and
ds
= 2ω m x 2 + d 2 .
dt
So
dx ds
=
dt dt
FIG. P35.60
x2 + d2
x2 + d2
= 2ω m
.
d
d
For polystyrene surrounded by air, internal reflection requires
θ 3 = sin −1
FG 1.00 IJ = 42.2° .
H 1.49 K
Then from geometry,
θ 2 = 90.0°−θ 3 = 47.8° .
From Snell’s law,
sin θ 1 = 1.49 sin 47.8° = 1.10 .
This has no solution.
Therefore, total internal reflection
(b)
(c)
always happens .
FIG. P35.61
FG 1.33 IJ = 63.2°
H 1.49 K
For polystyrene surrounded by water,
θ 3 = sin −1
and
θ 2 = 26.8° .
From Snell’s law,
θ 1 = 30.3° .
No internal refraction is possible
since the beam is initially traveling in a medium of lower index of refraction.
337
338
*P35.62
The Nature of Light and the Laws of Geometric Optics
θ1
The picture illustrates optical sunrise. At the center of the earth,
cos φ =
δ
6
6.37 × 10 m
6.37 × 10 6 m + 8 614
1
φ = 2.98°
θ 2 = 90 − 2.98° = 87.0°
2
θ2
φ
FIG. P35.62
At the top of the atmosphere
n1 sin θ 1 = n 2 sin θ 2
1 sin θ 1 = 1.000 293 sin 87.0°
θ 1 = 87.4°
Deviation upon entry is
δ = θ1 −θ 2
δ = 87.364°−87.022° = 0.342°
Sunrise of the optical day is before geometric sunrise by 0.342°
FG 86 400 s IJ = 82.2 s. Optical sunset
H 360° K
occurs later too, so the optical day is longer by 164 s .
P35.63
tan θ 1 =
4.00 cm
h
and
tan θ 2 =
2.00 cm
h
b
tan 2 θ 1 = 2.00 tan θ 2
sin 2 θ 1
1 − sin 2 θ 1
Snell’s law in this case is:
= 4.00
g
2
= 4.00 tan 2 θ 2
F sin θ I .
GH 1 − sin θ JK
2
2
2
(1)
2
n1 sin θ 1 = n 2 sin θ 2
sin θ 1 = 1.333 sin θ 2 .
Squaring both sides,
Substituting (2) into (1),
sin 2 θ 1 = 1.777 sin 2 θ 2 .
1.777 sin 2 θ 2
1 − 1.777 sin 2 θ 2
= 4.00
(2)
F sin θ I .
GH 1 − sin θ JK
Defining x = sin 2 θ ,
0.444
1
=
.
1 − 1.777 x 1 − x
Solving for x,
0.444 − 0.444x = 1 − 1.777 x
2
2
2
and
From x we can solve for θ 2 : θ 2 = sin −1 0.417 = 40.2° .
Thus, the height is
h=
FIG. P35.63
2
2.00 cm 2.00 cm
=
= 2.36 cm .
tan θ 2
tan 40.2°
x = 0.417 .
Chapter 35
P35.64
δ = θ 1 − θ 2 = 10.0°
and
n1 sin θ 1 = n 2 sin θ 2
with
n1 = 1 , n 2 =
b
339
4
.
3
g
b
g
θ 1 = sin −1 n 2 sin θ 2 = sin −1 n 2 sin θ 1 − 10.0° .
Thus,
(You can use a calculator to home in on an approximate solution to this equation, testing different
values of θ 1 until you find that θ 1 = 36.5° . Alternatively, you can solve for θ 1 exactly, as shown
below.)
P35.65
b
g
4
sin θ 1 − 10.0° .
3
We are given that
sin θ 1 =
This is the sine of a difference, so
3
sin θ 1 = sin θ 1 cos 10.0°− cos θ 1 sin 10.0° .
4
Rearranging,
sin 10.0° cos θ 1 = cos 10.0°−
sin 10.0°
= tan θ 1 and
cos 10.0°−0.750
θ 1 = tan −1
FG
H
a0.740f =
IJ
K
3
sin θ 1
4
36.5° .
To derive the law of reflection, locate point O so that the time of travel
from point A to point B will be minimum.
The total light path is L = a sec θ 1 + b sec θ 2 .
The time of travel is t =
FG 1 IJ ba sec θ
H vK
1
g
+ b sec θ 2 .
FIG. P35.65
If point O is displaced by dx, then
dt =
FG 1 IJ ba secθ
H vK
1
g
tan θ 1 dθ 1 + b sec θ 2 tan θ 2 dθ 2 = 0
(1)
(since for minimum time dt = 0 ).
Also,
c + d = a tan θ 1 + b tan θ 2 = constant
so,
a sec 2 θ 1 dθ 1 + b sec 2 θ 2 dθ 2 = 0 .
Divide equations (1) and (2) to find θ 1 = θ 2 .
(2)
340
P35.66
The Nature of Light and the Laws of Geometric Optics
Observe in the sketch that the angle of incidence at point P is γ, and
using triangle OPQ:
sin γ =
L
.
R
R 2 − L2
.
R
cos γ = 1 − sin 2 γ =
Also,
Applying Snell’s law at point P, 1.00 sin γ = n sin φ .
sin γ
L
=
n
nR
Thus,
sin φ =
and
cos φ = 1 − sin 2 φ =
FIG. P35.66
a
n 2 R 2 − L2
.
nR
f b
g
From triangle OPS, φ + α + 90.0° + 90.0°−γ = 180° or the angle of incidence at point S is α = γ − φ .
Then, applying Snell’s law at point S
b
1.00 sin θ = n sin α = n sin γ − φ
gives
LMF L I n R − L
MNGH R JK nR
L F
sin θ =
n R −L − R −L I
K
R H
LL
O
θ = sin M FH n R − L − R − L IK P .
NR
Q
2
sin θ = n sin γ cos φ − cos γ sin φ = n
or
2
2
−1
and
P35.67
g
2
2
2
2
2
2
2
2
2
−
FG
H
R 2 − L2 L
R
nR
IJ OP
K PQ
2
2
2
As shown in the sketch, the angle of incidence at point A is:
θ = sin −1
FG d 2 IJ = sin FG 1.00 m IJ = 30.0° .
H 2.00 m K
HRK
−1
If the emerging ray is to be parallel to the incident ray, the path
must be symmetric about the centerline CB of the cylinder. In
the isosceles triangle ABC,
γ =α
and
Therefore,
α + β + γ = 180°
becomes
2α + 180°−θ = 180°
or
α=
β = 180°−θ .
θ
= 15.0° .
2
Then, applying Snell’s law at point A,
n sin α = 1.00 sin θ
or
n=
sin θ sin 30.0°
=
= 1.93 .
sin α sin 15.0°
FIG. P35.67
Chapter 35
*P35.68
(a)
The apparent radius of the glowing sphere
is R3 as shown. For it
θ2
R
sin θ 1 = 1
R2
R
sin θ 2 = 3
R2
n sin θ 1 = 1 sin θ 2
n
(b)
P35.69
(a)
R1 R3
=
R 2 R2
θ1
R2
R3 = nR1
FIG. P35.68(a)
FIG. P35.68(b)
B
1
.
n
Since sin θ c =
Solving for d,
air
n
θc
Squaring the last equation gives:
1
, this becomes
n
d 4
sin θ c
d
=
.
or
cos θ c 4t
t
tan θ c =
sin 2 θ c
sin 2 θ c
FG IJ
H K
d
=
=
2
2
4t
cos θ c 1 − sin θ c
FG IJ
H K
1
d
=
4t
n2 − 1
d=
Thus, if n = 1.52 and t = 0.600 cm , d =
(c)
R3
At the boundary of the air and glass, the critical angle is given by
Consider the critical ray PBB′ :
(b)
R3
R1
If nR1 > R 2 , then sin θ 2 cannot be equal to
nR1
. The ray considered in part (a)
R2
undergoes total internal reflection. In this
case a ray escaping the atmosphere as
shown here is responsible for the apparent
radius of the glowing sphere and
R3 = R 2 .
sin θ c =
341
4t
n2 − 1
2
or n = 1 +
P
d /4
d
2
.
FG 4t IJ
HdK
FIG. P35.69
2
.
.
a
f=
a1.52f − 1
4 0.600 cm
2
B'
t
2.10 cm .
Since violet light has a larger index of refraction, it will lead to a smaller critical angle and
the inner edge of the white halo will be tinged with violet light.
342
P35.70
The Nature of Light and the Laws of Geometric Optics
From the sketch, observe that the angle of
incidence at point A is the same as the
prism angle θ at point O. Given that
θ = 60.0° , application of Snell’s law at
point A gives
1.50 sin β = 1.00 sin 60.0° or β = 35.3° .
From triangle AOB, we calculate the angle
of incidence (and reflection) at point B.
b
g b
FIG. P35.70
g
θ = 90.0°− β + 90.0°−γ = 180° so
γ = θ − β = 60.0°−35.3° = 24.7° .
Thus the angle of incidence at point C is
b90.0°−γ g + b90.0°−δ g + a90.0°−θ f = 180° .
δ = a90.0°−θ f − γ = 30.0°−24.7° = 5.30° .
Finally, Snell’s law applied at point C gives
1.00 sin φ = 1.50 sin 5.30°
or
φ = sin −1 1.50 sin 5.30° = 7.96° .
Now, using triangle BCQ:
P35.71
(a)
a
f
Given that θ 1 = 45.0° and θ 2 = 76.0° .
Snell’s law at the first surface gives
n sin α = 1.00 sin 45.0°
(1)
Observe that the angle of incidence at the second
surface is
β = 90.0°−α .
Thus, Snell’s law at the second surface yields
a
f
n sin β = n sin 90.0°−α = 1.00 sin 76.0°
or
(b)
n cos α = sin 76.0° .
FIG. P35.71
(2)
sin 45.0°
= 0.729
sin 76.0°
Dividing Equation (1) by Equation (2),
tan α =
or
α = 36.1° .
Then, from Equation (1),
n=
sin 45.0° sin 45.0°
=
= 1.20 .
sin α
sin 36.1°
From the sketch, observe that the distance the light travels in the plastic is d =
the speed of light in the plastic is v =
∆t =
a
f
c
, so the time required to travel through the plastic is
n
1.20 0.500 m
d
nL
=
=
= 3.40 × 10 −9 s = 3.40 ns .
v c sin α
3.00 × 10 8 m s sin 36.1°
e
j
L
. Also,
sin α
Chapter 35
P35.72
sin θ 1
sin θ 2
0.174
0.131
343
sin θ 1
sin θ 2
1.330 4
0.342
0.261
1.312 9
0.500
0.379
1.317 7
0.643
0.480
1.338 5
0.766
0.576
1.328 9
0.866
0.647
1.339 0
0.940
0.711
1.322 0
0.985
0.740
1.331 5
The straightness of the graph line demonstrates Snell’s
proportionality.
The slope of the line is
n = 1.327 6 ± 0.01
and
n = 1.328 ± 0.8% .
FIG. P35.72
ANSWERS TO EVEN PROBLEMS
P35.2
227 Mm s
P35.30
(a) 41.5° ; (b) 18.5°; (c) 27.6°; (d) 42.6°
P35.4
(a) see the solution; (b) 300 Mm s
P35.32
(a) see the solution; (b) 37.2°; (c) 37.3° ;
(d) 37.3°
P35.6
(a) 1.94 m; (b) 50.0° above the horizontal :
antiparallel to the incident ray
P35.34
sin −1
P35.36
(a) 24.4° ; (b) 37.0° ; (c) 49.8°
P35.38
67.2
FH
n 2 − 1 sin Φ − cos Φ
IK
P35.8
five times by the right-hand mirror and
six times by the left-hand mirror
P35.10
25.5°; 442 nm
P35.12
(a) 474 THz ; (b) 422 nm; (c) 200 Mm s
P35.40
(a)
P35.14
22.5°
P35.42
P35.16
(a) 181 Mm s ; (b) 225 Mm s ;
(c) 136 Mm s
(a) 10.7°; (b) air; (c) Sound falling on the
wall from most directions is 100%
reflected.
P35.44
54.8° east of north
P35.46
(a)
P35.48
see the solution
P35.50
see the solution
P35.52
(a) 45.0°; (b) yes; see the solution
P35.54
3.79 m
P35.18
3.39 m
−1
P35.20
θ 1 = tan n
P35.22
106 ps
P35.24
23.1°
P35.26
(a) 58.9° ; (b) Only if θ 1 = θ 2 = 0
P35.28
see the solution
nd
; (b) yes; (c) 350 µm
n −1
FG
H
IJ
K
h n +1
n +1
; (b) larger by
times
2
2
c
344
P35.56
P35.58
The Nature of Light and the Laws of Geometric Optics
(a) 0.042 6 ; (b) no difference
θ = sin −1
P35.68
(a) nR1 ; (b) R 2
P35.70
7.96°
P35.72
see the solution; n = 1.328 ± 0.8%
0.706
P35.60
(a) 2ω m R ; (b) 2ω m
P35.62
164 s
P35.64
36.5°
x2 + d2
d
LM L F
NR H
P35.66
2
n 2 R 2 − L2 − R 2 − L2
IK OP
Q
36
Image Formation
CHAPTER OUTLINE
Images Formed by Flat
Mirrors
36.2 Images Formed by Spherical
Mirrors
36.3 Images Formed by
Refraction
36.4 Thin Lenses
36.5 Lens Aberrations
36.6 The Camera
36.7 The Eye
36.8 The Simple Magnifier
36.9 The Compound Microscope
36.10 The Telescope
ANSWERS TO QUESTIONS
36.1
Q36.1
The mirror shown in the textbook picture produces an inverted
image. It actually reverses top and bottom. It is not true in the
same sense that “Most mirrors reverse left and right.” Mirrors
don’t actually flip images side to side—we just assign the labels
“left” and “right” to images as if they were real people
mimicking us. If you stand face to face with a real person and
raise your left hand, then he or she would have to raise his or
her right hand to “mirror” your movement. Try this while
facing a mirror. For sake of argument, let’s assume you are
facing north and wear a watch on your left hand, which is on
the western side. If you raise your left hand, you might say that
your image raises its right hand, based on the labels we assign
to other people. But your image raises its western-side hand,
which is the hand with the watch.
Q36.2
With a concave spherical mirror, for objects beyond the focal length the image will be real and
inverted. For objects inside the focal length, the image will be virtual, upright, and magnified. Try a
shaving or makeup mirror as an example.
Q36.3
With a convex spherical mirror, all images of real objects are upright, virtual and smaller than the
object. As seen in Question 36.2, you only get a change of orientation when you pass the focal
point—but the focal point of a convex mirror is on the non-reflecting side!
Q36.4
The mirror equation and the magnification equation apply to plane mirrors. A curved mirror is
made flat by increasing its radius of curvature without bound, so that its focal length goes to infinity.
1 1 1
1
1
From + = = 0 we have = − ; therefore, p = − q . The virtual image is as far behind the mirror
p q f
p
q
q p
as the object is in front. The magnification is M = − = = 1 . The image is right side up and actual
p p
size.
Q36.5
Stones at the bottom of a clear stream always appears closer to the surface because light is refracted
away from the normal at the surface. Example 36.8 in the textbook shows that its apparent depth is
three quarters of its actual depth.
345
346
Q36.6
Image Formation
For definiteness, we consider real objects (p > 0 ).
q
1 1 1
to be negative, q must be positive. This will happen in = − if p > f , if the
q f p
p
object is farther than the focal point.
(a)
For M = −
(b)
For M = −
(c)
For a real image, q must be positive.
As in part (a), it is sufficient for p to be larger than f.
(d)
For q < 0 we need p < f .
(e)
For M > 1 , we consider separately
q
If M = − < −1, we need
p
q
to be positive, q must be negative.
p
1 1 1
From = − we need p < f .
q f p
From
q
>1
p
1 1
<
q p
1 1 1
+ >
p p f
or
2 1
>
p f
or
p
<f
2
or
p<2f .
or
−q > p
or
q < −p
1 1 1
= −
p f q
with
1
>0
f
or
−p > q.
we may require q < 0 , since then
gives
q>p
or
1 1 1
+ = ,
p q f
Now if −
M < −1 and M > 1 .
q
>1
or
p
1
1
> − as required
p
q
1 1 1
= − we need
p< f.
q f p
Thus the overall condition for an enlarged image is simply p < 2 f .
For q < 0 in
(f)
Q36.7
For M < 1 , we have the reverse of part (e), requiring p > 2 f .
Using the same analysis as in Question 36.6 except f < 0 .
(a)
Never.
(b)
Always.
(c)
Never, for light rays passing through the lens will always diverge.
(d)
Always.
(e)
Never.
(f)
Always.
Chapter 36
Q36.8
347
We assume the lens has a refractive index higher than its surroundings. For the biconvex lens in
1
1
−
are positive and f > 0 . For the
Figure 36.27(a), R1 > 0 and R 2 < 0 . Then all terms in n − 1
R1 R 2
other two lenses in part (a) of the figure, R1 and R 2 are both positive but R1 is less than R 2 . Then
1
1
>
and the focal length is again positive.
R1 R 2
a fFGH
IJ
K
For the biconcave lens and the plano-concave lens in Figure 36.27(b), R1 < 0 and R 2 > 0 . Then
1
1
−
and the focal length is negative. For the middle lens in part (b)
both terms are negative in
R1 R 2
1
1
of the figure, R1 and R 2 are both positive but R1 is greater than R 2 . Then
<
and the focal
R1 R 2
length is again negative.
Q36.9
Both words are inverted. However OXIDE has up-down symmetry whereas LEAD does not.
Q36.10
An infinite number. In general, an infinite number
of rays leave each point of any object and travel in
all directions. Note that the three principal rays
that we use for imaging are just a subset of the
infinite number of rays. All three principal rays can
be drawn in a ray diagram, provided that we
extend the plane of the lens as shown in
Figure Q36.10.
O
F
F
I
FIG. Q36.10
Q36.11
In this case, the index of refraction of the lens material is less than that of the surrounding medium.
Under these conditions, a biconvex lens will be diverging.
Q36.12
Chromatic aberration arises because a material medium’s refractive index can be frequency
dependent. A mirror changes the direction of light by reflection, not refraction. Light of all
wavelengths follows the same path according to the law of reflection, so no chromatic aberration
happens.
Q36.13
This is a convex mirror. The mirror gives the driver a wide field of view and an upright image with
the possible disadvantage of having objects appear diminished. Your brain can then interpret them
as farther away than the objects really are.
Q36.14
As pointed out in Question 36.11, if the converging lens is immersed in a liquid with an index of
refraction significantly greater than that of the lens itself, it will make light from a distant source
diverge. This is not the case with a converging (concave) mirror, as the law of reflection has nothing
to do with the indices of refraction.
348
Q36.15
Image Formation
As in the diagram, let the center of curvature C of the fishbowl
and the bottom of the fish define the optical axis, intersecting
the fishbowl at vertex V. A ray from the top of the fish that
reaches the bowl surface along a radial line through C has
angle of incidence zero and angle of refraction zero. This ray
exits from the bowl unchanged in direction. A ray from the top
of the fish to V is refracted to bend away from the normal. Its
extension back inside the fishbowl determines the location of
the image and the characteristics of the image. The image is
upright, virtual, and enlarged.
C
V
O I
FIG. Q36.15
Q36.16
Because when you look at the
in your rear view mirror, the apparent left-right
inversion clearly displays the name of the AMBULANCE behind you. Do not jam on your brakes
when a MIAMI city bus is right behind you.
Q36.17
The entire image is visible, but only at half the intensity. Each point on the object is a source of rays
that travel in all directions. Thus, light from all parts of the object goes through all unblocked parts
of the lens and forms an image. If you block part of the lens, you are blocking some of the rays, but
the remaining ones still come from all parts of the object.
Q36.18
With the meniscus design, when you direct your gaze near the outer circumference of the lens you
receive a ray that has passed through glass with more nearly parallel surfaces of entry and exit.
Thus, the lens minimally distorts the direction to the object you are looking at. If you wear glasses,
turn them around and look through them the wrong way to maximize this distortion.
Q36.19
The eyeglasses on the left are diverging lenses that correct for nearsightedness. If you look carefully
at the edge of the person’s face through the lens, you will see that everything viewed through these
glasses is reduced in size. The eyeglasses on the right are converging lenses, which correct for
farsightedness. These lenses make everything that is viewed through them look larger.
Q36.20
The eyeglass wearer’s eye is at an object distance from the lens that is quite small—the eye is on the
order of 10 −2 meter from the lens. The focal length of an eyeglass lens is several decimeters, positive
or negative. Therefore the image distance will be similar in magnitude to the object distance. The
onlooker sees a sharp image of the eye behind the lens. Look closely at the left side of Figure Q36.19
and notice that the wearer’s eyes seem not only to be smaller, but also positioned a bit behind the
plane of his face—namely where they would be if he was not wearing glasses. Similarly, in the right
half of Figure Q36.19, his eyes seem to be in front of the plane of his face and magnified. We as
observers take this light information coming from the object through the lens and perceive or
photograph the image as if it were an object.
Q36.21
In the diagram, only two of the three principal rays have
been used to locate images to reduce the amount of visual
clutter. The upright shaded arrows are the objects, and the
correspondingly numbered inverted arrows are the images.
As you can see, object 2 is closer to the focal point than
object 1, and image 2 is farther to the left than image 1.
O1
O2
F
C
V
I2 I1
FIG. Q36.21
Chapter 36
349
Q36.22
Absolutely. Only absorbed light, not transmitted light, contributes internal energy to a transparent
object. A clear lens can stay ice-cold and solid as megajoules of light energy pass through it.
Q36.23
One can change the f number either by changing the focal length (if using a “zoom” lens) or by
changing the aperture of the camera lens. As the f number increases, the exposure time required
increases also, as both increasing the focal length or decreasing the aperture decreases the light
intensity reaching the film.
Q36.24
Make the mirror an efficient reflector (shiny). Make it reflect to the image even rays far from the axis,
by giving it a parabolic shape. Most important, make it large in diameter to intercept a lot of solar
power. And you get higher temperature if the image is smaller, as you get with shorter focal length;
and if the furnace enclosure is an efficient absorber (black).
Q36.25
For the explanation, we ignore the lens and consider two objects. Hold your two thumbs parallel
and extended upward in front of you, at different distances from your nose. Alternately close your
left eye and your right eye. You see both thumbs jump back and forth against the background of
more distant objects. Parallax by definition is this apparent motion of a stationary object (one
thumb) caused by motion of the observer (jumping from right eye to left eye). Your nearer thumb
jumps by a larger angle against the background than your farther thumb does. They will jump by
the same amount only if they are equally distant from your face. The method of parallax for
adjusting one object so that it is the same distance away from you as another object will work even if
one ’object’ is an image.
Q36.26
The artist’s statements are accurate, perceptive, and eloquent. The image you see is “almost one’s
whole surroundings,” including things behind you and things farther in front of you than the globe
is, but nothing eclipsed by the opaque globe or by your head. For example, we cannot see Escher’s
index and middle fingers or their reflections in the globe.
The point halfway between your eyes is indeed the focus in a figurative sense, but it is not an
optical focus. The principal axis will always lie in a line that runs through the center of the sphere
and the bridge of your nose. Outside the globe, you are at the center of your observable universe. If
you wink at the ball, the center of the looking-glass world hops over to the location of the image of
your open eye.
Q36.27
The three mirrors, two of which are shown as M
and N in the figure to the right, reflect any incident
ray back parallel to its original direction. When you
look into the corner you see image I 3 of yourself.
FIG. Q36.21
350
Q36.28
Image Formation
You have likely seen a Fresnel mirror for sound.
The diagram represents first a side view of a band
shell. It is a concave mirror for sound, designed to
channel sound into a beam toward the audience in
front of the band shell. Sections of its surface can
be kept at the right orientations as they are pushed
around inside a rectangular box to form an
auditorium with good diffusion of sound from
stage to audience, with a floor plan suggested by
the second part of the diagram.
FIG. Q36.28
SOLUTIONS TO PROBLEMS
Section 36.1
P36.1
Images Formed by Flat Mirrors
I stand 40 cm from my bathroom mirror. I scatter light, which travels to the mirror and back to me in
time
0.8 m
~ 10 −9 s
8
3 × 10 m s
showing me a view of myself as I was at that look-back time. I’m no Dorian Gray!
P36.2
The virtual image is as far behind the mirror as the choir is in front
of the mirror. Thus, the image is 5.30 m behind the mirror. The
image of the choir is 0.800 m + 5.30 m = 6.10 m from the organist.
Using similar triangles:
View Looking Down
South
image of choir
mirror
h′
6.10 m
=
0.600 m 0.800 m
or
a
6.10 m I
fFGH 0.800
J=
mK
h′ = 0.600 m
0.600 m
4.58 m .
Organist
0.800 m
5.30 m
FIG. P36.2
h'
Chapter 36
P36.3
351
The flatness of the mirror is described
R=∞, f =∞
1
=0.
f
by
and
By our general mirror equation,
1 1 1
+ = =0
p q f
q = −p .
or
FIG. P36.3
Thus, the image is as far behind the mirror as the person is in front. The magnification is then
M=
−q
h′
=1=
p
h
h′ = h = 70.0 inches .
so
The required height of the mirror is defined by the triangle from the person’s eyes to the top and
bottom of his image, as shown. From the geometry of the triangle, we see that the mirror height
must be:
h′
F p I = h ′F p I = h ′ .
GH p − q JK GH 2p JK 2
Thus, the mirror must be at least 35.0 inches high .
P36.4
A graphical construction produces 5 images, with images I 1 and I 2
directly into the mirrors from the object O,
and
and
bO, I , I g
bI , I , I g
2
3
4
1
5
forming the vertices of equilateral triangles.
FIG. P36.4
P36.5
(1)
The first image in the left mirror is 5.00 ft behind the mirror, or 10.0 ft from the position of
the person.
(2)
The first image in the right mirror is located 10.0 ft behind the right mirror, but this location
is 25.0 ft from the left mirror. Thus, the second image in the left mirror is 25.0 ft behind the
mirror, or 30.0 ft from the person.
(3)
The first image in the left mirror forms an image in the right mirror. This first image is 20.0 ft
from the right mirror, and, thus, an image 20.0 ft behind the right mirror is formed. This
image in the right mirror also forms an image in the left mirror. The distance from this
image in the right mirror to the left mirror is 35.0 ft. The third image in the left mirror is,
thus, 35.0 ft behind the mirror, or 40.0 ft from the person.
352
*P36.6
Image Formation
(a)
The flat mirrors have
R→∞
and
f → ∞.
The upper mirror M 1
produces a virtual, actual
sized image I 1 according
to
1
1 1 1
+
= = =0
p1 q 1 f ∞
q1 = − p 1
with M 1 = −
q1
= +1 .
p1
As shown, this image is
above the upper mirror. It
is the object for mirror
M 2 , at object distance
p 2 = p1 + h .
The lower mirror
produces a virtual, actualsize, right-side-up image
according to
1
1
+
=0
p 2 q2
b
q 2 = − p 2 = − p1 + h
with M 2 = −
FIG. P36.6
g
q2
= +1 and M overall = M 1 M 2 = 1.
p2
Thus the final image is at distance p1 + h behind the lower mirror.
(b)
(c)
It is virtual .
Upright
(d)
With magnification +1 .
(e)
It does not appear to be reversed left and right. In a top view of the periscope, parallel
rays from the right and left sides of the object stay parallel and on the right and left.
Chapter 36
Section 36.2
P36.7
Images Formed by Spherical Mirrors
For a concave mirror, both R and f are positive.
f=
We also know that
(a)
R
= 10.0 cm .
2
1 1 1
1
1
3
= − =
−
=
q f p 10.0 cm 40.0 cm 40.0 cm
q = 13.3 cm
and
M=
q
13.3 cm
=−
= −0.333 .
40.0 cm
p
The image is 13.3 cm in front of the mirror, real, and inverted .
(b)
1 1 1
1
1
1
= − =
−
=
q f p 10.0 cm 20.0 cm 20.0 cm
q = 20.0 cm
and
M=
q
20.0 cm
=−
= −1.00 .
20.0 cm
p
The image is 20.0 cm in front of the mirror, real, and inverted .
(c)
1 1 1
1
1
= − =
−
=0
q f p 10.0 cm 10.0 cm
Thus,
q = infinity.
No image is formed . The rays are reflected parallel to each other.
P36.8
1 1 1
1
1
= − =−
−
q f p
0.275 m 10.0 m
gives
q = −0.267 m .
and
diminished
Thus, the image is virtual .
M=
−q
−0.267
=−
= 0.026 7
10.0 m
p
a f
Thus, the image is upright +M
c M < 1h .
353
354
P36.9
Image Formation
(a)
1 1 2
+ =
p q R
gives
1
1
2
+ =
30.0 cm q
−40.0 cm
1
2
1
=−
−
= −0.083 3 cm −1
40.0 cm 30.0 cm
q
so
q = −12.0 cm
a
M=
(b)
P36.10
f
−12.0 cm
−q
=−
= 0.400 .
p
30.0 cm
1 1 2
+ =
p q R
gives
1
1
2
+ =
−40.0 cm
60.0 cm q
1
2
1
=−
−
= −0.066 6 cm −1
40.0 cm 60.0 cm
q
so
q = −15.0 cm
a
M=
(c)
a
f
a
f
f
−15.0 cm
−q
=−
= 0.250 .
p
60.0 cm
Since M > 0 , the images are upright .
With radius 2.50 m, the cylindrical wall is a highly efficient mirror for sound, with focal length
f=
R
= 1.25 m .
2
In a vertical plane the sound disperses as usual, but that radiated in a horizontal plane is
concentrated in a sound image at distance q from the back of the niche, where
1 1 1
+ =
p q f
so
1
1
1
+ =
2.00 m q 1.25 m
q = 3.33 m .
P36.11
(a)
(b)
(c)
1 1 2
+ =
p q R
becomes
1
2
1
=
−
q 60.0 cm 90.0 cm
q = 45.0 cm
and
M=
1 1 2
+ =
p q R
becomes
1
2
1
=
−
q 60.0 cm 20.0 cm
q = −60.0 cm
and
M=
−q
45.0 cm
=−
= −0.500 .
90.0 cm
p
a
a
f
f
−60.0 cm
−q
=−
= 3.00 .
p
20.0 cm
The image (a) is real, inverted and diminished. That of (b) is
virtual, upright, and enlarged. The ray diagrams are similar
to Figure 36.15(a) and 36.15(b) in the text, respectively.
FIG. P36.11
Chapter 36
P36.12
355
For a concave mirror, R and f are positive. Also, for an erect image, M is positive. Therefore,
q
M = − = 4 and q = −4p .
p
1 1 1
1
1 1
3
= + becomes
= −
=
; from which, p = 30.0 cm .
f p q
40.0 cm p 4p 4p
*P36.13
The ball is a convex mirror with R = −4. 25 cm and
R
f = = −2.125 cm. We have
2
M=
q
3
=−
4
p
O
3
p
4
1 1 1
+ =
p q f
I F
q=−
FIG. P36.13
1
1
1
+
=
p − 3 4 p −2.125 cm
b g
3
4
1
−
=
3 p 3 p −2.125 cm
3 p = 2.125 cm
p = 0.708 cm in front of the sphere.
The image is upright, virtual, and diminished.
*P36.14
(a)
(b)
M = −4 = −
q
p
q = 4p
q − p = 0.60 m = 4p − p
p = 0.2 m
1 1 1
1
1
= + =
+
f p q 0.2 m 0.8 m
f = 160 mm
M=+
q
1
=−
2
p
p = −2 q
q + p = 0.20 m = − q + p = − q − 2 q
q = −66.7 mm
p = 133 mm
1 1 2
1
1
+ = =
+
p q R 0.133 m −0.066 7 m
R = −267 mm
q = 0.8 m
C
356
* P36.15
Image Formation
M=−
q
p
a
f
q = − Mp = −0.013 30 cm = −0.39 cm
1 1 1 2
+ = =
p q f R
1
1
2
+
=
30 cm −0.39 cm R
2
R=
= −0.790 cm
−2.53 m −1
FIG. P36.15
The cornea is convex, with radius of curvature 0.790 cm .
*P36.16
With
q
h ′ +4.00 cm
=
= +0.400 = −
10.0 cm
h
p
q = −0.400 p
M=
the image must be virtual.
(a)
It is a convex mirror that produces a diminished upright virtual image.
(b)
We must have
p + q = 42.0 cm = p − q
p = 42.0 cm + q
p = 42.0 cm − 0. 400 p
p=
42.0 cm
= 30.0 cm
1.40
The mirror is at the 30.0 cm mark .
(c)
1 1 1
1
1
1
+ = =
+
= = − 0.050 0 cm
p q f 30 cm −0.4 30 cm
f
a
f = −20.0 cm
f
The ray diagram looks like Figure 36.15(c) in the text.
P36.17
(a)
b
g
q = p + 5.00 m and, since the image must be real,
M=−
q
= −5
p
or
p + 5.00 m = 5 p
Therefore,
or
p = 1.25 m
From
1 1 2
+ = ,
p q R
q = 5p .
and
q = 6.25 m .
R=
a fa f
2 pq 2 1.25 6.25
=
1. 25 + 6.25
p+q
a
= 2.08 m concave
(b)
FIG. P36.17
f
From part (a), p = 1.25 m ; the mirror should be 1.25 m in front of the object.
Chapter 36
P36.18
Assume that the object distance is the same in both cases (i.e., her face is the same distance from the
hubcap regardless of which way it is turned). Also realize that the near image ( q = −10.0 cm ) occurs
when using the convex side of the hubcap. Applying the mirror equation to both cases gives:
(concave side: R = R ,
a
(convex side: R = − R ,
(1)
f
q = −10.0 cm )
1
1
2
−
=−
p 10.0
R
2 p − 10.0 cm
=
.
R
10.0 cm p
or
(a)
q = −30.0 cm )
1
1
2
−
=
p 30.0 R
2 30.0 cm − p
=
R
30.0 cm p
or
a
(2)
f
Equating Equations (1) and (2) gives:
or
30.0 cm − p
= p − 10.0 cm
3.00
p = 15.0 cm .
Thus, her face is 15.0 cm from the hubcap.
(b)
Using the above result ( p = 15.0 cm ) in Equation (1) gives:
2
30.0 cm − 15.0 cm
=
R
30.0 cm 15.0 cm
2
1
=
R 30.0 cm
a
or
and
fa
f
R = 60.0 cm .
The radius of the hubcap is 60.0 cm .
*P36.19
357
(a)
The flat mirror produces an image according to
1 1 1 2
+ = =
p q f R
1
1 1
+ = =0
24 cm q ∞
q = −24.0 m.
The image is 24.0 m behind the mirror, distant from your eyes by
1.55 m + 24.0 m = 25.6 m .
(b)
The image is the same size as the object, so
θ=
h 1.50 m
=
= 0.058 7 rad .
d 25.6 m
(c)
1 1 2
+ =
p q R
q=
1
= −0.960 m
− 1 1 m − 1 24 m
1
1
2
+ =
−2 m
24 m q
a
This image is distant from your eyes by
continued on next page
f
b
g b
g
1.55 m + 0.960 m = 2.51 m .
358
Image Formation
(d)
The image size is given by M =
q
h′
=−
h
p
h′ = − h
θ′ =
So its angular size at your eye is
(e)
(a)
h ′ 0.06 m
=
= 0.023 9 rad .
d 2.51 m
Your brain assumes that the car is 1.50 m high and calculate its distance as
d′ =
P36.20
IJ
K
FG
H
q
−0.960 m
= 0.060 0 m.
= −1.50 m
p
24 m
h 1.50 m
=
= 62.8 m .
θ ′ 0.023 9
The image starts from a point whose height above the mirror vertex is given by
1 1 1 2
+ = =
p q f R
1
1
1
+ =
.
3.00 m q 0.500 m
Therefore,
q = 0.600 m .
As the ball falls, p decreases and q increases. Ball and image pass when q1 = p1 . When this is
true,
1
1
1
2
+
=
=
p1 p1 0.500 m p1
p1 = 1.00 m.
or
As the ball passes the focal point, the image switches from infinitely far above the mirror to
infinitely far below the mirror. As the ball approaches the mirror from above, the virtual
image approaches the mirror from below, reaching it together when p 2 = q 2 = 0 .
(b)
The falling ball passes its real image when it has fallen
3.00 m − 1.00 m = 2.00 m =
a
f
2 2.00 m
1 2
gt , or when t =
= 0.639 s .
2
9.80 m s 2
The ball reaches its virtual image when it has traversed
3.00 m − 0 = 3.00 m =
Section 36.3
P36.21
a
f
2 3.00 m
1 2
gt , or at t =
= 0.782 s .
2
9.80 m s 2
Images Formed by Refraction
n1 n 2 n 2 − n 1
+
=
= 0 and R → ∞
p
q
R
q=−
a
f
n2
1
50.0 cm = −38.2 cm
p=−
1.309
n1
Thus, the virtual image of the dust speck is 38.2 cm below the top surface of the ice.
Chapter 36
P36.22
359
When R → ∞ , the equation describing image formation at a single refracting surface becomes
FG n IJ . We use this to locate the final images of the two surfaces of the glass plate. First, find
Hn K
the image the glass forms of the bottom of the plate.
F 1.33 IJ a8.00 cmf = −6.41 cm
q = −G
H 1.66 K
q = −p
2
1
B1
This virtual image is 6.41 cm below the top surface of the glass of 18.41 cm below the water surface.
Next, use this image as an object and locate the image the water forms of the bottom of the plate.
1.00
18.41 cm = −13.84 cm
qB2 = −
or
13.84 cm below the water surface.
1.33
Now find image the water forms of the top surface of the glass.
1
q3 = −
or
9.02 cm below the water surface.
12.0 cm = −9.02 cm
1.33
FG IJ a
H K
FG IJ a
H K
f
f
Therefore, the apparent thickness of the glass is ∆t = 13.84 cm − 9.02 cm = 4.82 cm .
P36.23
From Equation 36.8
n1 n 2 n 2 − n 1
+
=
.
p
q
R
Solve for q to find
q=
In this case,
n1 = 1.50 , n 2 = 1.00 , R = −15.0 cm
and
p = 10.0 cm .
So
q=
n1 n 2 n 2 − n1
+
=
p
q
R
b
a1.00fa−15.0 cmfa10.0 cmf
a10.0 cmfa1.00 − 1.50f − a1.50fa−15.0 cmf = −8.57 cm .
so
1.00
1.40
1.40 − 1.00
+
=
∞
21.0 mm 6.00 mm
0.066 7 = 0.066 7 .
and
They agree.
P36.25
g
apparent depth is 8.57 cm .
Therefore, the
P36.24
n 2 Rp
.
p n 2 − n 1 − n1 R
The image is inverted, real and diminished.
n1 n 2 n 2 − n 1
+
=
p
q
R
becomes
1.00 1.50 1.50 − 1.00
1
+
=
=
p
q
6.00 cm
12.0 cm
1.50
(a)
1.00
1.50
1
+
=
20.0 cm
q
12.0 cm
or
q=
b1.00 12.0 cmg − b1.00 20.0 cmg =
(b)
1.00
1.50
1
+
=
10.0 cm
q
12.0 cm
or
q=
b
(c)
1.00
1.50
1
+
=
3.0 cm
q
12.0 cm
or
q=
1.50
g b
g=
1.00 12.0 cm − 1.00 10.0 cm
1.50
b1.00 12.0 cmg − b1.00 3.0 cmg =
45.0 cm
−90.0 cm
−6.00 cm
360
P36.26
Image Formation
p = ∞ and q = +2 R
1.00 n 2 n 2 − 1.00
+
=
p
q
R
0+
n 2 n 2 − 1.00
=
R
2R
n 2 = 2.00
so
FIG. P36.26
P36.27
n p
n1 n 2 n 2 − n 1
+
=
becomes q = − 2 .
p
q
R
n1
For a plane surface,
Thus, the magnitudes of the rate of change in the image and object positions are related by
dq n 2 dp
.
=
dt n1 dt
If the fish swims toward the wall with a speed of 2.00 cm s , the speed of the image is given by
v image =
Section 36.4
P36.28
dq 1.00
=
2.00 cm s = 1.50 cm s .
dt 1.33
b
g
Thin Lenses
Let R1 = outer radius and R 2 = inner radius
OP a
Q
a fLM
N
fLMN
OP
Q
1
1
1
1
1
= n−1
−
= 1.50 − 1
−
= 0.050 0 cm −1
2.00 m 2.50 cm
f
R1 R 2
so
P36.29
(a)
f = 20.0 cm .
a fLM
N
OP a
Q
fLM
N
OP
fQ
1
1
1
1
1
= n−1
−
= 0.440
−
12.0 cm −18.0 cm
f
R1 R 2
a
f = 16.4 cm
(b)
a
fLM
N
OP
fQ
1
1
1
= 0.440
−
f
18.0 cm −12.0 cm
f = 16.4 cm
a
FIG. P36.29
Chapter 36
P36.30
For a converging lens, f is positive. We use
(a)
1 1 1
+ = .
p q f
1 1 1
1
1
1
= − =
−
=
q f p 20.0 cm 40.0 cm 40.0 cm
M=−
q = 40.0 cm
q
40.0
=−
= −1.00
40.0
p
The image is real, inverted , and located 40.0 cm past the lens.
(b)
1 1 1
1
1
= − =
−
=0
q f p 20.0 cm 20.0 cm
q = infinity
No image is formed. The rays emerging from the lens are parallel to each other.
(c)
1 1 1
1
1
1
= − =
−
=−
q f p 20.0 cm 10.0 cm
20.0 cm
M=−
a
q = −20.0 cm
f
−20.0
q
=−
= 2.00
p
10.0
The image is upright, virtual and 20.0 cm in front of the lens.
P36.31
(a)
1 1 1
1
1
= − =
−
q f p 25.0 cm 26.0 cm
q = 650 cm
The image is real, inverted, and enlarged .
(b)
1 1 1
1
1
= − =
−
q f p 25.0 cm 24.0 cm
q = −600 cm
The image is virtual, upright, and enlarged .
P36.32
(a)
1 1 1
+ = :
p q f
so
1
1
1
+
=
32.0 cm 8.00 cm f
f = 6.40 cm
q
8.00 cm
=−
= −0.250
32.0 cm
p
(b)
M=−
(c)
Since f > 0 , the lens is converging .
361
362
P36.33
Image Formation
We are looking at an enlarged, upright, virtual image:
M=
q
h′
=2=−
h
p
1 1 1
+ =
p q f
a
f
−2.84 cm
q
=−
= +1. 42 cm
2
2
so
p=−
gives
1
1
1
+
=
1.42 cm −2.84 cm
f
a
f
f = 2.84 cm .
FIG. P36.33
P36.34
(a)
(b)
1 1 1
+ = :
p q f
1
1
1
+
=
p −30.0 cm 12.5 cm
p = 8.82 cm
M=−
a
f
−30.0
q
=−
= 3.40 , upright
p
8.82
See the figure to the right.
FIG. P36.34(b)
P36.35
1 1 1
+ = :
p q f
p −1 + q −1 = constant
We may differentiate through with respect to p:
−1p −2 − 1q −2
dq
=0
dp
dq
q2
= − 2 = −M 2 .
dp
p
P36.36
P36.37
M=
The image is inverted:
−q
−1.80 m
h′
=
= −75.0 =
h 0.024 0 m
p
(b)
q + p = 3.00 m = 75.0 p + p
p = 39.5 mm
(a)
q = 2.96 m
1 1 1
1
1
= + =
+
f p q 0.039 5 m 2.96 m
(a)
1 1 1
+ =
p q f
1
1
1
+ =
−32.0 cm
20.0 cm q
so
q=−
a
FG 1 + 1 IJ
H 20.0 32.0 K
q = 75.0 p .
f = 39.0 mm
f
−1
= −12.3 cm
The image is 12.3 cm to the left of the lens.
a
f
−12.3 cm
q
=−
= 0.615
p
20.0 cm
(b)
M=−
(c)
See the ray diagram to the right.
FIG. P36.37
Chapter 36
*P36.38
In
1 1 1
+ =
p q f
p −1 + q −1 = constant,
we differentiate with respect to time
dp
dq
−1 p −2
− 1 q −2
=0
dt
dt
dq − q 2 dp
.
= 2
dt
p dt
e j
e j
We must find the momentary image location q:
1
1
1
+ =
20 m q 0.3 m
q = 0.305 m .
Now
*P36.39
a
f
2
0.305 m
dq
=−
5 m s = −0.001 16 m s = 1.16 mm s toward the lens .
2
dt
20 m
a
(a)
1 1 1
+ =
p q f
(b)
M=
f
q
h′
=−
h
p
1
1
1
+ =
480 cm q 7.00 cm
h′ =
a
q = 7.10 cm
fa
f
− hq − 5.00 mm 7.10 cm
=
= −0.074 0 mm
p
480 cm
diameter of illuminated spot = 74.0 µm
P36.40
af
0.100 W 4
P
P4
=
=
2
A πd
π 74.0 × 10 −6 m
= 2.33 × 10 7 W m 2
(c)
I=
(a)
1
1
1
1
1
= n−1
−
= 1.50 − 1
−
15.0 cm −12.0 cm
f
R1 R 2
e
a fLM
N
OP a
Q
j
2
fLM
N
a
(b)
The square is imaged as a trapezoid.
FIG. P36.40(b)
continued on next page
OP
fQ
or
f = 13.3 cm
363
364
Image Formation
(c)
To find the area, first find q R and qL , along with the heights hR′ and hL′ , using the thin lens
equation.
1
1
1
+
=
pR qR f
becomes
1
1
1
+
=
20.0 cm q R 13.3 cm
or
q R = 40.0 cm
F − q I = a10.0 cmfa−2.00f = −20.0 cm
GH p JK
hR′ = hM R = h
R
R
1
1
1
+
=
30.0 cm qL 13.3 cm
a
or
fa
qL = 24.0 cm
f
hL′ = hM L = 10.0 cm −0.800 = −8.00 cm
Thus, the area of the image is:
P36.41
(a)
Area = q R − qL hL′ +
q=d−p.
The image distance is:
1 1 1
+ =
p q f
Thus,
1
q R − q L hR′ − hL′ = 224 cm 2 .
2
becomes
This reduces to a quadratic equation:
1
1
1
+
= .
p d−p f
a f
p 2 + −d p + fd = 0
p=
which yields:
d ± d 2 − 4 fd
2
=
d
d2
±
− fd .
2
4
d
, both solutions are meaningful and the two solutions are not equal to each
4
other. Thus, there are two distinct lens positions that form an image on the screen.
Since f <
(b)
The smaller solution for p gives a larger value for q, with a real, enlarged, inverted image .
The larger solution for p describes a real, diminished, inverted image .
P36.42
To properly focus the image of a distant object, the lens must be at a distance equal to the focal
length from the film (q1 = 65.0 mm). For the closer object:
1
1 1
+
=
p2 q2 f
1
1
1
+
=
becomes
2 000 mm q 2 65.0 mm
I.
fFGH 2 0002 000
− 65.0 JK
a
q 2 = 65.0 mm
and
The lens must be moved away from the film by a distance
a
I − 65.0 mm =
fFGH 2 0002 000
− 65.0 JK
D = q 2 − q1 = 65.0 mm
2.18 mm .
Chapter 36
*P36.43
365
In the first arrangement the lens is used as a magnifying glass, producing an upright, virtual
enlarged image:
q
h ′ 120 cm
= 33.3 = −
M= =
h 3.6 cm
p
a
f
q = −33.3 p = −33.3 20 cm = −667 cm
For the lens,
1 1 1
1
1
1
+ = =
+
=
p q f 20 cm −667 cm f
f = 20.62 cm
In the second arrangement the lens us used as a projection lens to produce a real inverted enlarged
image:
q
120 cm
−
= −33.3 = − 2
q 2 = 33.3 p 2
3.6 cm
p2
1
1
1
34.3
1
=
+
=
p 2 = 21.24 cm
33.3 p 2 20.62 cm
p 2 33.3 p 2 20.62 cm
The lens was moved 21.24 cm − 20.0 cm = 1.24 cm .
Section 36.5
P36.44
(a)
Lens Aberrations
The focal length of the lens is given by
a fFGH
IJ a
K
fFGH
1
1
1
1
1
= n −1
−
= 1.53 − 1.00
−
−32.5 cm 42.5 cm
f
R1 R 2
IJ
K
f = −34.7 cm
Note that R1 is negative because the center of curvature of the
first surface is on the virtual image side.
p=∞
When
the thin lens equation gives q = f .
Thus, the violet image of a very distant object is formed
(b)
at
q = −34.7 cm .
The image is
virtual, upright and diminshed .
The same ray diagram and image characteristics apply for red light.
Again,
q= f
and now
1
1
1
= 1.51 − 1.00
−
−32.5 cm 42.5 cm
f
giving
f = −36.1 cm .
a
fFGH
IJ
K
FIG. P36.44
366
P36.45
Image Formation
Ray h1 is undeviated at the plane surface and strikes the second
surface at angle of incidence given by
FG h IJ = sin FG 0.500 cm IJ = 1.43° .
H 20.0 cm K
H RK
F 0.500 IJ
1.00 sin θ = 1.60 sin θ = a1.60fG
H 20.0 cm K
θ 1 = sin −1
Then,
−1
1
2
1
FIG. P36.45
so
θ 2 = 2.29° .
The angle this emerging ray makes with the horizontal is
θ 2 − θ 1 = 0.860° .
It crosses the axis at a point farther out by f1
f1 =
where
h1
b
tan θ 2 − θ 1
g
=
0.500 cm
= 33.3 cm .
tan 0.860°
a
f
The point of exit for this ray is distant axially from the lens vertex by
a20.0 cmf − a0.500 cmf
2
20.0 cm −
2
= 0.006 25 cm
so ray h1 crosses the axis at this distance from the vertex:
x1 = 33.3 cm − 0.006 25 cm = 33.3 cm .
Now we repeat this calculation for ray h2 :
FG 12.0 cm IJ = 36.9°
H 20.0 cm K
F 12.00 IJ
1.00 sin θ = 1.60 sin θ = a1.60fG
H 20.0 K
θ = sin −1
2
f2 =
h2
b
tan θ 1 − θ 2
fFH
a
g
=
12.0 cm
= 16.0 cm
tan 36.8°
x 2 = 16.0 cm 20.0 cm −
*P36.46
a20.0 cmf − a12.0 cmf IK = 12.0 cm .
2
2
∆x = 33.3 cm − 12.0 cm = 21.3 cm .
Now
Section 36.6
θ 2 = 73.7°
1
The Camera
The same light intensity is received from the subject, and the same light energy on the film is
required:
IA1 ∆t1 = IA 2 ∆t 2
π d12
π d 22
∆t 1 =
∆t 2
4
4
2
f
1
1
s = d 22
s
4
16
128
FG IJ FG
H KH
d2 =
IJ
K
FG
H
f
128 f
=
16 4
1. 41
IJ
K
Chapter 36
Section 36.7
367
The Eye
1 1 1 1
1
= + = −
= −4.00 diopters = −4.00 diopters, a diverging lens
f p q ∞ 0. 250 m
P36.47
P=
P36.48
For starlight going through Nick’s glasses,
1 1 1
+ =
p q f
1
1
1
+
= = −1.25 diopters .
∞ −0.800 m
f
a
1
1
+
= −1.25 m −1 , so p = 23.2 cm .
p −0.180 m
For a nearby object,
P36.49
f
a
f
Consider an object at infinity, imaged at the person’s far point:
1 1 1
+ =
p q f
1 1
+ = −4.00 m −1
∞ q
q = −25.0 cm .
The person’s far point is 25.0 cm + 2.00 cm = 27.0 cm from his eyes. For the contact lenses we want
1
1
1
+
= = −3.70 diopters .
∞ −0.270 m
f
a
f
Section 36.8
The Simple Magnifier
Section 36.9
The Compound Microscope
Section 36.10
The Telescope
P36.50
P36.51
(a)
From the thin lens equation:
(b)
M=−
1
1
1
+
=
or p = 4.17 cm .
p −25.0 cm 5.00 cm
a
q
25.0 cm
25.0 cm
=1+
=1+
= 6.00
p
f
5.00 cm
FG L IJ FG 25.0 cm IJ = −FG 23.0 cm IJ FG 25.0 cm IJ =
H f K H f K H 0.400 cm K H 2.50 cm K
F 25.0 cm IJ ⇒ f = FG M IJ a25.0 cmf = FG −12.0 IJ a25.0 cmf =
=M G
H −140 K
HMK
H f K
Using Equation 36.24, M ≈ −
o
P36.52
P36.53
f
M = M om e
f o = 20.0 m
o
e
e
−575 .
e
o
2.14 cm
f e = 0.025 0 m
(a)
The angular magnification produced by this telescope is: m = −
(b)
Since m < 0 , the image is inverted .
fo
= −800 .
fe
368
P36.54
Image Formation
(a)
The lensmaker’s equation
1 1 1
+ =
p q f
gives
q=
Then,
M=
1
1
=
1 f −1 p
p− f
b
h′ = −
fp
.
p− f
hf
.
p
For p >> f , f − p ≈ − p . Then,
(c)
Suppose the telescope observes the space station at the zenith:
(b)
=
hf
.
f −p
(b)
h′ = −
P36.55
fp
q
f
h′
=− =−
h
p
p− f
h′ =
gives
g
a
fa
f
108.6 m 4.00 m
hf
=−
= −1.07 mm .
p
407 × 10 3 m
Call the focal length of the objective f o and that of the eyepiece − f e . The distance between
the lenses is f o − f e . The objective forms a real diminished inverted image of a very distant
object at q1 = f o . This image is a virtual object for the eyepiece at p 2 = − f e .
For it
becomes
1
1
1
1
=0
+ =
,
q
− fe q − fe
2
q2 = ∞ .
and
(a)
1 1 1
+ =
p q f
The user views the image as virtual . Letting
h′ represent the height of the first image,
h′
h′
and θ =
θo =
. The angular
fo
fe
θ0
F0
θ0
F0
magnification is
m=
(c)
I
L1
f
θ h′ fe
=
= 0 .
θ o h′ f o
fe
f
Here, f o − f e = 10.0 cm and o = 3.00 .
fe
Thus,
fe =
fo
3.00
and
Fe
f e = −5.00 cm
O
L2
FIG. P36.55
and
Fe
θ
2
f o = 10.0 cm .
3
f o = 15.0 cm
f e = 5.00 cm
h’
369
Chapter 36
P36.56
Let I 0 represent the intensity of the light from the nebula and θ 0 its angular diameter. With the
h′
as h′ = −θ o 2 000 mm .
first telescope, the image diameter h′ on the film is given by θ o = −
fo
b
g
L π a200 mmf OP , and the light
The light power captured by the telescope aperture is P = I A = I M
MN 4 PQ
L π a200 mmf OPa1.50 minf .
energy focused on the film during the exposure is E = P ∆t = I M
MN 4 PQ
2
1
0
1
0
2
1
1
1
0
Likewise, the light power captured by the aperture of the second telescope is
LM π a60.0 mmf OP and the light energy is E
MN 4 PQ
2
= I0
the same light energy per unit area, it is necessary that
a
f
I 0 π 60.0 mm
a
2
f
π θ o 900 mm
4 ∆t 2
2
4
=
a
2
f 4 a1.50 minf
.
b2 000 mmg 4
I 0 π 200 mm
π θo
LM π a60.0 mmf OP∆t . Therefore, to have
MN 4 PQ
2
2
P2 = I 0 A 2 = I 0
2
2
The required exposure time with the second telescope is
a200 mmf a900 mmf a1.50 minf =
a60.0 mmf b2 000 mmg
2
∆t 2 =
2
2
2
3.38 min .
Additional Problems
P36.57
Only a diverging lens gives an upright diminished image. The image is virtual and
d = p− q = p+q:
p=
f=
P36.58
M=−
a
f
−M + 1 1 − M
1 1 1 1
1
+ = = +
=
=
− Mp
p q f p − Mp
− Md
d
:
1−M
− Md
a1 − Mf
q
so q = − Mp and d = p − Mp
p
2
=
a
fa
f=
− 0.500 20.0 cm
a1 − 0.500f
2
−40.0 cm .
If M < 1 , the lens is diverging and the image is virtual.
M=−
p=
q
p
d
:
1−M
so
2
q = − Mp
d =p− q = p+q
d = p − Mp
and
a
f a
f
1−M
−M + 1
1 1 1 1
1
+ = = +
=
=
p q f p − Mp
− Mp
− Md
b
g
2
f=
− Md
a1 − M f
2
.
2
.
If M > 1 , the lens is converging and the image is still virtual.
Now
d = −q − p .
We obtain in this case
f=
Md
a M − 1f
370
P36.59
Image Formation
(a)
a fFGH
IJ
K
F 1 − 1 IJ
1
= a1.66 − 1fG
−65.0 cm
H 50.0 cm R K
A
1
1
1
= n−1
−
f
R1 R 2
(b)
B
C
R2
R2 = 23.1 cm
so
FIG. P36.59
The distance along the axis from B to A is
a
f
R1 − R12 − 2.00 cm
2
2.00 cm
R1
2
1
1
1
=
+
R 2 50.0 cm 42.9 cm
D
a50.0 cmf − a2.00 cmf
2
= 50.0 cm −
2
= 0.040 0 cm .
Similarly, the axial distance from C to D is
23.1 cm −
a23.1 cmf − a2.00 cmf
2
2
= 0.086 8 cm .
Then, AD = 0.100 cm − 0.040 0 cm + 0.086 8 cm = 0.147 cm .
*P36.60
We consider light entering the rod. The surface of entry is convex
to the object rays, so R1 = +4.50 cm
n1 n 2 n 2 − n 1
+
=
p 1 q1
R1
1.33
1.50 1.50 − 1.33
+
=
100 cm q1
4.50 cm
1.50
= 0.037 8 cm − 0.013 3 cm = 0.024 5 cm
q1
q1 = 61.3 cm
O1
V
I2
O2
The first image is real, inverted and diminished. To find its
magnification we can use two similar triangles in the ray diagram
with their vertices meeting at the center of curvature:
h1′
h1
=
100 cm + 4.5 cm 61.3 cm − 4.5 cm
I1
C
C
V
FIG. P36.60
h1′
= −0.543 .
h1
Now the first image is a real object for the second surface at object distance from its vertex
75.0 cm + 4.50 cm + 4.50 cm − 61.3 cm = 22.7 cm
1.50
1.33 1.33 − 1.50
+
=
22.7 cm q 2
−4.50 cm
1.33
= 0.037 8 cm − 0.066 0 cm = −0.028 2 cm
q2
q 2 = −47.1 cm
(a)
The final image is inside the rod, 47.1 cm from the second surface .
(b)
It is virtual, inverted, and enlarged . Again by similar triangles meeting at C we have
h2
h2′
=
22.7 cm − 4.5 cm 47.1 cm − 4.5 cm
h2′
= 2.34 .
h2
Since h2 = h1′ , the overall magnification is M 1 M 2 =
a
fa f
h1′ h2′ h2′
=
= −0.543 2.34 = −1.27 .
h1 h2 h1
Chapter 36
*P36.61
(a)
1
1
1
1
1
= −
=
−
q1 f1 p1 5 cm 7.5 cm
q
15 cm
= −2
M1 = − 1 = −
7.5 cm
p1
∴ q1 = 15 cm
a f
M = M1 M 2
∴M2 = −
371
∴ 1 = −2 M 2
q
1
=− 2
2
p2
1
1
1
+
=
p 2 q2 f2
∴
∴ p 2 = 2 q2
1
1
1
+
=
2 q 2 q 2 10 cm
∴ q 2 = 15 cm, p 2 = 30 cm
p1 + q1 + p 2 + q 2 = 7.5 cm + 15 cm + 30 cm + 15 cm = 67.5 cm
(b)
1
1
1
1
+
=
=
p1′ q1′
f1 5 cm
Solve for q1′ in terms of p1′ : q1′ =
5 p1′
p1′ − 5
(1)
q1′
5
, using (1).
=−
p1′
p1′ − 5
q′
3
M′
∴ M 2′ =
= − p1′ − 5 = − 2
M ′ = M 1′ M 2′
5
M 1′
p 2′
3
∴ q ′2 = p 2′ p1′ − 5
(2)
5
1
1
1
1
+
=
=
Substitute (2) into the lens equation
and obtain p ′2 in terms of p1′ :
p ′2 q 2′
f 2 10 cm
10 3 p1′ − 10
.
(3)
p 2′ =
3 p1′ − 5
Substituting (3) in (2), obtain q 2′ in terms of p1′ :
q ′2 = 2 3 p1′ − 10 .
(4)
Now, p1′ + q1′ + p ′2 + q 2′ = a constant.
Using (1), (3) and (4), and the value obtained in (a):
10 3 p1′ − 10
5 p1′
+
p1′ +
+ 2 3 p1′ − 10 = 67.5 .
p1′ − 5
3 p′ − 5
This reduces to the quadratic equation
21p1′ 2 − 322.5 p1′ + 1 212.5 = 0 ,
which has solutions p1′ = 8.784 cm and 6.573 cm.
Case 1:
p1′ = 8.784 cm
∴ p1′ − p1 = 8.784 cm − 7.5 cm = 1.28 cm.
From (4):
q ′2 = 32.7 cm
∴ q ′2 − q 2 = 32.7 cm − 15 cm = 17.7 cm.
Case 2:
p1′ = 6.573 cm
∴ p1′ − p1 = 6.573 cm − 7.5 cm = −0.927 cm .
From (4):
q 2′ = 19. 44 cm
∴ q ′2 = q 2 = 19.44 cm − 15 cm = 4.44 cm.
From these results it is concluded that:
The lenses can be displaced in two ways. The first lens can be moved 1.28 cm farther from
M 1′ = −
b
b
b
b
b
g
g
g
g
b
g
b
g b
g
g
the object and the second lens 17.7 cm toward the object. Alternatively, the first lens can
be moved 0.927 cm toward the object and the second lens 4.44 cm toward the object.
372
P36.62
Image Formation
1
1
1
1
1
= −
=
−
q1 f1 p1 10.0 cm 12.5 cm
so
q1 = 50.0 cm (to left of mirror).
This serves as an object for the lens (a virtual object), so
1
1
1
1
1
=
−
=
−
and q 2 = −50.3 cm,
q2 f2 p 2
−16.7 cm
−25.0 cm
a
f a
f
meaning 50.3 cm to the right of the lens. Thus, the final image is located
25.3 cm to right of mirror .
M1 = −
q1
50.0 cm
=−
= −4.00
p1
12.5 cm
M2 = −
−50.3 cm
q2
=−
= −2.01
p2
−25.0 cm
a
a
f
f
M = M 1 M 2 = 8.05
Thus, the final image is virtual, upright , 8.05 times the size of object, and 25.3 cm to right of the
mirror.
P36.63
We first find the focal length of the mirror.
1 1 1
1
1
9
= + =
+
=
f p q 10.0 cm 8.00 cm 40.0 cm
*P36.64
and
f = 4. 44 cm .
Hence, if p = 20.0 cm ,
1 1 1
1
1
15.56
= − =
−
=
.
q f p 4. 44 cm 20.0 cm 88.8 cm
Thus,
q = 5.71 cm , real.
A telescope with an eyepiece decreases the diameter of a
beam of parallel rays. When light is sent through the same
device in the opposite direction, the beam expands. Send
the light first through the diverging lens. It will then be
diverging from a virtual image found like this:
1 1 1
+ =
p q f
1 1
1
+ =
∞ q −12 cm
FIG. P36.64
q = −12 cm .
Use this image as a real object for the converging lens, placing it at the focal point on the object side
of the lens, at p = 21 cm . Then
1 1 1
+ =
p q f
1
1
1
+ =
21 cm q 21 cm
q=∞.
The exiting rays will be parallel. The lenses must be 21.0 cm − 12.0 cm = 9.00 cm apart.
By similar triangles,
d 2 21 cm
=
= 1.75 times .
d 1 12 cm
Chapter 36
P36.65
A hemisphere is too thick to be described as a thin
lens. The light is undeviated on entry into the flat
face. We next consider the light’s exit from the
second surface, for which R = −6.00 cm .
The incident rays are parallel, so p = ∞ .
Then,
n1 n 2 n 2 − n 1
+
=
p
q
R
becomes
0+
1 1.00 − 1.56
=
q −6.00 cm
q = 10.7 cm .
and
P36.66
FIG. P36.65
(a)
I=
P
4.50 W
=
2
4π r
4π 1.60 × 10 −2 m
(b)
I=
P
4.50 W
=
2
4π r
4π 7.20 m
(c)
1 1 1
+ = :
p q f
1
1
1
+ =
7.20 m q 0.350 m
so
q = 0.368 m
and
M=
e
a
j
f
2
2
= 1.40 kW m 2
= 6.91 mW m 2
q
h′
0.368 m
=− =−
3. 20 cm
p
7.20 m
h′ = 0.164 cm
(d)
e
jLMN π4 a0.150 mf OPQ
e
j a
f
The lens intercepts power given by
P = IA = 6.91 × 10 −3 W m 2
and puts it all onto the image where
I=
2
f
6.91 × 10 −3 W m 2 π 15.0 cm
P
=
2
A
π 0.164 cm 4
I = 58.1 W m 2 .
a
2
4
373
374
P36.67
Image Formation
a
fa
a
f
f
−6.00 cm 12.0 cm
f1 p1
=
= −4.00 cm .
p1 − f1 12.0 cm − −6.00 cm
From the thin lens equation,
q1 =
When we require that q 2 → ∞ ,
the thin lens equation becomes p 2 = f 2 .
In this case,
p 2 = d − −4.00 cm .
Therefore,
d + 4.00 cm = f 2 = 12.0 cm
a
f
and
d = 8.00 cm .
FIG. P36.67
*P36.68
The inverted real image is formed by the lens operating on light directly from the object, on light
that has not reflected from the mirror.
q
p
q = 1.50 p
For this we have
M = −1.50 = −
1 1 1
+ =
p q f
1
1
1
2.50
+
=
=
p 1.50 p 10 cm 1.50 p
FG 2.5 IJ = 16.7 cm
H 1.5 K
p = 10 cm
40.0 cm − 16.7 cm = 23.3 cm .
Then the object is distant from the mirror by
The second image seen by the person is formed by light that first reflects from the mirror and then
goes through the lens. For it to be in the same position as the inverted image, the lens must be
receiving light from an image formed by the mirror at the same location as the physical object. The
formation of this image is described by
1 1 1
+ =
p q f
P36.69
1
1
1
+
=
23.3 cm 23.3 cm f
f = 11.7 cm .
R
= +1.50 m . In addition, because the distance to the Sun is so much larger than
2
any other distances, we can take p = ∞ .
For the mirror, f =
The mirror equation,
1 1 1
+ = , then gives
p q f
q = f = 1.50 m .
M=−
Now, in
q h′
=
p h
the magnification is nearly zero, but we can be more precise:
Thus, the image diameter is
h′ = −
fFGH
IJ a
K
h
is the angular diameter of the object.
p
hq
π rad
= −0.533°
1.50 m = −0.014 0 m = −1.40 cm .
180°
p
a
f
Chapter 36
P36.70
(a)
For the light the mirror intercepts,
P = I 0 A = I 0 π R a2
e
j
350 W = 1 000 W m 2 π R a2
Ra = 0.334 m or larger .
and
(b)
In
we have
so
h′ = − q
so
where
1 1 1 2
+ = =
p q f R
p→∞
R
q=
2
q
h′
M= =−
h
p
F h I = −FG R IJ L0.533° FG π rad IJ O = −FG R IJ a9.30 m radf
GH p JK H 2 K MN H 180° K PQ H 2 K
h
is the angle the Sun subtends. The intensity at the image is
p
then
I=
4I 0 π R a2
P
=
=
πh ′ 2 4
π h′ 2
R2
4I 0 R a2
b g e9.30 × 10
16e1 000 W m jR
=
R e9.30 × 10 radj
2
2
120 × 10 3 W m 2
so
P36.71
−3
2
rad
j
2
2
a
2
Ra
= 0.025 5 or larger .
R
In the original situation,
p1 + q1 = 1.50 m .
In the final situation,
p 2 = p1 + 0.900 m
and
q 2 = q1 − 0.900 m = 0.600 m − p1 .
1
1 1
1
1
+
= =
+
.
p1 q 1 f p 2 q 2
Our lens equation is
−3
Substituting, we have
1
1
1
1
+
=
+
.
p1 1.50 m − p1 p1 + 0.900 0.600 − p1
Adding the fractions,
1.50 m − p1 + p1
0.600 − p1 + p1 + 0.900
=
.
p1 1.50 m − p1
p1 + 0.900 0.600 − p1
Simplified, this becomes
b
g b
p b1.50 m − p g = b p
(a)
p1 =
(b)
(c)
Thus,
1
1
1
gb
g
+ 0.900gb0.600 − p g .
1
0.540
m = 0.300 m
1.80
1
1
1
=
+
f 0.300 m 1.50 m − 0.300 m
and
p 2 = p1 + 0.900 = 1.20 m
f = 0.240 m
The second image is real, inverted, and diminished
with
M=−
q2
= −0.250 .
p2
FIG. P36.71
375
376
P36.72
Image Formation
(a)
becomes:
(b)
a fFGH
1
1
1
= n−1
+
f
R1 R 2
The lens makers’ equation,
a fLM
N
OP giving n =
fQ
1
1
1
= n−1
−
9.00 cm −11.0 cm
5.00 cm
a
IJ
K
1.99 .
As the light passes through the lens for the first time, the thin lens equation
1
1 1
+
=
p 1 q1 f
1
1
1
+
=
8.00 cm q1 5.00 cm
becomes:
or
q1 = 13.3 cm ,
M1 = −
and
q1
13.3 cm
=−
= −1.67 .
8.00 cm
p1
This image becomes the object for the concave mirror with:
p m = 20.0 cm − q1 = 20.0 cm − 13.3 cm = 6.67 cm
R
= +4.00 cm .
2
1
1
1
+
=
6.67 cm qm 4.00 cm
f=
and
The mirror equation becomes:
giving
qm = 10.0 cm
and
M2 = −
qm
10.0 cm
=−
= −1.50 .
6.67 cm
pm
The image formed by the mirror serves as a real object for the lens on the second pass of the
light through the lens with:
p 3 = 20.0 cm − q m = +10.0 cm .
The thin lens equation yields:
1
1
1
+
=
10.0 cm q3 5.00 cm
or
q3 = 10.0 cm
and
M3 = −
10.0 cm to the left of the lens .
The final image is a real image located
M total = M1 M 2 M 3 = −2.50 .
The overall magnification is
(c)
P36.73
q3
10.0 cm
=−
= −1.00 .
10.0 cm
p3
Since the total magnification is negative, this final image is inverted .
For the objective:
1 1 1
+ = becomes
p q f
1
1
1
+ =
so q = 25.5 mm .
3.40 mm q 3.00 mm
q
25.5 mm
=−
= −7.50 .
3.40 mm
p
The objective produces magnification
M1 = −
For the eyepiece as a simple magnifier,
me =
and overall
M = M1 m e = −75.0 .
25.0 cm 25.0 cm
=
= 10.0
f
2.50 cm
Chapter 36
P36.74
(a)
377
Start with the second lens: This lens must form a virtual image located 19.0 cm to the left of
it (i.e., q 2 = −19.0 cm ). The required object distance for this lens is then
p2 =
a
fa
f
−19.0 cm 20.0 cm 380 cm
q2 f2
=
.
=
39.0
q2 − f2
−19.0 cm − 20.0 cm
The image formed by the first lens serves as the object for the second lens. Therefore, the
image distance for the first lens is
q1 = 50.0 cm − p 2 = 50.0 cm −
380 cm 1 570 cm
=
.
39.0
39.0
The distance the original object must be located to the left of the first lens is then given by
1
1
1
1
39.0
157 − 39.0
118
= −
=
−
=
=
p1 f1 q1 10.0 cm 1 570 cm 1 570 cm 1 570 cm
(b)
P36.75
P36.76
or
p1 =
1 570 cm
= 13.3 cm .
118
F q I F − q I = LMFG 1 570 cm IJ F 118 I OPL a−19.0 cmfa39.0f O =
GH p JK GH p JK MNH 39.0 K GH 1 570 cm JK PQMN 380 cm PQ
M = M1 M 2 = −
1
2
1
2
(c)
Since M < 0 , the final image is inverted .
(a)
P=
1 1 1
1
1
= + =
+ = 44.6 diopters
f p q
0.022 4 m ∞
(b)
P=
1 1 1
1
1
= + =
+ = 3.03 diopters
0.330 m ∞
f p q
b
a
−5.90
g
f
The object is located at the focal point of the upper mirror. Thus, the
upper mirror creates an image at infinity (i.e., parallel rays leave this
mirror).
The lower mirror focuses these parallel rays at its focal point, located at
the hole in the upper mirror.
Thus, the
image is real, inverted, and actual size .
For the upper mirror:
1 1 1
+ = :
p q f
1
1
1
+
=
7.50 cm q1 7.50 cm
q1 = ∞.
For the lower mirror:
1 1
1
+
=
∞ q 2 7.50 cm
q 2 = 7.50 cm.
Light directed into the hole in the upper mirror reflects as shown, to
behave as if it were reflecting from the hole.
FIG. P36.76
378
P36.77
Image Formation
(a)
For lens one, as shown in the first
figure,
1
1
1
+
=
40.0 cm q1 30.0 cm
q1 = 120 cm
M1 = −
q1
120 cm
=−
= −3.00
p1
40.0 cm
This real image I 1 = O 2 is a virtual
object for the second lens. That is, it is
behind the lens, as shown in the second
figure. The object distance is
p 2 = 110 cm − 120 cm = −10.0 cm
1
1
1
+
=
:
−10.0 cm q 2 −20.0 cm
q 2 = 20.0 cm
M2 = −
q2
20.0 cm
=−
= +2.00
−10.0 cm
p2
a
f
M overall = M 1 M 2 = −6.00
(b)
M overall < 0 , so final image is
inverted .
(c)
If lens two is a converging lens (third
figure):
1
1
1
+
=
−10.0 cm q 2 20.0 cm
q 2 = 6.67 cm
M2 = −
a
6.67 cm
= +0.667
−10.0 cm
f
M overall = M 1 M 2 = −2.00
Again, M overall < 0 and the final image is inverted .
FIG. P36.77
Chapter 36
*P36.78
379
The first lens has focal length described by
gFGH
IJ b
K
gFGH
IJ b
K
gFGH
IJ
K
n −1
1
1
1
1 1
= n1 − 1
−
= n1 − 1
−
=− 1
.
∞ R
f1
R11 R12
R
b
For the second lens
gFGH
IJ
K
b
g
2 n2 − 1
1
1
1
1
1
= n2 − 1
−
=+
= n2 − 1
−
.
+R −R
f2
R 21 R 22
R
b
Let an object be placed at any distance p1 large compared to the thickness of the doublet. The first
lens forms an image according to
1
1
1
+
=
p1 q1 f1
1 −n 1 + 1 1
=
−
.
q1
R
p1
b
g
This virtual q1 < 0 image is a real object for the second lens at distance p 2 = − q1 . For the second lens
1
1
1
+
=
p 2 q2 f2
2n − 2 1 2n 2 − 2 −n1 + 1 1 2n 2 − n1 − 1 1
1 2n 2 − 2 1
=
−
= 2
+
=
+
−
=
− .
q2
R
p2
R
q1
R
R
p1
R
p1
Then
1
1 2n 2 − n1 − 1
1 2n − n 1 − 1
+
=
so the doublet behaves like a single lens with = 2
.
p1 q 2
R
f
R
ANSWERS TO EVEN PROBLEMS
P36.2
4.58 m
P36.22
4.82 cm
P36.4
see the solution
P36.24
P36.6
(a) p1 + h ; (b) virtual; (c) upright; (d) +1;
(e) No
see the solution;
real, inverted, diminished
P36.26
2.00
P36.28
20.0 cm
P36.30
(a) q = 40.0 cm real, inverted, actual size
M = −1.00 ;
(b) q = ∞ , M = ∞ , no image is formed;
(c) q = −20.0 cm upright, virtual , enlarged
M = +2.00
P36.8
at q = −0.267 m virtual upright and
diminished with M = 0.026 7
P36.10
at 3.33 m from the deepest point of the
niche
P36.12
30.0 cm
P36.14
(a) 160 mm; (b) R = −267 mm
P36.32
(a) 6.40 cm; (b) −0. 250 ; (c) converging
P36.16
(a) convex; (b) At the 30.0 cm mark;
(c) –20.0 cm
P36.34
(a) 3.40 , upright; (b) see the solution
P36.18
(a) 15.0 cm; (b) 60.0 cm
P36.36
(a) 39.0 mm; (b) 39.5 mm
P36.20
(a) see the solution;
(b) at 0.639 s and at 0.782 s
P36.38
1.16 mm s toward the lens
380
Image Formation
P36.40
(a) 13.3 cm;
(b) see the solution; a trapezoid;
(c) 224 cm 2
P36.60
(a) inside the rod, 47.1 cm from the
second surface ;
(b) virtual, inverted, and enlarged
P36.42
2.18 mm away from the film
P36.62
P36.44
(a) at q = −34.7 cm
virtual, upright and diminshed;
(b) at q = −36.1 cm
virtual, upright and diminshed
25.3 cm to right of mirror ,
virtual, upright , enlarged 8.05 times
P36.64
place the lenses 9.00 cm apart and let light
pass through the diverging lens first.
1.75 times
(a) 1.40 kW m 2 ; (b) 6.91 mW m 2 ;
P36.46
f
1.41
P36.66
P36.48
23. 2 cm
P36.68
11.7 cm
P36.50
(a) at 4.17 cm; (b) 6.00
P36.70
P36.52
2.14 cm
(a) 0.334 m or larger ;
R
(b) a = 0.025 5 or larger
R
P36.54
(a) see the solution; (b) h ′ = −
P36.72
(a) 1.99 ;
(b) 10.0 cm to the left of the lens ; −2.50 ;
(c) inverted
P36.74
(a) 13.3 cm; (b) −5.90 ; (c) inverted
P36.76
see the solution;
real, inverted, and actual size
P36.78
see the solution
(c) 0.164 cm; (d) 58.1 W m 2
hf
;
p
(c) −1.07 mm
P36.56
3.38 min
P36.58
if M < 1 , f =
− Md
a1 − M f
Md
if M > 1 , f =
a M − 1f
2
2
,
37
Interference of Light Waves
CHAPTER OUTLINE
37.1
37.2
37.3
37.4
37.5
37.6
37.7
Conditions for Interference
Young’s Double-Slit
Experiment
Intensity Distribution of the
Double-Slit Interference
Pattern
Phasor Addition of Waves
Change of Phase Due to
Reflection
Interference in Thin Films
The Michelson
Interferometer
ANSWERS TO QUESTIONS
Q37.1
(a)
Two waves interfere constructively if their path
difference is zero, or an integral multiple of the
wavelength, according to δ = mλ , with
m = 0 , 1, 2 , 3 , ….
(b)
Two waves interfere destructively if their path
difference is a half wavelength, or an odd multiple of
λ
1
, described by δ = m + λ , with m = 0 , 1, 2 , 3 , ….
2
2
FG
H
Q37.2
IJ
K
The light from the flashlights consists of many different
wavelengths (that’s why it’s white) with random time
differences between the light waves. There is no coherence
between the two sources. The light from the two flashlights
does not maintain a constant phase relationship over time.
These three equivalent statements mean no possibility of an
interference pattern.
λ air
. Since the positions of light
n water
and dark bands are proportional to λ, (according to Equations 37.2 and 37.3), the underwater fringe
separations will decrease.
Q37.3
Underwater, the wavelength of the light would decrease, λ water =
Q37.4
Every color produces its own pattern, with a spacing between the maxima that is characteristic of the
wavelength. With several colors, the patterns are superimposed and it can be difficult to pick out a
single maximum. Using monochromatic light can eliminate this problem.
Q37.5
The threads that are woven together to make the cloth have small meshes between them. These bits
of space act as pinholes through which the light diffracts. Since the cloth is a grid of such pinholes,
an interference pattern is formed, as when you look through a diffraction grating.
Q37.6
If the oil film is brightest where it is thinnest, then n air < n oil < n water . With this condition, light
reflecting from both the top and the bottom surface of the oil film will undergo phase reversal. Then
these two beams will be in phase with each other where the film is very thin. This is the condition
for constructive interference as the thickness of the oil film decreases toward zero.
381
382
Interference of Light Waves
Q37.7
As water evaporates from the ‘soap’ bubble, the thickness of the bubble wall approaches zero. Since
light reflecting from the front of the water surface is phase-shifted 180° and light reflecting from the
back of the soap film is phase-shifted 0°, the reflected light meets the conditions for a minimum.
Thus the soap film appears black, as in the illustration accompanying textbook Example 37.5,
“Interference in a Wedge-Shaped Film.”
Q37.8
If the film is more than a few wavelengths thick, the interference fringes are so close together that
you cannot resolve them.
Q37.9
If R is large, light reflecting from the lower surface of the lens can interfere with light reflecting from
the upper surface of the flat. The latter undergoes phase reversal on reflection while the former does
not. Where there is negligible distance between the surfaces, at the center of the pattern you will see
a dark spot because of the destructive interference associated with the 180° phase shift. Colored
rings surround the dark spot. If the lens is a perfect sphere the rings are perfect circles. Distorted
rings reveal bumps or hollows on the fine scale of the wavelength of visible light.
Q37.10
A camera lens will have more than one element, to correct (at least) for chromatic aberration. It will
have several surfaces, each of which would reflect some fraction of the incident light. To maximize
light throughput the surfaces need antireflective coatings. The coating thickness is chosen to
produce destructive interference for reflected light of some wavelength.
Q37.11
To do Young’s double-slit interference experiment with light from an ordinary source, you must first
pass the light through a prism or diffraction grating to disperse different colors into different
directions. With a single narrow slit you select a single color and make that light diffract to cover
both of the slits for the interference experiment. Thus you may have trouble lining things up and
you will generally have low light power reaching the screen. The laser light is already
monochromatic and coherent across the width of the beam.
Q37.12
Suppose the coating is intermediate in index of refraction between vacuum and the glass. When the
coating is very thin, light reflected from its top and bottom surfaces will interfere constructively, so
you see the surface white and brighter. As the thickness reaches one quarter of the wavelength of
violet light in the coating, destructive interference for violet will make the surface look red or
perhaps orange. Next to interfere destructively are blue, green, yellow, orange, and red, making the
surface look red, purple, and then blue. As the coating gets still thicker, we can get constructive
interference for violet and then for other colors in spectral order. Still thicker coating will give
constructive and destructive interference for several visible wavelengths, so the reflected light will
start to look white again.
Q37.13
Assume the film is higher in refractive index than the medium on both sides of it. The condition for
λ
. The
2
ray that reflects through the film undergoes phase reversal both at the bottom and at the top surface.
λ
Then this ray should also travel an extra distance of . Since this ray passes through two extra
2
λ
thicknesses of film, the thickness should be . This is different from the condition for destructive
4
interference of light reflected from the film, but it is the same as the condition for constructive
interference of reflected light. The energy of the extra reflected light is energy diverted from light
otherwise transmitted.
destructive interference of the two transmitted beams is that the waves be out of phase by
Chapter 37
Q37.14
383
The metal body of the airplane is reflecting radio waves broadcast by the television station. The
reflected wave that your antenna receives has traveled an extra distance compared to the stronger
signal that came straight from the transmitter tower. You receive it with a short time delay. On the
television screen you see a faint image offset to the side.
SOLUTIONS TO PROBLEMS
Section 37.1
Conditions for Interference
Section 37.2
Young’s Double-Slit Experiment
P37.1
∆y bright =
P37.2
y bright =
ja f
e
−9
5.00
λL 632.8 × 10
=
m = 1.58 cm
−4
d
2.00 × 10
λL
m
d
λ=
For m = 1 ,
P37.3
e
je
j
3.40 × 10 −3 m 5.00 × 10 −4 m
yd
=
= 515 nm
3.30 m
L
Note, with the conditions given, the small angle approximation does not
work well. That is, sin θ , tan θ , and θ are significantly different. We treat
the interference as a Fraunhofer pattern.
(a)
At the m = 2 maximum, tan θ =
400 m
= 0.400
1 000 m
400 m
θ = 21.8°
λ=
so
(b)
300 m
a
f
300 m sin 21.8°
d sin θ
=
= 55.7 m .
2
m
The next minimum encountered is the m = 2 minimum;
FG
H
IJ
K
1
λ
2
and at that point,
d sin θ = m +
which becomes
d sin θ =
or
sin θ =
and
θ = 27.7°
so
y = 1 000 m tan 27.7° = 524 m .
b
5
λ
2
FG
H
IJ
K
5 λ 5 55.7 m
=
= 0.464
2 d 2 300 m
g
Therefore, the car must travel an additional 124 m .
If we considered Fresnel interference, we would more precisely find
1
550 2 + 1 000 2 − 250 2 + 1 000 2 = 55.2 m and (b) 123 m.
(a) λ =
2
FH
IK
1 000 m
FIG. P37.3
384
P37.4
Interference of Light Waves
λ=
v 354 m s
=
= 0.177 m
f 2 000 s −1
a0.300 mf sinθ = 1a0.177 mf
so
d sin 36.2° = 1b0.030 0 mg
d sin θ = mλ
e1.00 × 10 mj sin 36.2° = a1fλ
d sin θ = mλ
(a)
(b)
−6
(c)
f=
P37.5
c
λ
=
3.00 × 10 8 m s
FG
H
d sin θ = m +
The first minimum is described by
m=0
and the tenth by m = 9 :
sin θ =
sin θ ≈ tan θ .
Thus,
d=
λ=
ja
7.26 × 10
m
d = 5.08 cm
so
λ = 590 nm
f = 1.54 × 10
IJ
K
y
L
IJ
K
Source
y
L
but for small θ,
−3
and
1
λ.
2
FG
H
tan θ =
e
θ = 36.2°
λ
1
9+
.
2
d
Also,
9.5 5 890 × 10 −10 m 2.00 m
and
= 508 THz
5.90 × 10 −7 m
In the equation
d=
P37.6
so
FIG. P37.5
9.5 λ 9.5 λL
=
sin θ
y
−3
m = 1.54 mm .
340 m s
= 0.170 m
2 000 Hz
Maxima are at
d sin θ = mλ :
m=0
gives
θ = 0°
m=1
gives
sin θ =
λ 0.170 m
=
d 0.350 m
θ = 29.1°
m=2
gives
sin θ =
2λ
= 0.971
d
θ = 76.3°
m=3
gives
sin θ = 1. 46
Minima are at
d sin θ = m +
m=0
gives
sin θ =
m=1
gives
sin θ =
m=2
gives
sin θ = 1.21
FG
H
IJ
K
No solution.
1
λ:
2
λ
= 0.243
θ = 14.1°
3λ
= 0.729
2d
θ = 46.8°
2d
No solution.
So we have maxima at 0° , 29.1° , and 76.3° ; minima at 14.1° and 46.8° .
d
Chapter 37
P37.7
(a)
For the bright fringe,
y bright =
y=
y
mλ L
where m = 1
d
e546.1 × 10
−9
ja
f = 2.62 × 10
m 1.20 m
0. 250 × 10
−3
m
y 2 − y1
−3
m = 2.62 mm .
Source
IJ
K
λL LF 1 I F
1 I O λL
=
1 + J − G0 + JP =
a1f
G
M
2K H
2KQ d
d NH
e546.1 × 10 mja1.20 mf
=
For the dark bands, y dark =
(b)
L = 1.20 m
FG
H
λL
1
; m = 0 , 1, 2 , 3 , …
m+
2
d
d = 0.250 m
bright bright bright
dark
dark
dark
−9
0.250 × 10 −3 m
y1
∆y = 2.62 mm .
y2
FIG. P37.7
P37.8
Taking m = 0 and y = 0.200 mm in Equation 37.6 gives
L≈
2dy
=
e
je
Bright
j = 0.362 m
2 0.400 × 10 −3 m 0.200 × 10 −3 m
λ
L ≈ 36.2 cm
442 × 10
−9
m
Bright
0.2 mm
Geometric optics incorrectly predicts bright regions opposite the
slits and darkness in between. But, as this example shows,
interference can produce just the opposite.
P37.9
Dark
0.2 mm
Dark
L
FIG. P37.7
Location of A = central maximum,
Location of B = first minimum.
P37.10
FG IJ
H K
a3.00 mfa150 mf =
=
So,
∆y = y min − y max =
Thus,
d=
a
f
40.0 m
11.3 m .
d sin θ = mλ
At 30.0° ,
e3.20 × 10
λL
2 20.0 m
1
1 λL
λL
0+
−0=
= 20.0 m .
2
2 d
d
−4
j
e
j
m sin 30.0° = m 500 × 10 −9 m
so
m = 320
There are 320 maxima to the right, 320 to the left, and one for m = 0 straight ahead.
There are 641 maxima .
Bright
385
386
*P37.11
Interference of Light Waves
Observe that the pilot must not only home in on the airport, but must be headed in the right
direction when she arrives at the end of the runway.
8
c 3 × 10 m s
=
= 10.0 m
f 30 × 10 6 s −1
(a)
λ=
(b)
The first side maximum is at an angle given by d sin θ = 1 λ .
af
θ = 14.5°
a40 mf sinθ = 10 m
y = L tan θ = b 2 000 mg tan 14.5° = 516 m
(c)
*P37.12
tan θ =
y
L
The signal of 10-m wavelength in parts (a) and (b) would show maxima at 0°, 14.5°, 30.0°,
48.6°, and 90°. A signal of wavelength 11.23-m would show maxima at 0°, 16.3°, 34.2°, and
57.3°. The only value in common is 0°. If λ 1 and λ 2 were related by a ratio of small integers
λ
n
(a just musical consonance!) in 1 = 1 , then the equations d sin θ = n 2 λ 1 and d sin θ = n1 λ 2
λ 2 n2
would both be satisfied for the same nonzero angle. The pilot could come flying in with that
inappropriate bearing, and run off the runway immediately after touchdown.
In d sin θ = mλ
d
y
= mλ
L
e
e
y=
mλ L
d
j
−9
dy mλ dL 1 633 × 10 m
=
=
3 m s = 6.33 mm s
dt
d dt
0.3 × 10 −3 m
P37.13
φ=
2π
λ
d sin θ =
j
2π
d
λ
FG y IJ
H LK
2π
1.20 × 10
je
(a)
φ=
(b)
φ=
(c)
If φ = 0.333 rad =
e
2π
e5.00 × 10
−7
e1.20 × 10
mj
2πd sin θ
λ
j a
f
−4
m sin 0.500° = 13.2 rad
−4
m
5.00 × 10 −7 m
× 10 m I
=
jFGH 5.001.20
m JK
θ = sin −1
−3
F λ φ I = sin
GH 2π d JK
6.28 rad
LM e5.00 × 10 mja0.333 radf OP
MN 2π e1.20 × 10 mj PQ
−7
−1
−4
θ = 1.27 × 10 −2 deg .
(d)
If d sin θ =
λ
4
θ = sin −1
FG λ IJ = sin
H 4d K
−1
LM 5 × 10 m OP
MN 4e1.20 × 10 mj PQ
θ = 5.97 × 10 −2 deg .
−7
−4
Chapter 37
P37.14
δ = d sin θ 1 − d sin θ 2 .
The path difference between rays 1 and 2 is:
For constructive interference, this path difference must be equal to an integral number of
wavelengths: d sin θ 1 − d sin θ 2 = mλ , or
b
g
d sin θ 1 − sin θ 2 = mλ .
P37.15
(a)
The path difference δ = d sin θ and when L >> y
δ=
e
je
j
1.80 × 10 −2 m 1.50 × 10 −4 m
yd
=
= 1.93 × 10 −6 m = 1.93 µm .
1.40 m
L
(b)
δ 1.93 × 10 −6 m
=
= 3.00 , or δ = 3.00λ
λ 6.43 × 10 −7 m
(c)
Point P will be a maximum since the path difference is an integer multiple of the
wavelength.
Section 37.3
P37.16
(a)
Intensity Distribution of the Double-Slit Interference Pattern
I
I max
= cos 2
FG φ IJ
H 2K
(Equation 37.11)
φ = 2 cos −1
Therefore,
(b)
P37.17
δ=
a
fa
I
I max
f
486 nm 1.29 rad
λφ
=
= 99.8 nm
2π
2π
I av = I max cos 2
FG π d sinθ IJ
H λ K
For small θ,
sin θ =
and
I av = 0.750 I max
y=
y
L
λL
cos −1
πd
I av
I max
e6.00 × 10 ja1.20 mf cos
π e 2.50 × 10 mj
−7
y=
P37.18
= 2 cos −1 0.640 = 1.29 rad .
−1
−3
FG π yd IJ
H λL K
L π e6.00 × 10 mje1.80 × 10 mj OP
= cos M
MN e656.3 × 10 mja0.800 mf PQ =
0.750 I max
= 48.0 µm
I max
I = I max cos 2
I
I max
−3
−4
2
−9
0.968
387
388
P37.19
Interference of Light Waves
(a)
From Equation 37.8,
φ=
2π d
λ
sin θ =
e
2π d
λ
y
⋅
y2 + D2
je
ja
j
−3
−3
2π yd 2π 0.850 × 10 m 2.50 × 10 m
φ≈
=
= 7.95 rad
λD
600 × 10 −9 m 2.80 m
(b)
I
I max
I
I max
P37.20
(a)
=
e
b
g
cos 2 π d λ sin θ
cos
2
= cos 2
=
f
b g
cos 2 φ 2
cos mπ
bπ d λ g sinθ
φ
F 7.95 rad IJ = 0.453
= cos G
H 2 K
2
2
max
2
The resultant amplitude is
b
g b
g
Er = E0 sin ω t + E0 sin ω t + φ + E0 ω t + 2φ ,
where
φ=
2π
λ
d sin θ .
b
g
E = E bsin ω t ge1 + cos φ + 2 cos φ − 1j + E bcos ω t gbsin φ + 2 sin φ cos φ g
E = E b1 + 2 cos φ gbsin ω t cos φ + cos ω t sin φ g = E b1 + 2 cos φ g sinbω t + φ g
F 1I
Then the intensity is
I ∝ E = E b1 + 2 cos φ g G J
H 2K
Er = E0 sin ω t + sin ω t cos φ + cos ω t sin φ + sin ω t cos 2φ + cos ω t sin 2φ
r
0
r
0
2
0
0
2
r
b
2
2
0
g
sin 2 ω t + φ is
where the time average of
1
.
2
From one slit alone we would get intensity I max ∝ E02
I = I max
(b)
LM1 + 2 cosFG 2π d sinθ IJ OP
H λ KQ
N
FG 1 IJ so
H 2K
2
.
Look at the N = 3 graph in Figure 37.14. Minimum intensity is zero, attained where
1
cos φ = − . One relative maximum occurs at cos φ = −1.00 , where I = I max .
2
The larger local maximum happens where cos φ = +1.00 , giving I = 9.00 I 0 .
The ratio of intensities at primary versus secondary maxima is 9.00 .
389
Chapter 37
Section 37.4
P37.21
(a)
Phasor Addition of Waves
FG A + B IJ cosFG A − B IJ to find the sum of the two sine functions
H 2 2K H 2 2K
We can use sin A + sin B = 2 sin
to be
f
b
g a
= b19.7 kN C g sina15 x − 4.5t + 35.0°f
E1 + E2 = 24.0 kN C sin 15 x − 4.5t + 35.0° cos 35.0°
E1 + E2
Thus, the total wave has amplitude 19.7 kN C and has a constant phase difference of
35.0° from the first wave.
(b)
In units of kN/C, the resultant phasor is
e j e
a16.1f + a11.3f at tan FGH 1116..31 IJK =
kx - ωt
y
j
E R = E 1 + E 2 = 12.0 i + 12.0 cos 70.0° i + 12.0 sin 70.0 j = 16.1 i + 11.3 j
2
ER =
2
−1
ER
19.7 kN C at 35.0°
E2
70.0°
α
x
E1
FIG. P37.21(b)
(c)
y
E R = 12.0 cos 70.0° i + 12.0 sin 70.0° j
+17.0 cos 160° i + 17.0 sin 160° j
P37.22
(a)
f
e
x
E3
The wave function of the total wave is
EP = 9.36 kN C sin 15 x − 4.5t + 169° .
g a
E2
ER
E R = −9.18 i + 1.83 j = 9.36 kN C at 169°
b
FIG. P37.21(c)
j e
E R = E0 i + i cos 20.0°+ j sin 20.0° + i cos 40.0°+ j sin 40.0°
j
y
ER
E R = E0 2.71i + 0.985 j = 2.88E0 at 20.0° = 2.88E0 at 0.349 rad
b
EP = 2.88E0 sin ω t + 0.349
kx - ωt
E1
+15.5 cos 80.0° i − 15.5 sin 80.0° j
g
E3
E2
x
E1
FIG. P37.22(a)
(b)
e
j e
E R = E0 i + i cos 60.0°+ j sin 60.0° + i cos 120°+ j sin 120°
E R = E0 1.00 i + 1.73 j = 2.00E0 at 60.0° = 2.00E0 at
FG
H
EP = 2.00E0 sin ω t +
π
3
IJ
K
j
y
E3
π
rad
3
ER
E2
E1
x
FIG. P37.22(b)
continued on next page
390
Interference of Light Waves
e
j e
E R = E0 i + i cos 120°+ j sin 120° + i cos 240°+ j sin 240°
(c)
j
y
E3
E R = E0 0 i + 0 j = 0
E2
EP = 0
x
E1
FIG. P37.22(c)
LM FG
N H
E R = E0 i + i cos
(d)
IJ e
K
3π
3π
+ i cos 3π + j sin 3π
+ j sin
2
2
jOPQ
y
E3
3π
rad
E R = E0 0 i − 1.00 j = E0 at 270° = E0 at
2
3π
EP = E0 sin ω t +
2
FG
H
P37.23
E R = 6.00 i + 8.00 j =
E2
IJ
K
a6.00f + a8.00f
2
2
at tan −1
FIG. P37.22(d)
FG 8.00 IJ
H 6.00 K
y
ER
E R = 10.0 at 53.1° = 10.0 at 0.927 rad
b
EP = 10.0 sin 100π t + 0.927
x
E1
g
8.00
α
π2
6.00
x
FIG. P37.23
P37.24
b
g
If E1 = E01 sin ω t and E2 = E02 sin ω t + φ , then by phasor addition,
the amplitude of E is
E0 =
bE
01
+ E02 cos φ
g + bE
2
02
sin φ
g
2
e
x
FIG. P37.24
j
y
E R = 21.0 i + 15.6 j = 26.2 at 36.6°
b
E 02
φ
θ
E 01
E R = 12.0 i + 18.0 cos 60.0° i + 18.0 sin 60.0° j
ER = 26.2 sin ω t + 36.6°
E0
2
2
+ 2E01 E02 cos φ + E02
E01
=
E sin φ
and the phase angle is found from sin θ = 02
.
E0
P37.25
y
ER
g
θ
18.0
60.0°
12.0
x
FIG. P37.25
P37.26
Constructive interference occurs where m = 0 , 1, 2 , 3 , … , for
FG 2π x − 2π ft + π IJ − FG 2π x
H λ
6K H λ
bx − x g + 1 − 1 = m
1
1
2
λ
12
16
2
− 2π ft +
IJ
K
π
= 2π m
8
b
g + FG π − π IJ = 2π m
H 6 8K
F 1I
= Gm − J λ
m = 0 , 1, 2 , 3 , …
H 48 K
2π x 1 − x 2
λ
x1 − x 2
.
Chapter 37
P37.27
See the figure to the right:
φ=
391
π /2
π
.
2
π /2
π /2
ωt
FIG. P37.27
P37.28
ER2 = E12 + E22 − 2E1 E2 cos β
where
ER
β = 180 − φ .
E2
φ = π /4
β
I ∝ E2
Since
E1
I R = I 1 + I 2 + 2 I 1 I 2 cos φ .
FIG. P37.28
P37.29
360°
where N defines the
N
number of coherent sources. Then,
Take φ =
ER =
y
The N = 6 case
φ=
N
∑ E0 sinbω t + mφ g = 0 .
360°
N
=
360°
6
= 60. 0°
m =1
In essence, the set of N electric field
components complete a full circle and
return to zero.
60.0°
x
FIG. P37.29
Section 37.5
Change of Phase Due to Reflection
Section 37.6
Interference in Thin Films
P37.30
Light reflecting from the first surface suffers phase reversal. Light reflecting from the second surface
does not, but passes twice through the thickness t of the film. So, for constructive interference, we
require
λn
+ 2t = λ n
2
where
λn =
λ
is the wavelength in the material.
n
Then
2t =
λn λ
=
2
2n
a fa
f
λ = 4nt = 4 1.33 115 nm = 612 nm .
392
P37.31
Interference of Light Waves
(a)
The light reflected from the top of the oil film undergoes phase
reversal. Since 1.45 > 1.33 , the light reflected from the bottom
undergoes no reversal. For constructive interference of reflected
light, we then have
FG
H
2nt = m +
or
λm =
IJ
K
1
λ
2
a fa
f
FIG. P37.31
2 1.45 280 nm
2nt
.
=
m+ 1 2
m+ 1 2
b g
Substituting for m gives:
b g
m=0 ,
λ 0 = 1 620 nm (infrared)
m=1,
λ 1 = 541 nm (green)
m=2,
λ 2 = 325 nm (ultraviolet).
Both infrared and ultraviolet light are invisible to the human eye, so the dominant color in
reflected light is green .
(b)
The dominant wavelengths in the transmitted light are those that produce destructive
interference in the reflected light. The condition for destructive interference upon reflection
is
2nt = mλ
or
λm =
2nt 812 nm
=
.
m
m
Substituting for m gives:
m=1,
λ 1 = 812 nm (near infrared)
m=2,
λ 2 = 406 nm (violet)
m=3 ,
λ 3 = 271 nm (ultraviolet).
Of these, the only wavelength visible to the human eye (and hence the dominate
wavelength observed in the transmitted light) is 406 nm. Thus, the dominant color in the
transmitted light is violet .
P37.32
Since 1 < 1.25 < 1.33 , light reflected both from the top and from the bottom surface of the oil suffers
phase reversal.
mλ cons
n
For constructive interference we require
2t =
and for destructive interference,
2t =
Then
λ cons
1
640 nm
=1+
=
= 1.25 and m = 2 .
2m 512 nm
λ dest
Therefore,
t=
b g
m + 1 2 λ des
n
a
f
.
2 640 nm
= 512 nm .
2 1.25
a f
Chapter 37
P37.33
393
Treating the anti-reflectance coating like a camera-lens coating,
FG
H
2t = m +
Let m = 0 :
t=
λ
4n
=
IJ
K
1 λ
.
2 n
3.00 cm
= 0.500 cm .
4 1.50
a f
This anti-reflectance coating could be easily countered by changing the wavelength of the radar—to
1.50 cm—now creating maximum reflection!
P37.34
FG
H
2nt = m +
IJ
K
1
λ
2
so
Minimum
P37.35
FG 1 IJ λ
H 2 K 2n
F 1 I a500 nmf =
t=G J
H 2 K 2a1.30f
t= m+
96.2 nm .
Since the light undergoes a 180° phase change at each surface of the film, the condition for
2nt
. The film thickness is
constructive interference is 2nt = mλ , or λ =
m
t = 1.00 × 10 −5 cm = 1.00 × 10 −7 m = 100 nm . Therefore, the wavelengths intensified in the reflected
light are
λ=
a fa
f
2 1.38 100 nm
276 nm
where m = 1, 2 , 3 , …
=
m
m
or λ 1 = 276 nm , λ 2 = 138 nm , . . . . All reflection maxima are in the ultraviolet and beyond.
No visible wavelengths are intensified.
P37.36
(a)
For maximum transmission, we want destructive interference in the light reflected from the
front and back surfaces of the film.
If the surrounding glass has refractive index greater than 1.378, light reflected from the front
surface suffers no phase reversal and light reflected from the back does undergo phase
reversal. This effect by itself would produce destructive interference, so we want the
distance down and back to be one whole wavelength in the film: 2t =
t=
=
656.3 nm
= 238 nm
2 1.378
a
f
(b)
The filter will expand. As t increases in 2nt = λ , so does λ increase .
(c)
Destructive interference for reflected light happens also for λ in 2nt = 2 λ ,
or
P37.37
λ
2n
λ
.
n
a
f
λ = 1.378 238 nm = 328 nm
anear ultravioletf .
If the path length difference ∆ = λ , the transmitted light will be bright. Since ∆ = 2d = λ ,
d min =
λ 580 nm
=
= 290 nm .
2
2
394
P37.38
Interference of Light Waves
The condition for bright fringes is
2t +
λ
λ
=m
2n
n
θ
m = 1, 2 , 3 , … .
R
From the sketch, observe that
f FGH
a
t = R 1 − cos θ ≈ R 1 − 1 +
P37.39
I
JK
FG IJ
H K
θ2
R r
=
2
2 R
2
=
r2
.
2R
t
2
FG
H
r
IJ
K
The condition for a bright fringe becomes
r
1 λ
= m−
.
R
2 n
Thus, for fixed m and λ,
nr 2 = constant .
Therefore, n liquid r f2 = n air ri2 and
n liquid = 1.00
cmf
a f aa11..5031 cm
f
FIG. P37.38
2
2
= 1.31 .
For destructive interference in the air,
2t = mλ .
For 30 dark fringes, including the one where the plates
meet,
t=
a
f
29 600 nm
= 8.70 × 10 −6 m .
2
Therefore, the radius of the wire is
r=
t 8.70 µm
=
= 4.35 µm .
2
2
FIG. P37.39
P37.40
For total darkness, we want destructive interference for reflected light for both 400 nm and 600 nm.
With phase reversal at just one reflecting surface, the condition for destructive interference is
2n air t = mλ
m = 0 , 1, 2 , … .
The least common multiple of these two wavelengths is 1 200 nm, so we get no reflected light at
2 1.00 t = 3 400 nm = 2 600 nm = 1 200 nm , so t = 600 nm at this second dark fringe.
a f a
f a
f
By similar triangles,
600 nm 0.050 0 mm
=
,
x
10.0 cm
or the distance from the contact point is
x = 600 × 10 −9 m
e
m I
J=
jFGH 5.000.100
× 10 m K
−5
1.20 mm .
Chapter 37
Section 37.7
P37.41
The Michelson Interferometer
When the mirror on one arm is displaced by ∆ , the path difference changes by 2∆ . A shift
resulting in the reversal between dark and bright fringes requires a path length change of one-half
mλ
, where in this case, m = 250 .
wavelength. Therefore, 2 ∆ =
2
∆ =m
P37.42
395
a fe
j
250 6.328 × 10 −7 m
λ
=
= 39.6 µm
4
4
e
j
Distance = 2 3.82 × 10 −4 m = 1 700 λ
λ = 4.49 × 10 −7 m = 449 nm
The light is blue .
P37.43
Counting light going both directions, the number of wavelengths originally in the cylinder is
2L
2L 2nL
=
m1 =
. It changes to m 2 =
as the cylinder is filled with gas. If N is the number of bright
λ
λn
λ
2L
fringes passing, N = m 2 − m1 =
n − 1 , or the index of refraction of the gas is
λ
n = 1+
a f
Nλ
2L
Additional Problems
*P37.44
(a)
Where fringes of the two colors coincide we have d sin θ = mλ = m ′λ ′ , requiring
(b)
λ = 430 nm , λ ′ = 510 nm
∴
λ m′
.
=
λ′ m
m ′ 430 nm 43
=
=
, which cannot be reduced any further. Then m = 51, m ′ = 43 .
m 510 nm 51
F mλ IJ = sin LM a51fe430 × 10 mj OP = 61.3°
= sin G
HdK
MN 0.025 × 10 m PQ
= L tan θ = a1.5 mf tan 61.3° = 2.74 m
−9
θm
ym
P37.45
−1
−1
−3
m
The wavelength is λ =
c 3.00 × 10 8 m s
=
= 5.00 m.
f
60.0 × 10 6 s −1
Along the line AB the two traveling waves going in opposite directions add to give a standing wave.
The two transmitters are exactly 2.00 wavelengths apart and the signal from B, when it arrives at A,
will always be in phase with transmitter B. Since B is 180° out of phase with A, the two signals
always interfere destructively at the position of A.
The first antinode (point of constructive interference) is located at distance
λ 5.00 m
=
= 1.25 m from the node at A.
4
4
396
*P37.46
Interference of Light Waves
Along the line of length d joining the source, two identical waves
moving in opposite directions add to give a standing wave. An
d λ
antinode is halfway between the sources. If > , there is space for
2 2
d
two more antinodes for a total of three. If > λ , there will be at least
2
d
five antinodes, and so on. To repeat, if > 0 , the number of antinodes
s
N
A
N
s
λ
4
FIG. P37.46
λ
d
> 1, the number of antinodes is 3 or more. If > 2 , the
λ
λ
number of antinodes is 5 or more. In general,
is 1 or more. If
d
d
.
λ
The number of antinodes is 1 plus 2 times the greatest integer less than or equal to
d λ
d λ
< , there will be no nodes. If > , there will be space for at least two nodes, as shown in the
2 4
2 4
d 3λ
d 5λ
picture. If >
, there will be at least four nodes. If >
six or more nodes will fit in, and so on.
2
4
2
4
To repeat, if 2d < λ the number of nodes is 0. If 2d > λ the number of nodes is 2 or more. If 2d > 3 λ
d 1
+
> 1,
the number of nodes is 4 or more. If 2d > 5 λ the number of nodes is 6 or more. Again, if
λ 2
d 1
d 1
the number of nodes is at least 2. If
+
> 2 , the number of nodes is at least 4. If
+
> 3 , the
λ 2
λ 2
number of nodes is at least 6. In general,
If
FG
H
IJ
K
FG
H
FG
H
the number of nodes is 2 times the greatest nonzero integer less than
IJ
K
IJ
K
FG d + 1 IJ .
H λ 2K
Next, we enumerate the zones of constructive interference. They are described by d sin θ = mλ ,
m = 0 , 1, 2 , … with θ counted as positive both left and right of the maximum at θ = 0 in the center.
d
The number of side maxima on each side is the greatest integer satisfying sin θ ≤ 1, d1 ≥ mλ , m ≤ .
λ
So the total
number of bright fringes is one plus 2 times the greatest integer less than or equal to
It is equal to the number of antinodes on the line joining the sources.
FG
H
The interference minima are to the left and right at angles described by d sin θ = m +
FG
H
IJ
K
d
.
λ
IJ
K
1
λ,
2
1
d 1
d 1
λ , m max < − or m max + 1 < + . Let n = 1, 2 , 3 , ….
λ 2
λ 2
2
d 1
Then the number of side minima is the greatest integer n less than + . Counting both left and
λ 2
d 1
right, the number of dark fringes is two times the greatest positive integer less than
+ . It is
λ 2
equal to the number of nodes in the standing wave between the sources.
m = 0 , 1, 2 , …. With sin θ < 1, d1 > m max +
FG
H
IJ
K
Chapter 37
P37.47
397
My middle finger has width d = 2 cm .
(a)
Two adjacent directions of constructive interference for 600-nm light are described by
d sin θ = mλ
θ0 = 0
e2 × 10
(b)
−2
j
e
m sin θ 1 = 1 6 × 10 −7 m
Thus,
θ 1 = 2 × 10 −3 degree
and
θ 1 − θ 0 ~ 10 −3 degree .
Choose
θ 1 = 20°
e2 × 10
−2
j
af
j
m sin 20° = 1 λ
λ = 7 mm
Millimeter waves are microwaves .
f=
P37.48
λ
f=
:
3 × 10 8 m s
7 × 10 −3 m
~ 10 11 Hz
If the center point on the screen is to be a dark spot rather than bright, passage through the plastic
must delay the light by one-half wavelength. Calling the thickness of the plastic t.
t
λ
P37.49
c
+
t
nt
1
=
=
2 λn λ
t=
or
λ
2 n −1
a f
where n is the index of refraction for the plastic.
No phase shift upon reflection from the upper surface (glass to air) of the film, but there will be a
λ
due to the reflection at the lower surface of the film (air to metal). The total phase
2
difference in the two reflected beams is
shift of
then
δ = 2nt +
λ
.
2
For constructive interference, δ = mλ
or
a f
2 1.00 t +
λ
= mλ .
2
Thus, the film thickness for the m th order bright fringe is
FG
H
tm = m −
FG IJ
H K
IJ
K
λ
λ
1 λ
=m
−
2 2
2
4
and the thickness for the m − 1 bright fringe is:
a
t m−1 = m − 1
fFGH λ2 IJK − λ4 .
Therefore, the change in thickness required to go from one bright fringe to the next is
∆t = t m − t m −1 =
continued on next page
λ
.
2
398
Interference of Light Waves
To go through 200 bright fringes, the change in thickness of the air film must be:
200
FG λ IJ = 100λ .
H 2K
Thus, the increase in the length of the rod is
e
j
∆L = 100 λ = 100 5.00 × 10 −7 m = 5.00 × 10 −5 m .
P37.50
From
∆L = Liα∆t
we have:
α=
∆L
5.00 × 10 −5 m
=
= 20.0 × 10 −6 ° C −1 .
Li ∆T
0.100 m 25.0° C
a
fa
f
Since 1 < 1.25 < 1.34 , light reflected from top and bottom surfaces of the oil undergoes phase reversal.
The path difference is then 2t, which must be equal to
mλ n =
mλ
n
for maximum reflection, with m = 1 for the given first-order condition and n = 1.25 . So
t=
a
f
mλ 1 500 nm
=
= 200 nm .
2n
2 1.25
a f
a
A=
P37.51
1.00 m3
e
200 10
−9
m
j
f
1.00 m3 = 200 nm A
The volume we assume to be constant:
= 5.00 × 10 6 m 2 = 5.00 km 2 .
One radio wave reaches the receiver R directly from the distant
source at an angle θ above the horizontal. The other wave
undergoes phase reversal as it reflects from the water at P.
Constructive interference first occurs for a path difference of
d=
λ
2
(1)
It is equally far from P to R as from P to R ′ , the mirror image of
the telescope.
The angles θ in the figure are equal because they each form
part of a right triangle with a shared angle at R ′ .
a
f
FIG. P37.51
a
f
So the path difference is
d = 2 20.0 m sin θ = 40.0 m sin θ .
The wavelength is
λ=
Substituting for d and λ in Equation (1),
a40.0 mf sinθ = 5.002 m .
Solving for the angle θ, sin θ =
8
c 3.00 × 10 m s
=
= 5.00 m .
f
60.0 × 10 6 Hz
5.00 m
and θ = 3.58° .
80.0 m
Chapter 37
P37.52
For destructive interference, the path length must differ by mλ . We may treat this problem as a
π
-phase shift at the mirror. The
2
second slit is the mirror image of the source, 1.00 cm below the mirror plane. Modifying
Equation 37.5,
double slit experiment if we remember the light undergoes a
y dark
P37.53
e
ja
f
−7
mλL 1 5.00 × 10 m 100 m
=
=
= 2.50 mm .
d
2.00 × 10 −2 m
e
j
a15.0 kmf + h = 30.175 km
a15.0 kmf + h = 227.63
2
2
2
2
2
h = 1.62 km
P37.54
FIG. P37.53
2nt = mλ
For dark fringes,
and at the edge of the wedge, t =
so
P37.55
a fa fa
f
2nt = mλ
When submerged in water,
m=
a
84 500 nm
.
2
f
2 1.33 42 500 nm
500 nm
FIG. P37.54
m + 1 = 113 dark fringes .
I
From Equation 37.13,
I max
I
Let λ 2 equal the wavelength for which
I max
λ2 =
Then
But
FG
H
π yd
I1
= λ 1 cos −1
L
I max
IJ
K
12
a
FG I IJ
HI K
max
→
F π yd I .
GH λ L JK
I2
= 0.640 .
I max
π yd L
cos
−1
a
bI
2
I max
g
12
.
f
= 600 nm cos −1 0.900 = 271 nm .
Substituting this value into the expression for λ 2 ,
Note that in this problem, cos −1
f
= cos 2
λ2 =
271 nm
cos
−1
e0.640 j
12
must be expressed in radians.
12
= 421 nm .
399
400
P37.56
Interference of Light Waves
At entrance, 1.00 sin 30.0° = 1.38 sin θ 2
θ 2 = 21.2°
Call t the unknown thickness. Then
t
t
cos 21.2° =
a=
a
cos 21.2°
c
tan 21.2° =
c = t tan 21.2°
t
b
sin θ 1 =
b = 2t tan 21.2° sin 30.0°
2c
The net shift for the second ray, including the phase reversal on reflection
of the first, is
FIG. P37.56
λ
2
where the factor n accounts for the shorter wavelength in the film. For constructive interference, we
require
λ
2 an − b − = mλ .
2
λ
2 an − b − = 0 .
The minimum thickness will be given by
2
λ
nt
= 2 an − b = 2
− 2t tan 21.2° sin 30.0°
2
cos 21. 2°
590 nm
2 × 1.38
t = 115 nm
=
− 2 tan 21.2° sin 30.0° t = 2.57t
cos 21.2°
2
2 an − b −
a
f
FG
H
P37.57
IJ
K
λ
2
where a and b are as shown in the ray diagram, n is the index of
λ
is due to phase reversal at the top
refraction, and the term
2
surface. For constructive interference, δ = mλ where m has
integer values. This condition becomes
1
2na − b = m + λ
(1)
2
t
From the figure’s geometry, a =
cos θ 2
t sin θ 2
c = a sin θ 2 =
cos θ 2
2t sin θ 2
b = 2 c sin φ 1 =
sin φ 1
cos θ 2
Also, from Snell’s law,
sin φ 1 = n sin θ 2 .
The shift between the two reflected waves is δ = 2na − b −
FG
H
Thus,
b=
IJ
K
FIG. P37.57
2nt sin 2 θ 2
.
cos θ 2
With these results, the condition for constructive interference given in Equation (1) becomes:
FG t IJ − 2nt sin θ = 2nt e1 − sin θ j = FG m + 1 IJ λ
H 2K
H cos θ K cos θ cosθ
F 1I
2nt cos θ = G m + J λ .
H 2K
2
2n
2
or
2
2
2
2
2
2
Chapter 37
P37.58
(a)
Minimum:
2nt = mλ 2
Maximum:
2nt = m ′ +
for λ 1 > λ 2 ,
FG m′ + 1 IJ < m
H 2K
so
m′ = m − 1 .
Then
2nt = mλ 2 = m −
FG
H
401
for m = 0 , 1, 2 , …
IJ
K
1
λ1
2
FG
H
for m ′ = 0 , 1, 2 , …
IJ
K
1
λ1
2
2mλ 2 = 2mλ 1 − λ 1
m=
so
(b)
m=
λ1
2 λ1 − λ 2
b
g
.
500
= 1.92 → 2 (wavelengths measured to ±5 nm )
2 500 − 370
a
f
Minimum:
2nt = mλ 2
a f a
f
1I
F
2nt = G m − 1 + J λ = 1.5 λ
H
2K
2a1.40ft = 1.5a500 nmf
2 1.40 t = 2 370 nm
Maximum:
t = 264 nm
t = 268 nm
Film thickness = 266 nm .
P37.59
From the sketch, observe that
x = h2 +
FG d IJ
H 2K
2
=
4h 2 + d 2
.
2
d
x
h
x
Including the phase reversal due to reflection from
the ground, the total shift between the two waves
is δ = 2 x − d −
λ
.
2
d/2
FIG. P37.59
(a)
For constructive interference, the total shift must be an integral number of wavelengths, or
δ = mλ where m = 0 , 1, 2 , 3 , … .
Thus,
FG
H
2x − d = m +
IJ
K
1
λ
2
or
λ=
4x − 2d
.
2m + 1
For the longest wavelength, m = 0 , giving λ = 4x − 2d = 2 4h 2 + d 2 − 2d .
(b)
For destructive interference,
Thus,
FG
H
IJ
K
1
λ where m = 0 , 1, 2 , 3 , … .
2
2x − d
λ=
.
or
2x − d = mλ
m
δ = m−
For the longest wavelength, m = 1 giving λ = 2 x − d =
4h 2 + d 2 − d .
402
P37.60
Interference of Light Waves
FG IJ
H K
FλI
2t = G J m .
H nK
F hI
t = G Jx .
HK
2t =
Bright fringes occur when
and dark fringes occur when
The thickness of the film at x is
Therefore, x bright =
P37.61
FG
H
λ
1
m+
2 hn
2
IJ
K
λ
1
m+
n
2
and x dark =
λ m
.
2 hn
FIG. P37.60
Call t the thickness of the film. The central maximum
corresponds to zero phase difference. Thus, the added
distance ∆r traveled by the light from the lower slit must
introduce a phase difference equal to that introduced by
the plastic film. The phase difference φ is
φ = 2π
FG t IJ an − 1f .
Hλ K
L
Thin
film
y'
d
θ
α
Screen
∆r
a
The corresponding difference in path length ∆r is
FIG. P37.61
Fλ I F t I Fλ I
∆r = φ G J = 2π G J an − 1fG J = tan − 1f .
H 2π K H λ K H 2π K
a
Zero
order
m=0
a
a
Note that the wavelength of the light does not appear in this equation. In the figure, the two rays
from the slits are essentially parallel.
P37.62
∆r y ′
=
.
d
L
Thus the angle θ may be expressed as
tan θ =
Eliminating ∆r by substitution,
t n −1 L
y′ t n − 1
=
gives y ′ =
.
L
d
d
a f
a f
The shift between the waves reflecting from the top and bottom
surfaces of the film at the point where the film has thickness t is
θ
λ
λ
, with the factor of
being due to a phase reversal
2
2
at one of the surfaces.
δ = 2tn film +
For the dark rings (destructive interference), the total shift should
1
be δ = m + λ with m = 0 , 1, 2 , 3 , … . This requires that
2
mλ
t=
.
2n film
FG
H
IJ
K
a f
To find t in terms of r and R,
R2 = r 2 + R − t
Since t is much smaller than R,
t 2 << 2 Rt
2
R
nfilm
t
r
FIG. P37.62
so
r 2 = 2 Rt + t 2 .
and
r 2 ≈ 2 Rt = 2 R
FG mλ IJ .
H 2n K
film
Thus, where m is an integer,
r≈
mλ R
.
n film
Chapter 37
P37.63
(a)
FG
H
Constructive interference in the reflected light requires 2t = m +
403
IJ
K
1
λ . The first bright ring
2
has m = 0 and the 55th has m = 54 , so at the edge of the lens
t=
e
j = 17.7 µm .
54.5 650 × 10 −9 m
2
Now from the geometry in Figure 37.18, the distance from the center of curvature down to
the flat side of the lens is
R 2 − r 2 = R − t or R 2 − r 2 = R 2 − 2 Rt + t 2
e
j e
j
2
5.00 × 10 −2 m + 1.77 × 10 −5 m
r 2 + t2
=
R=
2t
2 1.77 × 10 −5 m
(b)
*P37.64
a fFGH
e
IJ
K
j
FG
H
2
= 70.6 m
IJ
K
1
1
1
1
1
so f = 136 m
= 0.520
−
= n−1
−
∞ −70.6 m
f
R2 R 2
Light reflecting from the upper interface of the air layer
suffers no phase change, while light reflecting from the
lower interface is reversed 180°. Then there is indeed a
dark fringe at the outer circumference of the lens, and a
dark fringe wherever the air thickness t satisfies 2t = mλ ,
m = 0 , 1, 2 , ….
(a)
At the central dark spot m = 50 and
50λ
= 25 589 × 10 −9 m = 1.47 × 10 −5 m .
t0 =
2
(b)
In the right triangle,
e
a8 mf
2
e
j = r + a8 mf − 2a8 mfe1.47 × 10
2a8 mfe1.47 × 10 mj = 1.53 × 10 m
IJ a
K
2
2
2
−5
fFGH
1
1
1
1
1
= n −1
−
= 1.50 − 1
−
∞ 8.00 m
f
R1 R 2
f = −16.0 m
R=8m
FIG. P37.64
= r 2 + 8 m − 1.47 × 10 −5 m
a fFGH
r
j
last term is negligible. r =
(c)
t0
IJ
K
−2
−5
j
m + 2 × 10 −10 m 2 . The
404
P37.65
Interference of Light Waves
For bright rings the gap t between surfaces is given by
1
2t = m + λ . The first bright ring has m = 0 and the hundredth
2
has m = 99.
FG
H
So, t =
IJ
K
a fe
1
99.5 500 × 10 −9 m = 24.9 µm .
2
j
Call rb the ring radius. From the geometry of the figure at the right,
FH
t = r − r 2 − rb2 − R − R 2 − rb2
IK
Since rb << r , we can expand in series:
F
GH
FIG. P37.65
I − R + RF 1 − 1 I = 1 − 1
JK
GH 2 R JK 2 r 2 R
L 2t OP = LM 2e24.9 × 10 mj OP = 1.73 cm
r =M
N 1 r − 1 R Q MN 1 4.00 m − 1 12.0 m PQ
7π
4π O
L π
E = E + E + E = Mcos + 3.00 cos
+ 6.00 cos P i
2
3 Q
N 6
π
π
π
4
7
L
O
+ Msin + 3.00 sin
+ 6.00 sin P j
3 Q
2
N 6
t = r −r 1−
rb2
2
rb2
2
1
2r
rb2
12
−6
12
rb2
b
P37.66
R
1
2
y
3
a−2.13f + a−7.70f
2
2
b
at tan −1
Thus, EP = 7.99 sin ω t + 4.44 rad
P37.67
(a)
(b)
(c)
g
ER
E3
FG −7.70 IJ = 7.99 at 4.44 rad
H −2.13 K
FIG. P37.66
.
FG
H
Bright bands are observed when
2nt = m +
Hence, the first bright band ( m = 0 ) corresponds to
nt =
Since
x 1 t1
=
x2 t2
we have
x 2 = x1
t1 =
λ 1 420 nm
=
= 78.9 nm
4n 4 1.33
t2 =
λ 2 680 nm
=
= 128 nm
4n 4 1.33
1
1
a f
θ ≈ tan θ =
a f
t1 78.9 nm
=
= 2.63 × 10 −6 rad
x 1 3.00 cm
2
1
IJ
K
1
λ.
2
λ
.
4
FG t IJ = x FG λ IJ = a3.00 cmfFG 680 nmIJ =
H 420 nm K
Ht K Hλ K
2
x
E2
E R = −2.13 i − 7.70 j
ER =
E1
4.86 cm .
405
Chapter 37
*P37.68
Depth = one-quarter of the wavelength in plastic.
t=
P37.69
*P37.70
λ 780 nm
=
= 130 nm
4n 4 1.50
a f
F 1I
2 h sin θ = G m + J λ
H 2K
F ∆y I 1
2 hG J = λ
H 2L K 2
bright
a
fe
j
2.00 m 606 × 10 −9 m
Lλ
=
= 0.505 mm
h=
2 ∆y
2 1. 2 × 10 −3 m
so
e
j
Represent the light radiated from each slit to point P as a phasor. The two have
equal amplitudes E. Since intensity is proportional to amplitude squared, they
3E 2
add to amplitude 3E . Then cos θ =
, θ = 30° . Next, the obtuse angle
E
between the two phasors is 180 − 30 − 30 = 120°, and φ = 180 − 120° = 60° . The
phase difference between the two phasors is caused by the path difference
δ
φ
60° λ
,δ =λ
= . Then
δ = SS 2 − SS 1 according to =
λ 360°
360° 6
3E
θ
E
FIG. P37.70
λ
6
2 Lλ λ 2
+
L2 + d 2 = L2 +
6
36
L2 + d 2 − L =
The last term is negligible
d=
P37.71
FG 2Lλ IJ
H6K
12
=
a
f
2 1. 2 m 620 × 10 −9 m
= 0.498 mm .
6
Superposing the two vectors, ER = E1 + E 2
ER = E 1 + E 2 =
ER =
FG E
H
0
+
E0
cos φ
3
IJ + FG E sin φIJ
K H3 K
2
0
2
= E02 +
E2
E2
2 2
E0 cos φ + 0 cos 2 φ + 0 sin 2 φ
3
9
9
10 2 2 2
E0 + E0 cos φ
9
3
Since intensity is proportional to the square of the amplitude,
I=
10
2
I max + I max cos φ .
9
3
Using the trigonometric identity cos φ = 2 cos 2
I=
or
FG
H
IJ
K
φ
− 1 , this becomes
2
φ
φ
10
2
4
4
I max + I max 2 cos 2 − 1 = I max + I max cos 2 ,
9
3
2
9
3
2
I=
FG
H
φ
4
I max 1 + 3 cos 2
9
2
IJ
K
.
φ
E
406
Interference of Light Waves
ANSWERS TO EVEN PROBLEMS
P37.2
515 nm
P37.40
1.20 mm
P37.4
(a) 36.2°; (b) 5.08 cm ; (c) 508 THz
P37.42
449 nm; blue
P37.6
maxima at 0° , 29.1° , 76.3° ;
minima at 14.1° and 46.8°
P37.44
(a) see the solution; (b) 2.74 m
P37.8
36.2 cm
P37.46
P37.10
641
number of antinodes = number of
constructive interference zones
= 1 plus 2 times the greatest positive
d
integer ≤
P37.12
6.33 mm s
P37.14
see the solution
P37.16
(a) 1.29 rad; (b) 99.8 nm
P37.18
0.968
P37.20
(a) see the solution; (b) 9.00
P37.22
(a) 2.88E0 at 0.349 rad ;
(b) 2.00E0 at
(d) E0 at
P37.24
P37.26
λ
π
rad ; (c) 0;
3
3π
rad
2
see the solution
FG
H
IJ
K
1
λ where
48
m = 0 , 1, 2 , 3 , …
x1 − x 2 = m −
P37.28
see the solution
P37.30
612 nm
P37.32
512 nm
P37.34
96.2 nm
P37.36
(a) 238 nm; (b) λ increase ; (c) 328 nm
P37.38
1.31
number of nodes = number of destructive
interference zones = 2 times the greatest
d 1
positive integer <
+
λ 2
FG
H
IJ
K
P37.48
λ
2 n −1
P37.50
5.00 km 2
P37.52
2.50 mm
P37.54
113
P37.56
115 nm
P37.58
(a) see the solution; (b) 266 nm
P37.60
see the solution
P37.62
see the solution
P37.64
(a) 14.7 µm ; (b) 1.53 cm; (c) −16.0 m
P37.66
7.99 sin ω t + 4.44 rad
P37.68
130 nm
P37.70
0.498 mm
a f
b
g
38
Diffraction Patterns and Polarization
CHAPTER OUTLINE
38.1
38.2
38.3
38.4
38.5
38.6
Introduction to Diffraction
Patterns
Diffraction Patterns from
Narrow Slits
Resolution of Single-Slit and
Circular Apertures
The Diffraction Grating
Diffraction of X-Rays by
Crystals
Polarization of Light Waves
ANSWERS TO QUESTIONS
Q38.1
Audible sound has wavelengths on the order of meters or
centimeters, while visible light has a wavelength on the order
of half a micrometer. In this world of breadbox-sized objects,
is large for sound, and sound diffracts around behind walls
λ
a
λ
is a tiny fraction for visible light passing
a
ordinary-size objects or apertures, so light changes its direction
by only very small angles when it diffracts.
Another way of phrasing the answer: We can see by a
small angle around a small obstacle or around the edge of a
small opening. The side fringes in Figure 38.1 and the Arago
spot in the center of Figure 38.3 show this diffraction. We
cannot always hear around corners. Out-of-doors, away from
reflecting surfaces, have someone a few meters distant face
away from you and whisper. The high-frequency, shortwavelength, information-carrying components of the sound do
not diffract around his head enough for you to understand his
words.
with doorways. But
Q38.2
The wavelength of light is extremely small in comparison to the dimensions of your hand, so the
diffraction of light around an obstacle the size of your hand is totally negligible. However, sound
waves have wavelengths that are comparable to the dimensions of the hand or even larger.
Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.
Q38.3
If you are using an extended light source, the gray
area at the edge of the shadow is the penumbra. A
bug looking up from there would see the light
source partly but not entirely blocked by the book.
If you use a point source of light, hold it and the
book motionless, and look at very small angles out
from the geometrical edge of the shadow, you may
see a series of bright and dark bands produced by
diffraction of light at the straight edge, as shown in
the diagram.
FIG. Q38.3
407
408
Diffraction Patterns and Polarization
3 × 10 8 m s
Q38.4
An AM radio wave has wavelength on the order of
Q38.5
The intensity of the light coming through the slit decreases, as you would expect. The central
~ 300 m . This is large compared to
1 × 10 6 s −1
the width of the mouth of a tunnel, so the AM radio waves can reflect from the surrounding ground
as if the hole were not there. (In the same way, a metal screen forming the dish of a radio telescope
can reflect radio waves as if it were solid, and a hole-riddled screen in the door of a microwave oven
keeps the microwaves inside.) The wave does not “see” the hole. Very little of the radio wave energy
enters the tunnel, and the AM radio signal fades. An FM radio wave has wavelength a hundred
times smaller, on the order of a few meters. This is smaller than the size of the tunnel opening, so
the wave can readily enter the opening. (On the other hand, the long wavelength of AM radio
waves lets them diffract more around obstacles. Long-wavelength waves can change direction more
in passing hills or large buildings, so in some experiments FM fades more than AM.)
λ
for
a
destructive interference on each side of the central maximum, θ increases as a decreases.
maximum increases in width as the width of the slit decreases. In the condition sin θ =
Q38.6
It is shown in the correct orientation. If the horizontal width of the opening is equal to or less than
the wavelength of the sound, then the equation a sin θ = 1 λ has the solution θ = 90° , or has no
solution. The central diffraction maximum covers the whole seaward side. If the vertical height of
the opening is large compared to the wavelength, then the angle in a sin θ = 1 λ will be small, and
the central diffraction maximum will form a thin horizontal sheet.
af
af
Q38.7
Q38.8
Q38.9
The speaker is mounted incorrectly—it should be rotated by 90°. The speaker is mounted with its
narrower dimension vertical. That means that the sound will diffract more vertically than it does
horizontally. Mounting the speaker so that its thinner dimension is horizontal will give more
diffraction spreading in the horizontal plane, broadcasting “important” information to the troops,
instead of to the birds in the air and the worms in the ground, as the speaker was mounted in the
movie.
1.22λ
for the resolution of a circular aperture, the pupil of your eye.
D
Suppose your dark-adapted eye has pupil diameter D = 5 mm . An average wavelength for visible
light is λ = 550 nm . Suppose the headlights are 2 m apart and the car is a distance L away. Then
2m
= 1.22 × 1.1 × 10 −4 so L ~ 10 km. The actual distance is less than this because the variableθm =
L
temperature air between you and the car makes the light refract unpredictably. The headlights
twinkle like stars.
We apply the equation θ m =
Consider incident light nearly parallel to the horizontal ruler.
Suppose it scatters from bumps at distance d apart to produce a
diffraction pattern on a vertical wall a distance L away. At a
y
point of height y, where θ = gives the scattering angle θ, the
L
character of the interference is determined by the shift δ
between beams scattered by adjacent bumps, where
d
δ=
− d . Bright spots appear for δ = mλ , where
cos θ
0 , 1, 2 , 3 , …. For small θ, these equations combine and reduce
to mλ =
y m2 d
. Measurement of the heights y m of bright spots
2L2
allows calculation of the wavelength of the light.
y
θ
d
L
FIG. Q38.9
Chapter 38
409
Q38.10
Yes, but no diffraction effects are observed because the separation distance between adjacent ribs is
so much greater than the wavelength of x-rays. Diffraction does not limit the resolution of an x-ray
image. Diffraction might sometimes limit the resolution of an ultrasonogram.
Q38.11
Vertical. Glare, as usually encountered when driving or boating, is horizontally polarized. Reflected
light is polarized in the same plane as the reflecting surface. As unpolarized light hits a shiny
horizontal surface, the atoms on the surface absorb and then reemit the light energy as a reflection.
We can model the surface as containing conduction electrons free to vibrate easily along the surface,
but not to move easily out of surface. The light emitted from a vibrating electron is partially or
completely polarized along the plane of vibration, thus horizontally.
Q38.12
The earth has an atmosphere, while the moon does not. The nitrogen and oxygen molecules in the
earth’s atmosphere are of the right size to scatter short-wavelength (blue) light especially well, while
there is nothing surrounding the moon to scatter light.
Q38.13
The little particles of dust diffusely reflect light from the light beam. Note that this is not necessarily
scattering. Scattering is a resonance phenomenon—as when the O 2 and N 2 molecules in our
atmosphere scatter blue light more than red. In general, light is visible when it enters your eye. Your
eyes and brain are well prepared to make you think on a subconscious level that you can ‘see’ where
light is coming from or sometimes ‘see’ light going past you, but really you see only light entering
your eye.
Q38.14
Light from the sky is partially polarized. Light from the blue sky that is polarized at 90° to the
polarization axis of the glasses will be blocked, making the sky look darker as compared to the
clouds.
Q38.15
First think about the glass without a coin and about one particular point P on the screen. We can
divide up the area of the glass into ring-shaped zones centered on the line joining P and the light
source, with successive zones contributing alternately in-phase and out-of-phase with the light that
takes the straight-line path to P. These Fresnel zones have nearly equal areas. An outer zone
contributes only slightly less to the total wave disturbance at P than does the central circular zone.
Now insert the coin. If P is in line with its center, the coin will block off the light from some
particular number of zones. The first unblocked zone around its circumference will send light to P
with significant amplitude. Zones farther out will predominantly interfere destructively with each
other, and the Arago spot is bright. Slightly off the axis there is nearly complete destructive
interference, so most of the geometrical shadow is dark. A bug on the screen crawling out past the
edge of the geometrical shadow would in effect see the central few zones coming out of eclipse. As
the light from them interferes alternately constructively and destructively, the bug moves through
bright and dark fringes on the screen. The diffraction pattern is shown in Figure 38.3 in the text.
Q38.16
Since obsidian glass is opaque, a standard method of measuring incidence and refraction angles and
using Snell’s Law is ineffective. Reflect unpolarized light from the horizontal surface of the obsidian
through a vertically polarized filter. Change the angle of incidence until you observe that none of
the reflected light is transmitted through the filter. This means that the reflected light is completely
horizontally polarized, and that the incidence and reflection angles are the polarization angle. The
tangent of the polarization angle is the index of refraction of the obsidian.
410
Diffraction Patterns and Polarization
Q38.17
The fine hair blocks off light that would otherwise go through a fine slit and produce a diffraction
pattern on a distant screen. The width of the central maximum in the pattern is inversely
proportional to the distance across the slit. When the hair is in place, it subtracts the same diffraction
pattern from the projected disk of laser light. The hair produces a diffraction minimum that crosses
the bright circle on the screen. The width of the minimum is inversely proportional to the diameter
of the hair. The central minimum is flanked by narrower maxima and minima. Measure the width 2y
y λ
of the central minimum between the maxima bracketing it, and use Equation 38.1 in the form =
L a
to find the width a of the hair.
Q38.18
The condition for constructive interference is that the three radio signals arrive at the city in phase.
We know the speed of the waves (it is the speed of light c), the angular bearing θ of the city east of
north from the broadcast site, and the distance d between adjacent towers. The wave from the
westernmost tower must travel an extra distance 2d sin θ to reach the city, compared to the signal
from the eastern tower. For each cycle of the carrier wave, the western antenna would transmit first,
d sin θ
, and the eastern antenna after an additional equal time
the center antenna after a time delay
c
delay.
SOLUTIONS TO PROBLEMS
Section 38.1
Introduction to Diffraction Patterns
Section 38.2
Diffraction Patterns from Narrow Slits
P38.1
sin θ =
λ 6.328 × 10 −7
=
= 2.11 × 10 −3
−4
a
3.00 × 10
y
= tan θ ≈ sin θ = θ (for small θ)
1.00 m
2 y = 4.22 mm
P38.2
The positions of the first-order minima are
y
λ
≈ sin θ = ± . Thus, the spacing between these two
L
a
FG λ IJ L and the wavelength is
H aK
F ∆y I F a I F 4.10 × 10 m IJ FG 0.550 × 10 m IJ =
λ = G JG J = G
H 2 K H L K H 2 K H 2.06 m K
minima is ∆y = 2
−3
P38.3
y
mλ
≈ sin θ =
L
a
∆m = 3 − 1 = 2
−3
547 nm .
∆y = 3.00 × 10 −3 nm
and
a=
a=
∆mλL
∆y
e
ja
f=
2 690 × 10 −9 m 0.500 m
e3.00 × 10
−3
j
m
2.30 × 10 −4 m
Chapter 38
P38.4
For destructive interference,
sin θ = m
λ λ 5.00 cm
= =
= 0.139
a a 36.0 cm
θ = 7.98°
and
d
= tan θ
L
a
f
d = L tan θ = 6.50 m tan 7.98° = 0.912 m
gives
d = 91.2 cm .
P38.5
v 340 m s
=
= 0.523 m .
f
650 s −1
If the speed of sound is 340 m/s,
λ=
Diffraction minima occur at angles described by
a sin θ = mλ
a1.10 mf sinθ
a1.10 mf sinθ
a1.10 mf sinθ
1
2
3
a
f
= 2a0.523 mf
= 3a0.523 mf
θ 1 = 28.4°
= 1 0.523 m
θ 2 = 72.0°
θ 3 nonexistent
Maxima appear straight ahead at 0° and left and right at an angle given approximately by
a1.10 mf sinθ
x
a
f
θ x ≈ 46° .
= 1.5 0.523 m
There is no solution to a sin θ = 2.5λ , so our answer is already complete, with three sound
maxima.
P38.6
(a)
sin θ =
y mλ
=
L
a
Therefore, for first minimum, m = 1 and
L=
(b)
e
je
j
7.50 × 10 −4 m 8.50 × 10 −4 m
ay
=
= 1.09 m .
mλ
1 587.5 × 10 −9 m
a fe
j
w = 2 y1 yields y1 = 0.850 mm
e
j
w = 2 0.850 × 10 −3 m = 1.70 mm
411
412
*P38.7
Diffraction Patterns and Polarization
The rectangular patch on the wall is wider than it is tall. The aperture will be taller than it is wide.
For horizontal spreading we have
y width 0.110 m 2
=
= 0.012 2
L
4.5 m
a width sin θ width = 1λ
tan θ width =
a width =
632.8 × 10 −9 m
= 5.18 × 10 −5 m
0.012 2
For vertical spreading, similarly
0.006 m 2
= 0.000 667
4.5 m
1λ
632.8 × 10 −9 m
=
=
= 9.49 × 10 −4 m
0.000 667
sin θ h
tan θ height =
a height
P38.8
mλ
, where m = ±1, ± 2 , ± 3 , …. The
a
requirement for m = 1 is from an analysis of the extra path distance
traveled by ray 1 compared to ray 3 in Figure 38.5. This extra
Equation 38.1 states that sin θ =
λ
for destructive interference. When the
2
source rays approach the slit at an angle β, there is a distance added
a
to the path difference (of ray 1 compared to ray 3) of sin β Then,
2
for destructive interference,
distance must be equal to
FIG. P38.8
λ
λ
a
a
sin β + sin θ = so sin θ = − sin β .
2
2
2
a
Dividing the slit into 4 parts leads to the 2nd order minimum:
Dividing the slit into 6 parts gives the third order minimum:
Generalizing, we obtain the condition for the mth order minimum:
P38.9
sin θ ≈
y 4.10 × 10 −3 m
=
1.20 m
L
e
j F 4.10 × 10 m I = 7.86 rad
GH 1.20 m JK
L sina7.86f OP = 1.62 × 10
=M
N 7.86 Q
−4
β π a sin θ π 4.00 × 10 m
=
=
λ
2
546.1 × 10 −9 m
I
I max
=
LM sinbβ 2g OP
MN β 2 PQ
2
−3
2
−2
2λ
− sin β .
a
3λ
− sin β .
sin θ =
a
mλ
− sin β .
sin θ =
a
sin θ =
Chapter 38
38.10
413
Double-slit interference maxima are at angles given by d sin θ = mλ .
(a)
θ 0 = 0° .
For m = 0 ,
b
g
b
a
g
f
For m = 1 , 2.80 µm sin θ = 1 0.501 5 µm : θ 1 = sin −1 0.179 = 10.3° .
θ 2 = 21.0° , θ 3 = 32.5° , θ 4 = 45.8° ,
Similarly, for m = 2 , 3 , 4, 5 and 6,
a f
θ 5 = 63.6° , and θ 6 = sin −1 1.07 = nonexistent.
Thus, there are 5 + 5 + 1 = 11 directions for interference maxima .
We check for missing orders by looking for single-slit diffraction minima, at a sin θ = mλ .
(b)
For m = 1 ,
b0.700 µmg sinθ = 1b0.501 5 µmg
and
θ 1 = 45.8° .
Thus, there is no bright fringe at this angle. There are only nine bright fringes , at
θ = 0° , ± 10.3° , ± 21.0° , ± 32.5° , and ± 63.6° .
I = I max
(c)
LM sinbπ a sinθ λ g OP
MN π sinθ λ PQ
At θ = 0° ,
At θ = 10.3° ,
2
sin θ
θ
I max
b
I max
P38.11
I
→ 1.00 .
g
π a sin θ π 0.700 µm sin 10.3°
=
= 0.785 rad = 45.0°
0.501 5 µm
λ
I
Section 38.3
→ 1 and
=
LM sin 45.0° OP
N 0.785 Q
2
= 0.811 .
Similarly, at θ = 21.0° ,
π a sin θ
I
= 0.405 .
= 1.57 rad = 90.0° and
λ
I max
At θ = 32.5° ,
π a sin θ
I
= 0.090 1 .
= 2.36 rad = 135° and
I max
λ
At θ = 63.6° ,
π a sin θ
I
= 0.032 4 .
= 3.93 rad = 225° and
I max
λ
Resolution of Single-Slit and Circular Apertures
sin θ =
λ 5.00 × 10 −7 m
=
= 1.00 × 10 −3 rad
a
5.00 × 10 −4
414
P38.12
Diffraction Patterns and Polarization
θ min =
y
λ
= 1.22
L
D
a1.22fe5.00 × 10 jb0.030 0g =
y=
y = radius of star-image
L = length of eye
λ = 500 nm
D = pupil diameter
θ = half angle
−7
7.00 × 10 −3
P38.13
2.61 µm
Undergoing diffraction from a circular opening, the beam spreads into a cone of half-angle
θ min = 1.22
F
GH
I
JK
λ
632.8 × 10 −9 m
= 1.54 × 10 −4 rad .
= 1.22
D
0.005 00 m
The radius of the beam ten kilometers away is, from the definition of radian measure,
e
j
rbeam = θ min 1.00 × 10 4 m = 1.544 m
and its diameter is d beam = 2rbeam = 3.09 m .
*P38.14
When you are at the maximum range, the elves’ eyes will be resolved by Rayleigh’s criterion:
λ
d
= θ min = 1.22
L
D
0.100 m
660 × 10 −9 m
= 1.22
= 1.15 × 10 −4
−3
L
7 × 10 m
0.1 m
= 869 m
L=
1.15 × 10 −4
*P38.15
λ
d
= 1.22 , where θ min is the smallest angular separation of two
L
D
objects for which they are resolved by an aperture of diameter D, d is the separation of the two
objects, and L is the maximum distance of the aperture from the two objects at which they can be
resolved.
Two objects can be resolved if their angular separation is greater than θ min . Thus, θ min should
be as small as possible. Therefore, the smaller of the two given wavelengths is easier to resolve, i.e.
violet .
By Rayleigh’s criterion: θ min =
L=
e
je
j
5.20 × 10 −3 m 2.80 × 10 −2 m 1.193 × 10 −4 m 2
Dd
=
=
λ
1. 22λ
1.22λ
Thus L = 186 m for λ = 640 nm , and L = 271 m for λ = 440 nm . The viewer can resolve adjacent tubes
of violet in the range 186 m to 271 m , but cannot resolve adjacent tubes of red in this range.
P38.16
θ min = 1.22
λ d
=
D L
1.22
F 5.80 × 10
GH 4.00 × 10
−7
−3
I = d FG 1 mi IJ
J
m K 1.80 mi H 1 609 m K
m
d = 0.512 m
The shortening of the wavelength inside the patriot’s eye does not change the answer.
Chapter 38
P38.17
P38.18
*P38.19
415
By Rayleigh’s criterion, two dots separated center-to-center by 2.00 mm would overlap
when
θ min =
Thus,
L=
λ
D = 1.22
=
θ min
λ
d
= 1.22 .
L
D
e
je
j
2.00 × 10 −3 m 4.00 × 10 −3 m
dD
=
= 13.1 m .
1.22λ
1.22 500 × 10 −9 m
e
1.22 5.00 × 10 −7
1.00 × 10
−5
e
j m=
j
6.10 cm
The concave mirror of the spy satellite is probably about 2 m in diameter, and is surely not more
than 5 m in diameter. That is the size of the largest piece of glass successfully cast to a precise shape,
for the mirror of the Hale telescope on Mount Palomar. If the spy satellite had a larger mirror, its
manufacture could not be kept secret, and it would be visible from the ground. Outer space is
probably closer than your state capitol, but the satellite is surely above 200-km altitude, for
reasonably low air friction. We find the distance between barely resolvable objects at a distance of
200 km, seen in yellow light through a 5-m aperture:
y
λ
= 1.22
L
D
b
ga fFGH 6 × 105 m m IJK = 3 cm
−7
y = 200 000 m 1.22
(Considering atmospheric seeing caused by variations in air density and temperature, the distance
b
ga
between barely resolvable objects is more like 200 000 m 1 s
1° I F π rad I
fFGH 3 600
G J = 97 cm .) Thus the
s JK H 180° K
snooping spy satellite cannot see the difference between III and II or IV on a license plate. It cannot
count coins spilled on a sidewalk, much less read the date on them.
P38.20
1.22
λ
D
=
d
L
λ=
D = 2.10 m
d = 1.22
c
= 0.020 0 m
f
L = 9 000 m
b0.020 0 mgb9 000 mg =
2.10 m
λ
a2.00 mf =
a10.0 mf
105 m
P38.21
θ min = 1.22
P38.22
L = 88.6 × 10 9 m , D = 0.300 m , λ = 590 × 10 −9 m
D
= 1.22
λ
= θ min = 2.40 × 10 −6 rad
(a)
1.22
(b)
d = θ min L = 213 km
D
0.244 rad = 14.0°
416
Diffraction Patterns and Polarization
Section 38.4
P38.23
d=
The Diffraction Grating
1.00 cm 1.00 × 10 −2 m
=
= 5.00 µm
2 000
2 000
e
j
−9
mλ 1 640 × 10 m
=
= 0.128
sin θ =
d
5.00 × 10 −6 m
P38.24
θ = 7.35°
The principal maxima are defined by
d sin θ = mλ
m = 0 , 1, 2 , … .
For m = 1 ,
λ = d sin θ
θ
where θ is the angle between the central ( m = 0 ) and the first order
( m = 1 ) maxima. The value of θ can be determined from the information
given about the distance between maxima and the grating-to-screen
distance. From the figure,
tan θ =
1.72 m
0.488 m
= 0.284
1.72 m
so
θ = 15.8°
and
sin θ = 0.273 .
0.488 m
FIG. P38.24
The distance between grating “slits” equals the reciprocal of the number of grating lines per
centimeter
d=
The wavelength is
P38.25
1
5 310 cm −1
= 1.88 × 10 −4 cm = 1.88 × 10 3 nm .
ja
e
f
λ = d sin θ = 1.88 × 10 3 nm 0.273 = 514 nm .
The grating spacing is
d=
1.00 × 10 −2 m
= 2.22 × 10 −6 m .
4 500
In the 1st-order spectrum, diffraction angles are given by
sin θ =
λ
:
d
sin θ 1 =
656 × 10 −9 m
= 0.295
2.22 × 10 −6 m
so that for red
θ 1 = 17.17°
and for violet
sin θ 2 =
so that
θ 2 = 11.26° .
434 × 10 −9 m
2.22 × 10 −6 m
FIG. P38.25
= 0.195
The angular separation is in first-order,
∆θ = 17.17°−11.26° = 5.91° .
In the second-order spectrum,
∆θ = sin −1
Again, in the third order,
∆θ = sin −1
FG 2λ
Hd
FG 3λ
Hd
1
1
IJ − sin
K
IJ − sin
K
−1
−1
FG 2λ
Hd
FG 3λ
Hd
2
2
IJ =
K
IJ =
K
13.2° .
26.5° .
Since the red does not appear in the fourth-order spectrum, the answer is complete.
Chapter 38
P38.26
sin θ = 0.350 :
d=
λ
sin θ
=
417
632.8 nm
= 1.81 × 10 3 nm
0.350
Line spacing = 1.81 µm
P38.27
(a)
(b)
d=
1
= 2.732 × 10 −4 cm = 2.732 × 10 −6 m = 2 732 nm
3 660 lines cm
λ=
d sin θ
:
m
d=
λ
sin θ 1
At θ = 10.09°
λ = 478.7 nm
At θ = 13.71° ,
λ = 647.6 nm
At θ = 14.77° ,
λ = 696.6 nm
and
2 λ = d sin θ 2
sin θ 2 =
so
a
2λ
2λ
=
= 2 sin θ 1 .
d
λ sin θ 1
f
Therefore, if θ 1 = 10.09° then sin θ 2 = 2 sin 10.09° gives θ 2 = 20.51° .
Similarly, for θ 1 = 13.71° , θ 2 = 28.30°
P38.28
sin θ =
and for θ 1 = 14.77° , θ 2 = 30.66° .
mλ
d
Therefore, taking the ends of the visible spectrum to be λ v = 400 nm and λ r = 750 nm , the ends the
different order spectra are:
End of second order:
sin θ 2 r =
2 λ r 1 500 nm
=
.
d
d
Start of third order:
sin θ 3 v =
3 λ v 1 200 nm
=
.
d
d
Thus, it is seen that θ 2 r > θ 3 v and these orders must overlap regardless of the value of the grating
spacing d.
P38.29
(a)
In the 3rd order,
f
b
ga
R = a1fe1.20 × 10 linesj = 1.20 × 10
R = a 2fe1.20 × 10 linesj = 2.40 × 10
R = a3fe1.20 × 10 linesj = 3.60 × 10
From Equation 38.11,
R=
In the 3rd order,
∆λ =
From Equation 38.12, R = Nm where
In the 1st order,
In the 2nd order,
(b)
N = 3 000 lines cm 4.00 cm = 1.20 × 10 4 lines .
4
4
.
4
4
.
4
4
.
λ
:
∆λ
400 nm
λ
=
= 0.011 1 nm = 11.1 pm .
R 3.60 × 10 4
418
*
P38.30
Diffraction Patterns and Polarization
For a side maximum,
tan θ =
y 0.4 µm
=
L 6.9 µm
θ = 3.32°
a1fe780 × 10
d=
d sin θ = mλ
P38.32
λ
af
N1 =
Nm =
(b)
1.32 × 10 −2 m
= 4.72 µm
2 800
∆λ
1 × 10 −3 m
= 74. 2 .
13.5 × 10 -6 m
FIG. P38.30
531.7 nm
= 2 800
0.19 nm
(a)
d=
j = 13.5 µm.
m
sin 3.32°
The number of grooves per millimeter =
P38.31
−9
1
= 2.38 × 10 −6 m = 2 380 nm .
4 200 cm
d sin θ = mλ
Thus,
For m = 1 ,
For m = 2 ,
For m = 3 ,
or
FG mλ IJ and y = L tanθ = L tanLMsin FG mλ IJ OP .
HdK
N H d KQ
R L F mλ IJ O − tanLsin FG mλ IJ OUV .
∆y = L Stan Msin G
T N H d K PQ MN H d K PQW
R| L F 589.6 I O − tanLsin F 589 I OU| = 0.554 mm .
∆y = a 2.00 mfStan Msin G
P M
PV
T| MN H 2 380 JK PQ MN GH 2 380 JK PQW|
R| L F 2a589.6f I O − tanLsin F 2a589f I OU| = 1.54 mm .
∆y = a 2.00 mfStan Msin G
|T MN H 2 380 JK PPQ MMN GH 2 380 JK PPQV|W
R| L F 3a589.6f I O − tanLsin F 3a589f I OU| = 5.04 mm .
∆y = a 2.00 mfStan Msin G
|T MN H 2 380 JK PPQ MMN GH 2 380 JK PPQV|W
θ = sin −1
−1
−1
Thus, the observed order must be m = 2 .
2
−1
−1
1
−1
−1
−1
−1
−1
Chapter 38
P38.33
d=
(a)
1.00 × 10 −3 m mm
= 4.00 × 10 −6 m = 4 000 nm
250 lines mm
d sin θ max
λ
=
b4 000 nmg sin 90.0° = 5.71
700 nm
The highest order in which the violet end of the spectrum can be seen is:
m max =
d sin θ max
λ
=
b4 000 nmg sin 90.0° = 10.0
400 nm
10 orders in the short-wavelength region .
or
(a)
λ
5 orders is the maximum .
or
*P38.34
d sin θ
The number of times a complete order is seen is the same as the number of orders in which
the long wavelength limit is visible.
m max =
(b)
d sin θ = mλ ⇒ m =
419
The several narrow parallel slits make a diffraction
grating. The zeroth- and first- order maxima are
separated according to
af
d sin θ = 1 λ
sin θ =
b
λ 632.8 × 10 −9 m
=
d
1.2 × 10 −3 m
g
θ = sin −1 0.000 527 = 0.000 527 rad
a
fb
g
y = L tan θ = 1.40 m 0.000 527 = 0.738 mm .
(b)
FIG. P38.34
Many equally spaced transparent lines appear on the film. It is itself a diffraction grating.
When the same light is sent through the film, it produces interference maxima separated
according to
af
d sin θ = 1 λ
sin θ =
a
fb
g
λ 632.8 × 10 −9 m
=
= 0.000 857
d 0.738 × 10 −3 m
y = L tan θ = 1.40 m 0.000 857 = 1.20 mm
An image of the original set of slits appears on the screen. If the screen is removed, light
diverges from the real images with the same wave fronts reconstructed as the original slits
produced. Reasoning from the mathematics of Fourier transforms, Gabor showed that light
diverging from any object, not just a set of slits, could be used. In the picture, the slits or
maxima on the left are separated by 1.20 mm. The slits or maxima on the right are separated
by 0.738 mm. The length difference between any pair of lines is an integer number of
wavelengths. Light can be sent through equally well toward the right or toward the left.
Section 38.5
P38.35
Diffraction of X-Rays by Crystals
2d sin θ = mλ :
λ=
e
j
−9
2d sin θ 2 0.353 × 10 m sin 7.60°
=
= 9.34 × 10 −11 m = 0.093 4 nm
m
1
420
Diffraction Patterns and Polarization
P38.36
2d sin θ = mλ ⇒ d =
P38.37
2d sin θ = mλ :
sin θ =
f
a
f
e
e
j
j
−9
mλ 1 0.140 × 10 m
=
= 0.249
2d 2 0.281 × 10 −9 m
θ = 14.4°
and
P38.38
a fa
1 0.129 nm
mλ
=
= 0.455 nm
2 sin θ
2 sin 8.15°
mλ
:
2d
2λ
sin θ 2 =
= 2 0.218 so
2d
Three other orders appear:
sin θ m =
a
sin 12.6° =
f
θ 2 = 25.9°
θ 5 = sin −1
Figure 38.27 of the text shows the situation.
2d sin θ
or
λ=
2d sin θ = mλ
m
2 2.80 m sin 80.0°
m=1:
= 5.51 m
λ1 =
1
2 2.80 m sin 80.0°
m=2:
= 2.76 m
λ2 =
2
2 2.80 m sin 80.0°
= 1.84 m
λ3 =
m=3 :
3
a
a
a
Section 38.6
P38.40
f
f
f
Polarization of Light Waves
The average value of the cosine-squared function is one-half, so the first polarizer transmits
light. The second transmits cos 2 30.0° =
If =
P38.41
a
f
a4 × 0.218f = 60.8°
a5 × 0.218f = nonexistent
θ 3 = sin −1 3 × 0.218 = 40.9°
θ 4 = sin −1
P38.39
1λ
= 0.218
2d
3
.
4
1 3
3
× Ii =
Ii
2 4
8
I = I max cos 2 θ
⇒
θ = cos −1
I
I max
(a)
1
I
=
I max 3.00
⇒
θ = cos −1
1
= 54.7°
3.00
(b)
1
I
=
I max 5.00
⇒
θ = cos −1
1
= 63.4°
5.00
1
10.0
⇒
θ = cos −1
1
= 71.6°
10.0
(c)
I
I max
=
1
the
2
Chapter 38
P38.42
(a)
θ 1 = 20.0° , θ 2 = 40.0° , θ 3 = 60.0°
b
g
b
g b
g
= a10.0 unitsf cos a 20.0°f cos a 20.0°f cos a 20.0°f
I f = I i cos 2 θ 1 − 0° cos 2 θ 2 − θ 1 cos 2 θ 3 − θ 2
If
2
2
2
= 6.89 units
(b)
θ 1 = 0° , θ 2 = 30.0° , θ 3 = 60.0°
a
f
2
a f
2
a
f
2
a
FIG. P38.42
f
I f = 10.0 units cos 0° cos 30.0° cos 30.0° = 5.63 units
P38.43
*P38.44
a
(a)
f
n = tan θ p = tan 48.0° = 1.11 .
By Brewster’s law,
At incidence, n1 sin θ 1 = n 2 sin θ 2 and θ 1′ = θ 1 . For complete
polarization of the reflected light,
θ 1 θ 1'
b90 − θ ′ g + b90 − θ g = 90°
1
2
θ 1′ + θ 2 = 90 = θ 1 + θ 2
b
n1
n2
g
Then n1 sin θ 1 = n 2 sin 90 − θ 1 = n 2 cos θ 1
θ 3 θ 3'
θ2
sin θ 1 n 2
=
= tan θ 1
cos θ 1 n1
n1
At the bottom surface, θ 3 = θ 2 because the normals to the surfaces
of entry and exit are parallel.
Then n 2 sin θ 3 = n1 sin θ 4
n 2 sin θ 2 = n1 sin θ 4
FIG. P38.44(a)
θ ′3 = θ 3
θ 4 = θ1
and
and
θ4
The condition for complete polarization of the reflected light is
90 − θ ′3 + 90 − θ 4 = 90°
θ 2 + θ 1 = 90
This is the same as the condition for θ 1 to be Brewster’s angle at the top surface.
(b)
We consider light moving in a plane perpendicular to
the line where the surfaces of the prism meet at the
unknown angle Φ. We require
n1 sin θ 1 = n 2 sin θ 2
θ 1 + θ 2 = 90°
b
Φ
n1
n2
n3
g
So n1 sin 90 − θ 2 = n 2 sin θ 2
And n 2 sin θ 3 = n 3 sin θ 4
n 2 sin θ 3 = n 3 cos θ 3
θ1 θ1
θ2
n1
= tan θ 2
n2
θ 3 + θ 4 = 90°
n
tan θ 3 = 3
n2
θ3
θ4
FIG. P38.44(b)
In the triangle made by the faces of the prism and the ray in the prism,
b
g
Φ + 90 + θ 2 + 90 − θ 3 = 180 .
So one particular apex angle is required, and it is
Φ = θ 3 − θ 2 = tan −1
FG n IJ − tan FG n IJ
Hn K
Hn K
3
2
−1
1
.
2
Here a negative result is to be interpreted as meaning the same as a positive result.
θ3
421
422
P38.45
Diffraction Patterns and Polarization
sin θ c =
1
n
Also, tan θ p = n .
P38.46
sin θ c =
n=
Thus,
θ p = tan −1 n = tan −1 1.77 = 60.5° .
af
a f
1
and tan θ p = n
n
Thus, sin θ c =
P38.47
1
1
=
= 1.77 .
sin θ c sin 34.4°
or
1
or cot θ p = sin θ c .
tan θ p
tan θ p = 1.33
Complete polarization occurs at Brewster’s angle
θ p = 53.1° .
Thus, the Moon is 36.9° above the horizon.
Additional Problems
P38.48
For incident unpolarized light of intensity I max :
After transmitting 1st disk:
I=
1
I max .
2
After transmitting 2nd disk:
I=
1
I max cos 2 θ .
2
After transmitting 3rd disk:
I=
1
I max cos 2 θ cos 2 90°−θ .
2
a
f
FIG. P38.48
where the angle between the first and second disk is θ = ω t .
Using trigonometric identities cos 2 θ =
a
1
1 + cos 2θ
2
a
f
1
f
a1 − cos 2θ f
2
LM a1 + cos 2θ f OPLM a1 − cos 2θ f OP
N 2 QN 2 Q
e1 − cos 2θ j = 18 I FGH 12 IJK a1 − cos 4θ f .
and
cos 2 90°−θ = sin 2 θ =
we have
I=
1
I max
2
I=
1
I max
8
2
max
Since θ = ω t , the intensity of the emerging beam is given by I =
b
1
I max 1 − 4ω t
16
g
.
Chapter 38
P38.49
P38.50
423
Let the first sheet have its axis at angle θ to the original plane of polarization, and let each further
sheet have its axis turned by the same angle.
The first sheet passes intensity
I max cos 2 θ .
The second sheet passes
I max cos 4 θ
and the nth sheet lets through
I max cos 2n θ ≥ 0.90 I max
Try different integers to find
cos 2 × 5
(a)
So n = 6
(b)
θ = 7.50°
FG 45° IJ = 0.885
H5K
where θ =
cos 2 × 6
45°
.
n
FG 45° IJ = 0.902 .
H6K
Consider vocal sound moving at 340 m/s and of frequency 3 000 Hz. Its wavelength is
λ=
v 340 m s
=
= 0.113 m .
f 3 000 Hz
If your mouth, for horizontal dispersion, behaves similarly to a slit 6.00 cm wide, then a sin θ = mλ
predicts no diffraction minima. You are a nearly isotropic source of this sound. It spreads out from
you nearly equally in all directions. On the other hand, if you use a megaphone with width 60.0 cm
at its wide end, then a sin θ = mλ predicts the first diffraction minimum at
θ = sin −1
FG mλ IJ = sin FG 0.113 m IJ = 10.9° .
H 0.600 m K
HaK
−1
This suggests that the sound is radiated mostly toward the front into a diverging beam of angular
diameter only about 20°. With less sound energy wasted in other directions, more is available for
your intended auditors. We could check that a distant observer to the side or behind you receives
less sound when a megaphone is used.
P38.51
P38.52
af
The first minimum is at
a sin θ = 1 λ .
This has no solution if
λ
>1.
a
or if
a < λ = 632.8 nm .
x = 1.22
F
GH
Ie
JK
λ
5.00 × 10 −7 m
D = 1.22
250 × 10 3 m = 30.5 m
d
5.00 × 10 −3 m
j
D = 250 × 10 3 m
λ = 5.00 × 10 −7 m
d = 5.00 × 10 −3 m
424
P38.53
Diffraction Patterns and Polarization
d=
1
400 mm
= 2.50 × 10 −6 m
−1
(a)
d sin θ = mλ
(b)
λ=
(c)
d sin θ a = 2 λ
θ a = sin −1
541 × 10 −9 m
= 4.07 × 10 −7 m
1.33
θ b = sin −1
d sin θ b =
F 2 × 541 × 10 m I = 25.6°
GH 2.50 × 10 m JK
F 2 × 4.07 × 10 m I = 19.0°
GH 2.50 × 10 m JK
−9
−6
−7
−6
2λ
n
af
n sin θ b = 1 sin θ a
P38.54
(a)
λ=
v
:
f
λ=
θ min = 1.22
3.00 × 10 8 m s
1.40 × 10 9 s −1
= 0.214 m
FG
H
IJ
K
λ
0.214 m
= 7.26 µrad
: θ min = 1.22
D
3.60 × 10 4 m
FG 180 × 60 × 60 s IJ =
H π K
θ min = 7.26 µrad
P38.55
g
e
jb
F 500 × 10 m I = 50.8 µrad a10.5 seconds of arcf
λ
θ
θ
= 1.22
= 1.22G
D
H 12.0 × 10 m JK
d = θ L = e50.8 × 10 radja30.0 mf = 1.52 × 10 m = 1.52 mm
(b)
θ min =
(c)
min
(d)
1.50 arc seconds
d
:
L
d = θ min L = 7.26 × 10 −6 rad 26 000 ly = 0.189 ly
−9
min
−3
−6
min
−3
With a grazing angle of 36.0°, the angle of incidence is 54.0°
tan θ p = n = tan 54.0° = 1.38 .
In the liquid, λ n =
*P38.56
(a)
λ
n
=
750 nm
= 545 nm .
1.38
Bragg’s law applies to the space lattice of melanin rods. Consider the planes d = 0.25 µm
apart. For light at near-normal incidence, strong reflection happens for the wavelength
given by 2d sin θ = mλ . The longest wavelength reflected strongly corresponds to m = 1:
e
j
2 0.25 × 10 −6 m sin 90° = 1λ
λ = 500 nm . This is the blue-green color.
(b)
For light incident at grazing angle 60°, 2d sin θ = mλ gives
1λ = 2 0.25 × 10 −6 m sin 60° = 433 nm . This is violet.
(c)
Your two eyes receive light reflected from the feather at different angles, so they receive
light incident at different angles and containing different colors reinforced by constructive
interference.
e
continued on next page
j
Chapter 38
P38.57
425
(d)
The longest wavelength that can be reflected with extra strength by these melanin rods is
the one we computed first, 500 nm blue-green.
(e)
If the melanin rods were farther apart (say 0.32 µm ) they could reflect red with constructive
interference.
(a)
d sin θ = mλ
or d =
e
j
3 500 × 10 −9 m
mλ
=
= 2.83 µm
sin θ
sin 32.0°
Therefore, lines per unit length =
1
1
=
d 2.83 × 10 −6 m
or lines per unit length = 3.53 × 10 5 m −1 = 3.53 × 10 3 cm −1 .
sin θ =
(b)
P38.58
e
j a
−9
mλ m 500 × 10 m
=
= m 0.177
d
2.83 × 10 −6 m
f
a
f
For sin θ ≤ 1.00 , we must have
m 0.177 ≤ 1.00
or
m ≤ 5.66 .
Therefore, the highest order observed is
m=5 .
Total number of primary maxima observed is
2m + 1 = 11 .
For the air-to-water interface,
n water 1.33
=
1.00
n air
tan θ p =
θp
θ p = 53.1°
θ
a1.00f sinθ = a1.33f sinθ
F sin 53.1° IJ = 36.9° .
θ = sin G
H 1.33 K
and
p
2
θ2 θ3
2
−1
For the water-to-glass interface, tan θ p = tan θ 3 =
n glass
n water
=
1.50
so
1.33
FIG. P38.58
θ 3 = 48.4° .
The angle between surfaces is θ = θ 3 − θ 2 = 11.5° .
*P38.59
A central maximum and side maxima in seven orders of
interference appear. If the seventh order is just at 90°,
d sin θ = mλ
e
j
d1 = 7 654 × 10 −9 m
d = 4.58 µm .
If the seventh order is at less than 90°, the eighth order
might be nearly ready to appear according to
e
j
d1 = 8 654 × 10 −9 m
d = 5.23 µm .
Thus 4.58 µm < d < 5.23 µm .
FIG. P38.59
Air
Water
426
P38.60
Diffraction Patterns and Polarization
(a)
We require
D
=
radius of diffraction disk D
=
.
L
2L
D 2 = 2.44λL .
Then
e
ja
f
D = 2.44 500 × 10 −9 m 0.150 m = 428 µm
(b)
P38.61
λ
θ min = 1.22
The limiting resolution between lines θ min
e
e
j
j
550 × 10 −9 m
λ
= 1.22 = 1.22
= 1.34 × 10 −4 rad .
D
5.00 × 10 −3 m
Assuming a picture screen with vertical dimension , the minimum viewing distance for no visible
485
. The desired ratio is then
lines is found from θ min =
L
L
P38.62
(a)
=
1
1
=
= 15.4 .
485θ min 485 1.34 × 10 −4 rad
e
j
Applying Snell’s law gives n 2 sin φ = n1 sin θ . From the sketch, we also
see that:
b
g
θ + φ + β = π , or φ = π − θ + β .
which reduces to:
b g
sin φ = sinbθ + β g .
Applying the identity again:
sin φ = sin θ cos β + cos θ sin β .
Snell’s law then becomes:
n 2 sin θ cos β + cos θ sin β = n1 sin θ
or (after dividing by cos θ ):
Solving for tan θ gives:
(b)
(a)
b
g
n btan θ cos β + sin β g = n tan θ .
2
1
tan θ =
n 2 sin β
.
n1 − n 2 cos β
If β = 90.0° , n1 = 1.00 , and n 2 = n , the above result becomes:
tan θ =
P38.63
b
a f
n 1.00
, or n = tan θ , which is Brewster’s law.
1.00 − 0
From Equation 38.1,
θ = sin −1
In this case m = 1 and
λ=
Thus,
θ = sin −1
continued on next page
g
sin φ = sin π cos θ + β − cos π sin θ + β ,
Using the given identity:
FG mλ IJ .
HaK
8
c 3.00 × 10 m s
=
= 4.00 × 10 −2 m .
9
f
7.50 × 10 Hz
F 4.00 × 10
GH 6.00 × 10
−2
−2
I=
J
mK
m
41.8° .
FIG. P38.62(a)
Chapter 38
(b)
427
L sinbβ 2g OP where β = 2π a sinθ .
=M
λ
I
MN β 2 PQ
2π b0.060 0 mg sin 15.0°
β=
= 2.44 rad
2
I
From Equation 38.4,
max
When θ = 15.0° ,
0.040 0 m
I
and
(c)
sin θ =
I max
L sina1.22 radf OP
=M
N 1.22 rad Q
2
= 0.593 .
λ
so θ = 41.8° :
a
This is the minimum angle subtended by the two sources at the slit.
Let α be the half angle between the sources, each a distance
= 0.100 m from the center line and a distance L from the slit
plane. Then,
41.8°
= 0.262 m .
L = cot α = 0.100 m cot
2
a
f FGH
IJ
K
P38.64
I
1
1
= cos 2 45.0° cos 2 45.0° =
I max 2
8
P38.65
d sin θ = mλ
e
je
j
a
d cos θ dθ = mdλ
or
d 1 − sin 2 θ ∆θ ≈ m∆λ
d 1−
*P38.66
f
and, differentiating,
∆θ ≈
so
FIG. P38.63(c)
m 2 λ2
∆θ ≈ m∆λ
d2
∆λ
e
d
2
.
j
m 2 − λ2
af
(a)
The angles of bright beams diffracted from the grating are given by d sin θ = mλ . The
dθ
dθ
m
dθ
=m
=
angular dispersion is defined as the derivative
:
d cos θ
dλ
dλ d cos θ
dλ
(b)
For the average wavelength 578 nm,
af
d sin θ = mλ
θ = sin −1
0.02 m
sin θ = 2 578 × 10 −9 m
8 000
e
j
2 × 578 × 10 −9 m
= 27.5°
2.5 × 10 −6 m
The separation angle between the lines is
dθ
m
2
∆λ =
∆λ =
2.11 × 10 −9 m
−6
dλ
d cos θ
2.5 × 10 m cos 27.5°
180°
= 0.109°
= 0.001 90 = 0.001 90 rad = 0.001 90 rad
π rad
∆θ =
FG
H
IJ
K
428
*P38.67
Diffraction Patterns and Polarization
(a)
Constructive interference of light of wavelength λ on the screen is described by d sin θ = mλ
−1 2
y
y
= mλ . Differentiating with
where tan θ = so sin θ =
. Then d y L2 + y 2
2
2
L
L +y
afe
j
respect to y gives
a f FGH 12 IJK eL + y j b0 + 2yg = m ddyλ
adfy = m dλ = adfL + adfy − adfy
d
−
dy
eL + y j e L + y j
eL + y j
adfL
dλ
=
dy m L + y
e
j
e
d1 L2 + y 2
j
−1 2
2
+ d y −
2 −3 2
2
2
2 12
2
2
2 3 2
2
2
2 32
2
2
2 3 2
2
(b)
Here d sin θ = mλ gives
I
JK
F
GH
10 −2 m
0.55 × 10 −6 m
sin θ = 1 550 × 10 −9 m , θ = sin −1
= 26.1°
8 000
1.25 × 10 −6 m
e
j
y = L tan θ = 2. 40 m tan 26.1° = 1.18 m
Now
P38.68
dλ
dL2
=
dy m L2 + y 2
e
j
32
=
a
f
1ea 2.4 mf + a1.18 mf j
1.25 × 10 −6 m 2.40 m
2
2 32
2
= 3.77 × 10 −7 = 3.77 nm cm .
For a diffraction grating, the locations of the principal maxima for wavelength λ are given by
a
mλ y
where a is the width of the grating
≈ . The grating spacing may be expressed as d =
sin θ =
N
d
L
NLmλ
and N is the number of slits. Thus, the screen locations of the maxima become y =
. If two
a
nearly equal wavelengths are present, the difference in the screen locations of corresponding
maxima is
∆y =
b g.
NLm ∆λ
a
For a single slit of width a, the location of the first diffraction minimum is sin θ =
y=
λ y
≈ , or
a L
FG L IJ λ . If the two wavelengths are to be just resolved by Rayleigh’s criterion, y = ∆y from above.
H aK
Therefore,
FG L IJ λ = NLmb∆λ g
H aK
a
or the resolving power of the grating is
R≡
λ
= Nm .
∆λ
Chapter 38
P38.69
(a)
429
The E and O rays, in phase at the surface of the plate, will have a phase difference
θ=
FG 2π IJδ
HλK
after traveling distance d through the plate. Here δ is the difference in the optical path lengths
of these rays. The optical path length between two points is the product of the actual path
length d and the index of refraction. Therefore,
δ = dnO − dn E .
The absolute value is used since
θ=
(b)
P38.70
(a)
(b)
FG 2π IJ dn
HλK
O
− dn E =
FG 2π IJ d n
HλK
O
nO
may be more or less than unity. Therefore,
nE
− nE .
jb g
e
550 × 10 −9 m π 2
λθ
d=
=
= 1.53 × 10 −5 m = 15.3 µm
2π nO − n E
2π 1.544 − 1.553
I
From Equation 38.4,
I max
If we define
φ≡
β
2
this becomes
Therefore, when
1
I
=
I max 2
we must have
Let y1 = sin φ and y 2 =
I
I max
sin φ
φ
the right.
2π a sin θ
λ
gives
If
.
2
=
1
2
, or sin φ =
φ
.
2
π
is shown to
2
The solution to the transcendental equation is found to be
φ = 1.39 rad .
β=
2
φ
.
2
A plot of y1 and y 2 in the range 1.00 ≤ φ ≤
(c)
L sinbβ 2g OP
=M
MN β 2 PQ
L sin φ OP .
=M
Nφ Q
FIG. P38.70(b)
= 2φ
sin θ =
FG φ IJ λ = 0.443 λ .
Hπ K a
a
λ
λ
is small, then θ ≈ 0. 443 .
a
a
This gives the half-width, measured away from the maximum at θ = 0 . The pattern is
symmetric, so the full width is given by
∆θ = 0.443
FG
H
IJ
K
λ
λ
0.886 λ
− −0.443
=
.
a
a
a
430
P38.71
Diffraction Patterns and Polarization
φ
1
2
1.5
1.4
1.39
1.395
1.392
1.391 5
1.391 52
1.391 6
1.391 58
1.391 57
1.391 56
1.391 559
1.391 558
1.391 557
1.391 557 4
2 sin φ
1.19
1.29
1.41
1.394
1.391
1.392
1.391 7
1.391 54
1.391 55
1.391 568
1.391 563
1.391 561
1.391 558
1.391 557 8
1.391 557 5
1.391 557 3
1.391 557 4
bigger than φ
smaller than φ
smaller
bigger
smaller
bigger
bigger
smaller
We get the answer to seven digits after 17 steps. Clever guessing, like using the value of
the next guess for φ, could reduce this to around 13 steps.
P38.72
2 sin φ as
b g I LM bβ 2g cosbβ 2gb1 2g − sinbβ 2gb1 2g OP
JK M
PQ
bβ 2g
N
FβI
and require that it be zero. The possibility sinG J = 0 locates all of the minima and the central
H 2K
In I = I max
LM sinbβ 2g OP
MN β 2 PQ
2
find
F
GH
2 sin β 2
dI
= I max
β 2
dβ
2
maximum, according to
2π a sin θ
β
= 0 , π , 2π , … ;
2
β=
The side maxima are found from
β
β
β
β
β
− sin
= 0 , or tan
= .
cos
2
2
2
2
2
This has solutions
λ
FG IJ
H K
= 0 , 2π , 4π , … ; a sin θ = 0 , λ , 2 λ , … .
FG IJ
H K
FG IJ
H K
β
β
= 4.493 4 ,
= 7.725 3 , and others, giving
2
2
(a)
π a sin θ = 4.493 4λ
a sin θ = 1.430 3 λ
(b)
π a sin θ = 7.725 3 λ
a sin θ = 2.459 0λ
431
Chapter 38
P38.73
The first minimum in the single-slit diffraction pattern
occurs at
sin θ =
λ y min
≈
.
a
L
Thus, the slit width is given by
a=
λL
.
y min
For a minimum located at y min = 6.36 mm ± 0.08 mm,
the width is
a=
e632.8 × 10
−9
ja
f=
m 1.00 m
6.36 × 10
−3
m
FIG. P38.73
99.5 µm ± 1% .
ANSWERS TO EVEN PROBLEMS
P38.2
547 nm
P38.34
(a) 0.738 mm; (b) see the solution
P38.4
91.2 cm
P38.36
0.455 nm
P38.6
(a) 1.09 m ; (b) 1.70 mm
P38.38
3
P38.8
see the solution
P38.40
P38.10
(a) 0° , 10.3°, 21.0° , 32.5° , 45.8°, 63.6° ;
(b) nine bright fringes at 0° and on either
side at 10.3°, 21.0° , 32.5° , and 63.6° ;
(c) 1.00 , 0.811 , 0.405 , 0.090 1 , 0.032 4
3
8
P38.42
(a) 6.89 units ; (b) 5.63 units
P38.44
(a) see the solution; (b) For light confined
to a plane, yes. tan −1
FG n IJ − tan FG n IJ
Hn K
Hn K
3
−1
P38.12
2.61 µm
P38.14
869 m
P38.46
see the solution
P38.16
0.512 m
P38.48
see the solution
P38.18
6.10 cm
P38.50
see the solution
P38.20
105 m
P38.52
30.5 m
P38.22
(a) 2.40 µ rad ; (b) 213 km
P38.54
P38.24
514 nm
(a) 1.50 sec; (b) 0.189 ly; (c) 10.5 sec;
(d) 1.52 mm
P38.26
1.81 µm
P38.56
see the solution
P38.28
see the solution
P38.58
11.5°
P38.30
74.2 grooves/mm
P38.60
(a) see the solution; (b) 428 µm
P38.32
2
P38.62
see the solution
2
1
2
432
Diffraction Patterns and Polarization
P38.64
1
8
P38.70
(a) see the solution; (b) φ = 1.39 rad;
(c) see the solution
P38.66
(a) see the solution; (b) 0.109°
P38.72
(a) a sin θ = 1.430 3 λ ; (b) a sin θ = 2.459 0λ
P38.68
see the solution
39
Relativity
CHAPTER OUTLINE
The Principle of Galilean
Relativity
39.2 The Michelson-Morley
Experiment
39.3 Einstein’s Principle of
Relativity
39.4 Consequences of the
Special Theory of Relativity
39.5 The Lorentz Transformation
Equations
39.6 The Lorentz Velocity
Transformation Equations
39.7 Relativistic Linear Momentum
and the Relativistic Form of
Newton’s Laws
39.8 Relativistic Energy
39.9 Mass and Energy
39.10 The General Theory of
Relativity
ANSWERS TO QUESTIONS
39.1
Q39.1
The speed of light c and the speed v of their relative motion.
Q39.2
An ellipsoid. The dimension in the direction of motion would
be measured to be scrunched in.
Q39.3
No. The principle of relativity implies that nothing can travel
faster than the speed of light in a vacuum, which is 300 Mm/s.
The electron would emit light in a conical shock wave of
Cerenkov radiation.
Q39.4
The clock in orbit runs slower. No, they are not synchronized.
Although they both tick at the same rate after return, a time
difference has developed between the two clocks.
Q39.5
Suppose a railroad train is moving past you. One way to measure its length is this: You mark the
tracks at the cowcatcher forming the front of the moving engine at 9:00:00 AM, while your assistant
marks the tracks at the back of the caboose at the same time. Then you find the distance between the
marks on the tracks with a tape measure. You and your assistant must make the marks
simultaneously in your frame of reference, for otherwise the motion of the train would make its
length different from the distance between marks.
Q39.6
(a)
Yours does.
(b)
His does.
(c)
If the velocity of relative motion is constant, both observers have equally valid views.
Q39.7
Get a Mr. Tompkins book by George Gamow for a wonderful fictional exploration of this question.
Driving home in a hurry, you push on the gas pedal not to increase your speed by very much, but
rather to make the blocks get shorter. Big Doppler shifts in wave frequencies make red lights look
green as you approach them and make car horns and car radios useless. High-speed transportation
is very expensive, requiring huge fuel purchases. And it is dangerous, as a speeding car can knock
down a building. Having had breakfast at home, you return hungry for lunch, but you find you
have missed dinner. There is a five-day delay in transmission when you watch the Olympics in
Australia on live television. It takes ninety-five years for sunlight to reach Earth. We cannot see the
Milky Way; the fireball of the Big Bang surrounds us at the distance of Rigel or Deneb.
Q39.8
Nothing physically unusual. An observer riding on the clock does not think that you are really
strange, either.
433
434
Relativity
Q39.9
By a curved line. This can be seen in the middle of Speedo’s world-line in Figure 39.12, where he
turns around and begins his trip home.
Q39.10
According to p = γmu , doubling the speed u will make the momentum of an object increase by the
factor 2
LM c − u OP
N c − 4u Q
2
2
2
2
12
.
Q39.11
As the object approaches the speed of light, its kinetic energy grows without limit. It would take an
infinite investment of work to accelerate the object to the speed of light.
Q39.12
There is no upper limit on the momentum of an electron. As more energy E is fed into the object
E
without limit, its speed approaches the speed of light and its momentum approaches .
c
Q39.13
Recall that when a spring of force constant k is compressed or stretched from its relaxed position a
1
distance x, it stores elastic potential energy U = kx 2 . According to the special theory of relativity,
2
any change in the total energy of the system is equivalent to a change in the mass of the system.
Therefore, the mass of a compressed or stretched spring is greater than the mass of a relaxed spring
U
by an amount 2 . The fractional change is typically unobservably small for a mechanical spring.
c
Q39.14
You see no change in your reflection at any speed you can attain. You cannot attain the speed of
light, for that would take an infinite amount of energy.
Q39.15
Quasar light moves at three hundred million meters per second, just like the light from a firefly at rest.
Q39.16
A photon transports energy. The relativistic equivalence of mass and energy means that is enough to
give it momentum.
Q39.17
Any physical theory must agree with experimental measurements within some domain. Newtonian
mechanics agrees with experiment for objects moving slowly compared to the speed of light.
Relativistic mechanics agrees with experiment for objects at all speeds. Thus the two theories must
and do agree with each other for ordinary nonrelativistic objects. Both statements given in the
question are formally correct, but the first is clumsily phrased. It seems to suggest that relativistic
mechanics applies only to fast-moving objects.
Q39.18
The point of intersection moves to the right. To state the problem precisely, let us assume that each
of the two cards moves toward the other parallel to the long dimension of the picture, with velocity
2v
= 2 v cot φ , where φ is the
of magnitude v. The point of intersection moves to the right at speed
tan φ
small angle between the cards. As φ approaches zero, cot φ approaches infinity. Thus the point of
intersection can move with a speed faster than c if v is sufficiently large and φ sufficiently small. For
example, take v = 500 m s and φ = 0.000 19° . If you are worried about holding the cards steady
enough to be sure of the angle, cut the edge of one card along a curve so that the angle will
necessarily be sufficiently small at some place along the edge.
Let us assume the spinning flashlight is at the center of a grain elevator, forming a circular
screen of radius R. The linear speed of the spot on the screen is given by v = ω R , where ω is the
angular speed of rotation of the flashlight. With sufficiently large ω and R, the speed of the spot
moving on the screen can exceed c.
continued on next page
Chapter 39
435
Neither of these examples violates the principle of relativity. Both cases are describing a point
of intersection: in the first case, the intersection of two cards and in the second case, the intersection
of a light beam with a screen. A point of intersection is not made of matter so it has no mass, and
hence no energy. A bug momentarily at the intersection point could yelp, take a bite out of one card,
or reflect the light. None of these actions would result in communication reaching another bug so
soon as the intersection point reaches him. The second bug would have to wait for sound or light to
travel across the distance between the first bug and himself, to get the message.
As a child, the author used an Erector set to build a superluminal speed generator using the
intersecting-cards method. Can you get a visible dot to run across a computer screen faster than
light? Want’a see it again?
Q39.19
In this case, both the relativistic and Galilean treatments would yield the same result: it is that the
experimentally observed speed of one car with respect to the other is the sum of the speeds of the cars.
Q39.20
The hotter object has more energy per molecule than the cooler one. The equivalence of energy and
mass predicts that each molecule of the hotter object will, on average, have a larger mass than those
in the cooler object. This implies that given the same net applied force, the cooler object would have
a larger acceleration than the hotter object would experience. In a controlled experiment, the
difference will likely be too small to notice.
Q39.21
Special relativity describes inertial reference frames: that is, reference frames that are not
accelerating. General relativity describes all reference frames.
Q39.22
The downstairs clock runs more slowly because it is closer to the Earth and hence in a stronger
gravitational field than the upstairs clock.
Q39.23
The ants notice that they have a stronger sense of being pushed outward when they venture closer
to the rim of the merry-go-round. If they wish, they can call this the effect of a stronger gravitational
field produced by some mass concentration toward the edge of the disk. An ant named Albert
figures out that the strong gravitational field makes measuring rods contract when they are near the
rim of the disk. He shows that this effect precisely accounts for the discrepancy.
SOLUTIONS TO PROBLEMS
Section 39.1
P39.1
The Principle of Galilean Relativity
In the rest frame,
pi = m1 v1i + m 2 v 2i = 2 000 kg 20.0 m s + 1 500 kg 0 m s = 4.00 × 10 4 kg ⋅ m s
b
g
b
b
gb
g
g b
gb
g
p f = m1 + m 2 v f = 2 000 kg + 1 500 kg v f
Since pi = p f ,
vf =
4.00 × 10 4 kg ⋅ m s
= 11.429 m s .
2 000 kg + 1 500 kg
In the moving frame, these velocities are all reduced by +10.0 m/s.
b
g
v1′ i = v1i − v ′ = 20.0 m s − +10.0 m s = 10.0 m s
b
g
v ′2i = v 2i − v ′ = 0 m s − +10.0 m s = −10.0 m s
b
g
v ′f = 11.429 m s − +10.0 m s = 1.429 m s
Our initial momentum is then
b
gb
g b
b
gb
gb
g
pi′ = m1 v1′ i + m 2 v ′2i = 2 000 kg 10.0 m s + 1 500 kg −10.0 m s = 5 000 kg ⋅ m s
and our final momentum is
b
g
g
p ′f = 2 000 kg + 1 500 kg v ′f = 3 500 kg 1.429 m s = 5 000 kg ⋅ m s .
436
P39.2
P39.3
Relativity
(a)
v = vT + v B = 60.0 m s
(b)
v = vT − v B = 20.0 m s
(c)
v = vT2 + v B2 = 20 2 + 40 2 = 44.7 m s
The first observer watches some object accelerate under applied forces. Call the instantaneous
velocity of the object v 1 . The second observer has constant velocity v 21 relative to the first, and
measures the object to have velocity v 2 = v 1 − v 21 .
dv 2 dv 1
=
.
dt
dt
This is the same as that measured by the first observer. In this nonrelativistic case, they measure the
same forces as well. Thus, the second observer also confirms that ∑ F = ma .
a2 =
The second observer measures an acceleration of
P39.4
The laboratory observer notes Newton’s second law to hold:
F1 = ma 1
(where the subscript 1 implies the measurement was made in the laboratory frame of reference). The
observer in the accelerating frame measures the acceleration of the mass as a 2 = a 1 − a ′
(where the subscript 2 implies the measurement was made in the accelerating frame of reference,
and the primed acceleration term is the acceleration of the accelerated frame with respect to the
laboratory frame of reference). If Newton’s second law held for the accelerating frame, that observer
would then find valid the relation
F2 = ma 2
or
F1 = ma 2
(since F1 = F2 and the mass is unchanged in each). But, instead, the accelerating frame observer will
find that F2 = ma 2 − ma ′ which is not Newton’s second law.
Section 39.2
The Michelson-Morley Experiment
Section 39.3
Einstein’s Principle of Relativity
Section 39.4
Consequences of the Special Theory of Relativity
P39.5
L = Lp
Taking L =
P39.6
∆t =
v=c
p
Lp
2
where L p = 1.00 m gives
∆t p
b g
1− v c
2 12
so
v=c
∆t = 2 ∆t p
p
p
L F ∆t I OP
v = c M1 − G
MN H ∆t JK PQ
L F ∆t I OP
v = c M1 − G
MN H 2 ∆t JK PQ
2
= c 1−
1
= 0.866 c .
4
2 12
p
.
2 12
For
FLI
1−G J
HL K
F L 2I
1−G
H L JK
2
v2
1− 2
c
p
p
LM
N
= c 1−
1
4
OP
Q
12
= 0.866 c .
Chapter 39
P39.7
(a)
1
γ =
b g
1− v c
2
=
a
1
1 − 0.500
f
2
=
437
2
3
The time interval between pulses as measured by the Earth observer is
2 60.0 s
∆t = γ∆t p =
= 0.924 s .
3 75.0
60.0 s min
= 64.9 min .
Thus, the Earth observer records a pulse rate of
0.924 s
FG
H
(b)
IJ
K
At a relative speed v = 0.990 c , the relativistic factor γ increases to 7.09 and the pulse rate
recorded by the Earth observer decreases to 10.6 min . That is, the life span of the
astronaut (reckoned by the duration of the total number of his heartbeats) is much longer as
measured by an Earth clock than by a clock aboard the space vehicle.
*P39.8
(a)
The 0.8 c and the 20 ly are measured in the Earth frame,
x 20 ly 20 ly 1c
∆t = =
=
= 25.0 yr .
so in this frame,
v 0.8 c
0.8 c 1 ly yr
(b)
We see a clock on the meteoroid moving, so we do not measure proper time; that clock
measures proper time.
25.0 yr
∆t
∆t p =
=
= 25.0 yr 1 − 0.8 2 = 25.0 yr 0.6 = 15.0 yr
∆t = γ∆t p :
γ 1 1 - v2 c2
a f
(c)
Method one: We measure the 20 ly on a stick stationary in our frame, so it is proper length.
The tourist measures it to be contracted to
Lp
20 ly
20 ly
=
=
= 12.0 ly .
L=
γ
1 1 − 0.8 2 1.667
Method two: The tourist sees the Earth approaching at 0.8 c
0.8 ly yr 15 yr = 12.0 ly .
b
gb
g
Not only do distances and times differ between Earth and meteoroid reference frames, but
within the Earth frame apparent distances differ from actual distances. As we have
interpreted it, the 20-lightyear actual distance from the Earth to the meteoroid at the time of
discovery must be a calculated result, different from the distance measured directly. Because
of the finite maximum speed of information transfer, the astronomer sees the meteoroid as it
was years previously, when it was much farther away. Call its apparent distance d. The time
d
required for light to reach us from the newly-visible meteoroid is the lookback time t = .
c
The astronomer calculates that the meteoroid has approached to be 20 ly away as it moved
with constant velocity throughout the lookback time. We can work backwards to reconstruct
her calculation:
0.8 cd
d = 20 ly + 0.8 ct = 20 ly +
c
0.2d = 20 ly
d = 100 ly
Thus in terms of direct observation, the meteoroid we see covers 100 ly in only 25 years.
Such an apparent superluminal velocity is actually observed for some jets of material
emanating from quasars, because they happen to be pointed nearly toward the Earth. If we
can watch events unfold on the meteoroid, we see them slowed by relativistic time dilation,
but also greatly speeded up by the Doppler effect.
438
P39.9
P39.10
Relativity
∆t p
∆t = γ∆t p =
∆t p =
so
1 − v2 c2
F
GG
H
I F v I
J ∆t ≅ GH1 − 2 c JK ∆t
c JK
F v I ∆t .
=G
H 2c JK
1−
v2
2
2
2
2
and
∆t − ∆t p
If
v = 1 000 km h =
then
v
= 9.26 × 10 −7
c
and
e∆t − ∆t j = e4.28 × 10 jb3 600 sg = 1.54 × 10
For
2
1.00 × 10 6 m
= 277.8 m s
3 600 s
−13
p
−9
v
= 0.990 , γ = 7.09
c
(a)
The muon’s lifetime as measured in the Earth’s rest frame is
∆t =
4.60 km
0.990 c
and the lifetime measured in the muon’s rest frame is
∆t p =
(b)
P39.11
∆t
γ
=
L = Lp 1 −
LM
MN
1
4.60 × 10 3 m
7.09 0.990 3.00 × 10 8 m s
FG v IJ
H cK
2
e
=
Lp
γ
=
OP
=
j PQ
2.18 µs .
4.60 × 10 3 m
= 649 m
7.09
The spaceship is measured by the Earth observer to be length-contracted to
L = Lp 1 −
v2
c2
or
F
GH
L2 = L2p 1 −
v2
c
2
I.
JK
Also, the contracted length is related to the time required to pass overhead by:
L = vt
or
v2
L2 = v 2 t 2 =
v2
c
2
af
Equating these two expressions gives
L2p − L2p
or
L2p + ct
Using the given values:
L p = 300 m
this becomes
e1.41 × 10
giving
v = 0.800 c .
c2
af
= ct
v2
2
c
5
actf
2
m2
2
2
.
v2
c2
= L2p .
and
j vc
2
2
t = 7.50 × 10 −7 s
= 9.00 × 10 4 m 2
s = 1.54 ns .
Chapter 39
P39.12
(a)
439
The spaceship is measured by Earth observers to be of length L, where
L = Lp 1 −
v2
c2
v∆ t = L p 1 −
F
v G ∆t
H
2
Solving for v,
2
L2p
+
c
2
v2
c2
and
L = v∆t
and
v 2 ∆t 2 = L2p 1 −
F
GH
I =L
JK
2
p
The tanks move nonrelativistically, so we have v =
(c)
For the data in problem 11,
v=
a
f
c 300 m
e3 × 10
8
ms
j e0.75 × 10 sj + a300 mf
2
−6
cL p
v=
(b)
2
2
c 2 ∆t 2 + L2p
a
f
c 300 m
=
a f
m sj a75 sf + a300 mf
c 300 m
e3 × 10
8
2
2
2
225 2 + 300 2 m
a
c 300 m
=
.
300 m
= 4.00 m s .
75 s
in agreement with problem 11. For the data in part (b),
v=
I
JK
v2
.
c2
e2.25 × 10 j
10 2
f
= 0.800 c
= 1.33 × 10 −8 c = 4.00 m s
+ 300 2 m
in agreement with part (b).
P39.13
GMm
We find Cooper’s speed:
r2
=
mv 2
.
r
L GM OP = LM e6.67 × 10 je5.98 × 10 j OP
v=M
N aR + hf Q MN e6.37 × 10 + 0.160 × 10 j PQ
2π a R + hf 2π e6.53 × 10 j
T=
=
= 5. 25 × 10 s .
12
Solving,
−11
24
6
6
12
= 7.82 km s .
6
Then the time period of one orbit,
(a)
v
7.82 × 10
LF v I
The time difference for 22 orbits is ∆t − ∆t = bγ − 1g∆t = MG 1 − J
MNH c K
F 1 v − 1I a22T f = 1 F 7.82 × 10 m s I 22e5.25 × 10 sj =
∆t − ∆t ≈ G 1 +
2 GH 3 × 10 m s JK
H 2 c JK
2
p
p
P39.14
γ =
2
8
For one orbit, ∆t − ∆t p =
1
e
1 − v2 c2
j
= 1.01
p
3
2
(b)
3
3
2
−1 2
OP
PQa f
− 1 22T
2
3
39.2 µs .
39.2 µs
= 1.78 µs . The press report is accurate to one digit .
22
so
v = 0.140 c
440
P39.15
*P39.16
Relativity
(a)
Since your ship is identical to his, and you are at rest with respect to your own ship, its
length is 20.0 m .
(b)
His ship is in motion relative to you, so you measure its length contracted to 19.0 m .
(c)
We have
L = Lp 1 −
from which
L 19.0 m
v2
=
= 0.950 = 1 − 2 and v = 0.312 c .
L p 20.0 m
c
v2
c2
In the Earth frame, Speedo’s trip lasts for a time
∆t =
20.0 ly
∆x
=
= 21.05 yr .
v
0.950 ly yr
Speedo’s age advances only by the proper time interval
∆t p =
∆t
γ
= 21.05 yr 1 − 0.95 2 = 6.574 yr during his trip.
Similarly for Goslo,
∆t p =
20.0 ly
∆x
v2
1− 2 =
1 − 0.75 2 = 17.64 yr .
v
0.750 ly yr
c
While Speedo has landed on Planet X and is waiting for his brother, he ages by
20.0 ly
20.0 ly
−
= 5.614 yr .
0.750 ly yr 0.950 ly yr
b
g
Then Goslo ends up older by 17.64 yr − 6.574 yr + 5.614 yr = 5.45 yr .
P39.17
(a)
∆t = γ∆t p =
∆t p
=
15.0 yr
= 21.0 yr
(b)
b g 1 − a0.700f
d = va ∆t f = 0.700 c b 21.0 yr g = a0.700fb1.00 ly yr g b 21.0 yr g =
(c)
The astronauts see Earth flying out the back window at 0.700 c :
e j
1− v c
2
b
2
g a
fb
gb
14.7 ly
g
d = v ∆t p = 0.700 c 15.0 yr = 0.700 1.00 ly yr 15.0 yr = 10.5 ly
(d)
Mission control gets signals for 21.0 yr while the battery is operating, and then for 14.7 years
after the battery stops powering the transmitter, 14.7 ly away:
21.0 yr + 14.7 yr = 35.7 yr
Chapter 39
P39.18
The orbital speed of the Earth is as described by
v=
GmS
=
r
e6.67 × 10
−11
∑ F = ma :
je
N ⋅ m 2 kg 2 1.99 × 10 30 kg
1.496 × 10
11
m
GmS m E
r2
j = 2.98 × 10
4
=
441
mE v 2
r
m s.
The maximum frequency received by the extraterrestrials is
f obs = fsource
1+v c
= 57.0 × 10 6 Hz
1−v c
e
j 1 − ee2.98 × 10
j e3.00 × 10
m sj e3.00 × 10
1 + 2.98 × 10 4 m s
4
8
8
j = 57.005 66 × 10
m sj
6
Hz .
j = 56.994 34 × 10
m sj
6
Hz .
ms
The minimum frequency received is
f obs = fsource
1−v c
= 57.0 × 10 6 Hz
1+v c
e
j 1 + ee2.98 × 10
j e3.00 × 10
m sj e3.00 × 10
1 − 2.98 × 10 4 m s
4
8
8
ms
The difference, which lets them figure out the speed of our planet, is
b57.005 66 − 56.994 34g × 10
P39.19
(a)
6
Hz = 1.13 × 10 4 Hz .
Let f c be the frequency as seen by the car. Thus,
f c = f source
and, if f is the frequency of the reflected wave,
f = fc
(c)
which gives
a c + vf .
a c − vf
f a c − vf = f
a c + vf
b f − f gc = b f + f g v ≈ 2 f
The beat frequency is then
f beat = f − fsource =
Using the above result,
f beat
λ=
(d)
c+v
.
c−v
f = f source
Combining gives
(b)
c+v
c−v
v=
source
source
a2fb30.0 m sge10.0 × 10
=
8
3.00 × 10 m s
c
fsource
f beat λ
2
=
3.00 × 10 8 m s
10.0 × 10 9 Hz
so
∆v =
9
Hz
j = a2fb30.0 m sg = 2 000 Hz =
b0.030 0 mg
source
2 f source v
2v
.
=
λ
c
2.00 kHz
= 3.00 cm
a
fb
g
source v .
5 Hz 0.030 0 m
∆f beat λ
=
= 0.075 0 m s ≈ 0.2 mi h
2
2
442
P39.20
Relativity
(a)
When the source moves away from an observer, the observed frequency is
f obs = fsource
FG c − v IJ
Hc+v K
12
s
where v s = v source .
s
When v s << c , the binomial expansion gives
FG c − v IJ
Hc+v K
s
12
12
s
f obs ≈ fsource
So,
LM FG v IJ OP LM1 + FG v IJ OP
N H c KQ N H c KQ
FG 1 − v IJ .
H cK
s
= 1−
FG
H
−1 2
s
≈ 1−
vs
2c
IJ FG 1 − v IJ ≈ FG1 − v IJ .
K H 2c K H c K
s
s
s
The observed wavelength is found from c = λ obs fobs = λfsource :
λ obs =
λf source
λfsource
λ
≈
=
f obs
1 − vs c
f source 1 − v s c
b
∆λ = λ obs − λ = λ
Since 1 −
(b)
*P39.21
∆λ
vs
≈1,
c
v source = c
λ
s
c
I F v cI
JK GH 1 − v c JK
−1 = λ
s
s
v source
.
c
FG ∆λ IJ = cFG 20.0 nm IJ =
H λ K H 397 nm K
0.050 4c
For the light as observed
1+v c
1+v c
c
c
=
fobs =
fsource =
λ obs
1−v c
1 − v c λ source
1 + v c λ source 650 nm
=
=
λ obs
520 nm
1− v c
v
v
= 1.562 − 1.562
c
c
v = 0.220 c = 6.59 × 10 7 m s
1+
Section 39.5
*P39.22
≈
F 1
GH 1 − v
g
1+v c
= 1.25 2 = 1.562
1−v c
v 0.562
=
= 0. 220
c 2.562
The Lorentz Transformation Equations
Let Suzanne be fixed in reference from S and see the two light-emission events with coordinates
x 1 = 0 , t1 = 0 , x 2 = 0 , t 2 = 3 µs . Let Mark be fixed in reference frame S ′ and give the events
coordinate x 1′ = 0 , t1′ = 0 , t 2′ = 9 µs .
(a)
Then we have
v
t 2′ = γ t 2 − 2 x 2
c
FG
H
1
9 µs =
v2
c
(b)
2
=
IJ
K
1− v
2
c
2
b 3 µs − 0 g
8
9
b
1−
v2
c
2
=
1
3
v = 0.943 c
g e
x 2′ = γ x 2 − vt 2 = 3 0 − 0.943 c × 3 × 10 −6 s
F
jGH 3 × 10c
8
ms
I=
JK
2.55 × 10 3 m
443
Chapter 39
P39.23
1
γ =
1− v
2
c
2
1
=
= 10.0
1 − 0.995 2
We are also given: L1 = 2.00 m , and θ = 30.0° (both measured in a
reference frame moving relative to the rod).
a fa f
= a 2.00 mfa0.500 f = 1.00 m
Thus,
L1 x = L1 cos θ 1 = 2.00 m 0.867 = 1.73 m
and
L1 y = L1 sin θ 1
L 2 x is a proper length, related to L1 x by L1 x =
L2x
.
γ
= 10.0L1 x = 17.3 m
Therefore,
L2x
and
L 2 y = L1 y = 1.00 m .
FIG. P39.23
(Lengths perpendicular to the motion are unchanged).
*P39.24
bL g + eL j
(a)
L2 =
(b)
θ 2 = tan −1
2
2x
2
2y
L2y
L2x
gives
L 2 = 17.4 m
gives
θ 2 = 3.30°
Einstein’s reasoning about lightning striking the ends of a train shows that the moving observer sees
the event toward which she is moving, event B , as occurring first. The S-frame coordinates of the
events we may take as (x = 0 , y = 0 , z = 0, t = 0 ) and (x = 100 m , y = 0 , z = 0, t = 0 ). Then the
coordinates in S ′ are given by the Lorentz transformation. Event A is at ( x ′ = 0 , y ′ = 0 , z ′ = 0 , t ′ = 0).
The time of event B is
FG
H
t′ = γ t −
v
c
2
IJ
K
x =
1
1 − 0.8
2
FG 0 − 0.8 c a100 mfIJ = 1.667FG − 80 m IJ = −4.44 × 10
H c
K
H 3 × 10 m s K
2
8
−7
s.
The time elapsing before A occurs is 444 ns .
P39.25
(a)
From the Lorentz transformation, the separations between the blue-light and red-light
events are described by
a
∆x ′ = γ ∆x − v∆t
v=
(b)
2.00 m
8.00 × 10
−9
s
f
0 = γ 2.00 m − v 8.00 × 10 −9 s
= 2.50 × 10 8 m s
γ =
e
j
1
e
8
1 − 2.50 × 10 m s
a
f
j e3.00 × 10
2
8
ms
j
2
= 1.81 .
Again from the Lorentz transformation, x ′ = γ x − vt :
e
je
x ′ = 1.81 3.00 m − 2.50 × 10 8 m s 1.00 × 10 −9 s
j
x ′ = 4.97 m .
(c)
F v xIJ :
t′ = γ G t −
H c K
2
LM
t ′ = 1.81 1.00 × 10
MM
N
−9
t ′ = −1.33 × 10 −8 s
e2.50 × 10
s−
e3.00 × 10
8
8
j a3.00 mfOP
PP
m sj
Q
ms
2
444
Relativity
Section 39.6
P39.26
The Lorentz Velocity Transformation Equations
u x = Enterprise velocity
v = Klingon velocity
From Equation 39.16
u x′ =
ux − v
1 − ux v c
2
=
0.900 c − 0.800 c
= 0.357 c .
1 − 0.900 0.800
a
fa
f
FIG. P39.26
P39.27
u ′x =
ux − v
1 − ux v c 2
=
−0.750 c − 0.750 c
= −0.960 c
1 − −0.750 0.750
a
fa
f
FIG. P39.27
*P39.28
Let frame S be the Earth frame of reference. Then
The components of the velocity of the first spacecraft are
v = −0.7 c .
u x = 0.6 c cos 50° = 0.386 c
and
uy
a f
= a0.6 c f sin 50° = 0.459 c .
As measured from the S ′ frame of the second spacecraft,
The magnitude of u ′ is
a f = 1.086 c = 0.855 c
1.27
1−u v c
1 − a0.386 c fa −0.7 c f c
u
0.459 c 1 − a0.7 f
0.459 ca0.714f
=
=
= 0.258 c
u′ =
1.27
γ e1 − u v c j 1 − a0.386 fa −0.7f
a0.855 cf + a0.285cf = 0.893c
and its direction is at
tan −1
u ′x =
ux − v
x
2
0.386 c − −0.7 c
=
2
2
y
y
x
Section 39.7
P39.29
(a)
2
2
2
0. 258 c
= 16.8° above the x ′ -axis .
0.855 c
Relativistic Linear Momentum and the Relativistic Form of Newton’s Laws
p = γ mu ; for an electron moving at 0.010 0 c,
1
γ =
b g
1− u c
Thus,
e
2
1
=
p = 1.00 9.11 × 10 −31
p = 2.73 × 10 −24 kg ⋅ m s .
(b)
(c)
= 1.000 05 ≈ 1.00 .
b g
kg jb0.010 0 ge3.00 × 10
1 − 0.010 0
Following the same steps as used in part (a),
we find at 0.500 c, γ = 1.15 and
p = 1.58 × 10 −22 kg ⋅ m s .
At 0.900 c, γ = 2.29 and
p = 5.64 × 10 −22 kg ⋅ m s .
2
8
ms
j
Chapter 39
P39.30
p=
we find the difference ∆p from the classical momentum, mu:
∆p = γ mu − mu = γ − 1 mu .
b g
The difference is 1.00% when γ − 1 mu = 0.010 0γ mu :
thus 1 −
(b)
FG u IJ = a0.990f
H cK
2
2
γ =
b g
FG u IJ = a0.900f
H cK
2
2
γ =
γ −1=
1
−1 ≈1+
b g
p − mu 1 F 90.0 m s I
= G
2 H 3.00 × 10 m s JK
mu
1− u c
2
FG IJ
HK
mu
b g
1− u c
1−
becomes
2
P39.33
c2
1 = u2
which gives:
or c 2 = u 2
u2
Fm c
GH p
2 2
2
I
JK
+1
b g
1
1
=
0.990
1− u c
b g
1
1
=
0.900
1− u c
b g
−1=
2
2
FG IJ
HK
1 u
2 c
2
2
= 4.50 × 10 −14
m 2u 2
p2
Fm
GH p
2
2
+
1
c2
I
JK
c
u=
and
=
2
1 u
2 c
8
p=
2
u = 0.436 c .
and
p − mu γ mu − mu
=
=γ −1:
mu
mu
b g
1− u c
u = 0.141c .
, and
The difference is 10.0% when γ − 1 mu = 0.100γ mu :
thus 1 −
P39.32
= γ mu
Using the relativistic form,
(a)
P39.31
mu
445
em c
2 2
j
p2 + 1
.
Relativistic momentum of the system of fragments must be conserved. For total momentum to be
zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and
subscript 1 to the lighter, p 2 = p1
or
γ 2 m 2u 2 = γ 1 m1u1 =
a
1 − 0.893
e1.67 × 10 kgju = e4.960 × 10
1 − bu c g
F 1.67 × 10 u I = 1 − u
GH 4.960 × 10 c JK
c
−27
or
2.50 × 10 −28 kg
2
2
2
Proceeding to solve, we find
−27
−28
12.3
u22
c2
2
2
2
2
2
= 1 and u 2 = 0.285 c .
f
2
−28
a
× 0.893 c
j
kg c .
f
446
Relativity
Section 39.8
P39.34
Relativistic Energy
b
g
∆E = γ 1 − γ 2 mc 2
mc 2 = 0.511 MeV
For an electron,
F 1 − 1 I mc = 0.582 MeV
GH a1 − 0.810f a1 − 0.250f JK
F
I
1
1
(b)
∆E = G
−
GH 1 − a0.990f 1 − 0.810 JJK mc = 2.45 MeV
I F
F
I
1
1
J
G
JJ mc
G
− 1J mc −
∑W = K − K = G
G
GH 1 − dv ci JK H 1 − bv cg K
I
F
1
1
JJ mc
G
−
or ∑ W = G
GH 1 − dv ci 1 − bv cg JK
(a)
2
∆E =
2
2
P39.35
f
2
i
2
2
2
i
f
2
2
2
i
f
(a)
∑W =
F
GG
H
∑W =
(b)
P39.36
a
1
1 − 0.750
f
2
−
I
J e1.673 × 10
1 − a0.500 f JK
−27
kg 2.998 × 10 8 m s
−
I
J e1.673 × 10
1 − a0.500 f JK
−27
kg 2.998 × 10 8 m s
1
2
je
j
2
je
j
2
5.37 × 10 −11 J
F
H
∑ W = GG
1 − 0.995
∑W =
1.33 × 10 −9 J
a
1
f
2
1
2
The relativistic kinetic energy of an object of mass m and speed u is K r =
For u = 0.100 c ,
The classical equation K c =
different by
1
mu 2 gives
2
Kr =
F
GH
Kc =
1
m 0.100 c
2
1
1 − 0.010 0
a
f
2
F
GG
H
1 − u2 c 2
I
JK
= 0.005 000mc 2
I
JJ
K
− 1 mc 2 .
− 1 mc 2 = 0.005 038mc 2 .
0.005 038 − 0.005 000
= 0.751% .
0.005 038
For still smaller speeds the agreement will be still better.
1
Chapter 39
P39.37
E = γ mc 2 = 2mc 2 or
FG IJ
H K
1
u
Thus, = 1 −
γ
c
γ =2.
2
3
or
2
=
u=
c 3
.
2
F c 3 I = F mc I 3
GH 2 JK GH c JK
F 938.3 MeV IJ 3 = 1.63 × 10
p=G
H c K
2
p = γ mu = 2m
The momentum is then
MeV
.
c
3
P39.38
447
K=
(a)
Using the classical equation,
(b)
Using the relativistic equation, K =
LM
K=M
MMN
b
ge
1
1
mv 2 = 78.0 kg 1.06 × 10 5 m s
2
2
F
GG
H
1
b g
1− v c
2
j
2
= 4.38 × 10 11 J .
I
JJ
K
− 1 mc 2 .
OP
− 1Pb78.0 kg ge 2.998 × 10 m sj = 4.38 × 10 J
PPQ
1 − e1.06 × 10 2.998 × 10 j
LM1 − FG v IJ OP ≈ 1 + 1 FG v IJ .
v
When << 1 , the binomial series expansion gives
c
2 H cK
MN H c K PQ
LM1 − FG v IJ OP − 1 ≈ 1 FG v IJ .
Thus,
2H cK
MN H c K PQ
1 F vI
1
and the relativistic expression for kinetic energy becomes K ≈ G J mc = mv . That is, in
H
K
2 c
2
1
2
8
11
8 2
5
2 −1 2
2
2 −1 2
2
2
2
2
the limit of speeds much smaller than the speed of light, the relativistic and classical
expressions yield the same results.
P39.39
e
je
(a)
ER = mc 2 = 1.67 × 10 −27 kg 2.998 × 10 8 m s
(b)
E = γ mc 2 =
1.50 × 10 −10 J
b
1 − 0.950 c c
2
= 1.50 × 10 −10 J = 938 MeV
= 4.81 × 10 −10 J = 3.00 × 10 3 MeV
K = E − mc 2 = 4.81 × 10 −10 J − 1.50 × 10 −10 J = 3.31 × 10 −10 J = 2.07 × 10 3 MeV
(c)
P39.40
g
2 12
j
The relativistic density is
ER
2
c V
=
mc 2 m
= =
c 2V V
eL jeL jLMNL
p
p
m
8.00 g
=
O
1 − bu c g P a1.00 cmf
Q
2
p
3
a
1 − 0.900
f
2
= 18.4 g cm 3 .
448
P39.41
Relativity
We must conserve both energy and relativistic momentum of the
system of fragments. With subscript 1 referring to the 0.868 c
particle and subscript 2 to the 0.987 c particle,
1
1
γ1 =
= 2.01 and γ 2 =
= 6.22 .
2
2
1 − 0.868
1 − 0.987
a
a
f
f
Conservation of energy gives E1 + E 2 = Etotal
which is
γ 1 m1 c 2 + γ 2 m 2 c 2 = m total c 2
or
2.01m1 + 6. 22m 2 = 3.34 × 10 −27 kg .
This reduces to:
m1 + 3.09m 2 = 1.66 × 10 −27 kg .
(1)
Since the final momentum of the system must equal zero, p1 = p 2
γ 1 m1u1 = γ 2 m 2 u 2
gives
or
a2.01fa0.868cfm = a6.22fa0.987cfm
which becomes
m1 = 3.52m 2 .
1
(2)
Solving (1) and (2) simultaneously, m1 = 8.84 × 10
*P39.42
Energy conservation:
3 108 kg 1 −
1
1 − 02
1 452 kg 1 −
P39.43
a
1 − 0.85 2
900 kg 0.85 c
1 − 0.85
2
−28
900 kgc 2
1 400 kgc 2 +
v2
= M.
c2
Momentum conservation: 0 +
FIG. P39.41
2
f=
kg and m 2 = 2.51 × 10 −28 kg .
Mc 2
=
1 − v2 c2
Mv
1 − v2 c2
v 2 Mv
=
.
c
c2
v 1 452
=
= 0.467
c 3 108
v = 0.467 c .
(a)
Dividing gives
(b)
Now by substitution 3 108 kg 1 − 0.467 2 = M = 2.75 × 10 3 kg .
E = γ mc 2
e
p = γ mu
p = bγ mug
j
F u IF u I
= eγ mc j − bγ mug c = γ F emc j − amc f u I = emc j G 1 − J G 1 − J = emc j
H
K
H c KH c K
E 2 = γ mc
2
2 2
E −p c
2 2
2
2
2 2
2 2
2
2 2
2
2
2 2
2
2
2
2
Q.E.D.
P39.44
(a)
a f
Thus, γ =
(b)
b g
q ∆V = K = γ − 1 m e c 2
b g
1
b g
1− u c
2
=1+
a f e
a f from which
q ∆V
mec 2
je
u = 0.302 c .
j
K = γ − 1 m e c 2 = q ∆V = 1.60 × 10 −19 C 2.50 × 10 4 J C = 4.00 × 10 −15 J
−1
2 2
Chapter 39
P39.45
(a)
E = γ mc 2 = 20.0 GeV with mc 2 = 0.511 MeV for electrons. Thus,
γ =
*P39.46
449
20.0 × 10 9 eV
0.511 × 10 6 eV
1
= 3.91 × 10 4 .
= 3.91 × 10 4 from which u = 0.999 999 999 7 c
(b)
γ =
(c)
L = Lp 1 −
(a)
P=
(b)
The kinetic energy of one electron with v = 0.999 9 c is
b g
1− u c
2
FG u IJ
H cK
2
=
Lp
γ
=
3.00 × 10 3 m
= 7.67 × 10 −2 m = 7.67 cm
4
3.91 × 10
energy
2J
=
= 2.00 × 10 13 W
∆t
100 × 10 −15 s
F
bγ − 1gmc = GG
H
1
2
1 − 0.999 9
2
I
JJ
K
e
− 1 9.11 × 10 −31 kg 3 × 10 8 m s
j
2
e
= 69.7 8.20 × 10 −14 J
j
= 5.72 × 10 −12 J
Then we require
N=
P39.47
0.01
2 J = N 5.72 × 10 −12 J
100
e
j
2 × 10 −4 J
= 3.50 × 10 7 .
5.72 × 10 −12 J
Conserving total momentum of the decaying particle system,
p before decay = p after decay = 0
b
g
p v = p µ = γm µ u = γ 207 m e u .
Conservation of mass-energy for the system gives Eµ + E v = Eπ : γ m µ c 2 + p v c = mπ c 2
b g pc = 273m .
u
γ b 207m g + γ b 207m g = 273m
c
v
γ 207m e +
Substituting from the momentum equation above,
FG
H
IJ
K
u
273
=
= 1.32 :
c
207
1+u c
= 1.74
1−u c
or
γ 1+
Then,
K µ = γ − 1 m µ c 2 = γ − 1 207 m e c 2 :
b g
b g e
j
e
e
e
u
= 0.270 .
c
Kµ =
F
GG
H
a
1
1 − 0.270
f
2
e
I
JJ a
K
− 1 207 0.511 MeV
K µ = 4.08 MeV .
Also,
Ev = Eπ − Eµ :
b
g
Ev = mπ c 2 − γ m µ c 2 = 273 − 207γ m e c 2
F
GG
H
Ev = 273 −
I
J a0.511 MeVf
1 − a0.270 f JK
Ev = 29.6 MeV
207
2
f
450
*P39.48
Relativity
3c
3c
and u 2 = − . Let observer B be at
4
4
3c
v = u1 = .
4
Let observer A hold the unprimed reference frame, with u1 =
rest in the primed frame with u1′ = 0 =
(a)
Then u 2′ =
u2 − v
1 − u2 v c
speed = u2′ =
(b)
2
=
u1 − v
1 − u1 v c 2
− 3c 4 − 3c 4
−1.5 c
=
1 − − 3 c 4 + 3 c 4 1 + 9 16
b
gb
g
3 c 2 24
=
c = 0.960 c .
25 16 25
In the unprimed frame the objects, each of mass m, together have energy
mc 2
γ mc 2 + γ mc 2 = 2
= 3.02mc 2 .
2
1 − 0.75
mc 2
In the primed frame the energy is
2
mc 2
+
2
= 4.57mc 2 , greater by
1−0
1 − 0.96
4.57mc 2
= 1.51 times greater as measured by observer B .
3.02mc 2
Section 39.9
P39.49
P39.50
Mass and Energy
Let a 0.3-kg flag be run up a flagpole 7 m high.
e
j
We put into it energy
mgh = 0.3 kg 9.8 m s 2 7 m ≈ 20 J .
So we put into it extra mass
∆m =
for a fractional increase of
2 × 10 −16 kg
~ 10 −15 .
0.3 kg
E
c2
=
20 J
e3 × 10
8
ms
j
2
= 2 × 10 −16 kg
E = 2.86 × 10 5 J . Also, the mass-energy relation says that E = mc 2 .
Therefore,
m=
2.86 × 10 5 J
E
=
c2
3.00 × 10 8 m s
e
j
2
= 3.18 × 10 −12 kg .
No, a mass loss of this magnitude (out of a total of 9.00 g) could not be detected .
jb
e
ge
j
a f e
je
e
j
9
7
E P ∆t 0.800 1.00 × 10 J s 3.00 yr 3.16 × 10 s yr
=
=
= 0.842 kg
2
c2
c2
3.00 × 10 8 m s
P39.51
∆m =
P39.52
1 030 kg m3 1.40 × 10 9 10 3 m 4 186 J kg⋅° C 10.0 ° C
ρVc ∆T
E mc ∆T
∆m = 2 =
=
=
2
c
c2
c2
3.00 × 10 8 m s
a f
∆m = 6.71 × 10 8 kg
je
e
jb
ga
3
j
f
Chapter 39
P39.53
P=
3.77 × 10 26 J s
dm
=
= 4.19 × 10 9 kg s
2
8
dt
3.00 × 10 m s
e
j
2m e c 2 = 1.02 MeV
Section 39.10
*P39.55
e j
2
dE d mc
dm
=
= c2
= 3.77 × 10 26 W
dt
dt
dt
Thus,
P39.54
(a)
451
Eγ ≥ 1.02 MeV
The General Theory of Relativity
GM E m
∑ F = ma :
For the satellite
r2
=
FG
H
mv 2 m 2π r
=
r
r
T
GM ET 2 = 4π 2 r 3
F 6.67 × 10
r =G
GH
jb
e
−11
N ⋅ m 2 5.98 × 10 24 kg 43 080 s
kg 2 4π 2
g IJ
JK
2
IJ
K
2
13
r = 2.66 × 10 7 m
e
j
7
2π r 2π 2.66 × 10 m
=
= 3.87 × 10 3 m s
43 080 s
T
(b)
v=
(c)
The small fractional decrease in frequency received is equal in magnitude to the fractional
increase in period of the moving oscillator due to time dilation:
I
F
1
J
G
fractional change in f = −bγ − 1g = − G
− 1J
GH 1 − e3.87 × 10 3 × 10 j JK
F 1 L F 3.87 × 10 I OI
= 1 − G 1 − M− G
GH 2 MN H 3 × 10 JK PPQJJK = −8.34 × 10
3
3
8 2
2
−11
8
(d)
The orbit altitude is large compared to the radius of the Earth, so we must use
GM E m
.
Ug = −
r
∆U g = −
e
j
6.67 × 10 −11 Nm 2 5.98 × 10 24 kg m
= 4.76 × 10
2
7
kg 2.66 × 10 m
7
+
J kg m
∆f ∆U g 4.76 × 10 7 m 2 s 2
= +5. 29 × 10 −10
=
=
2
8
f
mc 2
3 × 10 m s
e
(e)
j
−8.34 × 10 −11 + 5.29 × 10 −10 = +4.46 × 10 −10
e
j
6.67 × 10 −11 Nm 5.98 × 10 24 kg m
6
kg 6.37 × 10 m
452
Relativity
Additional Problems
P39.56
(a)
d earth = vt earth = vγ t astro
FG IJ e
H K
v2
v
1− 2 =
1.50 × 10 −5
c
c
v2
1 = 2 1 + 2.25 × 10 −10
c
v
= 1 − 1.12 × 10 −10
c
e
P39.57
F
GG
H
j
I
JJ
K
K=
(c)
6.00 × 10 27 J = 6.00 × 10 27
(a)
10 13 MeV = γ − 1 m p c 2
1− v
c
2
e
30.0 yr
j
−1 2
=1−
1
2.25 × 10 −10
2
e
F 2.00 × 10 yr − 1I b1 000gb1 000 kg g 3 × 10 m s
e
j
GH 30 yr
JK
F 13¢ IJ FG k IJ FG W ⋅ s IJ FG h IJ = $2.17 × 10
JG
H kWh K H 10 K H J K H 3 600 s K
6
8
b g
= 6.00 × 10 27 J
γ = 10 10
so
t′ =
t
γ
=
10 5 yr
10 10
= 10 −5 yr ~ 10 2 s
(b)
d ′ = ct ′ ~ 10 8 km
(a)
When K e = K p ,
mec 2 γ e − 1 = mp c 2 γ p − 1 .
In this case,
m e c 2 = 0.511 MeV , m p c 2 = 938 MeV
and
γ e = 1 − 0.750
Substituting,
γ p =1+
but
γp=
b
2
g
e
a
f
2 −1 2
j
= 1.511 9 .
g = 1 + a0.511 MeVfb1.511 9 − 1g = 1.000 279
b
me c 2 γ e − 1
mp c
2
938 MeV
1
LM1 − eu cj OP
N
Q
2 12
.
p
(b)
Therefore,
u p = c 1 − γ p−2 = 0.023 6 c .
When p e = p p
γ p m pu p = γ e m eu e or γ pu p =
Thus,
γ p up =
γ e m eu e
.
mp
b1.511 9ge0.511 MeV c ja0.750 cf = 6.177 2 × 10
2
and
which yields
up
c
j
20
3
vp ≈ c
P39.58
c2
c2
v
= 1 + 2.25 × 10 −10
c
so
2
−10
2
c2
− 1 mc 2 =
(b)
2
6
2
j
1
e2.00 × 10 yrjc = v 1 − v1
v e 2.25 × 10 j
v
=
1−
so
938 MeV c
= 6.177 2 × 10 −4 1 −
Fu I
GH c JK
p
2
2
u p = 6.18 × 10 −4 c = 185 km s .
−4
c
Chapter 39
P39.59
(a)
Since Mary is in the same reference frame, S ′ , as Ted, she measures the ball to have the
same speed Ted observes, namely u ′x = 0.800 c .
(b)
∆t ′ =
(c)
L = Lp 1 −
Lp
u x′
=
1.80 × 10 12 m
e
8
0.800 3.00 × 10 m s
v2
c2
e
j
453
= 7.50 × 10 3 s
a0.600cf
j
= 1.80 × 10 12 m 1 −
2
= 1.44 × 10 12 m
c2
Since v = 0.600 c and u ′x = −0.800 c , the velocity Jim measures for the ball is
ux =
(d)
(a)
(b)
P39.61
*P39.62
1 + u ′x v c
2
=
a−0.800 cf + a0.600 cf =
1 + a −0.800 fa0.600f
−0.385 c .
Jim measures the ball and Mary to be initially separated by 1.44 × 10 12 m . Mary’s motion at
0.600c and the ball’s motion at 0.385c nibble into this distance from both ends. The gap closes
at the rate 0.600 c + 0.385 c = 0.985 c , so the ball and catcher meet after a time
∆t =
*P39.60
u ′x + v
1.44 × 10 12 m
e
0.985 3.00 × 10 8 m s
j
= 4.88 × 10 3 s .
b
fb
ga
g
The charged battery stores energy
E = P t = 1.20 J s 50 min 60 s min = 3 600 J
so its mass excess is
∆m =
−14
kg
∆m 4.00 × 10
=
= 1.60 × 10 −12
−3
m
25 × 10 kg
a
f
3 600 J
E
=
2
c
3 × 10 8 m s
e
j
2
= 4.00 × 10 −14 kg .
too small to measure.
∆mc 2 4 938.78 MeV − 3 728.4 MeV
=
× 100% = 0.712%
4 938.78 MeV
mc 2
a
f
e
The energy of the first fragment is given by E12 = p12 c 2 + m1 c 2
E1 = 2.02 MeV .
a
For the second, E22 = 2.00 MeV
(a)
2
2
2
2
2
E2 = 2.50 MeV .
Energy is conserved, so the unstable object had E = 4.52 MeV . Each component of
momentum is conserved, so the original object moved with
2
2
1.75 MeV
2.00 MeV
+
. Then for it
p 2 = p x2 + p y2 =
c
c
2
3.65 MeV
2
2
2
4.52 MeV = 1.75 MeV + 2.00 MeV + mc 2
m=
.
c2
a
(b)
f + a1.50 MeVf
j = a1.75 MeVf + a1.00 MeVf
FG
H
f a
IJ
K
f
FG
H
a
Now E = γ mc 2 gives 4.52 MeV =
IJ
K
f e j
1
1 − v2 c2
3.65 MeV
1−
v2
= 0.654 , v = 0.589 c .
c2
454
P39.63
Relativity
(a)
Take the spaceship as the primed frame, moving toward the right at v = +0.600 c .
Then u x′ = +0.800 c , and
Lp
ux =
u ′x + v
b g
1 + u′x v c
b
=
2
0.800 c + 0.600 c
= 0.946 c .
1 + 0.800 0.600
a
fa
g 1 − a0.600f
2
f
(b)
L=
(c)
The aliens observe the 0.160-ly distance closing because the probe nibbles into it from one
end at 0.800c and the Earth reduces it at the other end at 0.600c.
L = 0.200 ly
:
γ
time =
Thus,
(d)
K=
F
GG
H
1
2
1−u c
2
I
JJ
K
− 1 mc 2 :
K=
F
GG
H
= 0.160 ly
0.160 ly
= 0.114 yr .
0.800 c + 0.600 c
a
1
1 − 0.946
f
2
I
JJ e
K
je
− 1 4.00 × 10 5 kg 3.00 × 10 8 m s
j
2
K = 7.50 × 10 22 J
P39.64
In this case, the proper time is T0 (the time measured by the students on a clock at rest relative to
∆t = γT0
them). The dilated time measured by the professor is:
where ∆t = T + t Here T is the time she waits before sending a signal and t is the time required for
the signal to reach the students.
T + t = γ T0 .
Thus, we have:
(1)
To determine the travel time t, realize that the distance the students will have moved beyond the
professor before the signal reaches them is:
d = v T+t .
The time required for the signal to travel this distance is:
Solving for t gives:
Substituting this into equation (1) yields:
a f
d F vI
t = = G J aT + t f .
c H cK
bv cgT .
t=
1 − b v cg
bv cgT = γ T
T+
1 − bv cg
0
T
= γ T0 .
1− v c
or
Then T = T0
b g =T
1 − ev c j
1− v c
2
2
0
b g
1 + b v c g 1 − b v cg
1− v c
= T0
b g
1 + b v cg
1− v c
.
Chapter 39
P39.65
455
Look at the situation from the instructors’ viewpoint since they are at rest relative to the clock, and
hence measure the proper time. The Earth moves with velocity v = −0.280 c relative to the
instructors while the students move with a velocity u ′ = −0.600 c relative to Earth. Using the velocity
addition equation, the velocity of the students relative to the instructors (and hence the clock) is:
u=
a
a
f a
fa
f
f
−0. 280 c − 0.600 c
v + u′
=
= −0.753 c (students relative to clock).
1 + vu ′ c 2 1 + −0. 280 c −0.600 c c 2
(a)
With a proper time interval of ∆t p = 50.0 min , the time interval measured by the students is:
∆t = γ ∆t p
with
γ =
1
a
1 − 0.753 c
f
2
c2
= 1.52 .
a
f
Thus, the students measure the exam to last T = 1.52 50.0 min = 76.0 minutes .
(b)
The duration of the exam as measured by observers on Earth is:
∆t = γ ∆t p with γ =
P39.66
1
a
1 − 0. 280 c
f
2
c2
a
f
so T = 1.04 50.0 min = 52.1 minutes .
The energy which arrives in one year is
e
je
j
E = P ∆t = 1.79 × 10 17 J s 3.16 × 10 7 s = 5.66 × 10 24 J .
Thus,
P39.67
m=
5.66 × 10 24 J
E
=
c2
3.00 × 10 8 m s
e
j
2
= 6.28 × 10 7 kg .
The observer measures the proper length of the tunnel, 50.0 m, but measures the train contracted to
length
L = Lp 1 −
shorter than the tunnel by
P39.68
v2
= 100 m 1 − 0.950
c2
a
f
2
= 31.2 m
50.0 − 31.2 = 18.8 m so it is completely within the tunnel. .
If the energy required to remove a mass m from the surface is equal to its rest energy mc 2 ,
then
and
GM s m
= mc 2
Rg
Rg =
GM s
c
2
e6.67 × 10
=
−11
je
N ⋅ m 2 kg 2 1.99 × 10 30 kg
e3.00 × 10
R g = 1.47 × 10 3 m = 1.47 km .
8
ms
j
2
j
456
P39.69
Relativity
(a)
At any speed, the momentum of the particle is given by
mu
p = γ mu =
2
−1 2
2
dp
:
dt
Since F = qE =
.
b g
OP
d L F
u I
qE = MmuG 1 − J
dt M H
PQ
c K
N
F u I du + 1 muF 1 − u I FG 2u IJ du .
qE = mG 1 − J
H c K dt 2 GH c JK H c K dt
L
O
qE du M 1 − u c + u c P
=
m dt M 1 − u c
j PPQ
MN e
du qE F
u I
=
a=
1− J
.
G
dt m H
c K
1− u c
2
−1 2
2
2
2
2
2
So
2
2
32
2
qE
. As u
m
approaches c, the acceleration approaches zero, so that the object can never reach the speed
of light.
(b)
For u small compared to c, the relativistic expression reduces to the classical a =
(c)
ze
0
du
1 − u2 c
j
2 3 2
z
t
x = udt = qEc
0
(a)
=
z
z
t
qE
dt
m
t =0
t
u=
tdt
2 2
2
x=
2 2
m c +q E t
0
qEct
2 2
m c + q2E2t 2
c
qE
FH
m 2 c 2 + q 2 E 2 t 2 − mc
IK
An observer at rest relative to the mirror sees the light travel a distance D = 2d − x , where
x = vtS is the distance the ship moves toward the mirror in time tS . Since this observer
agrees that the speed of light is c, the time for it to travel distance D is
tS =
(b)
2
2 3 2
2
u
P39.70
2
2
and
−3 2
D 2d − vtS
=
c
c
tS =
2d
.
c+v
The observer in the rocket measures a length-contracted initial distance to the mirror of
L = d 1−
v2
c2
and the mirror moving toward the ship at speed v. Thus, he measures the distance the light
vt
is the distance the mirror moves toward the ship before
travels as D = 2 L − y where y =
2
the light reflects from it. This observer also measures the speed of light to be c, so the time
for it to travel distance D is:
b
t=
LM
MN
g
OP
PQ a f
2d
D 2
v 2 vt
=
d 1− 2 −
so c + v t =
c
2
c c
c
ac + vfac − vf or t =
2d
c
c−v
.
c+v
Chapter 39
*P39.71
Take the two colliding protons as the system
E1 = K + mc 2
E2 = mc 2
E12 = p12 c 2 + m 2 c 4
p2 = 0 .
2
In the final state, E f = K f + Mc :
E 2f = p 2f c 2 + M 2 c 4 .
By energy conservation, E1 + E2 = E f , so
E12 + 2E1 E2 + E22 = E 2f
e
j
p12 c 2 + m 2 c 4 + 2 K + mc 2 mc 2 + m 2 c 4
= p 2f c 2 + M 2 c 4
By conservation of momentum,
p1 = p f .
Then
M 2 c 4 = 2Kmc 2 + 4m 2 c 4 =
Mc 2 = 2mc 2 1 +
K
2mc 2
.
4Km 2 c 4
+ 4m 2 c 4
2mc 2
FIG. P39.71
By contrast, for colliding beams we have
E1 = K + mc 2
In the original state,
E2 = K + mc 2 .
In the final state,
E f = Mc 2
E1 + E2 = E f :
K + mc 2 + K + mc 2 = Mc 2
FG
H
Mc 2 = 2mc 2 1 +
*P39.72
K
2mc 2
IJ .
K
Conservation of momentum γ mu :
mu
+
1 − u2 c 2
a f
m −u
3 1 − u2 c 2
Mv f
=
1 − v 2f c 2
=
2mu
3 1 − u2 c 2
.
Conservation of energy γ mc 2 :
mc 2
+
1 − u2 c 2
mc 2
3 1 − u2 c 2
Mc 2
=
1 − v 2f c 2
To start solving we can divide: v f =
M
1−u
M=
2
4c
2
=
4m
3 1−u
2
c
2
=
b1 2g
=
4mc 2
3 1 − u2 c 2
.
2u u
= . Then
4 2
M
4 − u2 c 2
2m 4 − u 2 c 2
3 1 − u2 c 2
Note that when v << c , this reduces to M =
4m
, in agreement with the classical result.
3
457
458
P39.73
Relativity
(a)
L20 = L20 x + L20 y and
L2 = L2x + L2y .
The motion is in the x direction: L y = L0 y = L0 sin θ 0
FG v IJ = bL cosθ g 1 − FG v IJ .
H cK
H cK
L F vI O
L F vI
L = L cos θ M1 − G J P + L sin θ = L M1 − G J
MN H c K PQ
MN H c K
L F vI
O .
L = L M1 − G J cos θ P
MN H c K
PQ
2
L x = L0 x 1 −
2
Thus,
2
0
2
0
P39.74
L0 y
2
0
0
0
Ly
0
2
2
or
2
2
0
2
0
2
cos 2 θ 0
OP
PQ
12
2
0
(b)
tan θ =
(b)
Consider a hermit who lives on an asteroid halfway between the Sun and Tau Ceti,
stationary with respect to both. Just as our spaceship is passing him, he also sees the blast
waves from both explosions. Judging both stars to be stationary, this observer concludes that
the two stars blew up simultaneously .
(a)
We in the spaceship moving past the hermit do not calculate the explosions to be
simultaneous. We measure the distance we have traveled from the Sun as
Lx
L = Lp 1 −
=
b g
L0 x 1 − v c
2
= γ tan θ 0
FG v IJ = b6.00 lyg 1 − a0.800f
H cK
2
2
= 3.60 ly .
We see the Sun flying away from us at 0.800c while the light from the Sun approaches at
1.00c. Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the time
we calculate to have elapsed since the Sun exploded is
3.60 ly
= 2.00 yr .
1.80 c
We see Tau Ceti as moving toward us at 0.800c, while its light approaches at 1.00c, only
0.200c faster. We measure the gap between that star and its blast wave as 3.60 ly and
growing at 0.200c. We calculate that it must have been opening for
3.60 ly
= 18.0 yr
0.200 c
and conclude that Tau Ceti exploded 16.0 years before the Sun .
Chapter 39
P39.75
Since the total momentum is zero before decay, it is necessary that after the decay
p nucleus = p photon =
Eγ
=
c
14.0 keV
.
c
e j with mc = 8.60 × 10 J = 53.8 GeV .
0 keV I
JK + 1 .
emc + K j = a14.0 keV f + emc j or FGH 1 + mcK IJK = FGH 14.mc
K
F 14.0 keV IJ ≈ 1 + 1 FG 14.0 keV IJ (Binomial Theorem)
= 1+G
1+
H mc K
K
2 H mc
mc
a14.0 keVf = e14.0 × 10 eVj = 1.82 × 10 eV .
K ≈=
2mc
2e53.8 × 10 eV j
Also, for the recoiling nucleus, E 2 = p 2 c 2 + mc 2
2
2
Thus,
2
2
2 2
2
2
P39.76
2
2
2
2
2
3
2
and
2
2
2
So
−9
2
−3
2
9
Take m = 1.00 kg .
Kc =
The classical kinetic energy is
and the actual kinetic energy is
af
u
c
0.000
Kc J
0.000
0.100 0.045 × 10 16
0.200 0.180 × 10 16
0.300 0.405 × 10 16
0.400 0.720 × 10 16
0.500
0.600
0.700
0.800
0.900
0.990
1.13 × 10 16
1.62 × 10 16
2.21 × 10 16
2.88 × 10 16
3.65 × 10 16
4.41 × 10 16
Kr
FG IJ = e4.50 × 10
HK
I
1
− 1J mc = e9.00 × 10
JK
1 − bu c g
1
1
u
mu 2 = mc 2
2
2
c
F
=G
GH
2
16
2
2
16
jFGH uc IJK
F 1
JjG
GH 1 − bu cg
2
J
af
Kr J
0.000
0.0453 × 10 16
0.186 × 10 16
0.435 × 10 16
0.820 × 10 16
1.39 × 10 16
2.25 × 10 16
3.60 × 10 16
FIG. P39.76
6.00 × 10 16
11.6 × 10 16
54.8 × 10 16
FG IJ
HK
1 u
K c = 0.990K r , when
2 c
2
LM
= 0.990 M
NM
OP
− 1P , yielding u =
1 − bu c g
QP
1
2
Similarly,
K c = 0.950K r
when
u = 0.257 c
and
K c = 0.500K r
when
u = 0.786 c .
0.115 c .
2
I
− 1J .
JK
459
460
Relativity
ANSWERS TO EVEN PROBLEMS
P39.2
(a) 60.0 m s ; (b) 20.0 m s ; (c) 44.7 m s
P39.44
(a) 0.302 c ; (b) 4.00 fJ
P39.4
see the solution
P39.46
(a) 20.0 TW; (b) 3.50 × 10 7 electrons
P39.6
0.866 c
P39.48
P39.8
(a) 25.0 yr; (b) 15.0 yr; (c) 12.0 ly
(a) 0.960c; (b) 1.51 times greater as
measured by B.
P39.10
(a) 2.18 µs ; (b) The moon sees the planet
surface moving 649 m up toward it.
P39.50
3.18 × 10 −12 kg , not detectable
P39.52
6.71 × 10 8 kg
P39.54
1.02 MeV
P39.56
(a)
P39.12
(a)
cL p
2
c ∆t
2
+ L2p
; (b) 4.00 m s ;
(c) see the solution
v
= 1 − 1.12 × 10 −10 ; (b) 6.00 × 10 27 J ;
c
(c) $2.17 × 10 20
P39.14
v = 0.140 c
P39.16
5.45 yr, Goslo is older
P39.58
(a) 0.023 6 c ; (b) 6.18 × 10 −4 c
P39.18
11.3 kHz
P39.60
(a) 4.00 × 10 −14 kg ; (b) 1.60 × 10 −12
P39.20
(a) see the solution; (b) 0.050 4 c
P39.62
(a)
P39.22
(a) 0.943 c ; (b) 2.55 km
P39.64
see the solution
P39.24
B occurred 444 ns before A
P39.66
6.28 × 10 7 kg
P39.68
1.47 km
P39.70
(a)
P39.72
M=
P39.74
(a) Tau Ceti exploded 16.0 yr before the
Sun; (b) they exploded simultaneously
P39.76
see the solution, 0.115 c , 0.257 c , 0.786 c
P39.26
0.357 c
P39.28
0.893 c at 16.8° above the x ′ -axis
P39.30
(a) 0.141c ; (b) 0.436 c
P39.32
see the solution
P39.34
(a) 0.582 MeV ; (b) 2.45 MeV
P39.36
see the solution
P39.38
(a) 438 GJ ; (b) 438 GJ
P39.40
18.4 g cm 3
P39.42
(a) 0.467c; (b) 2.75 × 10 3 kg
3.65 MeV
; (b) v = 0.589 c
c2
2d
2d
; (b)
c+v
c
2m
3
c−v
c+v
4 − u2 c 2
1 − u2 c2
40
Introduction to Quantum Physics
CHAPTER OUTLINE
40.1
40.2
40.3
40.4
40.5
40.6
40.7
40.8
Blackbody Radiation and
Planck’s Hypothesis
The Photoelectric Effect
The Compton Effect
Photons and
Electromagnetic Waves
The Wave Properties of
Particles
The Quantum Particle
The Double-Slit Experiment
Revisited
The Uncertainty Principle
ANSWERS TO QUESTIONS
Q40.1
Planck made two new assumptions: (1) molecular energy is
quantized and (2) molecules emit or absorb energy in discrete
irreducible packets. These assumptions contradict the classical
idea of energy as continuously divisible. They also imply that
an atom must have a definite structure—it cannot just be a
soup of electrons orbiting the nucleus.
Q40.2
The first flaw is that the Rayleigh–Jeans law predicts that the
intensity of short wavelength radiation emitted by a blackbody
approaches infinity as the wavelength decreases. This is known
as the ultraviolet catastrophe. The second flaw is the prediction
much more power output from a black-body than is shown
experimentally. The intensity of radiation from the blackbody
is given by the area under the red I λ , T vs. λ curve in
Figure 40.5 in the text, not by the area under the blue curve.
Planck’s Law dealt with both of these issues and brought
the theory into agreement with the experimental data by
adding an exponential term to the denominator that depends
1
on . This both keeps the predicted intensity from
b g
λ
approaching infinity as the wavelength decreases and keeps
the area under the curve finite.
Q40.3
Our eyes are not able to detect all frequencies of energy. For example, all objects that are above 0 K
in temperature emit electromagnetic radiation in the infrared region. This describes everything in a
dark room. We are only able to see objects that emit or reflect electromagnetic radiation in the visible
portion of the spectrum.
Q40.4
Most stars radiate nearly as blackbodies. Vega has a higher surface temperature than Arcturus. Vega
radiates most intensely at shorter wavelengths.
Q40.5
No. The second metal may have a larger work function than the first, in which case the incident
photons may not have enough energy to eject photoelectrons.
Q40.6
Comparing Equation 40.9 with the slope-intercept form of the equation for a straight line, y = mx + b ,
we see that the slope in Figure 40.11 in the text is Planck’s constant h and that the y intercept is −φ ,
the negative of the work function. If a different metal were used, the slope would remain the same
but the work function would be different, Thus, data for different metals appear as parallel lines on
the graph.
461
462
Introduction to Quantum Physics
Q40.7
Wave theory predicts that the photoelectric effect should occur at any frequency, provided the light
intensity is high enough. However, as seen in the photoelectric experiments, the light must have a
sufficiently high frequency for the effect to occur.
Q40.8
The stopping voltage measures the kinetic energy of the most energetic photoelectrons. Each of
them has gotten its energy from a single photon. According to Planck’s E = hf , the photon energy
depends on the frequency of the light. The intensity controls only the number of photons reaching a
unit area in a unit time.
Q40.9
Let’s do some quick calculations and see: 1.62 MHz is the highest frequency in the commercial AM
band. From the relationship between the energy and the frequency, E = hf , the energy available
from such a wave would be 1.07 × 10 −27 J , or 6.68 neV. That is 9 orders of magnitude too small to
eject electrons from the metal. The only thing this student could gain from this experiment is a hefty
fine and a long jail term from the FCC. To get on the order of a few eV from this experiment, she
would have to broadcast at a minimum frequency of 250 Thz, which is in the infrared region.
Q40.10
No. If an electron breaks free from an atom absorbing a photon, we say the atom is ionized.
Ionization typically requires energy of several eV. As with the photoelectric effect in a solid metal,
the light must have a sufficiently high frequency for a photon energy that is large enough. The gas
can absorb energy from longer-wavelength light as it gains more internal energy of random motion
of whole molecules.
Q40.11
Ultraviolet light has shorter wavelength and higher photon energy than visible light.
Q40.12
(c) UV light has the highest frequency of the three, and hence each photon delivers more energy to a
skin cell. This explains why you can become sunburned on a cloudy day: clouds block visible light
and infrared, but not much ultraviolet. You usually do not become sunburned through window
glass, even though you can see the visible light from the Sun coming through the window, because
the glass absorbs much of the ultraviolet and reemits it as infrared.
Q40.13
The Compton effect describes the scattering of photons from electrons, while the photoelectric effect
predicts the ejection of electrons due to the absorption of photons by a material.
Q40.14
In developing a theory in accord with experimental evidence, Compton assumed that photons
exhibited clear particle-like behavior, and that both energy and momentum are conserved in
electron-photon interactions. Photons had previously been thought of as bits of waves.
Q40.15
The x-ray photon transfers some of its energy to the electron. Thus, its frequency must decrease.
Q40.16
A few photons would only give a few dots of exposure, apparently randomly scattered.
Q40.17
Light has both classical-wave and classical-particle characteristics. In single- and double-slit
experiments light behaves like a wave. In the photoelectric effect light behaves like a particle. Light
may be characterized as an electromagnetic wave with a particular wavelength or frequency, yet at
the same time light may be characterized as a stream of photons, each carrying a discrete energy, hf.
Since light displays both wave and particle characteristics, perhaps it would be fair to call light a
“wavicle”. It is customary to call a photon a quantum particle, different from a classical particle.
Chapter 40
463
Q40.18
An electron has both classical-wave and classical-particle characteristics. In single- and double-slit
diffraction and interference experiments, electrons behave like classical waves. An electron has mass
and charge. It carries kinetic energy and momentum in parcels of definite size, as classical particles
do. At the same time it has a particular wavelength and frequency. Since an electron displays
characteristics of both classical waves and classical particles, it is neither a classical wave nor a
classical particle. It is customary to call it a quantum particle, but another invented term, such as
“wavicle”, could serve equally well.
Q40.19
The discovery of electron diffraction by Davisson and Germer was a fundamental advance in our
understanding of the motion of material particles. Newton’s laws fail to properly describe the
motion of an object with small mass. It moves as a wave, not as a classical particle. Proceeding from
this recognition, the development of quantum mechanics made possible describing the motion of
electrons in atoms; understanding molecular structure and the behavior of matter at the atomic
scale, including electronics, photonics, and engineered materials; accounting for the motion of
nucleons in nuclei; and studying elementary particles.
Q40.20
If we set
p2
= q∆V , which is the same for both particles, then we see that the electron has the
2m
h
smaller momentum and therefore the longer wavelength λ =
.
p
F
GH
I
JK
Q40.21
Any object of macroscopic size—including a grain of dust—has an undetectably small wavelength
and does not exhibit quantum behavior.
Q40.22
A particle is represented by a wave packet of nonzero width. The width necessarily introduces
uncertainty in the position of the particle. The width of the wave packet can be reduced toward zero
only by adding waves of all possible wavelengths together. Doing this, however, results in loss of all
information about the momentum and, therefore, the speed of the particle.
Q40.23
The intensity of electron waves in some small region of space determines the probability that an
electron will be found in that region.
Q40.24
The wavelength of violet light is on the order of
Q40.25
The spacing between repeating structures on the surface of the feathers or scales is on the order of
1/2 the wavelength of light. An optical microscope would not have the resolution to see such fine
detail, while an electron microscope can. The electrons can have much shorter wavelength.
Q40.26
(a)
1
µm , while the de Broglie wavelength of an
2
electron can be 4 orders of magnitude smaller. Would your height be measured more precisely with
1
an unruled meter stick or with one engraved with divisions down to
mm ?
10
The slot is blacker than any black material or pigment. Any radiation going in through the
hole will be absorbed by the walls or the contents of the box, perhaps after several
reflections. Essentially none of that energy will come out through the hole again. Figure 40.1
in the text shows this effect if you imagine the beam getting weaker at each reflection.
continued on next page
464
Introduction to Quantum Physics
(b)
The open slots between the glowing tubes are brightest. When you look into a slot, you
receive direct radiation emitted by the wall on the far side of a cavity enclosed by the fixture;
and you also receive radiation that was emitted by other sections of the cavity wall and has
bounced around a few or many times before escaping through the slot. In Figure 40.1 in the
text, reverse all of the arrowheads and imagine the beam getting stronger at each reflection.
Then the figure shows the extra efficiency of a cavity radiator. Here is the conclusion of
Kirchhoff’s thermodynamic argument: ... energy radiated. A poor reflector—a good
absorber—avoids rising in temperature by being an efficient emitter. Its emissivity is equal
to its absorptivity: e = a . The slot in the box in part (a) of the question is a black body with
reflectivity zero and absorptivity 1, so it must also be the most efficient possible radiator, to
avoid rising in temperature above its surroundings in thermal equilibrium. Its emissivity in
Stefan’s law is 100% = 1 , higher than perhaps 0.9 for black paper, 0.1 for light-colored paint,
or 0.04 for shiny metal. Only in this way can the material objects underneath these different
surfaces maintain equal temperatures after they come to thermal equilibrium and continue
to exchange energy by electromagnetic radiation. By considering one blackbody facing
another, Kirchhoff proved logically that the material forming the walls of the cavity made
no difference to the radiation. By thinking about inserting color filters between two cavity
radiators, he proved that the spectral distribution of blackbody radiation must be a universal
function of wavelength, the same for all materials and depending only on the temperature.
Blackbody radiation is a fundamental connection between the matter and the energy that
physicists had previously studied separately.
SOLUTIONS TO PROBLEMS
Section 40.1
Blackbody Radiation and Planck’s Hypothesis
2.898 × 10 −3 m ⋅ K
= 5.18 × 10 3 K
560 × 10 −9 m
P40.1
T=
P40.2
(a)
λ max =
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
~
~ 10 −7 m
T
10 4 K
(b)
λ max ~
2.898 × 10 −3 m ⋅ K
~ 10 −10 m
10 7 K
P40.3
ultraviolet
γ − ray
Planck’s radiation law gives intensity-per-wavelength. Taking E to be the photon energy and n to be
the number of photons emitted each second, we multiply by area and wavelength range to have
energy-per-time leaving the hole:
P=
n=
=
n=
bλ
b
gb g
2
2π hc 2 λ 2 − λ 1 π d 2
1
g FH e
+λ2 2
5
2 hc
bλ
b
1 +λ 2
gk T
B
IK
−1
g
= En = nhf
E = hf =
where
2 hc
λ1 + λ 2
8π 2 cd 2 λ 2 − λ 1
P
=
4
2 hc b λ 1 + λ 2 g k BT
E
λ1 + λ 2 e
−1
g FH
b
IK
e
je
j e1.00 × 10
8π 2 3.00 × 10 8 m s 5.00 × 10 −5 m
e1 001 × 10
−9
5.90 × 10 16 s
e
j
e 3.84 − 1
j FGH e
m
4
e
2 6.626 ×10
−34
= 1.30 × 10 15 s
je
8
J⋅s 3.00 ×10 m s
2
j e1 001×10
−9
je
−9
m 1.38 × 10
j
m
−23
je
J K 7 .50 × 10 3 K
j
IJ
K
−1
Chapter 40
P40.4
e
je
b5 000 K g = 2.898 × 10
jb
(a)
P = eAσT 4 = 1 20.0 × 10 −4 m 2 5.67 × 10 −8 W m 2 ⋅ K 4 5 000 K
(b)
λ maxT = λ max
(c)
We compute:
e
−3
je
e
e
je
2π hc 2 A = 2π 6.626 × 10 −34 3.00 × 10 8
f
= 7.09 × 10 4 W
j
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 2.88 × 10 −6 m
−23
kBT
1.38 × 10
J K 5 000 K
2π hc 2 A
5
a
4
m ⋅ K ⇒ λ max = 580 nm
g
jb
The power per wavelength interval is P bλ g = AI bλ g =
λ
P 580 nm =
g
465
j e20.0 × 10 j = 7.50 × 10
2
−4
7.50 × 10 −19 J ⋅ m 4 s
e580 × 10
−9
j
m
5
b
b
g
exp 2.88 µm 0.580 µm − 1
g
exp hc λk B T − 1
=
−19
, and
J ⋅ m4 s
1.15 × 10 13 J m ⋅ s
e 4.973 − 1
= 7.99 × 10 10 W m
(d)–(i) The other values are computed similarly:
(d)
(e)
(f)
(c)
(g)
(h)
(i)
(j)
1.00 nm
7.96 × 10 1251
5.00 nm
576.5
2.40 × 10 250
400 nm
7.21
1347
580 nm
4.97
143.5
700 nm
4.12
60.4
10.0 cm
2.88 × 10
0.00289
−5
2.88 × 10 −5
bg
a5.44 + 7.99f × 10
10
a7.99 + 7.38f × 10
2
10
a
f
a
f
W m 580 − 400 × 10 −9 m
W m 700 − 580 × 10 −9 m
2
4
P = 2.13 × 10 W
so the power radiated as visible light is approximately 20 kW .
P = eAσT 4 , so
LM
P I
3.77 × 10 W
F
T=G
H eAσ JK = MM 1L4π e6.96 × 10 mj Oe5.67 × 10
PQ
MN MN
14
26
8
(b)
P ( λ ), W m
λ5
7.50 × 10 26 9.42 × 10 −1226
2.40 × 10 23 1.00 × 10 −227
7.32 × 10 13 5.44 × 10 10
1.15 × 10 13
7.99 × 10 10
12
4.46 × 10
7.38 × 10 10
7.50 × 10 −4 0.260
7.50 × 10 −14 2.60 × 10 −9
We approximate the area under the P λ versus λ curve, between 400 nm and 700 nm, as
two trapezoids:
+
(a)
e hc / λkBT − 1
1.00 mm 0.00288
P=
P40.5
2π hc 2 A
hc
λk B T
2882.6
λ
λ max =
2
−8
2
W m ⋅K
4
OP
P
j PPQ
14
= 5.75 × 10 3 K
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
=
= 5.04 × 10 −7 m = 504 nm
T
5.75 × 10 3 K
466
P40.6
Introduction to Quantum Physics
E = hf =
n=
P40.7
P40.8
hc
λ
e6.626 × 10
=
−34
je
J ⋅ s 3.00 × 10 8 m s
589.3 × 10
−9
m
j = 3.37 × 10
−19
J photon
10.0 J s
P
=
= 2.96 × 10 19 photons s
E 3.37 × 10 −19 J photon
e
je
e
je
e
je
jFGH 1.601.00× 10eV
jFGH 1.601.00× 10eV
jFGH 1.601.00× 10eV
IJ = 2.57 eV
JK
I = 1.28 × 10
J
JK
I = 1.91 × 10
J
JK
(a)
E = hf = 6.626 × 10 −34 J ⋅ s 620 × 10 12 s −1
(b)
E = hf = 6.626 × 10 −34 J ⋅ s 3.10 × 10 9 s −1
(c)
E = hf = 6.626 × 10 −34 J ⋅ s 46.0 × 10 6 s −1
(d)
λ=
c 3.00 × 10 8 m s
=
= 4.84 × 10 −7 m = 484 nm, visible light blue
12
f
620 × 10 Hz
λ=
c 3.00 × 10 8 m s
=
= 9.68 × 10 −2 m = 9.68 cm, radio wave
f
3.10 × 10 9 Hz
λ=
c 3.00 × 10 8 m s
=
= 6.52 m, radio wave
f
46.0 × 10 6 Hz
−19
−19
−19
−5
eV
−7
eV
a f
Energy of a single 500-nm photon:
Eγ = hf =
hc
λ
e6.626 × 10 J ⋅ sje3.00 × 10
e500 × 10 mj
−34
=
8
ms
−9
j = 3.98 × 10
−19
J.
The energy entering the eye each second
e
E = P∆t = IA∆t = 4.00 × 10 −11 W m 2
jLMN π4 e8.50 × 10
−3
j OPQa1.00 sf = 2.27 × 10
m
2
The number of photons required to yield this energy
n=
P40.9
2.27 × 10 −15 J
E
=
= 5.71 × 10 3 photons .
Eγ 3.98 × 10 −19 J photon
Each photon has an energy
This implies that there are
e
je
j
E = hf = 6.626 × 10 −34 99.7 × 10 6 = 6.61 × 10 −26 J .
150 × 10 3 J s
6.61 × 10
−26
J photon
= 2.27 × 10 30 photons s .
−15
J.
Chapter 40
P40.10
467
We take θ = 0.030 0 radians. Then the pendulum’s total energy is
a
f
E = b1.00 kg ge9.80 m s jb1.00 − 0.999 5g = 4.41 × 10
E = mgh = mg L − L cos θ
2
The frequency of oscillation is f =
ω
1
=
2π 2π
The energy is quantized,
E = nhf .
Therefore,
n=
−3
J
g
= 0.498 Hz .
L
4.41 × 10 −3 J
E
=
hf
6.626 × 10 −34 J ⋅ s 0.498 s −1
e
je
FIG. P40.10
j
= 1.34 × 10 31
P40.11
The radiation wavelength of λ ′ = 500 nm that is observed by observers on Earth is not the true
wavelength, λ, emitted by the star because of the Doppler effect. The true wavelength is related to
the observed wavelength using:
c
λ′
=
c
λ
b g:
1 + bv cg
1− v c
λ = λ′
P40.12
2.898 × 10 −3 m ⋅ K
λ max
T=
:
b g
Planck’s radiation law is
I λ, T =
Using the series expansion
ex = 1 + x +
Planck’s law reduces to
I λ, T =
b g
a
a
f
f
1 − 0.280
= 375 nm .
1 + 0.280
λ maxT = 2.898 × 10 −3 m ⋅ K :
The temperature of the star is given by
T=
b g = a500 nmf
1 + bv c g
1− v c
e
2.898 × 10 −3 m ⋅ K
= 7.73 × 10 3 K .
−9
375 × 10
2π hc 2
λ5 e hc λk BT − 1
j
.
x 2 x3
+
+ ….
2! 3!
2π hc 2
5
λ
b1 + hc λk T + …g − 1
B
≈
2π hc 2
5
b
λ hc λk B T
g
=
2π ck B T
λ4
which is the Rayleigh-Jeans law, for very long wavelengths.
Section 40.2
P40.13
(a)
The Photoelectric Effect
λc
e6.626 × 10 J ⋅ sje3.00 × 10 m sj =
=
=
φ
a4.20 eVfe1.60 × 10 J eVj
(b)
λ
8
−19
fc =
hc
−34
hc
c
λc
=
3.00 × 10 8 m s
296 × 10 −9 m
= 1.01 × 10 15 Hz
e6.626 × 10 je3.00 × 10 j = a4.20 eVfe1.60 × 10
−34
= φ + e∆VS :
Therefore,
296 nm
180 × 10 −9
∆VS = 2.71 V
8
−19
j e
j
J eV + 1.60 × 10 −19 ∆VS
468
P40.14
P40.15
Introduction to Quantum Physics
K max =
1
1
2
mv max
= 9.11 × 10 −31 4.60 × 10 5
2
2
(a)
φ = E − K max =
(b)
fc =
(a)
λc =
e
je
j
2
= 9.64 × 10 −20 J = 0.602 eV
1 240 eV ⋅ nm
− 0.602 nm = 1.38 eV
625 nm
I
JK
F
GH
φ
1.38 eV
1.60 × 10 −19 J
=
= 3.34 × 10 14 Hz
−34
1
eV
h 6.626 × 10
J⋅s
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 540 nm
a2.30 eVfe1.60 × 10 J eVj
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 318 nm
=
a3.90 eVfe1.60 × 10 J eVj
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 276 nm
=
a4.50 eVfe1.60 × 10 J eVj
−34
hc
φ
Li:
λc =
Be:
λc
Hg:
λc
8
−19
−34
8
−19
−34
8
−19
λ < λ c for photo current. Thus, only lithium will exhibit the photoelectric effect.
(b)
For lithium,
hc
λ
= φ + K max
e6.626 × 10
−34
je
J ⋅ s 3.00 × 10 8 m s
−9
400 × 10 m
= 1.29 × 10 −19 J = 0.808 eV
K max
P40.16
From condition (i),
and
b g
g = b∆V g + 1.48 V .
hf = e ∆VS1 + φ 1
b ∆V
S1
j = a2.30 eVfe1.60 × 10 j + K
−19
b
g
hf = e ∆VS 2 + φ 2
S2
Then
φ 2 − φ 1 = 1.48 eV .
From condition (ii),
hf c 1 = φ 1 = 0.600 hf c 2 = 0.600φ 2
φ 2 − 0.600φ 2 = 1.48 eV
φ 2 = 3.70 eV
P40.17
(a)
e∆VS =
(b)
e∆VS =
hc
λ
hc
λ
−φ →φ =
−φ =
φ 1 = 2.22 eV .
1 240 nm ⋅ eV
− 0.376 eV = 1.90 eV
546.1 nm
1 240 nm ⋅ eV
− 1.90 eV → ∆VS = 0. 216 V
587.5 nm
max
Chapter 40
P40.18
The energy needed is
E = 1.00 eV = 1.60 × 10 −19 J .
The energy absorbed in time interval ∆t is
E = P∆t = IA∆t
∆t =
so
E
1.60 × 10 −19 J
=
IA
500 J s ⋅ m 2 π 2.82 × 10 −15 m
= 1.28 × 10
O
j QP
jLNM e
e
7
2
469
s = 148 days .
The gross failure of the classical theory of the photoelectric effect contrasts with the success of
quantum mechanics.
P40.19
Ultraviolet photons will be absorbed to knock electrons out of the sphere with maximum kinetic
energy K max = hf − φ ,
or
K max
e6.626 × 10
=
−34
je
J ⋅ s 3.00 × 10 8 m s
200 × 10
−9
j F 1.00 eV I − 4.70 eV = 1.51 eV .
GH 1.60 × 10 J JK
−19
m
The sphere is left with positive charge and so with positive potential relative to V = 0 at r = ∞ . As its
potential approaches 1.51 V, no further electrons will be able to escape, but will fall back onto the
sphere. Its charge is then given by
V=
P40.20
k eQ
r
Q=
or
jb
e
g
5.00 × 10 −2 m 1.51 N ⋅ m C
rV
=
= 8.41 × 10 −12 C .
ke
8.99 × 10 9 N ⋅ m 2 C 2
(a)
By having the photon source move toward the metal, the incident photons are Doppler
shifted to higher frequencies, and hence, higher energy.
(b)
If v = 0.280 c ,
f′= f
Therefore,
φ = 6.626 × 10 −34 J ⋅ s 9.33 × 10 14 Hz = 6.18 × 10 −19 J = 3.87 eV .
At v = 0.900 c ,
f = 3.05 × 10 15 Hz
(c)
1+v c
= 7.00 × 10 14
1−v c
e
e
e
j
1.28
= 9.33 × 10 14 Hz .
0.720
je
j
je
and K max = hf − φ = 6.626 × 10 −34 J ⋅ s 3.05 × 10 15 Hz
Section 40.3
P40.21
E=
p=
jFGH 1.601.00× 10eV J IJK − 3.87 eV =
The Compton Effect
hc
λ
h
λ
=
=
e6.626 × 10
−34
je
J ⋅ s 3.00 × 10 8 m s
700 × 10
−9
m
j = 2.84 × 10
6.626 × 10 −34 J ⋅ s
= 9.47 × 10 −28 kg ⋅ m s
−9
700 × 10 m
−19
J = 1.78 eV
−19
8.78 eV .
470
P40.22
Introduction to Quantum Physics
(a)
∆λ =
(b)
E0 =
h
6.626 × 10 −34
1 − cos 37.0° = 4.88 × 10 −13 m
1 − cos θ : ∆λ =
mec
9.11 × 10 −31 3.00 × 10 8
a
hc
λ0
f
e
e
:
f
6.626 × 10 je3.00 × 10
j e
λ
−34
je
8
ms
j
0
λ ′ = λ 0 + ∆λ = 4.63 × 10 −12 m
e
je
j
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 4.30 × 10 −14 J = 268 keV
λ′
4.63 × 10 −12 m
K e = E0 − E ′ = 300 keV − 268.5 keV = 31.5 keV
(c)
With K e = E ′ , K e = E0 − E ′ gives
E0
hc
and λ ′ =
E′
2
a
E ′ = E0 − E ′
λ′ =
λ ′ = λ 0 + λ C 1 − cos θ
f
hc
hc
=2
= 2λ 0
E0 2
E0
a
2 λ 0 = λ 0 + λ C 1 − cos θ
1 − cos θ =
P40.24
a
300 × 10 3 eV 1.60 × 10 −19 J eV =
E′ =
E′ =
j
λ 0 = 4.14 × 10 −12 m
and
P40.23
je
f
λ 0 0.001 60
=
λ C 0.002 43
θ = 70.0°
This is Compton scattering through 180°:
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 11.3 keV
λ
e0.110 × 10 mje1.60 × 10 J eVj
h
∆λ =
a1 − cosθ f = e2.43 × 10 mja1 − cos 180°f = 4.86 × 10
m c
hc
E0 =
−34
=
8
−9
0
−19
−12
−12
m
FIG. P40.24
e
hc
= 10.8 keV .
λ′
h
h
By conservation of momentum for the photon-electron system,
i=
−i + pe i
λ0
λ′
λ ′ = λ 0 + ∆λ = 0.115 nm so
E′ =
e j
FG 1
Hλ
pe = h
and
e
p e = 6.626 × 10
−34
F e3.00 × 10
J ⋅ sjG
GH 1.60 × 10
8
j IJ FG 1
J eV JK H 0.110 × 10
ms c
−19
−9
m
+
−9
1
λ′
IJ
K
IJ =
mK
22.1 keV
.
c
11.3 keV = 10.8 keV + K e
By conservation of system energy,
K e = 478 eV .
so that
Check:
0
1
0.115 × 10
+
em c
E 2 = p 2 c 2 + m e2 c 4 or
a511 keV + 0.478 keVf = a22.1 keVf + a511 keVf
2
2.62 × 10 11 = 2.62 × 10 11
2
e
2
2
+ Ke
j = bpcg + em c j
2
2
e
2 2
Chapter 40
P40.25
(a)
Conservation of momentum in the x direction gives:
pγ = pγ′ cos θ + p e cos φ
or since θ = φ ,
h
h
= pe +
cos θ .
λ0
λ′
Conservation of momentum in the y direction gives:
0 = pγ′ sin θ − p e sin θ ,
which (neglecting the trivial solution θ = 0 ) gives:
p e = pγ′ =
IJ
K
FG
H
h
2h
=
cos θ , or
λ0 λ′
Substituting [2] into [1] gives:
[2]
λ ′ = 2 λ 0 cos θ .
[3]
a
h
1 − cos θ
mec
f
giving
2 λ 0 cos θ − λ 0 =
or
2 cos θ − 1 =
hc 1
1 − cos θ .
λ 0 mec 2
2 cos θ − 1 =
FE
GH m c
hc
λ0
, this may be written as:
or cos θ =
m e c 2 + Eγ
2
2m e c + Eγ
f
a
f
2
I a1 − cos θ f
JK
F 2 + E I cos θ = 1 + E
GH m c JK
m c
γ
=
γ
2
e
2
0.511 MeV + 0.880 MeV
= 0.732 so that θ = φ = 43.0° .
1.02 MeV + 0.880 MeV
Using Equation (3): Eγ′ =
Then,
a
h
1 − cos θ
mec
γ
e
e
(c)
h
.
λ′
λ′ − λ0 =
which reduces to:
(b)
[1]
Then the Compton equation is
Since Eγ =
471
pγ′ =
Eγ
hc
hc
0.880 MeV
=
=
=
= 0.602 MeV = 602 keV .
λ ′ λ 0 2 cos θ
2 cos 43.0°
2 cos θ
a
Eγ′
c
=
From Equation (2), p e = pγ′ =
f
0.602 MeV
= 3.21 × 10 −22 kg ⋅ m s .
c
0.602 MeV
= 3.21 × 10 −22 kg ⋅ m s .
c
From energy conservation: K e = Eγ − Eγ′ = 0.880 MeV − 0.602 MeV = 0.278 MeV = 278 keV .
472
P40.26
Introduction to Quantum Physics
The energy of the incident photon is E0 = pγ c =
(a)
hc
.
λ0
Conserving momentum in the x direction gives
p λ = p e cos φ + pγ′ cos θ , or since φ = θ ,
E0
= p e + pγ′ cos θ .
c
d
i
[1]
Conserving momentum in the y direction (with φ = 0 ) yields
0 = pγ′ sin θ − p e sin θ , or p e = pγ′ =
h
.
λ′
[2]
Substituting Equation [2] into Equation [1] gives
FG
H
IJ
K
E0
2 hc
h
h
=
+
cos θ .
cos θ , or λ ′ =
λ′ λ′
E0
c
By the Compton equation, λ ′ − λ 0 =
[3]
a
f
a
2 hc
2 hc
h
=
1 − cos θ
cos θ −
E0
E0
mec
h
1 − cos θ ,
mec
e 2m c
which reduces to
e
2
j
+ E0 cos θ = m e c 2 + E0 .
φ = θ = cos −1
Thus,
F m c +E I
GH 2m c + E JK
e
e
(b)
λ′ =
2 hc
2 hc m e c 2 + E0
cos θ =
.
E0
E0 2 m e c 2 + E0
Therefore,
Eγ′ =
hc
hc
=
2
λ ′ 2 hc E0 m e c + E0
pγ′ =
b
Eγ′
c
=
ge
F
GH
E0 2 m e c 2 + E0
2 c m e c 2 + E0
From conservation of energy, K e = E0 − Eγ′ = E0 −
or
Ke =
I
JK
j e 2m c
e
2
+ E0
j
=
F
GH
.
F
GH
E0 2 m e c 2 + E0
2 m e c 2 + E0
F
GH
I
JK
F
GH
2
0
.
0
E 0 2 m e c 2 + E0
2 m e c 2 + E0
I
JK
E 0 2 m e c 2 + 2 E0 − 2 m e c 2 − E 0
E02
=
2
2
m e c + E0
2 m e c 2 + E0
Finally, from Equation (2), p e = pγ′ =
2
I
JK
From Equation [3],
and
(c)
F
GH
f
E0 2 m e c 2 + E 0
2 c m e c 2 + E0
I
JK
.
e
j
.
I
JK
,
Chapter 40
*P40.27
The electron’s kinetic energy is
K=
1
1
mv 2 = 9.11 × 10 −31 kg 2.18 × 10 6 m s
2
2
e
j
2
= 2.16 × 10 −18 J .
This is the energy lost by the photon, hf 0 − hf ′
hc hc
−
= 2.16 × 10 −18 J. We also have
λ0 λ′
λ′ − λ0 =
h
6.63 × 10 −34 Js s
1 − cos θ =
1 − cos 17.4°
me c
9.11 × 10 −31 kg 3 × 10 8 m
a
f
e
j
a
f
λ ′ = λ 0 + 1.11 × 10 −13 m
(a)
Combining the equations by substitution,
1
λ0
−
1
2.16 × 10 −18 J s
=
= 1.09 × 10 7 m
λ 0 + 0.111 pm 6.63 × 10 −34 Js 3 × 10 8 m
e
j
λ 0 + 0.111 pm − λ 0
= 1.09 × 10 7 m
λ20 + λ 0 0.111 pm
b
e
g
j
0.111 pm = 1.09 × 10 7 m λ20 + 1.21 × 10 −6 λ 0
1.09 × 10 7 λ20 + 1.21 × 10 −6 mλ 0 − 1.11 × 10 −13 m 2 = 0
λ0 =
−1. 21 × 10 −6 m ±
e1.21 × 10
j − 4e1.09 × 10 je−1.11 × 10
2e1.09 × 10 j
−6
m
2
7
7
only the positive answer is physical: λ 0 = 1.01 × 10 −10 m .
(b)
Then λ ′ = 1.01 × 10 −10 m + 1.11 × 10 −13 m = 1.01 × 10 −10 m.
Conservation of momentum in the transverse direction:
0=
h
sin θ − γm e v sin φ
λ′
e
j
9.11 × 10 −31 kg 2.18 × 10 6 m s sin φ
6.63 × 10 −34 J ⋅ s
°
=
sin
17
.
4
2
1.01 × 10 −10 m
1 − 2.18 × 10 6 3 × 10 8
e
1.96 × 10 −24 = 1.99 × 10 −24 sin φ
φ = 81.1°
j
−13
m2
j
473
474
P40.28
Introduction to Quantum Physics
(a)
Thanks to Compton we have four equations in the unknowns φ, v, and λ ′ :
hc hc
=
+ γ me c 2 − me c 2
λ0 λ′
h
λ0
0=
=
(energy conservation)
h
cos 2φ + γ m e v cos φ
λ′
h
sin 2φ − γ m e v sin φ
λ′
λ′ − λ0 =
b
h
1 − cos 2φ
mec
g
[1]
(momentum in
x direction)
[2]
(momentum in
y direction)
[3]
(Compton equation).
[4]
2h
cos φ .
λ′
Using sin 2φ = 2 sin φ cos φ in Equation [3] gives γ m e v =
Substituting this into Equation [2] and using cos 2φ = 2 cos 2 φ − 1 yields
h
h
h
2h
=
2 cos 2 φ − 1 +
cos 2 φ =
4 cos 2 φ − 1 ,
λ0 λ′
λ′
λ′
e
or
j
e
j
λ ′ = 4λ 0 cos 2 φ − λ 0 .
[5]
Substituting the last result into the Compton equation gives
4λ 0 cos 2 φ − 2 λ 0 =
h
hc
1 − 2 cos 2 φ − 1 = 2
1 − cos 2 φ .
mec
mec 2
e
With the substitution λ 0 =
cos 2 φ =
For x =
m e c 2 + E0
2
2 m e c + E0
j
e
j
hc
, this reduces to
E0
=
E
1+x
where x ≡ 0 2 .
2+x
me c
1+ x
= 33.0° .
2+x
0.700 MeV
= 1.37 , this gives φ = cos −1
0.511 MeV
FIG. P40.28(a)
(b)
e
j
LM FG 1 + x IJ − 1OP = λ FG 2 + 3x IJ .
N H 2 + xK Q H 2 + x K
From Equation [5], λ ′ = λ 0 4 cos 2 φ − 1 = λ 0 4
0
Then, Equation [1] becomes
FG 2 + x IJ + γ m c − m c or E − E FG 2 + x IJ + 1 = γ .
λ
λ H 2 + 3x K
m c
m c H 2 + 3x K
F 2 + x IJ , and with x = 1.37 we get γ = 1.614 .
Thus, γ = 1 + x − xG
H 2 + 3x K
hc
=
0
Therefore,
hc
0
e
2
e
2
0
e
2
0
e
2
v
= 1 − γ −2 = 1 − 0.384 = 0.785 or v = 0.785 c .
c
Chapter 40
P40.29
λ′ − λ =
a
h
1 − cos θ
me c
475
f
a
f
h
1 − cos π − θ
mec
h
h
h
h
λ ′′ − λ =
−
−
cos π − θ +
cos θ
me c me c
mec mec
λ ′′ − λ ′ =
a
a
f
f
Now cos π − θ = − cos θ , so λ ′′ − λ = 2
P40.30
h
= 0.004 86 nm .
mec
FIG. P40.29
Maximum energy loss appears as maximum increase in wavelength, which occurs for scattering
2h
h
=
angle 180°. Then ∆λ = 1 − cos 180°
where m is the mass of the target particle. The
mc
mc
fractional energy loss is
fFGH IJK
a
2 h mc
E0 − E ′ hc λ 0 − hc λ ′ λ ′ − λ 0
∆λ
=
=
.
=
=
λ′
λ 0 + ∆λ λ 0 + 2 h mc
E0
hc λ 0
Further, λ 0 =
(a)
2 h mc
E − E′
2 E0
hc
=
=
, so 0
.
2
E0
E0
hc E0 + 2 h mc mc + 2E0
For scattering from a free electron, mc 2 = 0.511 MeV , so
a
f
2 0.511 MeV
E0 − E ′
=
= 0.667 .
E0
0.511 MeV + 2 0.511 MeV
(b)
a
f
For scattering from a free proton, mc 2 = 938 MeV , and
a
f
2 0.511 MeV
E0 − E ′
=
= 0.001 09 .
E0
938 MeV + 2 0.511 MeV
Section 40.4
*P40.31
a
f
Photons and Electromagnetic Waves
With photon energy 10.0 eV = hf
f=
e
10.0 1.6 × 10 −19 J
6.63 × 10
−34
J⋅s
j = 2.41 × 10
15
Hz .
Any electromagnetic wave with frequency higher than 2.41 × 10 15 Hz counts as ionizing radiation.
This includes far ultraviolet light, x-rays, and gamma rays.
476
*P40.32
Introduction to Quantum Physics
hc
e
Sav =
=
−9
je
beam is 2 × 10 18
af
j
0.628 W 4
P
=
π r 2 π 1.75 × 10 −3 m
vector I = Sav =
j = 3.14 × 10
−19
J . The power carried by the
λ
633 × 10 m
photons s 3.14 × 10 −19 J photon = 0.628 W . Its intensity is the average Poynting
The photon energy is E =
(a)
e
6.63 × 10 −34 J ⋅ s 3 × 10 8 m s
e
1
µ0
Erms Brms sin 90° =
b
Emax = 2 µ 0 cSav
g
12
j
2
= 2.61 × 10 5 W m 2 .
1 Emax Bmax
µ0
2
2
ee
. Also Emax = Bmax c . So Sav =
je
je
2
Emax
.
2µ 0 c
= 2 4π × 10 −7 Tm A 3 × 10 8 m s 2.61 × 10 5 W m 2
jj
12
= 1. 40 × 10 4 N C
Bmax =
(b)
1.40 × 10 4 N C
8
3 × 10 m s
P
E
. The beam transports momentum at the rate . It
c
c
imparts momentum to a perfectly reflecting surface at the rate
2 0.628 W
2P
= force =
= 4.19 × 10 −9 N .
c
3 × 10 8 m s
Each photon carries momentum
a
(c)
Section 40.5
P40.33
λ=
P40.34
(a)
f
The block of ice absorbs energy mL = P∆t melting
P∆t 0.628 W 1.5 × 3 600 s
m=
=
= 1.02 × 10 −2 kg .
L
3.33 × 10 5 J kg
b
g
The Wave Properties of Particles
h
h
6.626 × 10 −34 J ⋅ s
=
=
= 3.97 × 10 −13 m
p mv 1.67 × 10 −27 kg 1.00 × 10 6 m s
e
je
p2
= 50.0 1.60 × 10 −19 J
2m
p = 3.81 × 10 −24 kg ⋅ m s
a fe
λ=
(b)
= 4.68 × 10 −5 T
j
j
h
= 0.174 nm
p
p2
= 50.0 × 10 3 1.60 × 10 −19 J
2m
p = 1. 20 × 10 −22 kg ⋅ m s
e
je
j
h
= 5.49 × 10 −12 m
p
The relativistic answer is slightly more precise:
λ=
λ=
h
=
p
LMemc
N
hc
2
+K
j
2
− m2c4
OP
Q
12
= 5.37 × 10 −12 m .
477
Chapter 40
P40.35
(a)
Electron:
λ=
and
λ=
h
p
K=
and
h
2m e K
=
p2
m2v2
1
=
mev 2 = e
2
2m e
2m e
p = 2m e K
so
6.626 × 10 −34 J ⋅ s
ja fe
e
2 9.11 × 10 −31 kg 3.00 1.60 × 10 −19 J
j
λ = 7.09 × 10 −10 m = 0.709 nm .
(b)
P40.36
(a)
Photon:
λ=
c
f
and
λ=
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 4.14 × 10 −7 m = 414 nm .
−19
E
J
3 1.60 × 10
E = hf
and
e
je
e
f=
so
E
h
j
j
h
h
=
where the
p
2mK
kinetic energy K is in joules. If the neutron kinetic energy K n is given in electron volts, its
The wavelength of a non-relativistic particle of mass m is given by λ =
e
j
kinetic energy in joules is K = 1.60 × 10 −19 J eV K n and the equation for the wavelength
becomes
λ=
h
2mK
=
6.626 × 10 −34 J ⋅ s
e
je
j
2 1.67 × 10 −27 kg 1.60 × 10 −19 J eV K n
=
2.87 × 10 −11
Kn
m
where K n is expressed in electron volts.
P40.37
(b)
If K n = 1.00 keV = 1 000 eV , then
(a)
λ ~ 10 −14 m or less.
p=
λ=
h
λ
~
2.87 × 10 −11
6.6 × 10 −34 J ⋅ s
= 10 −19 kg ⋅ m s or more.
−14
m
10
The energy of the electron is E = p 2 c 2 + m e2 c 4 ~
(b)
m = 9.07 × 10 −13 m = 907 fm .
1 000
e10 j e3 × 10 j + e9 × 10 j e3 × 10 j
−19 2
8 2
−31 2
or
E ~ 10 −11 J ~ 10 8 eV or more,
so that
K = E − m e c 2 ~ 10 8 eV − 0.5 × 10 6 eV ~ 10 8 eV or more.
e
8 4
j
The electric potential energy of the electron-nucleus system would be
e
je
ja f
9 × 10 9 N ⋅ m 2 C 2 10 −19 C − e
k e q1 q 2
~
~ −10 5 eV .
Ue =
−14
r
10
m
With its K + U e >> 0 ,
the electron would immediately escape the nucleus .
478
P40.38
Introduction to Quantum Physics
From the condition for Bragg reflection,
mλ = 2d sin θ = 2d cos
FG φ IJ
H 2K
FG φ IJ .
H 2K
d = a sin
But
where a is the lattice spacing.
λ=
h
=
p
FG φ IJ cosFG φ IJ = a sin φ
H 2K H 2K
λ = 2 a sin
Thus, with m = 1,
h
6.626 × 10 −34 J ⋅ s
λ=
2m e K
e
2 9.11 × 10
Therefore, the lattice spacing is a =
P40.39
(a)
je
−31
kg 54.0 × 1.60 × 10
−19
J
j
= 1.67 × 10 −10 m .
λ
1.67 × 10 −10 m
=
= 2.18 × 10 −10 = 0.218 nm .
sin φ
sin 50.0°
E2 = p 2 c 2 + m 2c 4
with E = hf ,
(b)
FIG. P40.38
p=
h
λ
so
h2 f 2 =
For a photon
f 1
= .
c λ
The third term
h2 c 2
λ
2
+
h2 c 2
λ2C
and
mc =
and
FG f IJ
H cK
h
λC
2
=
1
1
+
λ2 λ2C
(Eq. 1).
1
in Equation 1 for electrons and other massive particles shows that
λC
they will always have a different frequency from photons of the same wavelength .
*P40.40
h
h
=
. For the photon (which we represent as γ),
p γ mv
λγ
chγ mv
γ v
c ch ch
ch
=
=
=
=
. Then the ratio is
.
E = K and λ γ = =
2
2
λm
f E K
γ − 1 mc
γ − 1 mc h γ − 1 c
b g
For the massive particle, K = γ − 1 mc 2 and λ m =
(a)
(b)
λγ
λm
λγ
λm
=
=
b g
1a0.9f
1 − 0.9 2
LMFH1
N
b g
IK OP =
Q
1 − 0.9 2 − 1
a
1 0.001
L
1 − a0.001f MF 1
NH
2
1.60
f
O=
1 − a0.001f I − 1P
K Q
2
2.00 × 10 3
(c)
As
λγ
γ
v
→ 1, γ → ∞ and γ − 1 becomes nearly equal to γ. Then
→ 1= 1 .
λm
γ
c
(d)
As
v
v2
→ 0, 1− 2
c
c
F
GH
I
JK
−1 2
FG 1 IJ v
H 2K c
−1≈1− −
2
2
−1 =
λγ
vc
2c
1 v2
→1
=
→ ∞ .
and
2
2
2
λm
2 c
v
12 v c
b ge
j
Chapter 40
P40.41
λ=
h
p
p=
(a)
h
λ
=
479
6.626 × 10 −34 J ⋅ s
= 6.63 × 10 −23 kg ⋅ m s
1.00 × 10 −11 m
e
j
2
6.63 × 10 −23
p2
=
Ke =
J = 15.1 keV
2m e 2 9.11 × 10 −31
electrons:
e
j
The relativistic answer is more precisely correct:
K e = p 2 c 2 + m e2 c 4 − m e c 2 = 14.9 keV .
P40.42
e
je
photons:
(a)
The wavelength of the student is λ =
then we need
Using ∆t =
h
h
=
. If w is the width of the diffracting aperture,
p mv
FG h IJ
H mv K
F 6.626 × 10 J ⋅ s I =
h
= 10.0G
v ≤ 10.0
mw
H b80.0 kg ga0.750 mf JK
w ≤ 10.0 λ = 10.0
−34
so that
(b)
j
Eγ = pc = 6.63 × 10 −23 3.00 × 10 8 = 124 keV
(b)
d
we get:
v
∆t ≥
1.10 × 10 −34 m s .
0.150 m
= 1.36 × 10 33 s .
−34
1.10 × 10
ms
No . The minimum time to pass through the door is over 10 15 times the age of the
(c)
Universe.
Section 40.6
*P40.43
The Quantum Particle
E=K =
h
1
.
mu 2 = hf and λ =
mu
2
v phase = fλ =
mu 2 h
u
=
= v phase .
2 h mu
2
This is different from the speed u at which the particle transports mass, energy, and momentum.
*P40.44
As a bonus, we begin by proving that the phase speed v p =
vp =
ω
is not the speed of the particle.
k
F
GH
I
JK
p2c2 + m2c4
γ 2m 2 v2 c 2 + m 2 c 4
ω
c2
c2
v2
c2
c2
=
=
= c 1+ 2 2 = c 1+ 2 1− 2 = c 1+ 2 −1 =
k
v
γ mv
γ v
v
c
v
γ 2m2 v2
In fact, the phase speed is larger than the speed of light. A point of constant phase in the wave
function carries no mass, no energy, and no information.
Now for the group speed:
continued on next page
480
Introduction to Quantum Physics
vg =
dω d ω dE d
=
=
=
m2c4 + p2c2
dk
d k dp dp
vg =
1 2 4
m c + p2c2
2
e
p2c4
j e0 + 2 pc j =
−1 2
2
p2c2 + m2c4
e
j
v2 1 − v2 c2
γ 2m 2 v 2
=c 2
=c
vg = c 2 2 2
γ m v + m2c2
v 1 − v2 c2 + c2
e
j
e
v2 1 − v2 c2
ev
2
2
+c −v
2
j
j e1 − v c j
2
2
=v
It is this speed at which mass, energy, and momentum are transported.
Section 40.7
P40.45
The Double-Slit Experiment Revisited
h
6.626 × 10 −34 J ⋅ s
=
= 9.92 × 10 −7 m
mv 1.67 × 10 −27 kg 0.400 m s
(a)
λ=
(b)
For destructive interference in a multiple-slit experiment, d sin θ = m +
jb
e
FG
H
the first minimum.
FG λ IJ = 0.028 4°
H 2d K
y = L tan θ = a10.0 mfbtan 0.028 4°g =
so
P40.46
y
= tan θ
L
4.96 mm .
We cannot say the neutron passed through one slit. We can only say it passed through the slits.
Consider the first bright band away from the center:
e6.00 × 10
d sin θ = mλ
λ=
h
so
me v
me v =
K=
and
h2
∆V =
2 em e λ2
*P40.47
IJ
K
1
λ , with m = 0 for
2
θ = sin −1
Then,
(c)
g
−8
j FGH
m sin tan −1
h
LM 0.400 OPIJ = a1fλ = 1.20 × 10
N 200 QK
m
λ
m2v2
1
h2
=
= e∆V
mev 2 = e
2
2m e
2 m e λ2
∆V =
−10
e
2 1.60 × 10 −19
e6.626 × 10
C je9.11 × 10
−34
−31
j
kg je1.20 × 10
J⋅s
2
−10
j
m
2
= 105 V .
We find the speed of each electron from energy conservation in the firing process:
1
0 = K f + U f = mv 2 − eV
2
v=
2 eV
=
m
9.11 × 10
The time of flight is ∆t =
apart is I =
a
2 × 1.6 × 10 −19 C 45 V
−31
kg
f = 3.98 × 10
6
ms
∆x
0.28 m
=
= 7.04 × 10 −8 s . The current when electrons are 28 cm
6
v
3.98 × 10 m s
q
e 1.6 × 10 −19 C
=
=
= 2.27 × 10 −12 A .
t ∆t 7.04 × 10 −8 s
Chapter 40
Section 40.8
P40.48
481
The Uncertainty Principle
∆v ≥
h
2π J ⋅ s
=
= 0. 250 m s .
4π m∆x 4π 2.00 kg 1.00 m
(a)
∆p∆x = m∆v∆x ≥
(b)
The duck might move by 0. 25 m s 5 s = 1.25 m . With original position uncertainty of
so
2
b
f
ga
ga f
b
1.00 m, we can think of ∆x growing to 1.00 m + 1.25 m = 2.25 m .
P40.49
J⋅s
6.626 × 10
h
=
= 1.16 mm .
−
32
4π ∆p 4π 4.56 × 10
kg ⋅ m s
b
e
gb
j
ge
j
h
= 5.28 × 10 −32 m .
4π ∆p
∆x =
∆y ∆p y
h
=
and
d∆p y ≥
.
x
px
4π
Eliminate ∆p y and solve for x.
b g
d
x = 4π p x ∆y :
h
e
x = 4π 1.00 × 10
The answer, x = 3.79 × 10
28
−3
jb
ge
kg 100 m s 1.00 × 10
−2
e2.00 × 10
mj
e6.626 × 10
j
−3
m
−34
J⋅s
j
m , is 190 times greater than the diameter of the Universe!
With ∆x = 2 × 10 −15 m m, the uncertainty principle requires ∆p x ≥
= 2.6 × 10 −20 kg ⋅ m s .
2 ∆x
The average momentum of the particle bound in a stationary nucleus is zero. The uncertainty in
momentum measures the root-mean-square momentum, so we take p rms ≈ 3 × 10 −20 kg ⋅ m s . For an
electron, the non-relativistic approximation p = m e v would predict v ≈ 3 × 10 10 m s , while v cannot
be greater than c.
Thus, a better solution would be
LMe
N
E = me c 2
γ ≈ 110 =
For a proton, v =
*P40.52
j
∆p = m∆v = 0.020 0 kg 500 m s 1.00 × 10 −4 = 1.00 × 10 −3 kg ⋅ m s
For the bullet,
P40.51
ge
−34
∆x =
P40.50
jb
e
∆p = m e ∆v = 9.11 × 10 −31 kg 500 m s 1.00 × 10 −4 = 4.56 × 10 −32 kg ⋅ m s
For the electron,
j + bpcg OPQ
2
2
12
1
≈ 56 MeV = γ m e c 2
so
1 − v2 c2
v ≈ 0.999 96 c .
p
gives v = 1.8 × 10 7 m s, less than one-tenth the speed of light.
m
a f
2
mv
p2
1
mv 2 =
=
2
2m
2m
(a)
K=
(b)
To find the minimum kinetic energy, think of the minimum momentum uncertainty, and
maximum position uncertainty of 10 −15 m = ∆x . We model the proton as moving along a
. The average momentum is zero. The average squared
, ∆p =
2 ∆x
2
momentum is equal to the squared uncertainty:
straight line with ∆p∆x =
b g
e
j
2
2
2
2
6.63 × 10 −34 J ⋅ s
∆p
p2
= 8.33 × 10 −13 J
K=
=
=
=
=
2
2
2
2
−
−
2
15
27
2m
2m
∆
∆
4 x 2m 32π
x m 32π 10
m 1.67 × 10
kg
= 5.21 MeV
a f
a f
e
j
482
P40.53
Introduction to Quantum Physics
(a)
At the top of the ladder, the woman holds a pellet inside a small region ∆xi . Thus, the
uncertainty principle requires her to release it with typical horizontal momentum
2H
1
, so
∆p x = m∆v x =
. It falls to the floor in a travel time given by H = 0 + gt 2 as t =
2
g
2 ∆x i
the total width of the impact points is
b g
∆x f = ∆xi + ∆v x t = ∆xi +
A=
where
2m
A
2H
= ∆x i +
g
∆x i
i
2H
.
g
d i =0
d b ∆x g
d ∆x f
To minimize ∆x f , we require
FG
IJ
m
x
2
∆
H
K
1−
or
i
A
=0
∆xi2
∆xi = A .
so
The minimum width of the impact points is
d∆x i = FGH ∆x + ∆Ax IJK
f
min
i
i
=2 A =
∆xi = A
LM 2e1.054 6 × 10 J ⋅ sj OP L 2a2.00 mf O
∆x
=
d i M 5.00 × 10 kg P MM 9.80 m s PP
Q
Q N
N
12
−34
(b)
F I
GH JK
2 2H
m g
f
−4
min
2
14
.
14
= 5.19 × 10 −16 m
Additional Problems
P40.54
∆VS =
FG h IJ f − φ
H eK e
From two points on the graph
FG h IJ e4.1 × 10 Hzj − φ
H eK
e
φ
F hI
3.3 V = G J e12 × 10 Hzj − .
H eK
e
14
0=
14
and
Combining these two expressions we find:
FIG. P40.54
(a)
φ = 1.7 eV
(b)
h
= 4. 2 × 10 −15 V ⋅ s
e
(c)
At the cutoff wavelength
e
je
hc
λc
=φ =
λ c = 4.2 × 10 −15 V ⋅ s 1.6 × 10 −19 C
FG h IJ ec
H eK λ
c
3 × 10 m s
j a1.7 eVe fe1.6 × 10 j J eVj =
8
−19
730 nm
Chapter 40
P40.55
483
We want an Einstein plot of K max versus f
λ , nm
588
505
445
399
f , 10 14 Hz K max , eV
5.10
0.67
5.94
0.98
6.74
1.35
7.52
1.63
0.402 eV
± 8%
10 14 Hz
(a)
slope =
(b)
e∆VS = hf − φ
a
h = 0.402
(c)
fFGH 1.60 ×1010
−19
J⋅s
14
I=
JK
6. 4 × 10 −34 J ⋅ s ± 8%
f (THz)
K max = 0
FIG. P40.55
at f ≈ 344 × 10 12 Hz
φ = hf = 2.32 × 10 −19 J = 1.4 eV
P40.56
From the path the electrons follow in the magnetic field, the maximum kinetic energy is seen to be:
K max =
P40.57
e2B2 R 2
.
2m e
From the photoelectric equation,
K max = hf − φ =
Thus, the work function is
φ=
∆λ =
e
hc
λ
− K max =
j
hc
λ
−φ .
hc
λ
−
e2B2 R2
.
2m e
6.626 × 10 −34 J ⋅ s
h
1 − cos θ =
0.234 = 3.09 × 10 −16 m
−27
8
mp c
kg 3.00 × 10 m s
1.67 × 10
a
e
f
e
je
je
j
j
a
f
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
λ0 =
=
= 6.20 × 10 −15 m
E0
200 MeV 1.60 × 10 −13 J MeV
a
fe
λ ′ = λ 0 + ∆λ = 6.51 × 10
−15
m
hc
= 191 MeV
λ′
(a)
Eγ =
(b)
K p = 9.20 MeV
j
484
P40.58
Introduction to Quantum Physics
Isolate the terms involving φ in Equations 40.13 and 40.14. Square and add to eliminate φ.
h2
Solve for
v2
b
=
:
2
c
b + c2
e
j
LM 1
Nλ
OP
Q
1
2 cos θ
−
= γ 2 m e2 v 2
2
λ 0λ ′
λ′
+
2
0
2
OP
Q
F h IJ LM 1 − 1 OP = γ = FG 1 − b IJ = c + b .
1+G
H m cKNλ λ′Q H b + c K
c
F h I LM 1
2 hc L 1
1O h L 1
1O
− P = c +G
− P+
c +
M
M
λ′ Q m Nλ
λ′Q
m Nλ
H m JK N λ
F h IJ 1 − cosθ .
λ′ − λ = G
H m cK
b=
LM
N
1
1
2 cos θ
h
+ 2 −
.
2
2
λ 0λ ′
me λ 0 λ ′
−1 2
Substitute into Eq. 40.12:
e
0
e
From this we get Eq. 40.11:
P40.59
0
2
2
e
0
2
2
2
2
Square each side:
2
2
0
2
2
e
2
0
+
e
Show that if all of the energy of a photon is transmitted to an electron, momentum will not be
conserved.
hc hc
hc
=0
=
+ K e = m e c 2 γ − 1 if
Energy:
λ0 λ′
λ′
h
h
=
+ γ m e v = γ m e v if λ ′ = ∞
Momentum:
λ0 λ′
h
γ =
+1
From (1),
λ 0mec
b g
v = c 1−
FG λ m c IJ
H h + λ m cK
Substitute (3) and (4) into (2) and show the inconsistency:
h
λ0
F h IJ m c
= G1 +
H λ m cK
0
e
e
F λ m c IJ
1−G
H h+ λ m cK
e
0
0
2
=
e
b
b
g
λ 0 m e c + h h h + 2λ 0 m e c
h
=
2
λ0
λ0
h + λ 0 me c
g
h + 2λ 0 m e c
.
h
This is impossible, so all of the energy of a photon cannot be transmitted to an electron.
P40.60
Begin with momentum expressions:
p=
Equating these expressions,
γ
h
λ
, and p = γ mv = γ mc
C
2
2
C
2
2
or
C
2
C
2
2
2
2
C
v=
2
C
2
giving
FG v IJ .
H cK
FG v IJ = FG h IJ 1 = λ .
H c K H mc K λ λ
bv cg = FG λ IJ
HλK
1 − b v cg
FG v IJ = FG λ IJ − FG λ IJ FG v IJ
H cK H λ K H λ K H cK
bλ λ g = 1
v
=
c
1 + bλ λ g
bλ λ g + 1
2
Thus,
(2)
(3)
(4)
e
0
(1)
2
e
0
OP
Q
1
2 cos θ
−
.
λ 0λ ′
λ ′2
C
c
b
1 + λ λC
g
2
.
Chapter 40
P40.61
(a)
b g
I λ, T =
Starting with Planck’s law,
2π hc 2
λ5 e hc λk BT − 1
zb g
∞
the total power radiated per unit area
I λ , T dλ =
0
z
∞
0
2π hc 2
λ5 e hc λk BT − 1
hc
λk B T
hcdλ
and
dx = −
.
k B Tλ2
Note that as λ varies from 0 → ∞ , x varies from ∞ → 0 .
∞
2π k B4 T 4
Then
I λ , T dλ = −
h3 c 2
0
dλ .
x=
Change variables by letting
zb g
zb
∞
Therefore,
g
I λ , T dλ =
0
(b)
485
σ=
From part (a),
2π 5 k B4
15 h 3 c 2
=
ze
0
∞
F
I
GH 15h c JK T
2π 5 k B4
3 2
x3
x
e −1
4
j
dx =
F I
GH JK
2πk B4 T 4 π 4
.
15
h3 c 2
= σ T4 .
e
2π 5 1.38 × 10 −23 J K
e
15 6.626 × 10 −34 J ⋅ s
j
4
j e3.00 × 10
3
8
ms
j
2
σ = 5.67 × 10 −8 W m 2 ⋅ K 4 .
P40.62
b g
Planck’s law states I λ , T =
2π hc 2
λ5 e hc λk BT − 1
= 2π hc 2 λ−5 e hc λk BT − 1
−1
.
To find the wavelength at which this distribution has a maximum, compute
R|
S|
T
dI
= 2π hc 2 −5 λ−6 e hc λk BT − 1
dλ
2π hc 2
dI
=
dλ λ6 e hc λk BT − 1
−1
R| hc
S|−5 + λk T
T
B
− λ−5 e hc λk BT − 1
−2 hc λk T
B
e
U|
V=0
−1 |
W
F − hc I U|V = 0
GH λ k T JK |W
2
B
e hc λk BT
e hc λk BT
xe x
hc
= 5.
, the condition for a maximum becomes x
λk B T
e −1
We zero in on the solution to this transcendental equation by iterations as shown in the table below.
The solution is found to be
Letting x =
e
x
xe x e x − 1
4.000 00
4.500 00
5.000 00
4.900 00
4.950 00
4.975 00
4.963 00
4.969 00
4.966 00
4.074 629 4
4.550 552 1
5.033 918 3
4.936 762 0
4.985 313 0
5.009 609 0
4.997 945 2
5.003 776 7
5.000 860 9
x=
hc
= 4.965 115
λ max k B T
continued on next page
j
and
e
x
xe x e x − 1
4.964 50
4.965 50
4.965 00
4.965 25
4.965 13
4.965 07
4.965 10
4.965 115
4.999 403 0
5.000 374 9
4.999 889 0
5.000 132 0
5.000 015 3
4.999 957 0
4.999 986 2
5.000 000 8
λ maxT =
hc
.
4.965 115 k B
j
486
Introduction to Quantum Physics
Thus, λ max
e6.626 075 × 10
T=
−34
je
J ⋅ s 2.997 925 × 10 8 m s
e
4.965 115 1.380 658 × 10
−23
JK
j
j=
2.897 755 × 10 −3 m ⋅ K .
This result is very close to Wien’s experimental value of λ maxT = 2.898 × 10 −3 m ⋅ K for this constant.
P40.63
(a)
Planck’s radiation law predicts maximum intensity at a wavelength λ max we find from
RS
T
0 = 2π hc λ a −1f e b
−1
dI
d
hc λk B T g
2π hc 2 λ−5 e b
=0=
−1
dλ
dλ
2 −5
hc λkB T
g − 1 −2 eb hc λk T g F
7
λ kB
which reduces to
Define x =
5
hc
.
λk B T
GH
B
− hce b
or
UV
W
hc λkB T
I
JK
g
2
hc λk T g
−1
T eb
+
B
FG λk T IJ eb
H hc K
B
hc λkB T
a f
− hc
hc λk B T g
+ 2π hc 2 −5 λ−6 e b
−1
λ2 k B T
g −1
5
hc λk T g
−1
eb
6
λ
−1
=0
B
= eb
hc λk BT
g.
Then we require 5 e x − 5 = xe x .
Numerical solution of this transcendental equation gives x = 4.965 to four digits. So
hc
λ max =
, in agreement with Wien’s law.
4.965 k B T
zb g
∞
0
0
z
0
2π hc 2 dλ
λ5 e b hc λk BT g − 1
.
hc
hc
hc
so λ =
and dλ = − 2
dx .
xk B T
λk B T
x kBT
Again, define x =
Then, A + B =
z
∞
The intensity radiated over all wavelengths is I λ , T dλ = A + B =
−2π hc 2 x 5 k B5 T 5 hcdx
x =∞
e
j
h5 c 5 x 2 kBT e x − 1
The integral is tabulated as
=
2π k B4 T 4
h3 c 2
ze
∞
0
x 3 dx
ex − 1
j
.
2π 5 k B4 T 4
π4
.
, so (in agreement with Stefan’s law) A + B =
15
15 h 3 c 2
The intensity radiated over wavelengths shorter than λ max is
zb g
z
λ max
λ max
0
0
I λ , T dλ = A =
With x =
2π hc 2 dλ
λ5 e b hc λkBT g − 1
.
2π k B4 T 4
hc
, this similarly becomes A =
λk B T
h3 c 2
z
∞
x 3 dx
.
x
4.965 e − 1
So the fraction of power or of intensity radiated at wavelengths shorter than λ max is
A
=
A+B
e 2π k T
continued on next page
4
B
4
L
jMMN
h 3 c 2 π 4 15 −
z
4.965
e
O
j PPQ
x 3 dx e x − 1
0
2π 5 k B4 T 4 15 h 3 c 2
= 1−
15
π4
z
4.965
0
x 3 dx
.
ex − 1
Chapter 40
(b)
Here are some sample values of the integrand, along with a sketch of the curve:
e
j
x3 ex − 1
x
0.000
0.00
−1
0.100
9.51 × 10 −3
0.200
3.61 × 10 −2
1.00
0.582
2.00
1.25
3.00
1.42
4.00
1.19
FIG. P40.63(b)
4.90
0.883
4.965
0.860
Approximating the integral by trapezoids gives
P40.64
jb
e
a
ge
p = mv = 2mE = 2 1.67 × 10 −27 kg 0.040 0 eV 1.60 × 10 −19 J eV
λ=
f
A
15
≈ 1 − 4 4.870 = 0. 250 1 .
A+B
π
j
h
= 1.43 × 10 −10 m = 0.143 nm
mv
This is of the same order of magnitude as the spacing between atoms in a crystal so diffraction
should appear.
P40.65
λC =
p
λ C h me c
=
=
;
λ
hp
mec
h
h
and λ = :
mec
p
e
E2 = c 2 p 2 + me c 2
j
2
p=
:
λC
1
E2
=
− mec
λ
me c c 2
b g
P40.66
(a)
mgyi =
2
E2
− mec
c2
b g
2
LM E − bm cg OP = F E I
bm c g N c
Q GH m c JK
2
1
=
2
e
e
2
2
e
2
2
−1
1
mv 2f
2
ja
e
f
v f = 2 gyi = 2 9.80 m s 2 50.0 m = 31.3 m s
λ=
(b)
h
6.626 × 10 −34 J ⋅ s
=
= 2.82 × 10 −37 m
mv
75.0 kg 31.3 m s
b
∆E∆t ≥
so ∆E ≥
(c)
gb
g
anot observablef
2
6.626 × 10 −34 J ⋅ s
e
4π 5.00 × 10
−3
s
j
= 1.06 × 10 −32 J
∆E
1.06 × 10 −32 J
=
= 2.87 × 10 −35%
E
75.0 kg 9.80 m s 2 50.0 m
b
ge
ja
f
487
488
P40.67
Introduction to Quantum Physics
From the uncertainty principle
∆E∆t ≥
or
∆ mc 2 ∆t =
Therefore,
∆m
h
h
=
=
2
m
4π c ∆t m 4π ∆t ER
2
e j
2
.
a f
a f
∆m
6.626 × 10 −34 J ⋅ s
=
m
4π 8.70 × 10 −17 s 135 MeV
ja
e
P40.68
∆λ =
a
−13
2.81 × 10 −8 .
f
h
1 − cos θ = λ ′ − λ 0
mec
LM
N
a
hc
hc
h
=
= hc λ 0 +
E′ =
1 − cos θ
mec
λ ′ λ 0 + ∆λ
LM1 + hc a1 − cos θ fOP
E′ =
λ N m c λ
Q
O
hc L
hc
E′ =
1+
1 − cos θ fP
a
M
λ N m c λ
Q
hc
0
0
P40.69
FG 1 MeV IJ =
f H 1.60 × 10 J K
e
e
2
2
−1
0
0
−1
fOP
Q
−1
L E a1 − cosθ fOP
= E M1 +
N mc
Q
0
0
e
−1
2
(a)
The light is unpolarized. It contains both horizontal and vertical field oscillations.
(b)
The interference pattern appears, but with diminished overall intensity.
(c)
The results are the same in each case.
(d)
The interference pattern appears and disappears as the polarizer turns, with alternately
increasing and decreasing contrast between the bright and dark fringes. The intensity on the
screen is precisely zero at the center of a dark fringe four times in each revolution, when the
filter axis has turned by 45°, 135°, 225°, and 315° from the vertical.
(e)
Looking at the overall light energy arriving at the screen, we see a low-contrast interference
pattern. After we sort out the individual photon runs into those for trial 1, those for trial 2,
and those for trial 3, we have the original results replicated: The runs for trials 1 and 2 form
the two blue graphs in Figure 40.24 in the text, and the runs for trial 3 build up the red
graph.
Chapter 40
P40.70
489
Let u ′ represent the final speed of the electron and let
I
JK
F
GH
−1 2
u′ 2
. We must eliminate β and u ′ from the
c2
three conservation equations:
γ ′ = 1−
hc
hc
+ γ mec 2 =
+ γ ′m e c 2
λ0
λ′
h
h
+ γm eu − cos θ = γ ′m eu ′ cos β
λ0
λ′
h
sin θ = γ ′m eu ′ sin β
λ′
Square Equations [2] and [3] and add:
h2
λ20
h
2
λ20
+ γ 2 m e2 u 2 +
+
h
2
λ ′2
h2
λ ′2
+
[2]
FIG. P40.70
[3]
2 hγ m e u
−
2 h 2 cos θ 2 hγ m eu cos θ
−
= γ ′ 2 m e2 u ′ 2
λ 0λ ′
λ′
2 hγ m e u
−
2 hγ m eu cos θ 2 h 2 cos θ
m e2 u ′ 2
−
=
λ′
λ 0λ ′
1 − u′ 2 c 2
λ0
+ γ 2 m e2 u 2 +
[1]
λ0
Call the left-hand side b. Then b −
bu ′ 2
b
c 2b
2 2
2
m
u
u
=
′
′
=
=
and
.
e
c2
m e2 + b c 2 m e2 c 2 + b
Now square Equation [1] and substitute to eliminate γ ′ :
h2
λ20
+ γ 2 m e2 c 2 +
So we have
h2
λ20
h2
2
λ′
+
2 hγ m e c
+
λ0
h2
h2
λ20
Multiply through by
λ 0 λ ′γ 2 +
F
GH
2h2
λ 0λ ′
+ γ 2 m e2 c 2 +
λ ′2
= me c 2 +
−
+
h2
−
2 hγ m e c
m e2 c 2
=
= m e2 c 2 + b .
λ′
1 − u′ 2 c 2
2 hγ m e c
λ0
+ γ 2 m e2 u 2 +
λ ′2
−
2 hγ m e c 2 h 2
−
λ′
λ 0λ ′
2 hγ m e u
λ0
−
2 hγ m e u cos θ 2 h 2 cos θ
−
λ′
λ 0λ ′
λ 0λ ′
m e2 c 2
λ λ ′γ 2 u 2 2 hλ ′uγ 2 hγλ 0u cos θ 2 h 2 cos θ
2 hλ ′γ 2 hλ 0γ
2h2
−
− 2 2 = λ 0λ ′ + 0
+
−
−
mec
me c
me c
mec 2
me c 2
mec 2
m e2 c 2
λ 0λ ′ γ 2 − 1 −
I
JK
FG
H
IJ
K
FG
H
IJ
K
2 hγλ 0
γ 2u 2
u
u cos θ
2 hγλ ′
2h2
+
=
+ 2 2 1 − cos θ
1−
1−
2
mec
c
me c
c
c
me c
The first term is zero. Then
Since
this result may be written as
a
f
F 1 − au cosθ f c I + hγ F 1 I a1 − cosθ f .
λ′ = λ G
H 1 − u c JK m c GH 1 − u c JK
F uI F uI F uI
γ = 1 − G J = G1 − J G1 + J
H c K H c KH c K
F 1 − au cos θ f c I + h 1 + u c a1 − cosθ f .
λ′ = λ G
H 1 − u c JK m c 1 − u c
−1
0
e
2
−1
0
e
It shows a specific combination of what looks like a Doppler shift and a Compton shift. This problem
is about the same as the first problem in Albert Messiah’s graduate text on quantum mechanics.
490
Introduction to Quantum Physics
ANSWERS TO EVEN PROBLEMS
P40.2
(a) ~ 10 −7 m ultraviolet ;
(b) ~ 10 −10 m gamma ray
P40.4
(a) 70.9 kW; (b) 580 nm;
(c) 7.99 × 10 10 W m;
(d) 9.42 × 10 −1226 W m ;
P40.30
(a) 0.667 ; (b) 0.001 09
P40.32
(a) 14.0 kV/m, 46.8 µT ; (b) 4.19 nN;
(c) 10.2 g
P40.34
(a) 0.174 nm; (b) 5.37 pm or 5.49 pm
ignoring relativistic correction
P40.36
(a) see the solution; (b) 907 fm
P40.38
0.218 nm
P40.40
(a) 1.60; (b) 2.00 × 10 3 ; (c) 1; (d) ∞
P40.42
(a) 1.10 × 10 −34 m s ; (b) 1.36 × 10 33 s ; (c) no
(e) 1.00 × 10 −227 W m ; (f) 5.44 × 10 10 W m ;
(g) 7.38 × 10 10 W m ; (h) 0.260 W m;
(i) 2.60 × 10
19
−9
W m ; (j) 20 kW
P40.6
2.96 × 10
P40.8
5.71 × 10 3 photons
P40.10
1.34 × 10 31
P40.44
see the solution
P40.12
see the solution
P40.46
105 V
P40.14
(a) 1.38 eV ; (b) 334 THz
P40.48
(a) 0.250 m s ; (b) 2.25 m
P40.16
Metal one: 2.22 eV , Metal two: 3.70 eV
P40.50
P40.18
148 d, the classical theory is a gross failure
3.79 × 10 28 m , much larger than the
diameter of the observable Universe
P40.20
(a) The incident photons are Doppler
shifted to higher frequencies, and hence,
higher energy; (b) 3.87 eV ; (c) 8.78 eV
P40.52
(a) see the solution; (b) 5.21 MeV
P40.54
(a) 1.7 eV ; (b) 4.2 × 10 −15 V ⋅ s ; (c) 730 nm
P40.22
(a) 488 fm ; (b) 268 keV ; (c) 31.5 keV
P40.56
P40.24
22.1 keV
; K e = 478 eV
pe =
c
P40.58
see the solution
P40.26
(a) cos −1
P40.60
see the solution
P40.62
see the solution
P40.64
0.143 nm, comparable to the distance
between atoms in a crystal, so diffraction
can be observed
P40.66
(a) 2.82 × 10 −37 m; (b) 1.06 × 10 −32 J ;
(c) 2.87 × 10 −35%
P40.68
see the solution
P40.70
see the solution
photons s
F m c +E I;
GH 2m c + E JK
E F 2m c + E I
(b) E ′ =
,
2 GH m c + E JK
E F 2m c + E I
p′ =
;
2 c GH m c + E JK
2
e
e
0
e
0
γ
0
2
e
γ
e
0
e
(c) K e =
pe
P40.28
2
0
2
0
2
2
0
j
E F 2m c + E I
=
2 c GH m c + E JK
2 m e c 2 + E0
e
0
e
2
2
λ
−
e2B2 R 2
2m e
0
E02
e
hc
0
0
(a) 33.0° ; (b) 0.785 c
,
41
Quantum Mechanics
CHAPTER OUTLINE
41.1
41.2
41.3
41.4
41.5
41.6
41.7
41.8
An Interpretation of Quantum
Mechanics
A Particle in a Box
The Particle Under Boundary
Conditions
The Schrödinger Equation
A Particle in a Well of Finite
Height
Tunneling Through a
Potential Energy Barrier
The Scanning Tunneling
Microscope
The Simple Harmonic
Oscillator
ANSWERS TO QUESTIONS
Q41.1
A particle’s wave function represents its state, containing all the
information there is about its location and motion. The squared
absolute value of its wave function tells where we would
2
classically think of the particle as a spending most its time. Ψ
is the probability distribution function for the position of the
particle.
Q41.2
The motion of the quantum particle does not consist of moving
through successive points. The particle has no definite position.
It can sometimes be found on one side of a node and
sometimes on the other side, but never at the node itself. There
is no contradiction here, for the quantum particle is moving as
a wave. It is not a classical particle. In particular, the particle
does not speed up to infinite speed to cross the node.
Q41.3
Consider a particle bound to a restricted region of space. If its minimum energy were zero, then the
particle could have zero momentum and zero uncertainty in its momentum. At the same time, the
uncertainty in its position would not be infinite, but equal to the width of the region. In such a case,
the uncertainty product ∆x∆p x would be zero, violating the uncertainty principle. This contradiction
proves that the minimum energy of the particle is not zero.
Q41.4
The reflected amplitude decreases as U decreases. The amplitude of the reflected wave is
proportional to the reflection coefficient, R, which is 1 − T , where T is the transmission coefficient as
given in equation 41.20. As U decreases, C decreases as predicted by equation 41.21, T increases, and
R decreases.
Q41.5
Consider the Heisenberg uncertainty principle. It implies that electrons initially moving at the same
speed and accelerated by an electric field through the same distance need not all have the same
measured speed after being accelerated. Perhaps the philosopher could have said “it is necessary for
the very existence of science that the same conditions always produce the same results within the
uncertainty of the measurements.”
Q41.6
In quantum mechanics, particles are treated as wave functions, not classical particles. In classical
mechanics, the kinetic energy is never negative. That implies that E ≥ U . Treating the particle as a
wave, the Schrödinger equation predicts that there is a nonzero probability that a particle can tunnel
through a barrier—a region in which E < U .
491
492
Quantum Mechanics
Q41.7
Consider Figure 41.8, (a) and (b) in the text. In the square well with infinitely high walls, the
particle’s simplest wave function has strict nodes separated by the length L of the well. The particle’s
p2
h
h2
=
. Now in the well with walls of only
, and its energy
wavelength is 2L, its momentum
2m 8mL2
2L
finite height, the wave function has nonzero amplitude at the walls. The wavelength is longer. The
particle’s momentum in its ground state is smaller, and its energy is less.
Q41.8
Quantum mechanically, the lowest kinetic energy possible for any bound particle is greater than
zero. The following is a proof: If its minimum energy were zero, then the particle could have zero
momentum and zero uncertainty in its momentum. At the same time, the uncertainty in its position
would not be infinite, but equal to the width of the region in which it is restricted to stay. In such a
case, the uncertainty product ∆x∆p x would be zero, violating the uncertainty principle. This
contradiction proves that the minimum energy of the particle is not zero. Any harmonic oscillator
can be modeled as a particle or collection of particles in motion; thus it cannot have zero energy.
Q41.9
As Newton’s laws are the rules which a particle of large mass follows in its motion, so the
Schrödinger equation describes the motion of a quantum particle, a particle of small or large mass. In
particular, the states of atomic electrons are confined-wave states with wave functions that are
solutions to the Schrödinger equation.
SOLUTIONS TO PROBLEMS
Section 41.1
P41.1
(a)
An Interpretation of Quantum Mechanics
ψ x = Ae e
af
i 5.00 ×10 10 x
j = A cose5 × 10 10 xj + Ai sine5 × 1010 xj = A cosa kxf + Ai sinakx f goes through
a full cycle when x changes by λ and when kx changes by 2π . Then kλ = 2π where
2π m
2π
= 1.26 × 10 −10 m .
k = 5.00 × 10 10 m −1 =
. Then λ =
10
λ
5.00 × 10
e
h
6.626 × 10 −34 J ⋅ s
= 5.27 × 10 −24 kg ⋅ m s
1.26 × 10 −10 m
(b)
p=
(c)
m e = 9.11 × 10 −31 kg
λ
=
j
e
j
j
5.27 × 10 −24 kg ⋅ m s
m2v2 p2
=
=
K= e
2m e
2m
2 × 9.11 × 10 −31 kg
P41.2
Probability
P=
z
a
af
ψ x
2
−a
P=
1
π
e
= 1.52 × 10 −17 J =
z π ex + a j FGH πa IJK FGH 1a IJK tan FGH xa IJK
1 Lπ F π I O
1
1 − tan a−1f = M − G − J P =
H
K
π N4
4 Q
2
=
a
−a
tan −1
2
a
2
−1
2
dx =
−1
a
−a
1.52 × 10 −17 J
= 95.5 eV
1.60 × 10 −19 J eV
Chapter 41
Section 41.2
P41.3
A Particle in a Box
E1 = 2.00 eV = 3.20 × 10 −19 J
E1 =
For the ground-state,
P41.4
h
= 4.34 × 10 −10 m = 0.434 nm
(a)
L=
(b)
∆E = E2 − E1 = 4
8m e E1
h2
.
8m e L2
F h I −F h I =
GH 8m L JK GH 8m L JK
2
e
2
2
e
6.00 eV
For an electron wave to “fit” into an infinitely deep potential well, an integral
number of half-wavelengths must equal the width of the well.
nλ
= 1.00 × 10 −9 m
2
e
so
λ=
j
h 2 λ2
p2
h2
n2
=
=
2m e
2m e
2m e 2 × 10 −9
2.00 × 10 −9 h
=
n
p
e
j
= 0.377n 2 eV
Since
K=
For
K ≈ 6 eV
n=4
(b)
With
n = 4,
K = 6.03 eV
(a)
We can draw a diagram that parallels our treatment of standing
mechanical waves. In each state, we measure the distance d
from one node to another (N to N), and base our solution upon
that:
(a)
P41.5
2
Since
e
d N to N =
p=
j
2
λ
h
and λ =
p
2
h
h
=
.
λ 2d
LM e6.626 × 10
MM 8e9.11 × 10
N
−34
Next,
Evaluating,
K=
In state 1,
d = 1.00 × 10 −10 m
K 1 = 37.7 eV .
In state 2,
d = 5.00 × 10 −11 m
K 2 = 151 eV .
In state 3,
d = 3.33 × 10 −11 m
K 3 = 339 eV .
In state 4,
d = 2.50 × 10 −11 m
K 4 = 603 eV .
6.02 × 10 −38 J ⋅ m 2
d2
K=
j OP .
kg j P
PQ
J⋅s
p2
1
h2
=
= 2
K=
2
2m e 8m e d
d
continued on next page
FIG. P41.4
−31
2
3.77 × 10 −19 eV ⋅ m 2
.
d2
FIG. P41.5
493
494
Quantum Mechanics
(b)
When the electron falls from state 2 to state 1, it puts out energy
hc
E = 151 eV − 37.7 eV = 113 eV = hf =
λ
into emitting a photon of wavelength
e
je
j
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
λ= =
= 11.0 nm .
E
113 eV 1.60 × 10 −19 J eV
a
fe
j
The wavelengths of the other spectral lines we find similarly:
4→3
264
4.71
Transition
E eV
λ nm
a f
a f
P41.6
λ = 2D
4→ 2
452
2.75
4→1
565
2.20
3→2
188
6.60
e
j
e
P41.7
h
λ
∆E =
L=
P41.8
∆E =
so
P41.9
=
hc
λ
2→1
113
11.0
for the lowest energy state
2
6.626 × 10 −34 J ⋅ s
p2
h2
h2
K=
=
=
=
2m 2mλ2 8mD 2 8 4 1.66 × 10 −27 kg 1.00 × 10 −14 m
p=
3→1
302
4.12
je
j
2
= 8.27 × 10 −14 J = 0.517 MeV
6.626 × 10 −34 J ⋅ s
h
=
= 3.31 × 10 −20 kg ⋅ m s
2D 2 1.00 × 10 −14 m
e
=
j
F h I2
GH 8m L JK
2
e
2
2
− 12 =
3h2
8m e L2
3 hλ
= 7.93 × 10 −10 m = 0.793 nm
8m e c
hc
λ
=
L=
F h I2
GH 8m L JK
2
e
2
2
− 12 =
3h2
8m e L2
3 hλ
8m e c
The confined proton can be described in the same way as a standing
wave on a string. At level 1, the node-to-node distance of the standing
wave is 1.00 × 10 −14 m , so the wavelength is twice this distance:
h
= 2.00 × 10 −14 m .
p
The proton’s kinetic energy is
e
j
2
6.626 × 10 −34 J ⋅ s
p2
1
h2
K = mv 2 =
=
=
2
2m 2mλ2 2 1.67 × 10 −27 kg 2.00 × 10 −14 m
e
3. 29 × 10 −13 J
=
= 2.05 MeV
1.60 × 10 −19 J eV
continued on next page
je
j
2
FIG. P41.9
Chapter 41
495
In the first excited state, level 2, the node-to-node distance is half as long as in state 1. The
momentum is two times larger and the energy is four times larger: K = 8.22 MeV .
The proton has mass, has charge, moves slowly compared to light in a standing wave state, and
stays inside the nucleus. When it falls from level 2 to level 1, its energy change is
2.05 MeV − 8.22 MeV = −6.16 MeV .
Therefore, we know that a photon (a traveling wave with no mass and no charge) is emitted at the
speed of light, and that it has an energy of +6.16 MeV .
e
je
j
Its frequency is
6.16 × 10 6 eV 1.60 × 10 −19 J eV
E
= 1. 49 × 10 21 Hz .
f= =
h
6.626 × 10 −34 J ⋅ s
And its wavelength is
λ=
c 3.00 × 10 8 m s
=
= 2.02 × 10 −13 m .
f 1. 49 × 10 21 s −1
This is a gamma ray, according to the electromagnetic spectrum chart in Chapter 34.
P41.10
The ground state energy of a particle (mass m) in a 1-dimensional box of width L is E1 =
e
j
e6.626 × 10 J ⋅ sj
=
8e1.67 × 10
kg je 2.00 × 10
mj
For a proton m = 1.67 × 10 −27 kg in a 0.200-nm wide box:
(a)
2
−34
E1
−27
−10
e
= 8.22 × 10 −22 J = 5.13 × 10 −3 eV .
j
e6.626 × 10 J ⋅ sj
=
8e9.11 × 10
kg je 2.00 × 10
2
−34
E1
(c)
−31
−10
m
j
2
= 1.51 × 10 −18 J = 9.41 eV .
The electron has a much higher energy because it is much less massive.
F h In
GH 8mL JK
e6.626 × 10 J ⋅ sj
=
8e1.67 × 10
kg je 2.00 × 10
En =
2
2
2
−34
E1
2
For an electron m = 9.11 × 10 −31 kg in the same size box:
(b)
P41.11
h2
.
8mL2
−27
E1 = 0.513 MeV
2
−14
m
j
2
= 8.21 × 10 −14 J
E2 = 4E1 = 2.05 MeV
E3 = 9E1 = 4.62 MeV
Yes , the energy differences are ~ 1 MeV , which is a typical energy for a γ-ray photon.
496
*P41.12
Quantum Mechanics
The energies of the confined electron are En =
(a)
jump from state 1 to state 4 is
h2
n 2 . Its energy gain in the quantum
8m e L2
h2
4 2 − 1 2 and this is the photon energy:
8m e L2
e
j
FG
H
h 2 15
hc
15 hλ
= hf = . Then 8m e cL2 = 15 hλ and L =
2
λ
8m e c
8m e L
12
.
12 h 2
hc
h2
h2
2
2
4
2
.
=
−
=
λ ′ 8m e L2
8m e L2
8m e L2
Let λ ′ represent the wavelength of the photon emitted:
(b)
Then
e
j
2
2
5
hc λ ′ h 15 8m e L
=
= and λ ′ = 1. 25λ .
λ hc 8m e L2 12 h 2
4
Section 41.3
The Particle Under Boundary Conditions
Section 41.4
The Schrödinger Equation
P41.13
IJ
K
ψ = Ae ib kx −ω t g
We have
∂ 2ψ
= − k 2ψ .
∂x 2
and
∂ 2ψ
2m
= − k 2ψ = − 2 E − U ψ .
2
=
∂x
a
Schrödinger’s equation:
Since
k
2
a 2π f = b 2π p g
=
2
λ2
h2
2
=
p2
E −U =
and
=2
f
p2
.
2m
Thus this equation balances.
P41.14
ψ x = A cos kx + B sin kx
∂ψ
= − kA sin kx + kB cos kx
∂x
∂ 2ψ
= − k 2 A cos kx − k 2 B sin kx
∂x 2
−
af
a
f
a
2m
2mE
E − U ψ = − 2 A cos kx + B sin kx
=
=
Therefore the Schrödinger equation is satisfied if
FG
H
IJ a
K
∂ 2ψ
2m
= − 2 E − U ψ or
=
∂x 2
f FGH
a
f
− k 2 A cos kx + B sin kx = −
This is true as an identity (functional equality) for all x if E =
*P41.15
(a)
af
=2
IJ a A cos kx + B sin kxf .
K
=2k2
.
2m
a f
With ψ x = A sin kx
d
A sin kx = Ak cos kx
dx
and
d2
ψ = − Ak 2 sin kx .
2
dx
e j
e ja f
h 2 4π 2
p2
= 2 d 2ψ
=2k2
m2v2
1
ψ
ψ
ψ = mv 2ψ = Kψ .
−
=+
=
=
A sin kx =
2
2
2
2m dx
2m
2
m
2
m
2
4π λ 2m
Then
(b)
2mE
f
af
FG 2π x IJ = A sin kx , the proof given in part (a) applies again.
H λK
With ψ x = A sin
Chapter 41
z
L
P41.16
x = x
(a)
0
FG
H
IJ
K
IJ
K
LM 4π x sin 4π x + cos 4π x OP
L Q
NL L
z FGH
L
2π x
4π x
2
2
1 1
sin 2
dx =
x − cos
dx
L
L
L0 2 2
L
1 x2
x =
L 2
L
−
0
1 L2
L 16π 2
FG
H
z
0.510 L
(b)
Probability=
Probability
IJ
K
L
=
0
L
2
LM
N
2π x
4π x
2
1
1 L
sin 2
sin
dx =
x−
L
L
L
L 4π
L
0. 490 L
Probability= 0.020 −
(c)
a
OP
Q
0.510 L
0. 490 L
f
1
sin 2.04π − sin 1.96π = 5.26 × 10 −5
4π
LM x − 1 sin 4π x OP
N L 4π L Q
0 . 260 L
= 3.99 × 10 −2
0 . 240 L
In the n = 2 graph in Figure 41.4 (b), it is more probable to find the particle either near x =
(d)
or x =
3L
than at the center, where the probability density is zero.
4
Nevertheless, the symmetry of the distribution means that the average position is
P41.17
z
L
2
ψ dx = 1
or
z
L
A 2 sin 2
0
FG nπ x IJ dx = A FG L IJ = 1
H 2K
H LK
2
or
z
L4
The desired probability is
P=
where
sin 2 θ =
Thus,
P=
In 0 ≤ x ≤ L , the argument
2
ψ dx =
0
P41.19
A 2 sin 2
0
all space
P41.18
L
.
2
Normalization requires
z
2
L
FG nπ x IJ dx = 1
H LK
A=
2
.
L
z
FG 2π x IJ dx
H LK
L4
sin 2
0
1 − cos 2θ
.
2
FG x − 1 sin 4π x IJ
H L 4π L K
L4
=
0
FG 1 − 0 − 0 + 0IJ =
H4
K
0.250 .
2π x
of the sine function ranges from 0 to 2π . The probability density
L
FG 2 IJ sin FG 2π x IJ reaches maxima at sinθ = 1 and sinθ = −1 at
H LK H L K
2
2π x π
2π x 3π
= and
=
.
2
L
2
L
∴
497
The most probable positions of the particle are at at x =
L
3L
.
and x =
4
4
L
4
498
Quantum Mechanics
z
z FGH IJK
1L
L
F 2π x IJ OP =
= Mx −
sinG
H L KQ
2π
LN
A
*P41.20
(a)
2
= ψ 1 dx =
Probability
0
A
0
(b)
FG 2π x IJ OPdx
H L KQ
A
1
F 2π A IJ
−
sinG
HLK
L 2π
A
πx
2
1
dx =
sin 2
L0
L
L
z LMN
A
1 − cos
0
Probability Curve for an Infinite
Potential Well
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
A
L
1.5
FIG. P41.20(b)
(c)
The probability of finding the particle between x = 0 and x = A is
x = L is
z
A
1
.
3
2
Thus, ψ 1 dx =
0
∴
FG
H
2
3
IJ
K
2π A
A
1
2
sin
−
= ,
L 2π
L
3
This equation for
being
2
, and between x = A and
3
or
u−
1
2
sin 2π u = .
2π
3
A
can be solved by homing in on the solution with a calculator, the result
L
A
= 0.585 , or A = 0.585L to three digits.
L
Chapter 41
z
z FGH IJK z FGH 12 − 12 cos 2πL x IJK dx
F x 1 sin 2π x IJ = FG 1 − 1 sin 2π IJ = FG 1 − 3 IJ = 0.196
P=G −
H L 2π L K H 3 2π 3 K H 3 4π K
L3
P41.21
(a)
L3
2
ψ dx =
P=
The probability is
0
0
πx
2
2
dx =
sin 2
L
L
L
499
L3
0
L3
.
0
(b)
L
.
2
Thus, the probability of finding the particle between
2L
x=
and x = L is the same 0.196. Therefore, the
3
L
2L
is
probability of finding it in the range ≤ x ≤
3
3
P = 1.00 − 2 0.196 = 0.609 .
The probability density is symmetric about x =
a
P41.22
FIG. P41.21(b)
f
(c)
Classically, the electron moves back and forth with constant speed between the walls, and
the probability of finding the electron is the same for all points between the walls. Thus, the
classical probability of finding the electron in any range equal to one-third of the available
1
.
space is Pclassical =
3
(a)
ψ1 x =
FG IJ
H K
axf = L2 sinFGH 2πL x IJK ;
axf = L2 cosFGH 3πL x IJK ;
af
ψ2
ψ3
πx
2
cos
;
L
L
af
FG IJ
H K
axf = L2 sin FGH 2πL x IJK
axf = L2 cos FGH 3πL x IJK
af
P1 x = ψ 1 x
af
P2 x = ψ 2
af
P3 x = ψ 3
2
=
πx
2
cos 2
L
L
2
2
2
2
(b)
∞
∞
∞
∞
n=3
ψ
ψ
2
n=2
n=1
−
L
2
0
x
L
2
−
L
2
FIG. P41.22(b)
0
x
L
2
500
P41.23
Quantum Mechanics
Problem 43 in Chapter 16 helps students to understand how to draw conclusions from an identity
2 Ax
dψ
d 2ψ
2A
x2
ψ x = A 1− 2
=− 2
=−
(a)
2
L
dx
L
dx
L
a f FGH
I
JK
Schrödinger’s equation
d 2ψ
2m
= − 2 E −U ψ
2
=
dx
becomes
2 2
2
2
x2
2A
2m
2m − = x A 1 − x L
− 2 = − 2 EA 1 − 2 + 2
=
=
L
L
mL2 L2 − x 2
a
f
I
JK
F
GH
e
je
e
j
j
1
mE mEx 2 x 2
− 2 =− 2 + 2 2 − 4 .
L
=
= L
L
This will be true for all x if both
1 mE
=
L2 = 2
and
mE
1
− 4 =0
2 2
= L
L
E=
both these conditions are satisfied for a particle of energy
(b)
1=
For normalization,
z
L
−L
LM
N
(c)
P=
z
ψ
−L 3
P=
P41.24
(a)
2
2x 3
2
+
I
JK
2
dx = A 2
z FGH
L
−L
1−
I
JK
2x 2 x 4
+ 4 dx
L2
L
L
x5
2
4
2
−L
L3
4
3
4
2
2
2
−L 3
5
5
L3
−L 3
47
= 0.580
81
Setting the total energy E equal to zero and rearranging the Schrödinger equation to isolate
the potential energy function gives
a f FGH 2=m IJK ψ1 ddxψ .
ψ a xf = Axe
.
d ψ
e
= e 4 Ax − 6 AxL j
dx
L
d ψ e 4x − 6L j
ψ a xf
=
U x =
If
Then
or
and
(b)
x2
L2
OP = A LL − 2 L + L + L − 2 L + L O = A FG 16L IJ
15
A=
.
MN 3 5 3 5 PQ H 15 K
16L
3L
5L Q
F 1 − 2 x + x I dx = 15 LMx − 2 x + x OP = 30 LM L − 2L + L OP
15
dx =
GH L L JK 16L N 3L 5L Q 16L N 3 81 1 215 Q
z
16L
1 = A2 x −
L3
F
GH
A2 1 −
=2
.
L2 m
2
2
2
− x 2 L2
2
3
2
2
2
dx 2
af
U x =
− x 2 L2
2
4
2
L4
F
GH
=2
4x 2
−6
2mL2 L2
I
JK
.
See the figure to the right.
FIG. P41.24(b)
Chapter 41
Section 41.5
P41.25
A Particle in a Well of Finite Height
See figure to the right.
(a)
(b)
The wavelength of the transmitted wave
traveling to the left is the same as the original
wavelength, which equals 2L .
FIG. P41.25(a)
P41.26
FIG. P41.26
Section 41.6
P41.27
Tunneling Through a Potential Energy Barrier
T = e −2CL where C =
2CL =
P41.28
e
a
2m U − E
f
=
je
2 2 9.11 × 10 −31 8.00 × 10 −19
1.055 × 10
j
−34
e2.00 × 10 j = 4.58
−10
(a)
T = e −4.58 = 0.010 3 , a 1% chance of transmission.
(b)
R = 1 − T = 0.990 , a 99% chance of reflection.
C=
ja
e
fe
j
2 9.11 × 10 −31 5.00 − 4.50 1.60 × 10 −19 kg ⋅ m s
1.055 × 10
−34
e
J⋅s
je
j
FIG. P41.27
= 3.62 × 10 9 m −1
a
T = e −2CL = exp −2 3.62 × 10 9 m −1 950 × 10 −12 m = exp −6.88
f
T = 1.03 × 10 −3
P41.29
FIG. P41.28
From problem 28, C = 3.62 × 10 9 m −1
e
j
10 −6 = exp −2 3.62 × 10 9 m −1 L .
Taking logarithms,
e
j
−13.816 = −2 3.62 × 10 9 m −1 L .
New L = 1.91 nm
∆L = 1.91 nm − 0.950 nm = 0.959 nm .
501
502
*P41.30
Quantum Mechanics
The original tunneling probability is T = e −2CL where
c2maU − Efh
C=
12
=
=
a
e
f
2π 2 × 9.11 × 10 −31 kg 20 − 12 1.6 × 10 −19 J
j
12
6.626 × 10 −34 J ⋅ s
= 1.448 1 × 10 10 m −1 .
1 240 eV ⋅ nm
=
= 2.27 eV , to make the electron’s new kinetic energy
λ
546 nm
12 + 2.27 = 14.27 eV and its decay coefficient inside the barrier
hc
The photon energy is hf =
C′ =
a
e
f
2π 2 × 9.11 × 10 −31 kg 20 − 14.27 1.6 × 10 −19 J
j
12
6.626 × 10 −34 J ⋅ s
= 1.225 5 × 10 10 m −1 .
Now the factor of increase in transmission probability is
−9
−1
10
e −2C ′L
= e 2 LaC −C ′ f = e 2 ×10 m× 0. 223 ×10 m = e 4.45 = 85.9 .
−2CL
e
Section 41.7
P41.31
The Scanning Tunneling Microscope
With the wave function proportional to e −CL , the transmission coefficient and the tunneling current
2
are proportional to ψ , to e −2CL .
a
a
P41.32
With transmission coefficient e −2CL , the fractional change in transmission is
e
b
g
−2 10.0 nm L
−e
e
Section 41.8
P41.33
f
f
−2 10.0 nm ga 0.500 nm f
I 0.500 nm
e b
=
= e 20 .0 a0 .015 f = 1.35 .
−2 b10.0 nm ga 0.515 nm f
I 0.515 nm
e
Then,
b
gb
−2 10 .0 nm L + 0.002 00 nm
b
g
g
−2 10.0 nm L
=1−e
b
−20.0 0.002 00
g = 0.0392 =
3.92% .
The Simple Harmonic Oscillator
FG IJ x ψ + FG − mω IJψ .
H K
H =K
F mω IJ x ψ + FG − mω IJψ = −FG 2mE IJψ + FG mω IJ
Substituting into Equation 41.13 gives G
H=K
H =K H= K H=K
ψ = Be − bmω
g
2= x2
so
FG IJ
H K
dψ
mω
d 2ψ
mω
=−
=
xψ and
2
=
=
dx
dx
2
which is satisfied provided that E =
=ω
.
2
2
2
2
2
2
x 2ψ
Chapter 41
P41.34
Problem 43 in Chapter 16 helps students to understand how to draw conclusions from an identity.
ψ = Axe − bx so
2
2
dψ
= Ae − bx − 2bx 2 Ae − bx
dx
and
2
2
2
d 2ψ
= −2bxAe − bx − 4bxAe − bx + 4b 2 x 3 e − bx = −6bψ + 4b 2 x 2ψ .
2
dx
Substituting into Equation 41.13,
−6bψ + 4b 2 x 2ψ = −
2
FG 2mE IJψ + FG mω IJ
H = K H=K
2
x 2ψ .
For this to be true as an identity, it must be true for all values of x.
−6b = −
So we must have both
P41.35
FG IJ
H K
2mE
mω
and 4b 2 =
=
=2
Therefore
(b)
and
(c)
The wave function is that of the first excited state .
E=
2
.
mω
2=
b=
(a)
3b= 2
3
=ω .
=
m
2
The longest wavelength corresponds to minimum photon energy, which must be equal to the
spacing between energy levels of the oscillator:
F
jGH
9.11 × 10 −31 kg
k
m
= =ω = =
so λ = 2π c
= 2π 3.00 × 10 8 m s
m
λ
k
8.99 N m
hc
P41.36
503
(a)
e
With ψ = Be
b
g
− mω 2 = x 2
, the normalization condition
z
I
JK
12
= 600 nm .
2
ψ dx = 1
all x
becomes 1 =
z
∞
B2 e
b
g
−2 mω 2 = x 2
z
∞
dx = 2B 2 e
−∞
b
g
− mω = x 2
0
dx = 2B 2
1
2
π
mω =
where Table B.6 in Appendix B was used to evaluate the integral.
Thus, 1 = B
(b)
2
F I
GH JK
π=
mω
and B =
π=
mω
14
.
For small δ, the probability of finding the particle in the range −
z
δ 2
2
af
ψ dx = δ ψ 0
−δ 2
2
= δ B 2 e −0 = δ
F mω I
GH π = JK
12
.
δ
δ
< x < is
2
2
504
*P41.37
Quantum Mechanics
(a)
For the center of mass to be fixed, m1 v1 + m 2 v 2 = 0 . Then
m
m + m1
m2 v
m
. Similarly, v = 2 v 2 + v 2 and
v = v1 + v 2 = v1 + 1 v1 = 2
v1 and v1 =
m2
m2
m1 + m 2
m1
m1 v
. Then
v2 =
m1 + m 2
1
1
1
1 m1 m 22 v 2
1 m 2 m12 v 2
1
m1 v12 + m 2 v 22 + kx 2 =
+
+ kx 2
2
2
2
2
2
2 m1 + m 2
2 m1 + m 2
2
b
g
b
1 m m bm + m g
=
v
2 bm + m g
1
2
1
(b)
FG
H
1
2
2
2
g
2
+
1 2 1 2 1 2
kx = µv + kx
2
2
2
IJ
K
d 1
1
µ v 2 + kx 2 = 0 because energy is constant
dx 2
2
0=
1
dv 1
dx dv
dv
µ 2v + k2x = µ
+ kx = µ
+ kx .
2
dx 2
dt dx
dt
Then µ a = − kx , a = −
kx
. This is the condition for simple harmonic motion, that the
µ
acceleration of the equivalent particle be a negative constant times the excursion from
k
1 k
= 2π f and f =
equilibrium. By identification with a = −ω 2 x , ω =
.
µ
2π µ
P41.38
(a)
a f
With x = 0 and p x = 0, the average value of x 2 is ∆x
=
2
∆p x . Then ∆x ≥
requires
2 ∆p x
b g
E≥
(b)
2
and the average value of p x2 is
p x2 k = 2
p x2 k= 2
.
+
=
+
2m 2 4p x2
2m 8 p x2
1
1
dE
k= 2
=
0
=
+
−1 4 .
2
2m
8
dp x
px
a f
To minimize this as a function of p x2 , we require
Then
1
k= 2
=
8 p x4 2m
and
E≥
Emin =
=
2
= mk
= k =
k= 2 2
+
=
+
2 2m 8= mk 4 m 4
a f
=ω
k
=
.
m
2
p x2 =
so
k
m
F 2mk= I
GH 8 JK
2
12
=
= mk
2
Chapter 41
505
Additional Problems
P41.39
Suppose the marble has mass 20 g. Suppose the wall of the box is 12 cm high and 2 mm thick. While
it is inside the wall,
ge
ja f
1
1
E = K = mv = b0.02 kg gb0.8 m sg = 0.006 4 J .
2
2
2b0.02 kg gb0.017 1 Jg
2maU − Ef
C=
=
= 2.5 × 10
b
U = mgy = 0.02 kg 9.8 m s 2 0.12 m = 0.023 5 J
2
2
and
Then
1.055 × 10 −34 J ⋅ s
=
32
m −1
and the transmission coefficient is
e −2CL = e
P41.40
P41.41
e
je
−2 2 .5 × 10 32 2 × 10 −3
j = e −10 ×10
(a)
λ = 2L = 2.00 × 10 −10 m
(b)
p=
(c)
E=
(a)
h
λ
=
29
=e
e
−2 .30 4.3 × 10 29
j = 10 −4.3 ×10
29
= ~ 10 −10
30
.
6.626 × 10 −34 J ⋅ s
= 3.31 × 10 −24 kg ⋅ m s
2.00 × 10 −10 m
p2
= 0.172 eV
2m
See the figure.
(b)
FIG. P41.41(a)
See the figure.
FIG. P41.41(b)
(c)
ψ is continuous and ψ → 0 as x → ±∞ . The function can be normalized. It describes a
particle bound near x = 0 .
(d)
Since ψ is symmetric,
z
∞
2
z
∞
2
ψ dx = 2 ψ dx = 1
−∞
0
F 2 A I ee
GH −2α JK
z
∞
or
2
2 A 2 e −2α x dx =
0
−∞
j
− e0 = 1 .
This gives A = α .
(e)
Pb −1 2α g→b1 2α g = 2
e aj
2
z
1 2α
e −2α x dx =
x=0
FG 2α IJ ee
H −2α K
− 2α 2α
j e
j
− 1 = 1 − e −1 = 0.632
506
P41.42
Quantum Mechanics
(a)
Use Schrödinger’s equation
∂ 2ψ
2m
= − 2 E −U ψ
2
=
∂x
a
f
with solutions
ψ 1 = Aeik1 x + Be −ik1 x
[region I]
ψ 2 = Ce ik 2 x
[region II].
FIG. P41.42(a)
2mE
=
Where
k1 =
and
k2 =
a
2m E − U
=
f.
Then, matching functions and derivatives at x = 0
bψ g = bψ g
F dψ IJ = FG dψ IJ
and G
H dx K H dx K
1 0
2 0
1
2
0
A+B=C
gives
k1 A − B = k 2C .
0
a
f
Then
B=
1 − k 2 k1
A
1 + k 2 k1
and
C=
2
A.
1 + k 2 k1
Incident wave Ae
(b)
gives
ikx
reflects Be
− ikx
, with probability
b
b
1 − k 2 k1
B2
R= 2 =
A
1 + k 2 k1
With
E = 7.00 eV
and
U = 5.00 eV
k2
E −U
=
=
k1
E
a1 − 0.535f
a1 + 0.535f
g
g
2
2
=
bk
bk
− k2
1
2
2.00
= 0.535 .
7.00
2
The reflection probability is
R=
The probability of transmission is
T = 1 − R = 0.908 .
2
g
+k g
1
= 0.092 0 .
2
2
.
Chapter 41
P41.43
bk
R=
bk
g = b1 − k k g
+k g
b1 + k k g
1
− k2
1
2
2
2
1
2
1
2
507
2
2
=2k2
= E − U for constant U
2m
= 2 k12
= E since U = 0
2m
(1)
= 2 k 22
= E −U
2m
(2)
FIG. P41.43
k 22
Dividing (2) by (1),
k12
=1−
k
U
1
1 1
= 1 − = so 2 =
E
2 2
k1
2
e1 − 1 2 j = e
R=
e1 + 1 2 j e
2
and therefore,
P41.44
(a)
(b)
2
j
2 + 1j
2 −1
2
2
= 0.029 4 .
The wave functions and probability densities are the same as those shown in the two lower
curves in Figure 41.4 of the textbook.
FG π x IJ dx
H 1.00 nm K
L x 1.00 nm sinFG 2π x IJ OP
= b 2.00 nmgM −
N 2 4π H 1.00 nmK Q
F xI F 1 I
In the above result we used z sin axdx = G J − G J sina 2 ax f .
H 2 K H 4a K
L 1.00 nm sinFG 2π x IJ OP
Therefore, P = b1.00 nmgM x −
N 2π H 1.00 nm K Q
1.00 nm
R
U
sina0.700π f − sina0.300π f V = 0.200 .
P = b1.00 nmgS0.350 nm − 0.150 nm −
2
π
T
W
2
F 2π x IJ dx = 2.00LM x − 1.00 sinFG 4π x IJ OP
P =
sin G
H 1.00 K
1.00 z
N 2 8π H 1.00 K Q
L 1.00 sinFG 4π x IJ OP = 1.00RSa0.350 − 0.150f − 1.00 sina1.40π f − sina0.600π f UV
P = 1.00 M x −
4π
T
W
N 4π H 1.00 K Q
P1 =
z
0.350 nm
2
ψ 1 dx =
0.150 nm
FG 2 IJ
H 1.00 nmK
z
0 .350
sin 2
0 .150
0.350 nm
0 .150 nm
2
0.350 nm
1
0.150 nm
1
0.350
(c)
2
0 .350
2
0.150
0 .150
0.350
2
0.150
= 0.351
(d)
Using En =
n 2 h2
, we find that E1 = 0.377 eV and E2 = 1.51 eV .
8mL2
508
P41.45
*P41.46
Quantum Mechanics
a
f
F 1.60 × 10 J I =
j GH 1.00 eV JK
−19
(a)
f=
1.80 eV
E
=
h
6.626 × 10 −34 J ⋅ s
(b)
λ=
c 3.00 × 10 8 m s
=
= 6.91 × 10 −7 m = 691 nm
14
f 4.34 × 10 Hz
(c)
∆E∆t ≥
(a)
Taking L x = L y = L , we see that the expression for E becomes
e
4.34 × 10 14 Hz
=
=
6.626 × 10 −34 J ⋅ s
h
=
=
= 2.64 × 10 −29 J = 1.65 × 10 −10 eV
so ∆E ≥
2 ∆t 4π ∆t
2
4π 2.00 × 10 −6 s
E=
a f
e
j
h2
n x2 + n y2 .
2
8m e L
e
j
For a normalizable wave function describing a particle, neither n x nor n y can be zero. The
ground state, corresponding to n x = n y = 1, has an energy of
E1 , 1 =
h2
2
8m e L
e1
2
h2
j
+ 12 =
4m e L2
.
The first excited state, corresponding to either n x = 2 , n y = 1 or n x = 1 , n y = 2 , has an energy
E2 , 1 = E1 , 2 =
h2
8m e L2
e2
2
j
+ 12 =
5h2
8m e L2
.
The second excited state, corresponding to n x = 2 , n y = 2 has an energy of
E2 , 2 =
h2
8m e L2
e2
2
j
+ 22 =
h2
m e L2
.
Finally, the third excited state, corresponding to either n x = 1 , n y = 3 or n x = 3, n x = 1 , has an
energy
E1 , 3 = E3 , 1 =
(b)
5h2
h2
12 + 3 2 =
.
2
8m e L
4m e L2
e
j
The energy difference between the second
excited state and the ground state is given
by
∆E = E2 , 2 − E1 , 1 =
3h2
.
=
4m e L2
h2
m e L2
−
h2
4m e L2
E 1, 3 , E 3, 1
E 2, 2
energy
h2
m e L2
E 1, 2 , E 2, 1
E 1, 1
Energy level diagram
FIG. P41.46(b)
0
Chapter 41
P41.47
x2 =
z
∞
2
x 2 ψ dx
−∞
FG
H
z
∞
P41.48
(a)
FG
H
z
IJ
K
L
nπ x
2 2
L2
L2
x sin 2
dx =
− 2 2
L0
3 2n π
L
Thus, x 2 =
IJ
K
nπ x
2
sin
.
L
L
For a one-dimensional box of width L, ψ n =
(from integral tables).
2
ψ dx = 1 becomes
−∞
A2
z
L4
cos 2
−L 4
or A 2 =
(b)
FG 2π x IJ dx = A FG L IJ LM π x + 1 sinFG 4π x IJ OP
H 2π K N L 4 H L K Q
H LK
= A2
−L 4
z
FG L IJ LM π OP = 1
H 2π K N 2 Q
2
4
.
and A =
L
L
The probability of finding the particle between 0 and
L8
2
ψ dx = A
2
z
L8
cos 2
0
0
P41.49
L4
2
FG 2π x IJ dx = 1 + 1 =
H L K 4 2π
L
is
8
0.409 .
For a particle with wave function
af
ψ x =
2 −x a
e
a
for x > 0
for x < 0 .
and
0
(a)
ψ x
(b)
Prob x < 0 =
af
2
a
= 0, x < 0
f
z
0
af
ψ x
2
dx =
−∞
Normalization
2 −2 x a
e
, x>0
a
FIG. P41.49
za f
0
0 dx = 0
−∞
z
z
z z FGH IJK
∞
(c)
af
ψ2 x =
and
af
ψ x
2
dx =
−∞
0
0
z
∞
2
−∞
0dx +
−∞
∞
0
a
2
ψ dx + ψ dx = 1
0
2 −2 x a
e
dx = 0 − e −2 x a
a
f
z
a
2
Prob 0 < x < a = ψ dx =
0
z FGH IJK
a
0
∞
0
e
j
= − e −∞ − 1 = 1
2 −2 x a
e
dx = − e −2 x a
a
a
0
= 1 − e −2 = 0.865
509
510
P41.50
Quantum Mechanics
(a)
The requirement that
E=
bpcg + emc j
2 2
2
K n = En − mc 2 =
(b)
nλ
h nh
= L so p = =
is still valid.
λ 2L
2
⇒ En =
FG nhc IJ + emc j
H 2L K
2
FG nhc IJ + emc j
H 2L K
2
2 2
2 2
− mc 2
Taking L = 1.00 × 10 −12 m, m = 9.11 × 10 −31 kg , and n = 1, we find K 1 = 4.69 × 10 −14 J .
e
j
2
6.626 × 10 −34 J ⋅ s
h2
=
Nonrelativistic, E1 =
8mL2 8 9.11 × 10 −31 kg 1.00 × 10 −12 m
e
je
j
2
= 6.02 × 10 −14 J .
Comparing this to K 1 , we see that this value is too large by 28.6% .
P41.51
LM
N
FG
H
U=
(b)
From Equation 41.12,
(c)
E = U + K and
d=
g
a7fe18k e m j
2
K = 2E1 =
7kee2
dE
= 0 for a minimum:
dd
3h2
e
(d)
IJ a fOP b
K Q
− 7 3 e2
7k e 2
e2
1 1
1
−1 + − + −1 +
+ −1 =
= − e
4π ∈0 d
2 3
2
4π ∈0 d
3d
(a)
e
3d 2
e
−
2h2
e j
8m e 9d 2
j
je
2
6.626 × 10 −34
h2
=
=
42m e k e e 2
42 9.11 × 10 −31 8.99 × 10 9 1.60 × 10 −19 C
a fe
je
of one atom, we get:
13
h2
.
36m e d 2
h2
=0
18m e d 3
Since the lithium spacing is a, where Na 3 = V , and the density is
F Vm IJ
a=G
H Nm K
=
F m I
=G
H density JK
13
(5.62 times larger than c).
F 1.66 × 10 kg × 7 I
=G
H 530 kg JK
−27
j
2
= 0.049 9 nm .
Nm
, where m is the mass
V
13
m = 2.80 × 10 −10 m = 0. 280 nm
511
Chapter 41
P41.52
(a)
ψ = Bxe − bmω
g
2= x2
dψ
− mω
= Be b
dx
g
2= x 2
FG
H
IJ
K
mω
− mω
2 xe b
2=
+ Bx −
FG
H
IJ
K
d ψ
F mω IJ xe b
= −3BG
H=K
dx
d 2ψ
mω
− mω
= Bx −
xe b
=
dx 2
2
g
2= x 2
g
− mω 2 = x 2
2
g
2= x2
= Be
b
FG mω IJ 2 xe b
H=K
F mω IJ x e b
+ BG
H=K
g
− mω 2 = x 2
−B
2
FG mω IJ x e b g
H=K
F mω IJ x FG − mω IJ xe b
− BG
H=K H =K
g
− mω 2 = x 2
2
2 − mω 2 = x
−B
g
− mω 2 = x 2
2
g
2
3 − mω 2 = x
Substituting into the Schrödinger Equation (41.13), we have
FG mω IJ xe b
H=K
g
− mω 2 = x 2
−3B
This is true if −3ω = −
FG mω IJ
H=K
+B
2
x3 e
b
g
− mω 2 = x 2
=−
2mE
=
2
Bxe
We never find the particle at x = 0 because ψ = 0 there.
(c)
ψ is maximized if
(d)
We require
z
g
+
FG mω IJ
H=K
2
x 2 Bxe
b
g
− mω 2 = x 2
.
2E
3 =ω
; it is true if E =
.
=
2
(b)
∞
b
− mω 2 = x 2
FG IJ
H K
=
dψ
mω
= 0 = 1 − x2
, which is true at x = ±
.
=
dx
mω
2
ψ dx = 1 :
−∞
1=
z
∞
B2 x2 e
b
g
− mω = x 2
z
dx = 2B 2 x 2 e
−∞
FG IJ
H K
2 1 2 mω
Then B = 1 4
=
π
3 4
b
g
− mω = x 2
F 4m ω I
GH π = JK
3
=
3
dx = 2B 2
π
1
4
b mω = g
3
=
B2 π 1 2=3 2
.
2 mω 3 2
a f
14
.
3
FG IJ
H K
=
1
1
4=
= 2 =ω . This is larger than the
, the potential energy is mω 2 x 2 = mω 2
mω
2
2
mω
3 =ω
, so there is zero classical probability of finding the particle here.
total energy
2
(e)
At x = 2
(f)
Probability = ψ dx = Bxe
2
Probability = δ
2
π1 2
FH
b
g
− mω 2 = x 2
IK
2
FG mω IJ FG 4= IJ e b
H = K H mω K
32
δ = δB 2 x 2 e −b mω
gb
− mω = 4 = mω
g=
g
= x2
8δ
FG mω IJ
H =π K
12
e −4
512
Quantum Mechanics
z
L
P41.53
(a)
FG π x IJ + 16 sin FG 2π x IJ + 8 sinFG π x IJ sinFG 2π x IJ OPdx = 1
HLK
H L K H L K H L KQ
LMFG L IJ + 16FG L IJ + 8 sinFG π x IJ sinFG 2π x IJ dxOP = 1
MNH 2 K H 2 K z H L K H L K PQ
LM 17L + 16 sin FG π x IJ cosFG π x IJ dxOP = A LM 17L + 16L sin FG π x IJ OP = 1
MN 2 z H L K H L K PQ MN 2 3π H L K PQ
2
A2
ψ dx = 1 :
z LMN
L
sin 2
2
0
0
A2
L
0
A2
x=L
L
2
2
3
x=0
0
A2 =
z
a
(b)
2
2
, so the normalization constant is A =
.
17L
17L
z LMN
a
2
ψ dx = 1 :
2
A cos 2
−a
−a
FG π x IJ + B
H 2a K
2
2
sin 2
FG π x IJ + 2 A B cosFG π x IJ sinFG π x IJ OPdx = 1
HaK
H 2a K H a K Q
2
The first two terms are A a and B a . The third term is:
z
a
2 A B cos
−a
FG π x IJ LM2 sinFG π x IJ cosFG π x IJ OPdx = 4 A B z cos FG π x IJ sinFG π x IJ dx
H 2a K N H 2a K H 2a K Q
H 2a K H 2a K
8a A B
F π x IJ = 0
=
cos G
H 2a K
3π
a
2
−a
a
3
−a
e
2
so that a A + B
*P41.54
(a)
x
z
∞
=
0
x
−∞
FG a IJ
HπK
F 4a I
= z xG
H π JK
∞
(b)
(c)
x
x
1
12
01
3
2
j = 1 , giving
2
2
A +B =
1
.
a
2
e − ax dx = 0 , since the integrand is an odd function of x.
12
2
x 2 e − ax dx = 0 , since the integrand is an odd function of x.
−∞
=
z
∞
x
−∞
b
1
ψ 0 +ψ 1
2
g
2
dx =
1
x
2
0
+
1
x
2
+
1
z
∞
af af
xψ 0 x ψ 1 x dx
−∞
The first two terms are zero, from (a) and (b). Thus:
F aI
= z xG J
HπK
F 2a I
= 2G
H π JK
∞
x
01
14
−∞
2
=
1
2a
12
F 4a I xe dx = 2F 2a I
e
GH π JK
GH π JK
1F π I
G J , from Table B.6
4Ha K
3
− ax 2 2
12
3
14
− ax 2 2
2
1 2∞
z
0
2
x 2 e − ax dx
Chapter 41
P41.55
2
2
With one slit open
P1 = ψ 1
With both slits open,
P = ψ 1 +ψ 2 .
At a maximum, the wave functions are in phase
Pmax = ψ 1 + ψ 2
At a minimum, the wave functions are out of phase
Pmin = ψ 1 − ψ 2
ψ1
P
Now 1 =
P2 ψ 2
c
c
h = c5.00 ψ
h c5.00 ψ
c
h.
c
h.
2
2
ψ1
= 5.00
ψ2
= 25.0 , so
ψ1 + ψ2
P
and max =
Pmin
ψ1 − ψ2
or P2 = ψ 2 .
2
2
2
513
2
2
2
2
+ψ2
−ψ2
h = a6.00f
h a4.00f
2
2
2
2
=
36.0
= 2.25 .
16.0
ANSWERS TO EVEN PROBLEMS
P41.2
1
2
P41.4
(a) 4; (b) 6.03 eV
P41.6
0.517 MeV , 3.31 × 10 −20 kg ⋅ m s
P41.8
FG 3hλ IJ
H 8m c K
P41.22
af
af
af
af
af
12
e
P41.10
(a) 5.13 meV ; (b) 9.41 eV ; (c) The much
smaller mass of the electron requires it to
have much more energy to have the same
momentum.
P41.12
(a)
FG 15hλ IJ
H 8m c K
12
; (b) 1.25λ
P41.16
P41.18
P41.20
see the solution;
=2k2
2m
L
; (b) 5.26 × 10 −5 ; (c) 3.99 × 10 −2 ;
2
(d) see the solution
(a)
F
GH
I
JK
=2
4x 2
− 6 ; (b) see the solution
2mL2 L2
P41.24
(a)
P41.26
see the solution
P41.28
1.03 × 10 −3
P41.30
85.9
P41.32
3.92%
P41.34
(a) see the solution; b =
e
P41.14
FG IJ
H K
FG IJ
H K
FG IJ
H K
FG IJ
H K
FG IJ
H K
FG IJ
H K
πx
2
cos
;
L
L
πx
2
P1 x = cos 2
;
L
L
2π x
2
ψ2 x =
sin
;
L
L
2π x
2
;
P2 x = sin 2
L
L
3π x
2
ψ3 x =
cos
;
L
L
3π x
2
P3 x = cos 2
;
L
L
(b) see the solution
af
(a) ψ 1 x =
0.250
FG
H
IJ
K
2π A
A
1
sin
; (b) see the solution;
−
L 2π
L
(c) 0.585L
(a)
(c) first excited state
mω
3
; (b) E = =ω ;
2
2=
514
Quantum Mechanics
P41.36
F mω I
(a) B = G
H π = JK
P41.38
see the solution
P41.40
(a) 2.00 × 10 −10 m; (b) 3.31 × 10 −24 kg ⋅ m s ;
(c) 0.172 eV
P41.42
(a) see the solution; (b) 0.092 0 , 0.908
14
F mω I
; (b) δ G
H π = JK
12
P41.44
(a) see the solution; (b) 0.200 ; (c) 0.351 ;
(d) 0.377 eV , 1.51 eV
P41.46
(a)
h2
5h2
h2
5h2
,
,
,
;
4m e L2 8m e L2 m e L2 4m e L2
3h2
(b) see the solution,
4m e L2
P41.48
(a)
P41.50
(a)
2
L
; (b) 0.409
FG nhc IJ
H 2L K
2
+ m 2 c 4 − mc 2 ;
(b) 46.9 fJ ; 28.6%
P41.52
(a)
=
3 =ω
; (b) x = 0 ; (c) ±
;
mω
2
F 4m ω I
(d) G
H π = JK
3
3
3
P41.54
14
; (e) 0; (f) 8δ
a f
(a) 0; (b) 0; (c) 2 a
−1 2
FG mω IJ
H =π K
12
e −4
42
Atomic Physics
CHAPTER OUTLINE
42.1
42.2
42.3
Atomic Spectra of Gases
Early Models of the Atom
Bohr’s Model of the
Hydrogen Atom
42.4 The Quantum Model of the
Hydrogen Atom
42.5 The Wave Functions of
Hydrogen
42.6 Physical Interpretation of
the Quantum Numbers
42.7 The Exclusion Principle
and the Periodic Table
42.8 More on Atomic Spectra:
Visible and X-ray
42.9 Spontaneous and
Stimulated Transitions
42.10 Lasers
ANSWERS TO QUESTIONS
Q42.1
Neon signs emit light in a bright-line spectrum, rather than in a
continuous spectrum. There are many discrete wavelengths
which correspond to transitions among the various energy
levels of the neon atom. This also accounts for the particular
color of the light emitted from a neon sign. You can see the
separate colors if you look at a section of the sign through a
diffraction grating, or at its reflection in a compact disk. A
spectroscope lets you read their wavelengths.
Q42.2
One assumption is natural from the standpoint of classical
physics: The electron feels an electric force of attraction to the
nucleus, causing the centripetal acceleration to hold it in orbit.
The other assumptions are in sharp contrast to the behavior of
ordinary-size objects: The electron’s angular momentum must
be one of a set of certain special allowed values. During the
time when it is in one of these quantized orbits, the electron
emits no electromagnetic radiation. The atom radiates a photon
when the electron makes a quantum jump from one orbit to a
lower one.
Q42.3
If an electron moved like a hockey puck, it could have any arbitrary frequency of revolution around
an atomic nucleus. If it behaved like a charge in a radio antenna, it would radiate light with
frequency equal to its own frequency of oscillation. Thus, the electron in hydrogen atoms would
emit a continuous spectrum, electromagnetic waves of all frequencies smeared together.
Q42.4
(a)
Yes—provided that the energy of the photon is precisely enough to put the electron into one
of the allowed energy states. Strangely—more precisely non-classically—enough, if the
energy of the photon is not sufficient to put the electron into a particular excited energy
level, the photon will not interact with the atom at all!
(b)
Yes—a photon of any energy greater than 13.6 eV will ionize the atom. Any “extra” energy
will go into kinetic energy of the newly liberated electron.
Q42.5
An atomic electron does not possess enough kinetic energy to escape from its electrical attraction to
the nucleus. Positive ionization energy must be injected to pull the electron out to a very large
separation from the nucleus, a condition for which we define the energy of the atom to be zero. The
atom is a bound system. All this is summarized by saying that the total energy of an atom is
negative.
515
516
Q42.6
Atomic Physics
From Equations 42.7, 42.8 and 42.9, we have − E = −
kee2
k e2 k e2
= + e − e = K + U e . Then K = E and
r
2r
2r
U e = −2 E .
Q42.7
Bohr modeled the electron as moving in a perfect circle, with zero uncertainity in its radial
coordinate. Then its radial velocity is always zero with zero uncertainty. Bohr’s theory violates the
uncertainty principle by making the uncertainty product ∆r∆p r be zero, less than the minimum
allowable
2
.
Q42.8
Fundamentally, three quantum numbers describe an orbital wave function because we live in threedimensional space. They arise mathematically from boundary conditions on the wave function,
expressed as a product of a function of r, a function of θ, and a function of φ.
Q42.9
Bohr’s theory pictures the electron as moving in a flat circle like a classical particle described by
∑ F = ma . Schrödinger’s theory pictures the electron as a cloud of probability amplitude in the
three-dimensional space around the hydrogen nucleus, with its motion described by a wave
equation. In the Bohr model, the ground-state angular momentum is 1 ; in the Schrödinger model
the ground-state angular momentum is zero. Both models predict that the electron’s energy is
−13.606 eV
with n = 1, 2 , 3 .
limited to discrete energy levels, given by
n2
Q42.10
The term electron cloud refers to the unpredictable location of an electron around an atomic nucleus.
It is a cloud of probability amplitude. An electron in an s subshell has a spherically symmetric
probability distribution. Electrons in p, d, and f subshells have directionality to their distribution. The
shape of these electron clouds influences how atoms form molecules and chemical compounds.
Q42.11
The direction of the magnetic moment due to an orbiting charge is given by the right hand rule, but
assumes a positive charge. Since the electron is negatively charged, its magnetic moment is in the
opposite direction to its angular momentum.
Q42.12
Practically speaking, no. Ions have a net charge and the magnetic force q v × B would deflect the
beam, making it difficult to separate the atoms with different orientations of magnetic moments.
Q42.13
The deflecting force on an atom with a magnetic moment is proportional to the gradient of the
magnetic field. Thus, atoms with oppositely directed magnetic moments would be deflected in
opposite directions in an inhomogeneous magnetic field.
Q42.14
If the exclusion principle were not valid, the elements and their chemical behavior would be grossly
different because every electron would end up in the lowest energy level of the atom. All matter
would be nearly alike in its chemistry and composition, since the shell structures of all elements
would be identical. Most materials would have a much higher density. The spectra of atoms and
molecules would be very simple, and there would be very little color in the world.
Q42.15
The Stern-Gerlach experiment with hydrogen atoms shows that the component of an electron’s spin
angular momentum along an applied magnetic field can have only one of two allowed values. So
does electron spin resonance on atoms with one unpaired electron.
a
f
Chapter 42
517
Q42.16
The three elements have similar electronic configurations. Each has filled inner shells plus one
electron in an s orbital. Their single outer electrons largely determine their chemical interactions
with other atoms.
Q42.17
When a photon interacts with an atom, the atom’s orbital angular momentum changes, thus the
photon must carry orbital angular momentum. Since the allowed transitions of an atom are
restricted to a change in angular momentum of ∆ = ±1 , the photon must have spin 1.
Q42.18
In a neutral helium atom, one electron can be modeled as moving in an electric field created by the
nucleus and the other electron. According to Gauss’s law, if the electron is above the ground state it
moves in the electric field of a net charge of +2 e − 1e = +1e We say the nuclear charge is screened by
the inner electron. The electron in a He + ion moves in the field of the unscreened nuclear charge of
2 protons. Then the potential energy function for the electron is about double that of one electron in
the neutral atom.
Q42.19
At low density, the gas consists of essentially separate atoms. As the density increases, the atoms
interact with each other. This has the effect of giving different atoms levels at slightly different
energies, at any one instant. The collection of atoms can then emit photons in lines or bands,
narrower or wider, depending on the density.
Q42.20
An atom is a quantum system described by a wave function. The electric force of attraction to the
nucleus imposes a constraint on the electrons. The physical constraint implies mathematical
boundary conditions on the wave functions, with consequent quantization so that only certain wave
functions are allowed to exist. The Schrödinger equation assigns a definite energy to each allowed
wave function. Each wave function is spread out in space, describing an electron with no definite
position. If you like analogies, think of a classical standing wave on a string fixed at both ends. Its
position is spread out to fill the whole string, but its frequency is one of a certain set of quantized
values.
Q42.21
Each of the electrons must have at least one quantum number different
from the quantum numbers of each of the other electrons. They can differ
(in m s ) by being spin-up or spin-down. They can also differ (in ) in
angular momentum and in the general shape of the wave function. Those
electrons with = 1 can differ (in m ) in orientation of angular
momentum—look at Figure Q42.21.
FIG. Q42.21
Q42.22
The Mosely graph shows that the reciprocal square root of the wavelength of K α characteristic xrays is a linear function of atomic number. Then measuring this wavelength for a new chemical
element reveals its location on the graph, including its atomic number.
Q42.23
No. Laser light is collimated. The energy generally travels in the same direction. The intensity of a
laser beam stays remarkably constant, independent of the distance it has traveled.
Q42.24
Stimulated emission coerces atoms to emit photons along a specific axis, rather than in the random
directions of spontaneously emitted photons. The photons that are emitted through stimulation can
be made to accumulate over time. The fraction allowed to escape constitutes the intense, collimated,
and coherent laser beam. If this process relied solely on spontaneous emission, the emitted photons
would not exit the laser tube or crystal in the same direction. Neither would they be coherent with
one another.
518
Q42.25
Atomic Physics
(a)
The terms “I define” and “this part of the universe” seem vague, in contrast to the precision
of the rest of the statement. But the statement is true in the sense of being experimentally
verifiable. The way to test the orientation of the magnetic moment of an electron is to apply
a magnetic field to it. When that is done for any electron, it has precisely a 50% chance of
being either spin-up or spin-down. Its spin magnetic moment vector must make one of two
12
S
and
allowed angles with the applied magnetic field. They are given by cos θ = z =
S
3 2
cos θ =
−1 2
3 2
. You can calculate as many digits of the two angles allowed by “space
quantization” as you wish.
(b)
This statement may be true. There is no reason to suppose that an ant can comprehend the
cosmos, and no reason to suppose that a human can comprehend all of it. Our experience
with macroscopic objects does not prepare us to understand quantum particles. On the
other hand, what seems strange to us now may be the common knowledge of tomorrow.
Looking back at the past 150 years of physics, great strides in understanding the Universe—
from the quantum to the galactic scale—have been made. Think of trying to explain the
photoelectric effect using Newtonian mechanics. What seems strange sometimes just has an
underlying structure that has not yet been described fully. On the other hand still, it has
been demonstrated that a “hidden-variable” theory, that would model quantum uncertainty
as caused by some determinate but fluctuating quantity, cannot agree with experiment.
SOLUTIONS TO PROBLEMS
Section 42.1
P42.1
(a)
Atomic Spectra of Gases
Lyman series
1
λ
1
λ
(b)
Paschen series:
1
λ
F
GH
= R 1−
=
1
ni2
I
JK
ni = 2 , 3 , 4, …
jFGH
1
1
= 1.097 × 10 7 1 − 2
−9
ni
94.96 × 10
=R
e
F1 − 1I
GH 3 n JK
2
I
JK
ni = 5
ni = 4, 5 , 6 , …
2
i
The shortest wavelength for this series corresponds to ni = ∞ for ionization
F
GH
1
1 1
= 1.097 × 10 7
−
9 ni2
λ
I
JK
For ni = ∞ , this gives λ = 820 nm
This is larger than 94.96 nm, so this wave length
cannot be associated with the Paschen series .
Balmer series:
F1 − 1I
GH 2 n JK
λ
F1 1 I
1
= 1.097 × 10 G − J
λ
H4 n K
1
=R
2
2
i
7
2
i
ni = 3 , 4, 5 , …
with ni = ∞ for ionization, λ min = 365 nm
Once again the shorter given wavelength cannot be associated with the Balmer series .
Chapter 42
P42.2
(a)
(b)
Section 42.2
P42.3
(a)
hc
Emax
λ min =
Lyman (n f = 1 ):
λ min =
hc 1 240 eV ⋅ nm
=
= 91.2 nm
E1
13.6 eV
(Ultraviolet)
Balmer (n f = 2 ):
λ min =
hc 1 240 eV ⋅ nm
=
= 365 nm
E2
1 4 13.6 eV
(UV)
Paschen (n f = 3 ):
λ min
Bracket (n f = 4 ):
λ min
Emax =
2
821 nm
(Infrared)
2
1 460 nm
(IR)
hc
λ min
c= E h
3.40 eV c = E h
1.51 eV c = E h
0.850 eV c= E h
Lyman:
Emax = 13.6 eV
1
Balmer:
Emax =
2
Paschen:
Emax =
Brackett:
Emax =
3
4
Early Models of the Atom
For a classical atom, the centripetal acceleration is
a=
1
v2
e2
=
4π ∈0 r 2 m e
r
E=−
so
m v2
e2
e2
+ e =−
4π ∈0 r
2
8π ∈0 r
F
GH
−1 e 2 a 2
−e2
dE
e2
dr
e2
=
=
=
dt 8π ∈0 r 2 dt 6π ∈0 c 3
6π ∈0 c 3 4π ∈0 r 2 m e
−
z
0
12π
I
JK
2
.
dr
e4
=−
.
2 2 2 2 3
dt
12π ∈0 r m e c
Therefore,
(b)
b g
= … = 3 a91.2 nmf =
= … = 4 a91.2 nmf =
2
2.00 ×10 −10 m
∈02
r
2
m e2 c 3 dr
=e
4
z
T
0
dt
12π 2 ∈02 m e2 c 3 r 3
3
e4
FIG. P42.3
2.00 ×10 −10
= T = 8.46 × 10 −10 s
0
Since atoms last a lot longer than 0.8 ns, the classical laws (fortunately!) do not hold for
systems of atomic size.
519
520
P42.4
Atomic Physics
(a)
The point of closest approach is found when
k q q
E = K + U = 0 + e α Au
r
k e 2 e 79 e
or
rmin =
E
a fa f
e8.99 × 10 N ⋅ m C ja158fe1.60 × 10 Cj
=
a4.00 MeVfe1.60 × 10 J MeV j
9
rmin
(b)
2
−19
2
2
−13
= 5.68 × 10 −14 m .
The maximum force exerted on the alpha particle is
Fmax =
k e qα q Au
2
rmin
e8.99 × 10
=
9
ja fe
N ⋅ m 2 C 2 158 1.60 × 10 −19 C
e5.68 × 10
−14
j
m
2
j
2
= 11.3 N away from the
nucleus.
Section 42.3
P42.5
Bohr’s Model of the Hydrogen Atom
kee2
m e r1
v1 =
(a)
af
e8.99 × 10 N ⋅ m C je1.60 × 10 Cj
=
e9.11 × 10 kg je5.29 × 10 mj
2
r1 = 1 a 0 = 0.005 29 nm = 5.29 × 10 −11 m
where
9
v1
P42.6
2
−19
2
−31
−11
1
1
m e v12 = 9.11 × 10 −31 kg 2.19 × 10 6 m s
2
2
e
je
j
2
K1 =
(c)
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
kee2
=−
U1 = −
r1
5. 29 × 10 −11 m
a
∆E = 13.6 eV
= 2.19 × 10 6 m s
= 2.18 × 10 −18 J = 13.6 eV
(b)
e
2
je
j
2
= −4.35 × 10 −18 J = −27.2 eV
fFG n1 − n1 IJ
H
K
2
i
2
f
Where for ∆E > 0 we have absorption and for ∆E < 0 we have emission.
(i)
for ni = 2 and n f = 5 , ∆E = 2.86 eV (absorption)
(ii)
for ni = 5 and n f = 3 , ∆E = −0.967 eV (emission)
(iii)
for ni = 7 and n f = 4 , ∆E = −0.572 eV (emission)
(iv)
for ni = 4 and n f = 7 , ∆E = 0.572 eV (absorption)
(a)
E=
(b)
The atom gains most energy in transition i .
(c)
The atom loses energy in transitions ii and iii .
hc
λ
so the shortest wavelength is emitted in transition ii .
Chapter 42
P42.7
ga f
b
(a)
r22 = 0.052 9 nm 2
(b)
meke e2
me v2 =
=
r2
2
= 0.212 nm
e9.11 × 10
−31
je
je
kg 8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
j
2
0.212 × 10 −9 m
m e v 2 = 9.95 × 10 −25 kg ⋅ m s
P42.8
e
je
j
(c)
L 2 = m e v 2 r2 = 9.95 × 10 −25 kg ⋅ m s 0.212 × 10 −9 m = 2.11 × 10 −34 kg ⋅ m 2 s
(d)
me v2
1
K 2 = m e v 22 =
2
2m e
(e)
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
kee2
=−
U2 = −
r2
0.212 × 10 −9 m
(f)
E2 = K 2 + U 2 = 3.40 eV − 6.80 eV = −3.40 eV
b
g = e9.95 × 10 kg ⋅ m sj
kg j
2e9.11 × 10
−25
2
2
−31
e
= 5.43 × 10 −19 J = 3.40 eV
je
j
En =
We use
2
= −1.09 × 10 −18 J = −6.80 eV
−13.6 eV
.
n2
To ionize the atom when the electron is in the n th level,
it is necessary to add an amount of energy given by
P42.9
E = − En =
13.6 eV
.
n2
(a)
Thus, in the ground state where n = 1, we have E = 13.6 eV .
(b)
In the n = 3 level,
E=
13.6 eV
= 1.51 eV .
9
(a)
F 1 − 1 I = 1.097 × 10 m F 1 − 1 I so λ = 410 nm
jGH 2 6 JK
GH n n JK e
λ
J ⋅ sje3.00 × 10 m sj
hc e6.626 × 10
J = 3.03 eV
E=
=
= 4.85 × 10
(c)
f=
(b)
1
=R
2
f
7
2
i
−34
λ
2
=
2
8
410 × 10 −9 m
λ
c
−1
3.00 × 10 8
= 7.32 × 10 14 Hz
410 × 10 −9
−19
521
522
P42.10
P42.11
Atomic Physics
Starting with
k e2
1
mev2 = e
2
2r
we have
v2 =
kee2
mer
and using
rn =
n2 2
me ke e2
gives
vn2 =
or
vn =
ke e2
e
me n 2
2
me ke e2
j
kee2
.
n
Each atom gives up its kinetic energy in emitting a photon,
so
e
je
j
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
1
hc
2
mv =
=
= 1.63 × 10 −18 J
λ
2
1.216 × 10 −7 m
e
j
v = 4.42 × 10 4 m s .
P42.12
The batch of excited atoms must make these six transitions to get back to state one: 2 → 1 , and also
3 → 2 and 3 → 1 , and also 4 → 3 and 4 → 2 and 4 → 1 . Thus, the incoming light must have just
enough energy to produce the 1 → 4 transition. It must be the third line of the Lyman series in the
absorption spectrum of hydrogen. The absorbing atom changes from energy
Ei = −
13.6 eV
13.6 eV
= −13.6 eV to E f = −
= −0.850 eV ,
2
1
42
so the incoming photons have wavelength
e
je
a
j F 1.00 eV I = 9.75 × 10
f GH 1.60 × 10 J JK
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
λ=
=
−0.850 eV − −13.6 eV
E f − Ei
P42.13
(a)
−19
−8
m = 97.5 nm .
The energy levels of a hydrogen-like ion whose charge number
is Z are given by
a
En = −13.6 eV
f Zn
2
2
a
.
f
Thus for Helium Z = 2 , the energy levels are
En = −
(b)
54.4 eV
n2
n = 1, 2 , 3 , … .
For He + , Z = 2 , so we see that the ionization energy (the
energy required to take the electron from the n = 1 to the n = ∞
state) is
E = E∞ − E1 = 0 −
a−13.6 eVfa2f
a1f
2
2
= 54.4 eV .
FIG. P42.13
Chapter 42
*P42.14
(a)
1
= Z 2 RH
λ
523
F 1 − 1 I . The shortest wavelength, λ , corresponds to n = ∞ , and the longest
GH n n JK
2
f
s
2
i
i
wavelength, λ , to ni = n f + 1 .
1
λs
=
Z 2 RH
(1)
n 2f
L 1 1 OP Z R L F n I
=Z R M −
MM n dn + 1i PP = n MM1 − GH n + 1 JK
λ
N
N
Q
F n I
λ
=1−G
Divide (1) and (2):
λ
H n + 1 JK
1
2
2
H
2
f
2
2
f
f
f
H
f
2
OP
PQ
(2)
2
f
s
f
∴
nf
nf +1
= 1−
λs
22.8 nm
= 1−
= 0.800
63.3 nm
λ
n 2f
From (1): Z =
=
λ s RH
e22.8 × 10
∴n f = 4
42
−9
je
m 1.097 × 10 7 m −1
j
= 8.00 .
Hence the ion is O 7+ .
(b)
R|
λ = Se7.020 8 × 10
|T
8
m
−1
L
jMM 41 − a4 +1kf
N
2
2
OPU|
PQV|W
−1
, k = 1, 2 , 3 , …
Setting k = 2 , 3 , 4 gives λ = 41.0 nm, 33.8 nm, 30.4 nm .
P42.15
(a)
The speed of the moon in its orbit is v =
So,
(b)
j
je
je
j
L = mvr = 7.36 × 10 22 kg 1.02 × 10 3 m s 3.84 × 10 8 m = 2.89 × 10 34 kg ⋅ m 2 s .
We have L = n
or
(c)
e
e
8
2π r 2π 3.84 × 10 m
=
= 1.02 × 10 3 m s .
T
2.36 × 10 6 s
n=
L
=
2.89 × 10 34 kg ⋅ m 2 s
1.055 × 10 −34 J ⋅ s
We have n = L = mvr = m
so
r=
2
m 2 GM e
FG GM IJ
H r K
e
12
n 2 = Rn 2 and
which is approximately equal to
= 2.74 × 10 68 .
r,
a f
2
n + 1 R − n 2 R 2n + 1
∆r
=
=
r
n2R
n2
2
= 7.30 × 10 −69 .
n
524
Atomic Physics
Section 42.4
P42.16
The Quantum Model of the Hydrogen Atom
The reduced mass of positronium is less than hydrogen, so the photon energy will be less for
positronium than for hydrogen. This means that the wavelength of the emitted photon will be
longer than 656.3 nm. On the other hand, helium has about the same reduced mass but more charge
than hydrogen, so its transition energy will be larger, corresponding to a wavelength shorter than
656.3 nm.
All the factors in the given equation are constant for this problem except for the reduced mass
and the nuclear charge. Therefore, the wavelength corresponding to the energy difference for the
transition can be found simply from the ratio of mass and charge variables.
mpme
For hydrogen,
µ=
Its wavelength is
λ = 656.3 nm ,
(a)
mp + me
µ=
For positronium,
≈ me .
The photon energy is
∆E = E3 − E2 .
where
λ=
c hc
=
.
f ∆E
meme
m
= e
2
me + me
so the energy of each level is one half as large as in hydrogen, which we could call
“protonium”. The photon energy is inversely proportional to its wavelength , so for
positronium,
a
f
λ 32 = 2 656.3 nm = 1.31 µm (in the infrared region).
(b)
For He + ,
µ ≈ m e , q1 = e , and q 2 = 2 e ,
so the transition energy is 2 2 = 4 times larger than hydrogen.
λ 32 =
Then,
P42.17
(a)
∆x∆p ≥
(b)
Choosing ∆p ≈
U=
(c)
2
so if ∆x = r , ∆p ≥
r
,K=
FG 656 IJ nm =
H4K
2r
164 nm (in the ultraviolet region).
.
b g
2
2
∆p
p2
≈
=
2m e
2m e
2m e r 2
2
−kee2
ke e2
−
, so E = K + U ≈
.
r
r
2m e r 2
To minimize E,
2
2
k e2
dE
=−
+ e2 = 0 → r =
= a0
3
dr
mer
r
meke e2
Then, E =
2
2m e
Fm k e I
GH
JK
e e
2
2
2
− kee2
(the Bohr radius).
Fm k e I =−m k e
GH
JK 2
e e
2
2
2 4
e e
2
= −13.6 eV .
Chapter 42
Section 42.5
P42.18
The Wave Functions of Hydrogen
af
1
ψ 1s r =
e−r
a 03
π
4r 2 −2 r
e
a 03
af
P1s r =
a0
(Eq. 42.22)
a0
(Eq. 42.25)
FIG. P42.18
P42.19
z
(a)
z
∞
2
2
ψ dV = 4π ψ r 2 dr = 4π
0
z
Using integral tables,
F 1Ire
GH π a JK z
2 L
dV = −
Me
a MN
∞
3
0
2 −2 r a 0
2
ψ
dr
0
2
0
−2 r a 0
Fr
GH
a2
+ a0 r + 0
2
2
I OP = F − 2 I F − a I =
JK PQ GH a JK GH 2 JK
∞
2
0
2
0
0
1
so the wave function as given is normalized.
Pa0
(b)
2 → 3 a0 2
= 4π
z
3 a0 2
2
ψ r 2 dr = 4π
a0 2
F 1I
GH π a JK
3
0
Again, using integral tables,
Pa0 2 → 3 a0 2 = −
P42.20
ψ=
1
1
b g
3 2 a0
3 2
LMe
MN
2
a 02
r −r
e
a0
−2 r a 0
Fr
GH
2
z
3 a0 2
a2
+ a0 r + 0
2
I OP
JK PQ
r 2 −r
e
24a 05
.
r 2 e −2 r
a0
dr
a0 2
3 a0 2
=−
a0 2
2
a 02
LMe F 17a I − e F 5 a I OP =
MN GH 4 JK GH 4 JK PQ
−3
2 a0
so
Pr = 4π r 2 ψ 2 = 4π r 2
Set
dP
4π
4r 3 e − r
=
dr 24a 05
LM
MN
a0
a0
FG 1 IJ e
H aK
+ r4 −
− r a0
0
OP = 0 .
PQ
Solving for r, this is a maximum at r = 4a 0 .
P42.21
ψ=
d 2ψ
dr
2
1
π
=
a 03
e −r
1
π a 07
−2
2 dψ
2
ψ
=
e − r a0 = −
5
r dr r π a
ra 0
0
a0
e−r
a0
2
But
so
or
a0 =
−
=
1
a 02
ψ
−
2
2m e
F 1 − 2 Iψ − e ψ = Eψ
GH a ra JK 4π ∈ r
2
2
0
0
b4π ∈ g
0
mee
2
2
e
=E
8π ∈0 a 0
E=−
kee2
.
2 a0
This is true, so the Schrödinger equation is satisfied.
0
2
0
−1
2
0
0.497 .
525
526
P42.22
Atomic Physics
The hydrogen ground-state radial probability density is
af
P r = 4π r 2 ψ 1s
2
=
4r 2
a 03
FG 2r IJ .
H aK
exp −
0
The number of observations at 2 a 0 is, by proportion
b g = 1 000 b2 a g
N = 1 000
Pb a 2g
b a 2g
P 2 a0
0
0
Section 42.6
2
2
0
e −4 a 0 a 0
= 1 000 16 e −3 = 797 times .
− a0 a0
e
a f
Physical Interpretation of the Quantum Numbers
Note: Problems 17 and 25 in Chapter 29 and Problem 68 in Chapter 30 can be assigned with this section.
P42.23
(a)
In the 3d subshell, n = 3 and
we have n
= 2,
m
3
2
+2
3
2
+2
3
2
+1
3
2
+1
3
2
0
3
2
0
3
2
–1
3
2
–1
3
2
–2
3
2
–2
ms
+1/2
–1/2
+1/2
–1/2
+1/2
–1/2
+1/2
–1/2
+1/2
–1/2
(A total of 10 states)
(b)
In the 3p subshell, n = 3 and
we have n
=1,
m
3
1
+1
3
1
+1
3
1
+0
3
1
+0
3
1
–1
3
1
–1
ms
+1/2
–1/2
+1/2
–1/2
+1/2
–1/2
(A total of 6 states)
P42.24
P42.25
= 2.58 × 10 −34 J ⋅ s .
(a)
For the d state,
= 2,
L=
6
(b)
For the f state,
=3,
L=
a + 1f
L=
a + 1f
:
=
= 3.65 × 10 −34 J ⋅ s .
a + 1f FGH 6.6262×π10 IJK
a + 1f = e4.714 × 10 j a2π f = 1.998 × 10
e6.626 × 10 j
−34
4.714 × 10 −34 =
−4 2
−34 2
so
12
=4 .
2
1
a f
≈ 20 = 4 4 + 1
Chapter 42
P42.26
The 5th excited state has n = 6 , energy
The atom loses this much energy:
P42.27
−13.6 eV
= −0.378 eV .
36
e6.626 × 10 J ⋅ sje3.00 × 10
=
λ e1 090 × 10 mje1.60 × 10
−34
hc
−9
−0.378 eV − 1.14 eV = −1.52 eV
which is the energy in state 3:
−
(a)
33
n =1:
For n = 1 ,
= 0 , m = 0 , ms = ±
1
1
ms
0
0
–1/2
+1/2
0
0
af
Yields 2 sets; 2n 2 = 2 1
n=2:
m
2
j = 1.14 eV
J eV j
ms
= −1.51 eV .
a + 1f
can be as large as 2, giving angular momentum
n
(b)
13.6 eV
8
−19
to end up with energy
While n = 3 ,
527
=
6
.
1
2
= 2
For n = 2 ,
we have
n
2
2
2
2
0
1
1
1
m
ms
0
–1
0
1
±1/2
±1/2
±1/2
±1/2
af
2n 2 = 2 2
yields 8 sets;
2
= 8
Note that the number is twice the number of m values. Also, for each
different m values. Finally,
can take on values ranging from 0 to n − 1 .
So the general expression is
number =
n −1
∑ 2a 2
f
+1 .
0
The series is an arithmetic progression:
2 + 6 + 10 + 14…
the sum of which is
number =
n
2a + n − 1 d
2
where a = 2 , d = 4 :
number =
n
4 + n − 1 4 = 2n 2 .
2
(c)
n=3:
(d)
n=4:
(e)
n=5:
af af af
2a1f + 2a3f + 2a5f + 2a7 f = 32
32 + 2a9f = 32 + 18 = 50
2 1 + 2 3 + 2 5 = 2 + 6 + 10 = 18
af
= 2a 4f
= 2a 5 f
a f
a f
2n 2 = 2 3
2
= 18
2n 2
2
= 32
2
= 50
2n 2
a
f
there are 2 + 1
528
P42.28
Atomic Physics
For a 3d state,
n = 3 and
Therefore,
L=
m can have the values
–2, –1, 0, 1, and 2
we find the possible values of θ
1.67 × 10 −27 kg
m
=
V
4 3 π 1.00 × 10 −15 m
ρ=
(b)
Size of model electron:
F 3m I
r =G
H 4π ρ JK
(c)
Moment of inertia:
I=
b ge
v=
P42.30
.
F 3e9.11 × 10
=G
GH 4π e3.99 × 10
13
= 3.99 × 10 17 kg m3 .
j IJ
kg m j JK
−31
17
kg
13
= 8.17 × 10 −17 m .
3
e
e
2
=
je
j
2
= 2.43 × 10 −63 kg ⋅ m 2
Iv
.
r
je
j
6.626 × 10 −34 J ⋅ s 8.17 × 10 −17 m
r
=
= 1.77 × 10 12 m s .
−63
2
2I
2π 2 × 2. 43 × 10
kg ⋅ m
e
j
This is 5.91 × 10 3 times larger than the speed of light.
In the N shell, n = 4 . For n = 4 , can take on values of 0, 1, 2, and 3. For each value of , m can be
− to in integral steps. Thus, the maximum value for m is 3. Since L z = m , the maximum
value for L z is L z = 3
P42.31
j
3
2
2
mr 2 = 9.11 × 10 −31 kg 8.17 × 10 −17 m
5
5
L z = Iω =
Therefore,
and 2
145° , 114° , 90.0° , 65.9° , and 35.3° .
Density of a proton:
(d)
= 2.58 × 10 −34 J ⋅ s
6
Lz
L
cos θ =
Using the relation
(a)
=
L z can have the values − 2 , − , 0 ,
so
P42.29
a + 1f
= 2.
The 3d subshell has
.
= 2 , and n = 3 . Also, we have s = 1 .
Therefore, we can have n = 3 ,
= 2 ; m = −2 , − 1, 0 , 1, 2 ; s = 1; and m s = −1, 0 , 1
leading to the following table:
n
m
s
ms
3
2
–2
3
2
–2
3
2
–2
3
2
–1
3
2
–1
3
2
–1
3
2
0
3
2
0
3
2
0
3
2
1
3
2
1
3
2
1
3
2
2
3
2
2
3
2
2
1
–1
1
0
1
1
1
–1
1
0
1
1
1
–1
1
0
1
1
1
–1
1
0
1
1
1
–1
1
0
1
1
529
Chapter 42
Section 42.7
P42.32
The Exclusion Principle and the Periodic Table
1s 2 2 s 2 2 p 4
(a)
(b)
P42.33
For the 1s electrons,
n =1,
=0, m =0,
ms = +
1
2
and
−
1
.
2
For the two 2s electrons,
n=2,
=0, m =0,
ms = +
1
2
and
−
1
.
2
For the four 2p electrons,
n=2;
= 1 ; m = −1 , 0, or 1; and
ms = +
1
2
or
−
1
.
2
The 4s subshell fills first , for potassium and calcium, before the 3d subshell starts to fill for
scandium through zinc. Thus, we would first suppose that Ar 3d 4 4s 2 would have lower energy
than Ar 3d 5 4s 1 . But the latter has more unpaired spins, six instead of four, and Hund’s rule
suggests that this could give the latter configuration lower energy. In fact it must, for Ar 3d 5 4s 1 is
the ground state for chromium.
P42.34
Electronic configuration:
2
2
1s 2s 2 p
6
→
Na 11
+3s 2
→
Mg 12
+3s 2 3 p 1
→
Al 13
+3s 2 3 p 2
→
Si 14
+3s 2 3 p 3
→
P 15
+3s 2 3 p 4
→
S 16
+3s 2 3 p 5
→
Cl 17
+3s 2 3 p 6
→
Ar 18
→
K 19
+3s
1s 2 2s 2 2 p 6 3s 2 3 p 6 4s 1
*P42.35
Sodium to Argon
1
In the table of electronic configurations in the text, or on a periodic table, we look for the element
whose last electron is in a 3p state and which has three electrons outside a closed shell. Its electron
configuration then ends in 3s 2 3 p 1 . The element is aluminum .
530
P42.36
Atomic Physics
(a)
For electron one and also for electron two, n = 3 and
here in columns giving the other quantum numbers:
= 1 . The possible states are listed
electron m
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
one
1
2
1
1
2
0
1
2
0
1
2
–1
1
2
–1
1
−
2
1
1
−
2
0
1
−
2
0
1
−
2
–1
1
−
2
–1
1
2
1
1
2
1
1
2
0
1
2
–1
1
2
–1
1
2
−
1
2
1
2
−
1
2
1
2
−
1
2
1
2
−
1
2
1
2
−
0
–1
–1
–1
–1
–1
–1
–1
–1
–1
–1
1
2
–1
1
2
1
1
2
1
1
2
0
1
2
0
1
2
–1
1
2
1
2
ms
electron m
two
ms
electron m
−
1
2
0
electron m
1
2
1
two
1
2
one
ms
ms
−
1
2
−
0
0
1
2
1
−
−
1
2
1
2
1
2
0
−
0
1
2
–1
−
1
2
1
2
1
2
−
−
−
1
2
1
2
−
1
2
−
1
2
1
2
1
−
1
2
1
2
1
2
1
−
−
1
2
−
1
2
0
−
1
2
1
2
0
−
−
1
2
1
2
1
2
–1
−
1
2
There are thirty allowed states, since electron one can have any of three possible values for
m for both spin up and spin down, amounting to six states, and the second electron can
have any of the other five states.
(b)
Were it not for the exclusion principle, there would be 36 possible states, six for each
electron independently.
P42.37
(a)
n+
subshell
(b)
Z = 15 :
Z = 47 :
Z = 86 :
1
1s
2
2s
3
2p, 3s
5
3d, 4p, 5s
6
4d, 5p, 6s
7
4f, 5d, 6p, 7s
Valence subshell:
Prediction:
Element is phosphorus,
1s , 2s , 2 p , 3s
(12 electrons)
3 electrons in 3p subshell
Valence = +3 or – 5
Valence = +3 or – 5 (Prediction correct)
Filled subshells:
1s , 2s , 2 p , 3s , 3 p , 4s , 3 d , 4 p , 5 s
Outer subshell:
Prediction:
Element is silver,
(38 electrons)
9 electrons in 4d subshell
Valence = −1
(Prediction fails)
Valence is +1
Filled subshells:
Filled subshells:
Prediction
Element is radon, inert
P42.38
4
3p, 4s
1s , 2s , 2 p , 3s , 3 p , 4s , 3 d , 4 p , 5 s , 4d , 5 p , 6 s ,
4 f , 5d , 6 p
(86 electrons)
Outer subshell is full:
inert gas
(Prediction correct)
Listing subshells in the order of filling, we have for element 110,
1s 2 2s 2 2 p 6 3s 2 3 p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5 p 6 6s 2 4 f 14 5 d 10 6 p 6 7 s 2 5 f 14 6 d 8 .
In order of increasing principal quantum number, this is
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4p 6 4d 10 4 f 14 5s 2 5 p 6 5d 10 5 f 14 6s 2 6 p 6 6d 8 7 s 2 .
Chapter 42
*P42.39
In the ground state of sodium, the outermost electron is in an s state. This state is spherically
symmetric, so it generates no magnetic field by orbital motion, and has the same energy no matter
whether the electron is spin-up or spin-down. The energies of the states 3p and 3p above 3s are
hc
hc
hf1 =
and hf 2 =
.
A
λ
2 µ B B = hc
B=
so
B
λ2
The energy difference is
FG 1
Hλ
1
−
1
λ2
IJ
K
IJ e
K
FG
H
je
6.63 × 10 −34 J ⋅ s 3 × 10 8 m s
hc
1
1
−
=
2µ B λ 1 λ 2
2 9.27 × 10 −24 J T
e
j
j FG
1
H 588.995 × 10
−9
m
−
1
589.592 × 10
−9
IJ
mK
B = 18.4 T .
Section 42.8
P42.40
More on Atomic Spectra: Visible and X-ray
(a)
n = 3,
= 0, m = 0
n = 3,
= 1 , m = −1 , 0 , 1
For n = 3 ,
= 2 , m = −2 , − 1 , 0 , 1 , 2
ψ 300 corresponds to E300 = −
(b)
Z 2 E0
n
2
=−
a f=
2 2 13.6
32
−6.05 eV .
ψ 31 −1 , ψ 310 , ψ 311 have the same energy since n is the same.
ψ 32 − 2 , ψ 32 −1 , ψ 320 , ψ 321 , ψ 322 have the same energy since n is the same.
All states are degenerate.
P42.41
E=
hc
λ
e6.626 × 10 J ⋅ sje3.00 × 10
e10.0 × 10 mj
−34
= e∆ V :
8
ms
−9
j = e1.60 × 10 j∆V
−19
∆V = 124 V
P42.42
531
Some electrons can give all their kinetic energy K e = e∆V to the creation of a single photon of xradiation, with
hf =
λ=
hc
λ
= e∆ V
e
je
j
6.626 1 × 10 −34 J ⋅ s 2.997 9 × 10 8 m s
1 240 nm ⋅ V
hc
=
=
−19
∆V
e∆V
1.602 2 × 10
C ∆V
e
j
532
P42.43
Atomic Physics
Eγ =
Following Example 42.9
a
3
42 − 1
4
f a13.6 eVf = 1.71 × 10
2
4
eV = 2.74 × 10 −15 J
f = 4.14 × 10 18 Hz
λ = 0.072 5 nm .
and
P42.44
E=
For
hc
λ
=
1 240 eV ⋅ nm
=
1.240 keV ⋅ nm
λ
λ 1 = 0.018 5 nm ,
λ
E = 67.11 keV
λ 2 = 0.020 9 nm ,
E = 59.4 keV
λ 3 = 0.021 5 nm ,
E = 57.7 keV
The ionization energy for the K shell is 69.5 keV, so the ionization
energies for the other shells are:
L shell = 11.8 keV
P42.45
M shell = 10.1 keV
FIG. P42.44
N shell = 2.39 keV .
The K β x-rays are emitted when there is a vacancy in the ( n = 1 ) K shell and an electron from the
( n = 3 ) M shell falls down to fill it. Then this electron is shielded by nine electrons originally and by
one in its final state.
hc
λ
=−
a
13.6 Z − 9
32
f
2
eV +
a
13.6 Z − 1
f
2
12
eV
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = a13.6 eVfF − Z
GH 9
e0.152 × 10 mje1.60 × 10 J eVj
F 8 Z − 8I
8.17 × 10 eV = a13.6 eV fG
H 9 JK
−34
8
−9
2
−19
8Z 2
−8
9
so
601 =
and
Z = 26
Iron .
Section 42.9
Spontaneous and Stimulated Transitions
Section 42.10
Lasers
a
f
E4 − E3 = 20.66 − 18.70 eV = 1.96 eV =
The photon energy is
λ=
P42.47
I
JK
18Z 81
−
+ Z 2 − 2Z + 1
9
9
2
3
P42.46
+
F
GH
e6.626 × 10
−34
je
8
J ⋅ s 3.00 × 10 m s
e
j
1.96 1.60 × 10 −19 J
I FG 1 J IJ =
JK H 1 V ⋅ C K
f=
0.117 eV
1.60 × 10 −19 C
E
=
h 6.630 × 10 −34 J ⋅ s
e
λ=
c 3.00 × 10 8 m s
=
= 10.6 µm , infrared
f 2.82 × 10 13 s −1
hc
λ
j=
2.82 × 10 13 s −1
633 nm .
Chapter 42
(a)
e3.00 × 10 Jj
I=
e1.00 × 10 sjLMNπ e15.0 × 10
(b)
e3.00 × 10
533
−3
P42.48
P42.49
−9
e0.600 × 10 mj
Jj
e30.0 × 10 mj
−9
−3
e
2
2
−6
je
4.24 × 10 15 W m 2
=
mj O
PQ
2
= 1.20 × 10 −12 J = 7.50 MeV
j
E = P ∆t = 1.00 × 10 6 W 1.00 × 10 −8 s = 0.010 0 J
Eγ = hf =
N=
*P42.50
−6
(a)
hc
λ
e6.626 × 10 je3.00 × 10 j J = 2.86 × 10
−34
=
694.3 × 10
8
−9
−19
J
0.010 0
E
=
= 3.49 × 10 16 photons
Eγ 2.86 × 10 −19
− E3
N3 N g e
=
N 2 N g e − E2
b k ⋅300 K g
b k ⋅300 K g
B
B
=e
b
− E3 − E 2
g b k ⋅300 K g = e − hc λ b k ⋅300 K g
B
B
where λ is the wavelength of light radiated in the 3 → 2 transition.
N3
− 6.63 × 10 −34 J⋅s je 3 ×10 8 m s j
=e e
N2
N3
= e −75 .9 = 1.07 × 10 −33
N2
(b)
Nu
− E −E
=e b u
N
e632.8 ×10
−9
ja
je
m 1.38 × 10 −23 J K 300 K
f
gkT
B
where the subscript u refers to an upper energy state and the subscript
state.
Since Eu − E = Ephoton =
hc
λ
to a lower energy
Nu
= e − hc λk BT .
N
1.02 = e − hc λk BT
Thus, we require
a f
ln 1.02 = −
or
T=−
e6.63 × 10
e632.8 × 10
−34
−9
je
j
J K jT
J ⋅ s 3 × 10 8 m s
je
m 1.38 × 10
23
2.28 × 10 4
= −1.15 × 10 6 K .
ln 1.02
a f
A negative-temperature state is not achieved by cooling the system below 0 K, but by
heating it above T = ∞ , for as T → ∞ the populations of upper and lower states approach
equality.
(c)
Because Eu − E > 0 , and in any real equilibrium state T > 0 ,
e − bEu − E
g k T <1
B
and
Nu < N .
Thus, a population inversion cannot happen in thermal equilibrium.
534
*P42.51
Atomic Physics
(a)
The light in the cavity is incident perpendicularly on the
mirrors, although the diagram shows a large angle of
incidence for clarity. We ignore the variation of the index
of refraction with wavelength. To minimize reflection at
a vacuum wavelength of 632.8 nm, the net phase
difference between rays (1) and (2) should be 180°. There
is automatically a 180° shift in one of the two rays upon
reflection, so the extra distance traveled by ray (2) should
be one whole wavelength:
1
2
SiO2
FIG. P42.51
λ
n
λ 632.8 nm
t=
=
= 217 nm
2 1.458
2n
2t =
a
(b)
f
The total phase difference should be 360°, including contributions of 180° by reflection and
180° by extra distance traveled
λ
2n
λ 543 nm
t=
=
= 93.1 nm
4n 4 1.458
2t =
a
f
Additional Problems
*P42.52
(a)
Using the same procedure that was used in the Bohr model of the hydrogen atom, we apply
Newton’s second law to the Earth. We simply replace the Coulomb force by the
gravitational force exerted by the Sun on the Earth and find
G
MS M E
r2
= ME
v2
r
(1)
where v is the orbital speed of the Earth. Next, we apply the postulate that angular
momentum of the Earth is quantized in multiples of :
M E vr = n
bn = 1, 2, 3, …g .
Solving for v gives
v=
n
.
MEr
(2)
Substituting (2) into (1), we find
r=
continued on next page
n2 2
.
GM S M E2
(3)
Chapter 42
(b)
535
Solving (3) for n gives
n = GM S r
ME
.
(4)
Taking MS = 1.99 × 10 30 kg , and M E = 5.98 × 10 24 kg , r = 1.496 × 10 11 m ,
= 1.055 × 10 −34 Js , we find
G = 6.67 × 10 −11 Nm 2 kg 2 , and
n = 2.53 × 10 74 .
(c)
We can use (3) to determine the radii for the orbits corresponding to the quantum numbers
n and n + 1 :
a f
2
n +1 2
n2 2
=
r
and
.
+
n
1
GM S M E2
GM S M E2
Hence, the separation between these two orbits is
rn =
2
∆r =
GM S M E2
an + 1f
2
− n2 =
2
GM S M E2
a2n + 1f .
Since n is very large, we can neglect the number 1 in the parentheses and express the
separation as
∆r ≈
2
GM S M E2
a2nf =
1.18 × 10 −63 m .
e
j
This number is much smaller than the radius of an atomic nucleus ~ 10 −15 m , so the
distance between quantized orbits of the Earth is too small to observe.
*P42.53
(a)
e
ja
−19
C 6.63 × 10 −34 J ⋅ s 5.26 T
e B 1.60 × 10
∆E =
=
me
2π 9.11 × 10 −31 kg
e
j
f F N ⋅ s I F kg ⋅ m I = 9.75 × 10
GH T ⋅ C ⋅ m JK GH N ⋅ s JK
2
= 609 µeV
e
je
k BT = 1.38 × 10 −23 J K 80 × 10 −3 K = 1.10 × 10 −24 J = 6.90 µeV
(c)
f=
∆E
9.75 × 10 −23 J
=
= 1.47 × 10 11 Hz
h
6.63 × 10 -34 J ⋅ s
λ=
3 × 10 8 m s
c
=
= 2.04 × 10 −3 m
f 1.47 × 10 11 Hz
z
∞
*P42.54
j
(b)
(a)
LM F
MN GH
z
I
JK
4 ∞
2r ′ 2 2r ′
+
+ 1 e −2 r ′ a 0
Probability = P1 s r ′ dr ′ = 3 r ′ 2 e −2 r ′ a0 dr ′ = −
2
a
a
a
0
0
0 r
r
af
using integration by parts, or Example 42.5
=
continued on next page
F 2r
GH a
2
2
0
+
I
JK
2r
+ 1 e −2 r
a0
a0
OP
PQ
∞
,
r
−23
J
536
Atomic Physics
(b)
Probability Curve for Hydrogen
1.2
1
0.8
0.6
0.4
0.2
0
0
1
2
3
r /a0
4
5
FIG. P42.66
(c)
The probability of finding the electron inside or outside the sphere of radius r is
∴
F 2r
GH a
2
2
0
+
I
JK
2r
+ 1 e −2 r
a0
a0
=
1
.
2
2r
1
or z 2 + 2 z + 2 = e z where z =
a0
2
One can home in on a solution to this transcendental equation for r on a calculator, the
result being r = 1.34a0 to three digits.
P42.55
Let r represent the distance between the electron and the positron. The two move in a circle of
r
radius around their center of mass with opposite velocities. The total angular momentum of the
2
electron-positron system is quantized to according to
Ln =
mvr mvr
+
=n
2
2
n = 1, 2 , 3 , … .
where
For each particle,
kee2
∑ F = ma expands to
We can eliminate v =
r
2
=
mv 2
.
r 2
k e e 2 2mn 2 2
=
.
r
m 2r 2
n
to find
mr
r=
So the separation distances are
2n 2 2
= 2 a0n 2 =
mk e e 2
e1.06 × 10
−10
j
m n2 .
r
= a 0 n 2 , the same as for the electron in hydrogen.
2
k e2
1
1
The energy can be calculated from
E = K + U = mv 2 + mv 2 − e .
2
2
r
2
2
2
2
k e
k e
k e
k e
−ke e2
6.80 eV
.
= −
E= e − e =− e =
Since mv 2 = e ,
2
r
2r
2r
2r
n2
4a0n
The orbital radii are
P42.56
(a)
(b)
The energy difference between these two states is equal to the energy that is absorbed.
a−13.6 eVf − a−13.6 eVf = 10.2 eV =
Thus,
E = E2 − E1 =
3
E = k BT or
2
2 1.63 × 10 −18 J
2E
=
= 7.88 × 10 4 K .
T=
3 k B 3 1.38 × 10 −23 J K
e
4
e
1
j
j
1.63 × 10 −18 J .
Chapter 42
P42.57
f=
P42.58
F 1
GH an − 1f
2h
mk e F 2n − 1 I
GH an − 1f n JK
h
hf = ∆E =
2π 2
4π 2 mk e2 e 4
3
2
2 4
e
2
2
−
1
n2
I
JK
2
As n approaches infinity, we have f approaching
2π 2 mk e2 e 4 2
h3
n3
The classical frequency is
f=
where
r=
n 2 h2
4π mk e e 2
Using this equation to eliminate r from the expression for f,
f=
2π 2 mk e2 e 4 2
h3
n3
(a)
537
v
2π r
hc
The energy of the ground state is:
E1 = −
From the wavelength of the Lyman α line:
E2 − E1 =
λ series limit
hc
λ
=
=
=−
1
2π
ke e2 1
m r3 2
1 240 eV ⋅ nm
= −8.16 eV .
152.0 nm
1 240 nm ⋅ eV
= 6.12 eV
202.6 nm
E2 = E1 + 6.12 eV = −2.04 eV .
(b)
1 240 nm ⋅ eV
= 7.26 eV
170.9 nm
The wavelength of the Lyman β line gives:
E3 − E1 =
so
E3 = −0.902 eV .
Next, using the Lyman γ line gives:
E4 − E1 =
and
E4 = −0.508 eV .
From the Lyman δ line,
E5 − E1 =
so
E5 = −0.325 eV .
For the Balmer series,
For the α line, Ei = E3 and so
hc
λ
1 240 nm ⋅ eV
= 7.65 eV
162.1 nm
1 240 nm ⋅ eV
= 7.83 eV
158.3 nm
= Ei − E 2 , or λ =
λa =
a
1 240 nm ⋅ eV
.
Ei − E2
1 240 nm ⋅ eV
= 1 090 nm .
−0.902 eV − −2.04 eV
f a
f
Similarly, the wavelengths of the β line, γ line, and the short wavelength limit are found to
be: 811 nm , 724 nm , and 609 nm .
continued
538
Atomic Physics
(c)
Computing 60.0% of the wavelengths of the spectral lines shown on the energy-level
diagram gives:
a
f
0.600a158.3 nmf =
a
f
a
f
0.600 202.6 nm = 122 nm , 0.600 170.9 nm = 103 nm , 0.600 162.1 nm = 97.3 nm ,
a
f
95.0 nm , and 0.600 152.0 nm = 91.2 nm
These are seen to be the wavelengths of the α, β , γ, and δ lines as well as the short
wavelength limit for the Lyman series in Hydrogen.
(d)
The observed wavelengths could be the result of Doppler shift when the source moves away
from the Earth. The required speed of the source is found from
b g
b g
1− v c
f′ λ
=
=
= 0.600
λ′
f
1+ v c
P42.59
af
The wave function for the 2s state is given by Eq. 42.26: ψ 2 s r =
(a)
*P42.60
Taking
r = a 0 = 0.529 × 10 −10 m
we find
ψ 2s a0 =
b g
b g = e1.57 × 10
2
14
1
FG
2π H 0.529 × 10
1
4
m− 3 2
j
2
−10
IJ
mK
3 2
FG 1 IJ LM2 − r OPe
a Q
2π H a K N
3 2
1
4
0
− r 2 a0
.
0
2 − 1 e −1 2 = 1.57 × 10 14 m − 3 2 .
= 2.47 × 10 28 m −3
(b)
ψ 2s a0
(c)
Using Equation 42.24 and the results to (b) gives P2 s a0 = 4π a02 ψ 2 s a0
b g
b g
2
= 8.69 × 10 8 m −1 .
From Figure 42.20, a typical ionization energy is 8 eV. For internal energy to ionize most of the
atoms we require
2 × 8 1.60 × 10 −19 J
3
~ between 10 4 K and 10 5 K .
k BT = 8 eV :
T=
2
3 1.38 × 10 −23 J K
e
P42.61
v = 0.471 c .
yielding
(a)
e3.00 × 10
(b)
E=
hc
λ
N=
(c)
8
e
j
j
je
j
m s 14.0 × 10 −12 s = 4.20 mm
= 2.86 × 10 −19 J
3.00 J
= 1.05 × 10 19 photons
−19
2.86 × 10
J
a
f a
f
V = 4.20 mm π 3.00 mm
n=
2
= 119 mm3
1.05 × 10 19
= 8.82 × 10 16 mm −3
119
Chapter 42
P42.62
*P42.63
(a)
The length of the pulse is ∆L = c∆t .
(b)
The energy of each photon is
(c)
V = ∆Lπ
Eγ =
d2
4
n=
hc
so N =
λ
E
Eλ
=
.
Eγ
hc
F 4
GH c∆tπ d
N
=
V
I FG Eλ IJ
JK H hc K
2
The fermions are described by the exclusion principle. Two of them, one spin-up and one spindown, will be in the ground energy level, with
d NN = L =
1
h
h
λ , λ = 2L = , and p =
2
p
2L
K=
p2
1
h2
mv 2 =
=
.
2
2m 8mL2
K=
p2
h2
.
=
2m 2mL2
The third must be in the next higher level, with
d NN =
L λ
h
= , λ = L , and p =
L
2 2
h2
The total energy is then
P42.64
∆z =
dBz
dz
+
2
8mL
h2
2
8mL
+
h2
2
2mL
=
3h2
4mL2
.
FG IJ
b g FG ∆x IJ
e
µ =
and
K
H
2m
v
H K
kg je10 mje10 m s je 2 × 9.11 × 10
2m a ∆zfv b 2m g 2a108fe1.66 × 10
=
=
∆x e
e1.00 m je1.60 × 10 Cje1.05 × 10 J ⋅ sj
2
at 2 1 Fz 2 µ z dBz dz
=
t =
2
2 m0
2m 0
2
0
z
−27
−3
e
4
2
2
e
2
−19
2
−34
dBz
= 0.389 T m
dz
P42.65
By Equation 42.24,
(a)
a f 14 e2π a j FGH 2 − ar IJK e
1 F r IF
r I
Par f = 4π r ψ = G J G 2 − J
8 H a KH
a K
3 −1 2
0
ψ 2s r =
We use
2
− r 2 a0
2
2
.
0
3
0
2
e −r
a0
.
0
a f LM FG IJ − 2r FG 1 IJ FG 2 − r IJ − r FG 2 − r IJ FG 1 IJ OP e
MN H K a H a K H a K a H a K H a K PQ
1 F r IF
r IL F
r I 2r r F
r IO
− G 2 − J Pe
= 0.
or G J G 2 − J M 2G 2 − J −
8 H a KH
a K MN H
a K a
a H
a K PQ
dP r
r
1 2r
=
2−
dr
a0
8 a 03
3
0
The roots of
continued
2
2
3
0
2
2
0
3
0
0
0
− r a0
=0
0
− r a0
0
0
0
0
0
af
dP
= 0 at r = 0 , r = 2 a 0 and r = ∞ are minima with P r = 0 .
dr
−31
kg
j
539
540
Atomic Physics
F 6r I F r I
..... = 4 − G J + G J
Ha K Ha K
r = e3 ± 5 ja .
Therefore we require
0
with solutions
2
=0
0
0
af
We substitute the last two roots into P r to determine the most probable value:
0.051 9
.
a0
af
0.191
.
a0
j
Pr =
When r = 3 + 5 a 0 = 5.236 a 0 ,
e
j
Pr =
Therefore, the most probable value of r is
e3 + 5 ja
z
∞
(b)
af
e
When r = 3 − 5 a 0 = 0.763 9 a 0 ,
af
P r dr =
0
Let u =
z
∞
F
GH
z
∞
1 r2
8 a 03
0
IF2 − r I
JK GH a JK
0
= 5.236 a 0 .
2
e −r
a0
dr
0
r
, dr = a 0 du ,
a0
af
P r dr =
0
z
∞
z
∞
1 2
1 4
1
u 4 − 4u + u 2 e −u dr =
u − 4u 3 + 4u 2 e −u du = − u 4 + 4u 2 + 8u + 8 e −u
8
8
8
0
0
e
j
e
j
e
j
∞
=1
0
This is as required for normalization.
P42.66
E=
hc
λ
=
1 240 eV ⋅ nm
λ
λ 1 = 310 nm,
= ∆E
∆E1 = 4.00 eV
so
λ 2 = 400 nm,
∆E2 = 3.10 eV
λ 3 = 1 378 nm ,
∆E3 = 0.900 eV
and the ionization energy = 4.10 eV .
FIG. P42.66
The energy level diagram having the fewest levels and consistent with these energies is shown at
the right.
P42.67
With one vacancy in the K shell, excess energy
a
∆E ≈ − Z − 1
f a13.6 eVfFGH 21 − 11 IJK = 5.40 keV .
2
2
2
We suppose the outermost 4s electron is shielded by 22 electrons inside its orbit:
Eionization ≈
a
2 2 13.6 eV
4
2
f = 3.40 eV .
Note the experimental ionization energy is 6.76 eV.
K = ∆E − Eionization ≈ 5.39 keV .
Chapter 42
P42.68
The configuration we may model as SN
(a)
NS has higher energy than SN
541
SN . The
higher energy state has antiparallel magnetic moments, so it has parallel spins of the
oppositely charged particles.
P42.69
= 9.42 × 10 −25 J = 5.89 µeV
E=
(c)
∆E∆t ≈
z
∞
P=
2.50 a0
P=−
P42.70
hc
(b)
(a)
λ
4r 2 −2 r
e
a 03
a0
e
j
7
je
2 10 yr 3.16 × 10
7
−19
1.04 × 10 −30 eV
z
∞
∞
=−
5 .00
a
FG IJ b
H K
f
g
1
1
37
0 + 25.0 + 10.0 + 2.00 e −5 =
0.006 74 = 0.125
2
2
2
One molecule’s share of volume
3
3
F
GH
mass per molecule
27.0 g mol
=
density
6.02 × 10 23 molecules mol
I F 1.00 × 10 m I = 1.66 × 10
JK GH 2.70 g JK
−6
3
−29
m3
V = 2.55 × 10 −10 m~ 10 −1 nm .
U: V =
238 g
FG
I F 1.00 × 10 m I
H 6.02 × 10 molecules JK GH 18.9 g JK = 2.09 × 10
−6
3
23
−29
m3
V = 2.76 × 10 −10 m ~ 10 −1 nm .
The outermost electron in any atom sees the nuclear charge screened by all the electrons
below it. If we can visualize a single outermost electron, it moves in the electric field of net
charge, +Ze − Z − 1 e = + e , the charge of a single proton, as felt by the electron in hydrogen.
So the Bohr radius sets the scale for the outside diameter of every atom. An innermost
electron, on the other hand, sees the nuclear charge unscreened, and the scale size of its (Ka
shell) orbit is 0 .
Z
a
P42.71
e
F 1.00 eV I =
G
J
JK
s yr j H 1.60 × 10
1.055 × 10 −34 J ⋅ s
2r
1
z 2 e − z dz where z ≡
2 5.00
a0
dr =
1 2
z + 2z + 2 e−z
2
Al: V =
(b)
2
so ∆E ≈
f
∆E = 2 µ B B = hf
e
ja
f e
j
so
2 9.27 × 10 −24 J T 0.350 T = 6.626 × 10 −34 J ⋅ s f
and
f = 9.79 × 10 9 Hz .
542
P42.72
Atomic Physics
ψ
a f FGH a1 IJK FGH 2 − ar IJK e
IF 3 − r I
d ψ F Ae
=G
dr
H a JK GH 2 4a JK
25
1
= 2π
4
FG
H
3 2
−1 2
0
− r 2 a0
IJ
K
r −r
e
a0
= A 2−
0
dψ
= Ae − r
dr
2 a0
F− 2 + r I
GH a 2a JK
0
2
0
− r 2 a0
2
2
2
0
0
Substituting into Schrödinger’s equation and dividing by Ae − r
2
Now with a 0 =
2 a0
, we will have a solution if
2
2k e2
k e2
2 2
5 2
r
Er
+ e +
+
− e = 2E −
.
2
3
4 m e a0
a0
r
a0
8m e a0 m e a0 r
−
me e2 ke
, this reduces to
m e e 4 k e2
−
8
2
FG 2 − r IJ = EFG 2 − r IJ .
H aK H aK
0
0
This is true, so ψ 25 is a solution to the Schrödinger equation, provided E =
P42.73
2 a0
1
E1 = −3.40 eV .
4
Suppose the atoms move in the +x direction. The absorption of a photon by an atom is a
completely inelastic collision, described by
(a)
mvi i +
h
e− ij = mv i
λ
v f − vi = −
so
f
h
.
mλ
This happens promptly every time an atom has fallen back into the ground state, so it
happens every 10 −8 s = ∆t. Then,
a=
(b)
v f − vi
∆t
=−
6.626 × 10 −34 J ⋅ s
h
~−
~ −10 6 m s 2 .
−25
−9
−8
mλ∆t
10
kg 500 × 10 m 10 s
e
je
j
With constant average acceleration,
e
j + 2e−10 m s j∆x
e10 m sj ~ 1 m .
∆x ~
0 ~ 10 3 m s
v 2f = vi2 + 2 a∆x
3
so
z
∞
P42.74
je
1
4r 2 −2 r
=
e
3
r
0 a0
2
6
2
2
10 6 m s 2
a0
z
∞
1
4
4
1
− 2 a r
dr = 3 re b 0 g dr = 3
r
a0 0
a0 2 a0
b g
2
=
1
a0
1
1
2
=
=
, and find that the average reciprocal value is NOT the
3 a0 2 3 a0
r
reciprocal of the average value.
We compare this to
Chapter 42
543
ANSWERS TO EVEN PROBLEMS
(a) 91.2 nm, 365 nm, 821 nm, 1.46 µm ;
(b) 13.6 eV, 3.40 eV, 1.51 eV, 0.850 eV
P42.38
P42.4
(a) 56.8 fm; (b) 11.3 N away from the
nucleus
P42.40
P42.6
(a) ii; (b) i; (c) ii and iii
(a) = 0 with m = 0 ; = 1 with m = 1, 0,
or 1; and = 2 with m = −2 , –1, 0, 1, 2;
(b) –6.05 eV
P42.8
(a) 13.6 eV; (b)1.51 eV
P42.42
see the solution
P42.10
see the solution
P42.44
L shell 11.8 keV, M shell 10.1 keV, N shell
2.39 keV, see the solution
P42.12
97.5 nm
P42.46
see the solution
P42.14
(a) O 7+ ; (b) 41.0 nm, 33.8 nm, 30.4 nm
P42.48
(a) 4.24 PW m 2 ; (b) 1.20 pJ = 7.50 MeV
P42.16
(a) 1.31 µm ; (b) 164 nm
P42.50
P42.18
see the solution
(a) 1.07 × 10 −33 ; (b) −1.15 × 10 6 K ;
(c) negative temperatures do not describe
systems in thermal equilibrium
P42.20
4a0
P42.52
P42.22
797 times
(a) see the solution; (b) 2.53 × 10 74 ;
(c) 1.18 × 10 −63 m , unobservably small
P42.2
P42.24
4 f 14 5s 2 5 p 6 5d 10 5 f 14 6s 2 6 p 6 6d 8 7s 2
P42.54
−34
J ⋅ s;
(a) 6 = 2.58 × 10
(b) 12 = 3.65 × 10 −34 J ⋅ s
P42.26
6
P42.28
6 ; −2 , − , 0,
65.9°, 35.3°
P42.30
3
P42.32
(a) 1s 2 2s 2 2 p 4 ;
(b) n
m
ms
, 2 ; 145°, 114°, 90.0°,
1
1
2
2
2
2
2
2
0
0
0
0
1
1
1
0
1
2
0
1
−
2
0
1
2
0
1
−
2
1
1
2
1
1
−
2
0
1
2
1
−1
1
2
P42.34
see the solution
P42.36
(a) see the solution;
(b) 36 states instead of 30
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4p 6 4d 10
F 2r
GH a
2
I
JK
2r
+ 1 e −2 r a 0 ;
a0
(b) see the solution; (c) 1.34 a 0
(a) Probability =
2
0
+
P42.56
(a) 10.2 eV = 1.63 aJ ; (b) 7.88 × 10 4 K
P42.58
(a) –8.16 eV, –2.04 eV, –0.902 eV, –0.508 eV,
–0.325 eV;
(b) 1 090 nm, 811 nm, 724 nm, 609 nm;
(c) see the solution; (d) The spectrum
could be that of hydrogen, Doppler shifted
by motion away from us at speed 0.471c .
P42.60
between 10 4 K and 10 5 K
P42.62
(a) c∆t ; (b)
P42.64
0.389 T/m
P42.66
Energy levels at 0, –0.100 eV, –1.00 eV, and
–4.10 eV
Eλ
4Eλ
; (c)
hc
∆tπ d 2 hc 2
544
P42.68
Atomic Physics
(a) parallel spins; (b) 5.89 µeV ;
(c) 1.04 × 10
P42.70
−30
P42.72
see the solution
P42.74
1
, no
a0
eV
−1
(a) diameter ~ 10 nm for both;
(b) A K-shell electron moves in an orbit
a
with size on the order of 0
Z
43
Molecules and Solids
CHAPTER OUTLINE
43.1
43.2
43.3
43.4
43.5
43.6
43.7
43.8
Molecular Bonds
Energy States and Spectra
of Molecules
Bonding in Solids
Free-Electron Theory of
Metals
Band Theory of Solids
Electrical Conduction in
Metals, Insulators, and
Semiconductors
Semiconductor Devices
Superconductivity
ANSWERS TO QUESTIONS
Q43.1
Rotational, vibrational and electronic (as discussed in
Chapter 42) are the three major forms of excitation. Rotational
2
, where I is
2I
the moment of inertia of the molecule. A typical value for a
small molecule is on the order of 1 meV = 10 −3 eV . Vibrational
energy is on the order of hf, where f is the vibration frequency
of the molecule. A typical value is on the order of 0.1 eV.
Electronic energy depends on the state of an electron in the
molecule and is on the order of a few eV. The rotational energy
can be zero, but neither the vibrational nor the electronic
energy can be zero.
energy for a diatomic molecule is on the order of
Q43.2
The Pauli exclusion principle limits the number of electrons in
the valence band of a metal, as no two electrons can occupy the
same state. If the valence band is full, additional electrons must
be in the conduction band, and the material can be a good
conductor. For further discussion, see Q43.3.
Q43.3
The conductive properties of a material depend on the electron population of the conduction band
of the material. If the conduction band is empty and a full valence band lies below the conduction
band by an energy gap of a few eV, then the material will be an insulator. Electrons will be unable to
move easily through the material in response to an applied electric field. If the conduction band is
partly full, states are accessible to electrons accelerated by an electric field, and the material is a good
conductor. If the energy gap between a full valence band and an empty conduction band is
comparable to the thermal energy k B T , the material is a semiconductor.
Q43.4
Thermal excitation increases the vibrational energy of the molecules. It makes the crystal lattice less
orderly. We can expect it to increase the width of both the valence band and the conduction band, to
decrease the gap between them.
545
546
Molecules and Solids
Q43.5
First consider electric conduction in a metal. The number of conduction electrons is essentially fixed.
They conduct electricity by having drift motion in an applied electric field superposed on their
random thermal motion. At higher temperature, the ion cores vibrate more and scatter more
efficiently the conduction electrons flying among them. The mean time between collisions is
reduced. The electrons have time to develop only a lower drift speed. The electric current is
reduced, so we see the resistivity increasing with temperature.
Now consider an intrinsic semiconductor. At absolute zero its valence band is full and its
conduction band is empty. It is an insulator, with very high resistivity. As the temperature increases,
more electrons are promoted to the conduction band, leaving holes in the valence band. Then both
electrons and holes move in response to an applied electric field. Thus we see the resistivity
decreasing as temperature goes up.
Q43.6
In a metal, there is no energy gap between the valence and conduction bands, or the conduction
band is partly full even at absolute zero in temperature. Thus an applied electric field is able to inject
a tiny bit of energy into an electron to promote it to a state in which it is moving through the metal
as part of an electric current. In an insulator, there is a large energy gap between a full valence band
and an empty conduction band. An applied electric field is unable to give electrons in the valence
band enough energy to jump across the gap into the higher energy conduction band. In a
semiconductor, the energy gap between valence and conduction bands is smaller than in an
insulator. At absolute zero the valence band is full and the conduction band is empty, but at room
temperature thermal energy has promoted some electrons across the gap. Then there are some
mobile holes in the valence band as well as some mobile electrons in the conduction band.
Q43.7
Ionic bonds are ones between oppositely charged ions. A simple model of an ionic bond is the
electrostatic attraction of a negatively charged latex balloon to a positively charged Mylar balloon.
Covalent bonds are ones in which atoms share electrons. Classically, two children playing a
short-range game of catch with a ball models a covalent bond. On a quantum scale, the two atoms are
sharing a wave function, so perhaps a better model would be two children using a single hula hoop.
Van der Waals bonds are weak electrostatic forces: the dipole-dipole force is analogous to the
attraction between the opposite poles of two bar magnets, the dipole—induced dipole force is
similar to a bar magnet attracting an iron nail or paper clip, and the dispersion force is analogous to
an alternating-current electromagnet attracting a paper clip.
A hydrogen atom in a molecule is not ionized, but its electron can spend more time elsewhere
than it does in the hydrogen atom. The hydrogen atom can be a location of net positive charge, and
can weakly attract a zone of negative charge in another molecule.
Q43.8
Ionically bonded solids are generally poor electric conductors, as they have no free electrons. While
they are transparent in the visible spectrum, they absorb infrared radiation. Physically, they form
stable, hard crystals with high melting temperatures.
Q43.9
Covalently bonded solids are generally poor conductors, as they form structures in which the atoms
share several electrons in the outer shell, leaving no room for conducting electrons. Depending on
the structure of the solid, they are usually very hard and have high melting points.
Q43.10
Metals are good conductors, as the atoms have many free electrons in the conduction band. Metallic
bonds allow the mixing of different metals to form alloys. Metals are opaque to visible light, and can
be highly reflective. A metal can bend under stress instead of fracturing like ionically and covalently
bonded crystals. The physical properties vary greatly depending on the composition.
Chapter 43
547
Q43.11
The energy of the photon is given to the electron. The energy of a photon of visible light is sufficient
to promote the electron from the lower-energy valence band to the higher-energy conduction band.
This results in the additional electron in the conduction band and an additional hole—the energy
state that the electron used to occupy—in the valence band.
Q43.12
Along with arsenic (As), any other element in group V, such as phosphorus (P), antimony (Sb), and
bismuth (Bi), would make good donor atoms. Each has 5 valence electrons. Any element in group III
would make good acceptor atoms, such as boron (B), aluminum, (Al), gallium (Ga), and indium (In).
They all have only 3 valence electrons.
Q43.13
The two assumptions in the free-electron theory are that the conduction electrons are not bound to
any particular atom, and that the nuclei of the atoms are fixed in a lattice structure. In this model, it
is the “soup” of free electrons that are conducted through metals. The energy band model is more
comprehensive than the free-electron theory. The energy band model includes an account of the
more tightly bound electrons as well as the conduction electrons. It can be developed into a theory
of the structure of the crystal and its mechanical and thermal properties.
Q43.14
A molecule containing two atoms of 2 H, deuterium, has twice the mass of a molecule containing
two atoms of ordinary hydrogen 1 H . The atoms have the same electronic structure, so the molecules
have the same interatomic spacing, and the same spring constant. Then the moment of inertia of the
double-deuteron is twice as large and the rotational energies one-half as large as for ordinary
1
times that of H 2 .
hydrogen. Each vibrational energy level for D 2 is
2
Q43.15
Rotation of a diatomic molecule involves less energy than vibration. Absorption of microwave
photons, of frequency ~ 10 11 Hz , excites rotational motion, while absorption of infrared photons, of
frequency ~ 10 13 Hz , excites vibration in typical simple molecules.
Q43.16
Yes. A material can absorb a photon of energy greater than the energy gap, as an electron jumps into
a higher energy state. If the photon does not have enough energy to raise the energy of the electron
by the energy gap, then the photon will not be absorbed.
Q43.17
From the rotational spectrum of a molecule, one can easily calculate the moment of inertia of the
molecule using Equation 43.7 in the text. Note that with this method, only the spacing between
adjacent energy levels needs to be measured. From the moment of inertia, the size of the molecule
can be calculated, provided that the structure of the molecule is known.
SOLUTIONS TO PROBLEMS
Section 43.1
P43.1
Molecular Bonds
q2
e1.60 × 10 j e8.99 × 10 j N =
=
e5.00 × 10 j
−19 2
9
0.921 × 10 −9 N toward the other ion.
(a)
F=
(b)
1.60 × 10 −19 8.99 × 10 9
−q 2
=−
U=
J ≈ −2.88 eV
4π ∈0 r
5.00 × 10 −10
4π ∈0 r 2
−10 2
e
je
2
j
548
P43.2
Molecules and Solids
We are told
K + Cl + 0.7 eV → K + + Cl −
and
Cl + e − → Cl − + 3.6 eV
or
Cl − → Cl + e − − 3.6 eV .
By substitution,
K + Cl + 0.7 eV → K + + Cl+e − − 3.6 eV
K + 4.3 eV → K + +e −
or the ionization energy of potassium is 4.3 eV .
P43.3
(a)
Minimum energy of the molecule is found from
LM OP .
N Q
B O
L1 1 O B =
−
= −M − P
P
2 A B PQ
N4 2Q A
dU
2A
= −12 Ar −13 + 6Br −7 = 0 yielding r0 =
dr
B
(b)
E = U r =∞ − U r = r = 0 −
0
LM A
MN 4 A B
2
16
2
2
B2
4A
This is also the equal to the binding energy, the amount of energy given up by the two
atoms as they come together to form a molecule.
(c)
L 2e0.124 × 10
r =M
MN 1.488 × 10
e1.488 × 10
E=
4e0.124 × 10
−120
0
−60
−60
(a)
eV ⋅ m6
eV ⋅ m6
−120
*P43.4
eV ⋅ m12
j
j OP
PQ
16
= 7.42 × 10 −11 m = 74.2 pm
2
eV ⋅ m12
j
= 4.46 eV
We add the reactions K + 4.34 eV → K + + e −
and
I + e − → I − + 3.06 eV
to obtain
K + I → K + + I − + 4.34 − 3.06 eV .
a
f
The activation energy is 1.28 eV .
(b)
LM FG IJ
MN H K
σ
dU 4 ∈
=
−12
σ
dr
r
At r = r0 we have
σ
= 2 −1 6
r0
13
+6
FG σ IJ OP
H r K PQ
7
FG IJ
H K
dU
σ
= 0 . Here
dr
r0
a
13
=
FG IJ
H K
1 σ
2 r0
7
f
σ = 2 −1 6 0.305 nm = 0.272 nm = σ .
Then also
LF 2 r I F 2 r I OP
L1 1 O
−G
+ E = 4 ∈M − P + E
U br g = 4 ∈ MG
J
J
r
r
MNH
N4 2Q
K H
K PQ
∈= E − U br g = 1.28 eV + 3.37 eV = 4.65 eV =∈
−1 6
0
0
a
0
continued on next page
12
0
−1 6
0
6
0
a
a
= − ∈+ Ea
Chapter 43
(c)
af
F r =−
LM FG IJ
MN H K
σ
dU 4 ∈
=
12
σ
dr
r
13
−6
FG σ IJ OP
H r K PQ
7
To find the maximum force we calculate
σ
rrupture
Fmax =
=
FG 42 IJ
H 156 K
LM
MN
a
f LM FG IJ
MN H K
= −6.55 nN
13 6
−6
FG IJ
H K
σ
dF 4 ∈
=
−156
dr σ 2
r
16
4 4.65 eV
42
12
0.272 nm
156
549
14
+ 42
FG σ IJ OP = 0 when
H r K PQ
8
FG 42 IJ OP = −41.0 eV nm = −41.0 1.6 × 10
H 156 K PQ
10
76
−19
−9
Nm
m
Therefore the applied force required to rupture the molecule is +6.55 nN away from the
center.
(d)
b
g
LF
LMF σ I F σ I OP
I F I
GMH r + s JK − GH r + s JK P + E = 4 ∈ MMGH 2r + rs JK − GH r2 + s JK
N
Q
N
L 1 F s I 1 F s I OP
= 4 ∈M G1 + J
MN 4 H r K − 2 GH1 + r JK PQ + E
L 1 F s s − …I − 1 F1 − 6 s + 21 s − …I OP + E
= 4 ∈ M G 1 − 12 + 78
MN 4 H r r JK 2 GH r r JK PQ
12
U r0 + s = 4 ∈
6
0
−1 6
a
0
−12
0
where k =
P43.5
0
a
2
0
s
s2
s
s2
+ 78 ∈ 2 − 2 ∈+12 ∈ − 42 ∈ 2 + Ea + …
r0
r0
r0
r0
= − ∈+ E a + 0
g b g
0
OP
PQ + E
2
2
0
0
b
6
a
0
2
U r0 + s ≅ U r0 +
−1 6
−6
0
=∈−12 ∈
12
0
FG s IJ + 36 ∈ s
Hr K
r
2
2
0
0
+…
1 2
ks
2
a
f
72 ∈ 72 4.65 eV
=
= 3 599 eV nm 2 = 576 N m .
2
r02
0.305 nm
a
f
At the boiling or condensation temperature,
e
k BT ≈ 10 −3 eV = 10 −3 1.6 × 10 −19 J
T≈
1.6 × 10 −22 J
~ 10 K .
1.38 × 10 −23 J K
j
a
550
Molecules and Solids
Section 43.2
P43.6
µ=
Energy States and Spectra of Molecules
a
fe
132.9 126.9
m1 m 2
=
1.66 × 10 −27 kg = 1.08 × 10 −25 kg
m1 + m 2 132.9 + 126.9
e
j
je
I = µ r 2 = 1.08 × 10 −25 kg 0.127 × 10 −9 m
E=
(a)
a f
Iω
1 2
Iω =
2
2I
2
=
J = 0 gives E = 0
2
I
= 1.74 × 10 −45 kg ⋅ m 2
2
J J +1
2I
e6.626 × 10
=
4π e1.74 × 10
2
J = 1 gives E =
a f
j
2
j
2
−34
J⋅s
−45
kg ⋅ m 2
j
= 6.41 × 10 −24 J = 40.0 µeV
hf = 6.41 × 10 −24 J − 0 to f = 9.66 × 10 9 Hz
f=
(b)
P43.7
2
E1
h
=
=
∝ r −2
h
hI 4π 2 µ r 2
If r is 10% too small, f is 20% too large.
For the HCl molecule in the J = 1 rotational energy level, we are given
r0 = 0.127 5 nm.
Erot =
Taking J = 1 , we have
2
a f
J J +1
2I
Erot =
2
=
I
1 2
Iω or ω =
2
2 2
= 2 .
I
I2
The moment of inertia of the molecule is given by Equation 43.3.
m1 m 2
I = µ r02 =
r02
m1 + m 2
2
0
Therefore, ω = 2
P43.8
hf = ∆E =
I=
b
4 h 2π
2 hf
2
2I
g
a
fe
I
h
2
j
2π f
=
2
2I
a f
1 1+1 =
2
2I
a 4f
6.626 × 10 −34 J ⋅ s
2π
2
= 2.62 × 10 −47 kg ⋅ m 2 .
1.055 × 10 −34 J ⋅ s
= 5.69 × 10 12 rad s .
2
−47
2.62 × 10
kg ⋅ m
= 2
a f
=
je
= 0.972 u 1.66 × 10 −27 kg u 1.275 × 10 −10 m
2 2+1 −
2
FIG. P43.7
IJ
K
FG
H
L a1 ufa35 uf OPr
I=M
N 1 u + 35 u Q
2
e2.30 × 10
11
Hz
j
= 1.46 × 10 −46 kg ⋅ m 2
Chapter 43
P43.9
I = m1 r12 + m 2 r22
m1 r1 = m 2 r2
and
r1 + r2 = r .
Then r1 =
m 2 r2
m1
so
m 2 r2
+ r2 = r
m1
and
r2 =
m1 r
.
m1 + m 2
Also, r2 =
m1 r1
.
m2
Thus,
r1 +
m1 r1
=r
m2
and
r1 =
m2r
m1 + m 2
I = m1
P43.10
where
bm
m 22 r 2
+
m 2 m12 r 2
=
b
m1 m 2 r 2 m 2 + m1
g=m m r
1
2
g bm + m g
bm + m g
22.99a35.45f
µ=
a22.99 + 35.45f e1.66 × 10 kgj = 2.32 × 10 kg
I = µ r = e 2.32 × 10
kg je0.280 × 10 mj = 1.81 × 10
+ m2
1
2
1
2
2
1
2
2
m1 + m 2
−27
(a)
2
2
= µr2 .
−26
−26
2
2
−9
2
2
2
2
−45
kg ⋅ m 2
2
a f 2 I 1a1 + 1f = 3 I − I = 2 I = 42πh I
λ 2I
kg ⋅ m j
c 4π I e3.00 × 10 m sj4π e1.81 × 10
λ=
=
= 1.62 cm
2h
J ⋅ sj
2e6.626 × 10
F I
The energy of a rotational transition is ∆E = G J J where J is the rotational quantum number of the
HIK
hc
(b)
=
2 2+1 −
2
8
2
−45
2
2
−34
P43.11
551
2
higher energy state (see Equation 43.7). We do not know J from the data. However,
∆E =
hc
λ
=
e6.626 × 10
−34
je
J ⋅ s 3.00 × 10 8 m s
j F 1 eV I .
GH 1.60 × 10 J JK
−19
λ
For each observed wavelength,
λ (mm)
0.120 4
0.096 4
0.080 4
0.069 0
0.060 4
∆E (eV)
0.010 32
0.012 88
0.015 44
0.018 00
0.020 56
The ∆E’s consistently increase by 0.002 56 eV.
and I =
2
E1
e1.055 × 10 J ⋅ sj F 1 eV I =
=
b0.002 56 eVg GH 1.60 × 10 J JK
−34
E1 =
2
I
= 0.002 56 eV
2
−19
For the HCl molecule, the internuclear radius is
2.72 × 10 −47 kg ⋅ m 2 .
r=
I
µ
=
2.72 × 10 −47
m = 0.130 nm.
1.62 × 10 −27
552
P43.12
Molecules and Solids
(a)
Minimum amplitude of vibration of HI is
(b)
1 2 1
kA = hf :
2
2
A=
For HF,
A=
e6.626 × 10
hf
=
k
−34
je
J ⋅ s 6.69 × 10 13 s
320 N m
e6.626 × 10
−34
je
J ⋅ s 8.72 × 10 13 s
970 N m
j = 1.18 × 10
j = 7.72 × 10
−12
−11
m = 0.011 8 nm .
m = 0.007 72 nm .
Since HI has the smaller k, it is more weakly bound.
P43.13
µ=
m1 m 2
35
=
× 1.66 × 10 −27 kg = 1.61 × 10 −27 kg
m1 + m 2 36
∆Evib =
P43.14
(a)
k
= 1.055 × 10 −34
µ
e
480
= 5.74 × 10 −20 J = 0.358 eV
1.61 × 10 −27
j
a16 ufa16 uf
a16 uf + a16 uf = 8 u = 8e1.66 × 10 kg j = 1.33 × 10 kg .
The moment of inertia is then I = µ r = e1.33 × 10
kg je1.20 × 10 mj = 1.91 × 10
kg ⋅ m .
−27
The reduced mass of the O 2 is µ =
−26
2
(b)
2
−10
e1.055 × 10
=
J a J + 1f =
2I
2e1.91 × 10
= e 2.91 × 10
Jj J a J + 1f .
2
−26
j J a J + 1f .
kg ⋅ m j
−34
J⋅s
2
The rotational energies are
Erot
Thus
Erot
And for J = 0 , 1, 2 ,
Erot = 0 , 3.64 × 10 −4 eV, 1.09 × 10 −3 eV .
FG
H
1
2
FG
H
1
1 eV
3.14 × 10 −20 J
2
1.60 × 10 −19
Evib = v +
Evib = v +
IJ
K
IJ e
K
k
µ
FG
H
= v+
IJ e
K
−46
−23
j 8e1.166177× 10N mkgj
IJ = FG v + 1 IJ a0.196 eVf
2K
JK H
1
1.055 × 10 −34 J ⋅ s
2
jFGH
2
For v = 0 , 1, 2 , E vib = 0.098 2 eV, 0.295 eV, 0.491 eV .
−27
−46
2
Chapter 43
P43.15
553
In Benzene, the carbon atoms are each 0.110 nm from the axis and each hydrogen atom is
0.110 + 0.100 nm = 0.210 nm from the axis. Thus, I = ∑ mr 2 :
a
f
e
je
j + 6e1.67 × 10
I = 6 1.99 × 10 −26 kg 0.110 × 10 −9 m
2
−27
je
j
kg 0.210 × 10 −9 m
2
= 1.89 × 10 −45 kg ⋅ m 2 .
The allowed rotational energies are then
Erot
Erot
e1.055 × 10 J ⋅ sj Ja J + 1f = e2.95 × 10 Jj Ja J + 1f = e18.4 × 10
=
J a J + 1f =
2I
kg ⋅ m j
2e1.89 × 10
= b18.4 µeV g J a J + 1f where J = 0 , 1, 2 , 3 , …
2
−34
2
−24
−45
2
−6
ja f
eV J J + 1
The first five of these allowed energies are: Erot = 0 , 36.9 µeV, 111 µeV, 221 µeV, and 369 µeV .
*P43.16
We carry extra digits through the solution because part (c) involves the subtraction of two close
numbers. The longest wavelength corresponds to the smallest energy difference between the
rotational energy levels. It is between J = 0 and J = 1 , namely
λ=
2
I
hc
hc
4π 2 Ic
= 2 =
. If µ is the reduced mass, then
∆Emin
h
I
e
j = e1.624 605 × 10
m jµ e 2.997 925 × 10
I = µ r 2 = µ 0.127 46 × 10 −9 m
∴λ =
e
4π 2 1.624 605 × 10 −20
2
−20
2
8
ms
6.626 075 × 10 −34 J ⋅ s
µ 35 =
(a)
j
m2 µ
j = e2.901 830 × 10
b1.007 825ugb34.968 853ug = 0.979 593u = 1.626 653 × 10
−27
1.007 825u + 34.968 853 u
e
je
23
j
m kg µ
kg
j
From (1): λ 35 = 2.901 830 × 10 23 m kg 1.626 653 × 10 −27 kg = 472 µm
µ 37 =
(b)
b1.007 825ugb36.965 903ug = 0.981 077u = 1.629 118 × 10
−27
1.007 825u + 36.965 903 u
e
je
kg
j
From (1): λ 37 = 2.901 830 × 10 23 m kg 1.629 118 × 10 −27 kg = 473 µm
λ 37 − λ 35 = 472.742 4 µm − 472.027 0 µm = 0.715 µm
(c)
P43.17
hf =
h2
4π 2 I
J where the rotational transition is from J − 1 to J,
where f = 6.42 × 10 13 Hz and I = 1.46 × 10 −46 kg ⋅ m 2 from Example 43.1.
e
je
j
−46
2
kg ⋅ m 2 6.42 × 10 13 s
4π 2 If 4π 1.46 × 10
=
= 558
J=
h
6.626 × 10 −34 J ⋅ s
(1)
554
*P43.18
Molecules and Solids
We find an average spacing between peaks by counting 22 gaps between 7.96 × 10 13 Hz and
9.24 × 10 13 Hz :
∆f =
I=
*P43.19
a9.24 − 7.96f10
13
Hz
= 0.058 2 × 10 13 Hz =
22
h
4π 2 ∆f
=
F
GH
1 h2
h 4π 2 I
I
JK
6.63 × 10 −34 J ⋅ s
= 2.9 × 10 −47 kg ⋅ m 2
4π 2 5.82 × 10 11 s
We carry extra digits through the solution because the given wavelengths are close together.
(a)
FG
H
IJ
K
2
1
hf +
J J +1
2
2I
EvJ = v +
a f
1
hf ,
2
∴ E00 =
2
∴ E11 − E00 = hf +
∴ hf +
2
I
=
2
2
I
2
2
I
=
(b)
λ
je
−34
J ⋅ s 2.997 925 × 10 8 m s
2.211 2 × 10
−6
m
hc
λ
(1)
e6.626 075 × 10
=
−34
je
J ⋅ s 2.997 925 × 10 8 m s
2.405 4 × 10
−6
j
m
2
3
I
e
= 7.252 89 × 10 −21 J
3 1.054 573 × 10 −34 J ⋅ s
7.252 89 × 10 −21 J
(2)
j
2
= 4.60 × 10 −48 kg ⋅ m 2
From (1):
f=
e1.054 573 × 10 J ⋅ sj
−
J ⋅ s e 4.600 060 × 10
kg ⋅ m je6.626 075 × 10
−34
8.983 573 × 10 −20 J
6.626 075 × 10 −34
−48
2
= 1.32 × 10 14 Hz
(c)
j
= 8.258 284 × 10 −20 J
Subtract (2) from (1):
∴I =
e6.626 075 × 10
=
hc
= 8.983 573 × 10 −20 J
I
E11 − E02 = hf −
∴ hf −
2
3
1
3 2
, E02 = hf +
hf +
I
I
2
2
E11 =
I = µ r 2 , where µ is the reduced mass:
µ=
b
g
1
1
m H = 1.007 825u = 8.367 669 × 10 −28 kg .
2
2
So r =
I
µ
=
4.600 060 × 10 −48 kg ⋅ m 2
8.367 669 × 10 −28 kg
= 0.074 1 nm .
2
−34
J⋅s
j
Chapter 43
P43.20
The emission energies are the same as the absorption energies, but the final state must be below
v = 1, J = 0 . The transition must satisfy ∆J = ±1 , so it must end with J = 1 . To be lower in energy, it
b
g
b
g
must be v = 0 , J = 1 . The emitted photon energy is therefore
e
hf photon = Evib
v =1
+ Erot
J =0
j − eE
vib v = 0
+ Erot
J =1
j = eE
vib v =1
− Evib
v=0
j − eE
− Erot
rot J = 1
J =0
j
hf photon = hf vib − hfrot
Thus, fphoton = f vib − frot = 6.42 × 10 13 Hz − 1.15 × 10 11 Hz = 6.41 × 10 13 Hz .
P43.21
Ix =
The moment of inertia about the molecular axis is
2
2
2
4
mr 2 + mr 2 = m 2.00 × 10 −15 m .
5
5
5
e
The moment of inertia about a perpendicular axis is I y = m
The allowed rotational energies are Erot =
E1 =
E1 , x
E1 , y
Section 43.3
P43.22
2
I
FG R IJ
H 2K
2
+m
FG R IJ
H 2K
2
=
j
2
m
2.00 × 10 −10 m .
2
e
j
F I Ja J + 1f , so the energy of the first excited state is
GH 2 I JK
2
. The ratio is therefore
e
=
e
2
2
j = I = b1 2gme2.00 × 10
I j I
b4 5gme2.00 × 10
Ix
y
y
x
−10
−15
j
mj
m
2
2
=
5
10 5
8
e j
2
= 6.25 × 10 9 .
Bonding in Solids
Consider a cubical salt crystal of edge length 0.1 mm.
F 10 m I ~ 10 .
GH 0.261 × 10 m JK
e10 mj FGH 0.26110× 10m m IJK ~ 10
3
−4
The number of atoms is
17
−9
−4
This number of salt crystals would have volume
3
3
−4
−9
5
m3 .
If it is cubic, it has edge length 40 m.
P43.23
P43.24
555
FG
H
IJ b
K
j ee
j FG 1 − 1 IJ = −1.25 × 10
j H 8K
1.60 × 10 −19
α ke e2
1
9
= − 1.747 6 8.99 × 10
U=−
1−
r0
m
0.281 × 10 −9
ge
2
−18
J = −7.84 eV
Visualize a K + ion at the center of each shaded cube, a Cl − ion at the center
of each white one.
The distance ab is
Distance ac is
Distance ad is
a
f
2 0.314 nm = 0.444 nm .
a
f
2 0.314 nm = 0.628 nm .
22 +
e 2 j a0.314 nmf =
2
0.769 nm .
FIG. P43.24
556
P43.25
Molecules and Solids
ke e2 ke e2 ke e2 ke e2 ke e2 ke e2 ke e2 ke e2
−
+
+
−
−
+
+
−…
2r
2r
3r
3r
4r
4r
r
r
2k e 2
1 1 1
=− e
1 − + − +…
2 3 4
r
e –e e –e e –e
+ – + – + –
U=−
FG
IJ
H
K
x
x
x
But, lna1 + xf = 1 −
+
−
+…
2
3
4
2
so, U = −
3
2r
3r
4
2
FIG. P43.25
2
2kee
e
ln 2 , or U = − k eα
where α = 2 ln 2 .
r
r
Section 43.4
Free-Electron Theory of Metals
Section 43.5
Band Theory of Solids
P43.26
r
FG IJ
H K
23
h 2 3n e
EF =
2 m 8π
L
e6.626 × 10 J ⋅ sj
=M
MM 2e9.11 × 10 kg je1.60 × 10
N
2
−34
−31
e
−19
OPF 3 I
G J
J eV j PH 8π K
QP
23
n e2 3
j
EF = 3.65 × 10 −19 n e2 3 eV with n measured in electrons m3 .
P43.27
The density of conduction electrons n is given by EF =
or
FG
H
8π 2mEF
ne =
3
h2
IJ
K
32
ja fe
e
e
e
n Ag = 10.6 × 10 3 kg m3
So an average atom contributes
(a)
j
3 2
= 5.80 × 10 28 m −3 .
j
atom I F
1u
J
jFGH 1108
u K GH 1.66 × 10
−27
I = 5.91 × 10
J
kg K
28
m −3 .
5.80
= 0.981 electron to the conduction band .
5.91
1
mv 2 = 7.05 eV
2
v=
(b)
23
−31
kg 5.48 1.60 × 10 −19 J
8π 2 9.11 × 10
=
3
3
6.626 × 10 −34 J ⋅ s
The number-density of silver atoms is
P43.28
FG IJ
H K
h 2 3n e
2 m 8π
a
fe
2 7.05 eV 1.60 × 10 −19 J eV
9.11 × 10
−31
kg
j=
1.57 × 10 6 m s
Larger than 10 −4 m s by ten orders of magnitude. However, the energy of an electron at
room temperature is typically k BT =
1
eV .
40
Chapter 43
P43.29
557
For sodium, M = 23.0 g mol and ρ = 0.971 g cm3 .
e
je
6.02 × 10 23 electrons mol 0.971 g cm 3
NA ρ
=
ne =
23.0 g mol
M
(a)
j
n e = 2.54 × 10 22 electrons cm3 = 2.54 × 10 28 electrons m3
(b)
P43.30
EF
F h I FG 3n IJ
=G
H 2m JK H 8π K
2
e
23
e6.626 × 10
=
2e9.11 × 10
−34
−31
j LM 3e2.54 × 10
8π
kg j M
N
2
J⋅s
28
m −3
j OP
PQ
23
= 5.05 × 10 −19 J = 3.15 eV
The melting point of silver is 1 234 K. Its Fermi energy at 300 K is 5.48 eV. The approximate fraction
of electrons excited is
e
a
jb
g
j
1.38 × 10 −23 J K 1 234 K
k BT
=
≈ 2% .
EF
5.48 eV 1.60 × 10 −19 J eV
P43.31
fe
Taking EF = 5.48 eV for sodium at 800 K,
f = eb
g
E − EF k B T
e b E − EF g
k BT
=
+1
−1
= 0.950
1
− 1 = 0.052 6
0.950
E − EF
= ln 0.052 6 = −2.94
k BT
b
g
e1.38 × 10 ja800f J = −0.203 eV or
= −2.94
−23
E − EF
P43.32
1.60 × 10 −19 J eV
e
j
d = 1.00 mm , so
V = 1.00 × 10 −3 m
The density of states is
g E = CE 1 2 =
af
af
gE =
or
3
E = 5.28 eV .
= 1.00 × 10 −9 m 3 .
8 2π m 3 2
h
3
E1 2
e
8 2π 9.11 × 10 −31 kg
e6.626 × 10
−34
J⋅s
j
j
3
3 2
a4.00 eVfe1.60 × 10
−19
J eV
j
af
g E = 8.50 × 10 46 m −3 ⋅ J −1 = 1.36 × 10 28 m −3 ⋅ eV −1 .
So, the total number of electrons is
a fa f e
jb
ge
j
N = g E ∆E V = 1.36 × 10 28 m −3 ⋅ eV −1 0.025 0 eV 1.00 × 10 −9 m3 = 3.40 × 10 17 electrons .
558
P43.33
Molecules and Solids
Eav =
z
∞
af
1
EN E dE
ne 0
af
af
Since f E = 1 for E < EEF and f E = 0 for E > EF , we can take
Eav =
z
EF
1
ne
CE 3 2 dE =
0
z
EF
C
ne
2C 5 2
EF .
5n e
E 3 2 dE =
0
But from Equation 43.24,
P43.34
af
N aEf = CE
N E = 0 for E > EF ;.
At T = 0 ,
C 3 −3 2
= EF , so that
ne 2
Eav =
12
=
8 2π m e3 2
h3
FG 2 IJ FG 3 E IJ E
H 5KH 2 K
−3 2
F
5 2
F
E1 2
3
EF .
5
=
Consider first the wave function in x. At x = 0 and x = L , ψ = 0.
Therefore,
sin k x L = 0
and
k x L = π , 2π , 3π , ….
Similarly,
sin k y L = 0
and
k y L = π , 2π , 3π , …
sin k z L = 0
and
k z L = π , 2π , 3π , …
FG n π x IJ sinFG n π y IJ sinFG n π z IJ .
H L K H L K H L K
∂ ψ ∂ ψ ∂ ψ 2m
+
+
=
From
aU − Efψ ,
∂x
∂y
∂z
F − n π − n π − n π I ψ = 2m −E ψ
a f
GH L L L JK
ψ = A sin
y
x
2
2
2
2
x
2
2
2
2
y
2
2
2
z
2
2
z
2
e
2
2
2
we have inside the box, where U = 0,
e
E=
2
π2 2
n x + n y2 + n z2
2
2m e L
e
j
n x , n y , n z = 1, 2 , 3 , … .
Outside the box we require ψ = 0.
The minimum energy state inside the box is
P43.35
(a)
n x = n y = n z = 1, with E =
The density of states at energy E is
Hence, the required ratio is
(b)
3 2π 2
2m e L2
af
g a8.50 eV f C a8.50f
=
g a7.00 eV f C a7.00f
g E = CE 1 2 .
12
12
= 1.10 .
From Eq. 43.22, the number of occupied states having energy E is
Hence, the required ratio is
At T = 300 K , k BT = 4.14 × 10 −21 J = 0.025 9 eV ,
And
12
a f eb CEg + 1 .
N a8.50 eV f a8.50f L e a
=
M
N a7.00 eV f a7.00f MN e a
N a8.50 eV f a8.50f L
=
N a7.00 eV f a7.00f MN e a
N a8.50 eV f
= 1.55 × 10
N a7.00 eV f
NE =
E − EF k B T
f
f
OP .
+ 1 PQ
OP .
+ 1Q
12
7.00 − 7 .00 k BT
12
8 .50 − 7.00 kB T
12
12
2.00
f
1.50 0 .025 9
−25
+1
.
Comparing this result with that from part (a), we conclude that very few states with E > EF
are occupied.
Chapter 43
Section 43.6
P43.36
559
Electrical Conduction in Metals, Insulators, and Semiconductors
E g = 1.14 eV for Si
(a)
a
fe
j
hf = 1.14 eV = 1.14 eV 1.60 × 10 −19 J eV = 1.82 × 10 −19 J so f ≥ 2.75 × 10 14 Hz
c = λf ; λ =
(b)
P43.37
c 3.00 × 10 8 m s
=
= 1.09 × 10 −6 m = 1.09 µm (in the infrared region)
f 2.75 × 10 14 Hz
Photons of energy greater than 2.42 eV will be absorbed. This means wavelength shorter than
λ=
e
je
j
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 514 nm .
E
2.42 × 1.60 × 10 −19 J
All the hydrogen Balmer lines except for the red line at 656 nm will be absorbed.
hc
−34
je
J ⋅ s 3.00 × 10 8 m s
j J≈
Eg =
P43.39
If λ ≤ 1.00 × 10 −6 m, then photons of sunlight have energy
E≥
λ
=
e6.626 × 10
P43.38
650 × 10
e6.626 × 10
=
hc
λ max
−34
−9
m
je
J ⋅ s 3.00 × 10 8 m s
1.00 × 10
−6
1.91 eV
j F 1 eV I = 1.24 eV .
GH 1.60 × 10 J JK
−19
m
Thus, the energy gap for the collector material should be E g ≤ 1.24 eV . Since Si has an energy gap
E g ≈ 1.14 eV , it will absorb radiation of this energy and greater. Therefore, Si is acceptable as a
material for a solar collector.
P43.40
If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here,
bhf g
max
=
hc
λ min
λ min = 2.26 × 10
*P43.41
λ min =
= 5.5 eV :
−7
a′ =
2
me ke e2
2
je
a
fe
j
j
m = 226 nm .
In the Bohr model we replace k e by
a0 =
e
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
5.5 eV
5.5 eV 1.60 × 10 −19 J eV
ke
κ
and m e by m* . Then the radius of the first Bohr orbit,
in hydrogen, changes to
FG IJ
H K
FG
H
FG IJ
H K
IJ b
K
2
me
me
me
κ
κ
κ a0 =
=
=
11.7 0.052 9 nm = 2.81 nm .
0. 220m e
m∗ke e2
m∗
me ke e2
m∗
The energy levels are in hydrogen En = −
En′ = −
ke e2 1
and here
2 a0 n 2
F I
GH JK
ke e2 1
ke e2
m ∗ En
=−
=
−
2
κ 2a ′ n
me κ 2
κ 2 m e m ∗ κ a0
e
For n = 1 , E1′ = −0.220
j
13.6 eV
11.7 2
= −0.021 9 eV .
g
560
Molecules and Solids
Section 43.7
P43.42
Semiconductor Devices
I = I 0 e ea ∆V f
Thus,
e ea ∆V f
and
∆V =
k BT
I
ln 1 +
.
e
I0
At T = 300 K ,
∆V =
e
k BT
j
−1 .
k BT
IJ
K
FG
H
e1.38 × 10
=1+
−23
I
I0
ja
J K 300 K
1.60 × 10
−19
C
f lnF 1 + I I = a25.9 mVf lnF 1 + I I .
GH I JK
GH I JK
0
a
f a f
∆V = a 25.9 mV f lna0.100f =
0
∆V = 25.9 mV ln 10.0 = 59.5 mV .
(a)
If I = 9.00 I 0 ,
(b)
If I = −0.900 I 0 ,
−59.5 mV .
The basic idea behind a semiconductor device is that a large current or charge can be
controlled by a small control voltage.
P43.43
P43.44
The voltage across the diode is about 0.6 V. The voltage drop across the resistor is
0.025 A 150 Ω = 3.75 V . Thus, ε − 0.6 V − 3.8 V = 0 and ε = 4.4 V .
a
fa
f
First, we evaluate I 0 in I = I 0 e ea ∆V f
ja
a f e
e
e
k BT
j
− 1 , given that I = 200 mA when ∆V = 100 mV and T = 300 K .
f
f
1.60 × 10 −19 C 0.100 V
e ∆V
200 mA
I
= 3.86
= 4.28 mA
=
= 3.86 so I 0 = ea ∆V f k T
−
23
B
k BT
−1
−1 e
1.38 × 10
J K 300 K
e
If ∆V = −100 mV ,
I = I 0 e ea ∆V f
e
*P43.45
(a)
k BT
a f
ja
e ∆V
= −3.86 ; and the current will be
k BT
j a
fe
j
− 1 = 4.28 mA e −3.86 − 1 = −4.19 mA .
The currents to be plotted are
e
je
I D = 10 −6 A e ∆V
IW =
0.025 V
j
−1 ,
20
2.42 V − ∆V
745 Ω
je
Diode
Wire
10
The two graphs intersect at
∆V = 0.200 V . The currents
are then
e
Diode and Wire Currents
0
j
I D = 10 −6 A e 0. 200 V 0.025 V − 1
0
0.1
∆V(volts)
= 2.98 mA
FIG. P43.45
2.42 V − 0.200 V
= 2.98 mA . They agree to three digits.
IW =
745 Ω
∴ I D = I W = 2.98 mA
(b)
∆V
0.200 V
=
= 67.1 Ω
ID
2.98 × 10 −3 A
(c)
d ∆V
dI D
=
dI D
d ∆V
a f LM OP = LM 10 A e
N a f Q N 0.025 V
−1
−6
0 . 200 V 0.025 V
OP
Q
−1
= 8.39 Ω
0.2
0.3
Chapter 43
Section 43.8
P43.46
(a)
(b)
Superconductivity
See the figure at right.
For a surface current around the outside of the cylinder as shown,
B=
Nµ 0 I
or NI =
B
µ0
a0.540 Tfe2.50 × 10 mj =
=
e4π × 10 j T ⋅ m A
−2
−7
10.7 kA .
FIG. P43.46
P43.47
By Faraday’s law (Equation 32.1),
∆Φ B
∆I
∆B
=L
=A
.
∆t
∆t
∆t
Thus,
A ∆B π 0.010 0 m 0.020 0 T
∆I =
=
= 203 A .
L
3.10 × 10 −8 H
a f b
gb
g
2
The direction of the induced current is such as to maintain the B– field through the ring.
P43.48
(a)
∆V = IR
If R = 0 , then ∆V = 0 , even when I ≠ 0 .
(b)
The graph shows a direct proportionality.
Slope =
a
f
f
155 − 57.8 mA
∆I
1
=
=
= 43.1 Ω −1
∆
R
V
3.61 − 1.356 mV
a
R = 0.023 2 Ω
(c)
FIG. P43.48
Expulsion of magnetic flux and therefore fewer
current-carrying paths could explain the
decrease in current.
Additional Problems
P43.49
(a)
Since the interatomic potential is the same for both molecules, the spring constant is the
same.
Then f =
1
2π
k
µ
where µ 12 =
a12 ufa16 uf = 6.86 u and µ = a14 ufa16 uf = 7.47 u .
14
12 u + 16 u
14 u + 16 u
Therefore,
f14 =
1
2π
k
µ 14
continued on next page
=
1
2π
k
µ 12
FG µ IJ = f
Hµ K
12
14
12
µ 12
= 6.42 × 10 13 Hz
µ 14
e
j
6.86 u
= 6.15 × 10 13 Hz .
7.47 u
561
562
Molecules and Solids
(b)
The equilibrium distance is the same for both molecules.
FG µ IJ µ r = FG µ IJ I
Hµ K
Hµ K
F 7.47 u IJ e1.46 × 10 kg ⋅ m j = 1.59 × 10 kg ⋅ m
I =G
H 6.86 u K
The molecule can move to the b v = 1, J = 9g state or to the b v = 1, J = 11g state. The energy it
I 14 = µ 14 r 2 =
14
12
12
14
12
12
−46
14
(c)
2
−46
2
2
can absorb is either
LMFG 1 + 1 IJ hf
2K
λ NH
hc LF
1I
= MG 1 + J hf
∆E =
H
λ N
2K
∆E =
or
hc
=
a f 2 I OP − LMFGH 0 − 12 IJK hf + 10a10 + 1f 2 I OP ,
Q N
Q
OP − LMFG 0 + 1 IJ hf + 10a10 + 1f OP .
+ 11a11 + 1f
2 I Q NH
2K
2I Q
2
14 + 9 9 + 1
2
14
14
14
2
14
2
14
14
14
The wavelengths it can absorb are then
λ=
These are: λ =
λ=
and
P43.50
c
f14 − 10
b2π I g
or λ =
14
c
f14 + 11
b2π I g .
14
3.00 × 10 8 m s
e
6.15 × 10 13 Hz − 10 1.055 × 10 −34 J ⋅ s
j
3.00 × 10 8 m s
e
6.15 × 10 13 Hz + 11 1.055 × 10 −34 J ⋅ s
e
j
e
j
2π 1.59 × 10 −46 kg ⋅ m 2
j
2π 1.59 × 10 −46 kg ⋅ m 2
For the N 2 molecule, k = 2 297 N m, m = 2.32 × 10 −26 kg , r = 1.20 × 10 −10 m , µ =
k
ω=
µ
e
je
j
I = µ r 2 = 1.16 × 10 −26 kg 1.20 × 10 −10 m
= 4.45 × 10 14 rad s,
2
= 4.96 µm
= 4.79 µm .
m
2
= 1.67 × 10 −46 kg ⋅ m 2 .
For a rotational state sufficient to allow a transition to the first exited vibrational state,
2
2I
a f
a f
J J + 1 = ω so J J + 1 =
2 Iω
=
e
je
2 1.67 × 10 −46 4.45 × 10 14
1.055 × 10
−34
j = 1 410 .
Thus J = 37 .
P43.51
FG
H
∆Emax = 4.5 eV = v +
8.25 > 7.5
P43.52
1
2
IJ
K
a4.5 eVfe1.6 × 10 J eVj ≥ F v + 1 I
G J
e1.055 × 10 J ⋅ sje8.28 × 10 s j H 2 K
−19
ω
so
−34
14
−1
v=7
With 4 van der Waal bonds per atom pair or 2 electrons per atom, the total energy of the solid is
e
10 atoms I
jFGH 6.02 ×4.00
JK =
g
E = 2 1.74 × 10 −23 J atom
23
5. 23 J g .
563
Chapter 43
P43.53
The total potential energy is given by Equation 43.17: U total = −α
ke e2
B
+ m.
r
r
The total potential energy has its minimum value U 0 at the equilibrium spacing, r = r0 . At this point,
dU
= 0,
dr r = r0
dU
dr
or
P43.54
=
r = r0
F
GH
k e e 2 m −1
r0 .
m
Thus,
B=α
Substituting this value of B into U total ,
U 0 = −α
I
JK
k e2
d
B
−α e + m
dr
r
r
=α
r = r0
kee2
r02
F I
GH JK
−
mB
= 0.
r0m +1
FG
H
kee2
k e2
k e2
1
1
1−
+ α e r0m −1 m = −α e
r0
m
r0
m
r0
IJ
K
1
3
hf + 4. 48 eV = hf + 3.96 eV is the depth
2
2
of the well below the dissociation point. We see hf = 0.520 eV , so the depth of the well is
Suppose it is a harmonic-oscillator potential well. Then,
a
f
1
1
hf + 4.48 eV = 0.520 eV + 4.48 eV = 4.74 eV .
2
2
P43.55
.
(a)
For equilibrium,
dU
= 0:
dx
d
Ax −3 − Bx −1 = −3 Ax −4 + Bx −2 = 0
dx
e
j
x → ∞ describes one equilibrium position, but the stable equilibrium position is at
3 Ax 0−2 = B .
x0 =
(b)
e
The depth of the well is given by
U0 = U x= x = −
0
(c)
j
3 0.150 eV ⋅ nm3
3A
=
= 0.350 nm
3.68 eV ⋅ nm
B
Fx = −
U0 = U x= x =
0
a
e
f
A
B
AB 3 2
BB 1 2
−
=
−
x 03 x 0 3 3 2 A 3 2 3 1 2 A 1 2
3 2
2 3.68 eV ⋅ nm
2B 3 2
=−
32 12
3 A
3 3 2 0.150 eV ⋅ nm3
j
12
= −7.02 eV .
dU
= 3 Ax −4 − Bx −2
dx
To find the maximum force, we determine finite x m such that
dF
dx
= 0.
x = xm
FG 6 A IJ .
HBK
F B IJ − BFG B IJ = − B = − a3.68 eV ⋅ nmf
= 3 AG
H 6 A K H 6 A K 12 A 12e0.150 eV ⋅ nm j
F 1.60 × 10 J I FG 1 nm IJ = −1.20 × 10
= −7.52 eV nm G
H 1 eV JK H 10 m K
Thus,
−12 Ax −5 + 2Bx −3
Then
Fmax
or
Fmax
2
x = xm
12
= 0 so that x m =
2
2
3
−19
−9
−9
N = −1.20 nN .
564
P43.56
Molecules and Solids
(a)
For equilibrium,
dU
= 0:
dx
d
Ax −3 − Bx −1 = −3 Ax −4 + Bx −2 = 0
dx
e
j
x → ∞ describes one equilibrium position, but the stable equilibrium position is at
3 Ax 0−2 = B or x 0 =
3A
.
B
(b)
The depth of the well is given by U 0 = U x = x =
(c)
Fx = −
0
A
B
AB 3 2
BB 1 2
B3
−
= 3 2 3 2 − 1 2 1 2 = −2
.
3
27 A
3 A
x0 x0 3 A
dU
= 3 Ax −4 − Bx −2
dx
To find the maximum force, we determine finite x m such that
dFx
dx
P43.57
(a)
= −12 Ax −5 + 2Bx −3
x = xm
x = x0
= 0 then Fmax = 3 A
dU
dr
At equilibrium separation, r = re ,
FG B IJ
H 6AK
2
−B
FG B IJ =
H 6AK
−
−a r −r
− a r −r
= −2 aB e b e 0 g − 1 e b e 0 g = 0 .
r = re
We have neutral equilibrium as re → ∞ and stable equilibrium at
e − abre − r0 g = 1 ,
or
re = r0 .
(b)
At r = r0 , U = 0. As r → ∞ , U → B . The depth of the well is B .
(c)
We expand the potential in a Taylor series about the equilibrium point:
af b g
2
br − r g + 12 ddrU br − r g
1
U ar f ≈ 0 + 0 + a −2BafL− ae b g − ae b g e e b
MN
2
U r ≈ U r0 +
dU
dr
r = r0
0
−2 r − r0
0
2
2
r = r0
− r − r0
This is of the form
(d)
B2
.
12 A
−2 r − r0
g − 1 jO
QP br − r g ≈ Ba br − r g
1
1
kx = k br − r g
2
2
0
r = r0
2
2
2
0
for a simple harmonic oscillator with
k = 2Ba 2 .
Then the molecule vibrates with frequency
f=
The zero-point energy is
ha
1
1
ω = hf =
π
2
2
1
2π
k
µ
=
a
2π
0
2
2
2B
µ
=
a
π
B
.
2µ
B
.
8µ
Therefore, to dissociate the molecule in its ground state requires energy B −
ha
π
B
.
8µ
Chapter 43
T=0
P43.58
E
EF
0
0.500
0.600
0.700
0.800
0.900
1.00
1.10
1.20
1.30
1.40
1.50
T = 0.1TF
af
bE E g−1 bT T g
e
e −∞
e −∞
e −∞
e −∞
e −∞
e −∞
e0
e +∞
e +∞
e +∞
e +∞
e +∞
F
bE E g−1 bT T g
f E
F
T = 0.2TF
F
e
−10 .0
e
e −5.00
e −4.00
e −3.00
e −2.00
e −1.00
e0
e 1.00
e 2.00
e 3 .00
e 4.00
e 5 .00
1.00
1.00
1.00
1.00
1.00
1.00
0.500
0.00
0.00
0.00
0.00
0.00
F
af
f E
1.000
0.993
0.982
0.953
0.881
0.731
0.500
0.269
0.119
0.047 4
0.018 0
0.006 69
T = 0.5TF
bE E g−1 bT T g
F
e
−5.00
e
e −2.50
e −2.00
e −1.50
e −1.00
e −0.500
e0
e 0 .500
e 1.00
e 1.50
e 2.00
e 2.50
F
af
f E
bE E g−1 bT T g
F
e
−2
.
00
0.993
e
0.924
e −1.00
0.881
e −0.800
0.818
e −0.600
0.731
e −0. 400
0.622
e −0. 200
0.500
e0
0.378
e 0. 200
0.269
e 0. 400
0.182
e 0 .600
0.119
e 0 .800
0.075 9 e 1.00
FIG. P43.58
P43.59
(a)
There are 6 Cl − ions at distance r = r0 . The
contribution of these ions to the electrostatic
−6 k e e 2
potential energy is
.
r0
There are 12 Na + ions at distance r = 2 r0 .
Their contribution to the electrostatic potential
+12 k e e 2
energy is
. Next, there are 8 Cl − ions
2 r0
at distance r = 3 r0 . These contribute a term of
−8 k e e 2
3 r0
to the electrostatic potential energy.
FIG. P43.59
To three terms, the electrostatic potential energy is:
FG
H
U = −6 +
12
2
continued on next page
−
IJ k e
3K r
8
e
0
2
= −2.13
565
k e2
kee2
with α = 2.13 .
or U = −α e
r0
r0
F
af
f E
0.881
0.731
0.690
0.646
0.599
0.550
0.500
0.450
0.401
0.354
0.310
0.269
566
Molecules and Solids
(b)
The fourth term consists of 6 Na + at distance r = 2r0 . Thus, to four terms,
a
U = −2.13 + 3
f k re
e
2
= 0.866
0
kee2
.
r0
So we see that the electrostatic potential energy is not even attractive to 4 terms, and that the
infinite series does not converge rapidly when groups of atoms corresponding to nearest
neighbors, next-nearest neighbors, etc. are added together.
ANSWERS TO EVEN PROBLEMS
P43.2
4.3 eV
P43.30
2%
P43.4
(a) 1.28 eV; (b) σ = 0.272 nm, ∈= 4.65 eV ;
(c) 6.55 nN; (d) 576 N m
P43.32
3.40 × 10 17 electrons
P43.6
(a) 40.0 µeV , 9.66 GHz;
(b) If r is 10% too small, f is 20% too large.
P43.34
see the solution
P43.36
(a) 275 THz; (b) 1.09 µm
P43.8
1.46 × 10 −46 kg ⋅ m 2
P43.38
1.91 eV
P43.10
(a) 1.81 × 10 −45 kg ⋅ m 2 ; (b) 1.62 cm
P43.40
226 nm
P43.12
(a) 11.8 pm; (b) 7.72 pm; HI is more loosely
bound
P43.42
(a) 59.5 mV; (b) –59.5 mV
P43.44
4.19 mA
P43.46
(a) see the solution; (b) 10.7 kA
P43.48
see the solution
P43.50
37
P43.52
5.23 J g
P43.54
4.74 eV
P43.56
(a) x 0 =
P43.58
see the solution
P43.14
(a) 0, 364 µeV , 1.09 meV;
(b) 98.2 meV, 295 meV, 491 meV
P43.16
(a) 472 µm ; (b) 473 µm ; (c) 0.715 µm
P43.18
2.9 × 10 −47 kg ⋅ m 2
P43.20
only 64.1 THz
P43.22
(a) ~ 10 17 ; (b) ~ 10 5 m 3
P43.24
(a) 0.444 nm, 0.628 nm, 0.769 nm
P43.26
see the solution
P43.28
(a) 1.57 Mm s;
(b) larger by 10 orders of magnitude
B2
3A
B3
; (b) −2
; (c) −
B
27 A
12 A
44
Nuclear Structure
ANSWERS TO QUESTIONS
CHAPTER OUTLINE
44.1
44.2
44.3
44.4
44.5
44.6
44.7
44.8
Some Properties of Nuclei
Nuclear Binding Energy
Nuclear Models
Radioactivity
The Decay Processes
Natural Radioactivity
Nuclear Reactions
Nuclear Magnetic
Resonance and Magnetic
Resonance Imagining
Q44.1
Because of electrostatic repulsion between the positivelycharged nucleus and the +2e alpha particle. To drive the αparticle into the nucleus would require extremely high kinetic
energy.
Q44.2
There are 86 protons and 136 neutrons in the nucleus 222
86 Rn.
For the atom to be neutral, there must be 86 electrons orbiting
the nucleus—the same as the number of protons.
Q44.3
All of these isotopes have the same number of protons in the
nucleus. Neutral atoms have the same number of electrons.
Isotopes only differ in the number of neutrons in the nucleus.
Q44.4
Nuclei with more nucleons than bismuth-209 are unstable
because the electrical repulsion forces among all of the protons
is stronger than the nuclear attractive force between nucleons.
Q44.5
The nuclear force favors the formation of neutron-proton pairs, so a stable nucleus cannot be too far
away from having equal numbers of protons and neutrons. This effect sets the upper boundary of
the zone of stability on the neutron-proton diagram. All of the protons repel one another electrically,
so a stable nucleus cannot have too many protons. This effect sets the lower boundary of the zone of
stability.
Q44.6
Nucleus Y will be more unstable. The nucleus with the higher binding energy requires more energy
to be disassembled into its constituent parts.
Q44.7
Extra neutrons are required to overcome the increasing electrostatic repulsion of the protons. The
neutrons participate in the net attractive effect of the nuclear force, but feel no Coulomb repulsion.
Q44.8
In the liquid-drop model the nucleus is modeled as a drop of liquid. The nucleus is treated as a
whole to determine its binding energy and behavior. The shell model differs completely from the
liquid-drop model, as it utilizes quantum states of the individual nucleons to describe the structure
and behavior of the nucleus. Like the electrons that orbit the nucleus, each nucleon has a spin state
to which the Pauli exclusion principle applies. Unlike the electrons, for protons and neutrons the
spin and orbital motions are strongly coupled.
Q44.9
The liquid drop model gives a simpler account of a nuclear fission reaction, including the energy
released and the probable fission product nuclei. The shell model predicts magnetic moments by
necessarily describing the spin and orbital angular momentum states of the nucleons. Again, the
shell model wins when it comes to predicting the spectrum of an excited nucleus, as the quantum
model allows only quantized energy states, and thus only specific transitions.
567
568
Nuclear Structure
Q44.10
4
Q44.11
If one half the number of radioactive nuclei decay in one year, then one half the remaining number
will decay in the second year. Three quarters of the original nuclei will be gone, and one quarter will
remain.
Q44.12
The statement is false. Both patterns show monotonic decrease over time, but with very different
shapes. For radioactive decay, maximum activity occurs at time zero. Cohorts of people now living
will be dying most rapidly perhaps forty years from now. Everyone now living will be dead within
less than two centuries, while the mathematical model of radioactive decay tails off exponentially
forever. A radioactive nucleus never gets old. It has constant probability of decay however long it
has existed.
Q44.13
Since the samples are of the same radioactive isotope, their half-lives are the same. When prepared,
sample A has twice the activity (number of radioactive decays per second) of sample B. After 5 halflives, the activity of sample A is decreased by a factor of 2 5 , and after 5 half-lives the activity of
sample B is decreased by a factor of 2 5 . So after 5 half-lives, the ratio of activities is still 2:1.
Q44.14
After one half-life, one half the radioactive atoms have decayed. After the second half-life, one half
1 1 3
of the remaining atoms have decayed. Therefore + = of the original radioactive atoms have
2 4 4
decayed after two half-lives.
Q44.15
The motion of a molecule through space does not affect anything inside the nucleus of an atom of
the molecule. The half-life of a nucleus is based on nuclear stability which, as discussed in Questions
44.4 and Q44.5, is predominantly determined by Coulomb repulsion and nuclear forces, not
molecular motion.
Q44.16
Long-lived progenitors at the top of each of the three natural radioactive series are the sources of our
radium. As an example, thorium-232 with a half-life of 14 Gyr produces radium-228 and radium-224
at stages in its series of decays, shown in Figure 44.17.
Q44.17
A free neutron decays into a proton plus an electron and an antineutrino. This implies that a proton
is more stable than a neutron, and in particular the proton has lower mass. Therefore a proton
cannot decay into a neutron plus a positron and a neutrino. This reaction satisfies every
conservation law except for energy conservation.
Q44.18
A neutrino has spin
Q44.19
Let us consider the carbon-14 decay used in carbon dating. 146 C→ 147 N + e − + ν . The carbon-14 atom
has 6 protons in the nucleus. The nitrogen-14 atom has 7 protons in the nucleus, but the additional
+ charge from the extra proton is canceled by the – charge of the ejected electron. Since charge is
conserved in this (and every) reaction, the antineutrino must have zero charge.
Similarly, when nitrogen-12 decays into carbon-12, the nucleus of the carbon atom has one
fewer protons, but the change in charge of the nucleus is balanced by the positive charge of the
ejected positron. Again according to charge conservation, the neutrino must have no charge.
He ,
16
O,
40
Ca , and
208
Pb.
1
while a photon has spin 1. A neutrino interacts by the weak interaction while
2
a photon is a quantum of the electromagnetic interaction.
Chapter 44
Q44.20
569
1
. When a nucleus undergoes beta decay, an electron and
2
antineutrino are ejected. With all nucleons paired, in their ground states the carbon-14, nitrogen-14,
nitrogen-12, and carbon-12 nuclei have zero net angular momentum. Angular momentum is
conserved in any process in an isolated system and in particular in the beta-decays of carbon-14 and
1
nitrogen-12. Conclusion: the neutrino must have spin quantum number , so that its z-component
2
=
=
of angular momentum can be just or − . A proton and a neutron have spin quantum number 1.
2
2
For conservation of angular momentum in the beta-decay of a free neutron, an antineutrino must
1
have spin quantum number .
2
An electron has spin quantum number
Q44.21
The alpha particle and the daughter nucleus carry equal amounts of momentum in opposite
p2
directions. Since kinetic energy can be written as
, the small-mass alpha particle has much more
2m
of the decay energy than the recoiling nucleus.
Q44.22
Bullet and rifle carry equal amounts of momentum p. With a much smaller mass m, the bullet has
p2
much more kinetic energy K =
. The daughter nucleus and alpha particle have equal momenta
2m
and the massive daughter nucleus, like the rifle, has a very small share of the energy released.
Q44.23
Yes. The daughter nucleus can be left in its ground state or sometimes in one of a set of excited
states. If the energy carried by the alpha particle is mysteriously low, the daughter nucleus can
quickly emit the missing energy in a gamma ray.
Q44.24
In a heavy nucleus each nucleon is strongly bound to its momentary neighbors. Even if the nucleus
could step down in energy by shedding an individual proton or neutron, one individual nucleon is
never free to escape. Instead, the nucleus can decay when two protons and two neutrons, strongly
bound to one another but not to their neighbors, happen momentarily to have a lot of kinetic
energy, to lie at the surface of the nucleus, to be headed outward, and to tunnel successfully through
the potential energy barrier they encounter.
Q44.25
mv 2
, or qBr = mv , a charged particle fired into a magnetic field is deflected
r
into a path with radius proportional to its momentum. If they have equal kinetic energies K, the
much greater mass m of the alpha particle gives it more momentum mv = 2mK than an electron.
Thus the electron undergoes greater deflection. This conclusion remains true if one or both particles
are moving relativistically.
From
∑ F = ma , or qvB =
Q44.26
The alpha particle stops in the wood, while many beta particles can make it through to deposit some
or all of their energy in the film emulsion.
Q44.27
The reaction energy is the amount of energy released as a result of a nuclear reaction. Equation 44.28
in the text implies that the reaction energy is (initial mass – final mass) c 2 . The Q-value is taken as
positive for an exothermic reaction.
570
Nuclear Structure
Q44.28
Carbon-14 is produced when carbon-12 is bombarded by cosmic rays. Both carbon-12 and carbon-14
combine with oxygen to form the atmospheric CO 2 that plants absorb in respiration. When the
plant dies, the carbon-14 is no longer replenished and decays at a known rate. Since carbon-14 is a
beta-emitter, one only needs to compare the activity of a living plant to the activity of the sample to
determine its age, since the activity of a radioactive sample exponentially decreases in time.
Q44.29
The samples would have started with more carbon-14 than we first thought. We would increase our
estimates of their ages.
Q44.30
There are two factors that determine the uncertainty on dating an old sample. The first is the fact
that the activity level decreases exponentially in time. After a long enough period of time, the
activity will approach background radiation levels, making precise dating difficult. Secondly, the
ratio of carbon-12 to carbon-14 in the atomsphere can vary over long periods of time, and this effect
contributes additional uncertainty.
Q44.31
An α-particle is a helium nucleus: 24 He
A β-particle is an electron or a positron: either e − or e + .
A γ-ray is a high-energy photon emitted when a nucleus makes a downward transition
between two states.
Q44.32
I z may have 6 values for I =
I = 3.
1
3
5
5
5 3 1
, namely , , , − , − , and − . Seven I z values are possible for
2
2
2
2
2 2 2
Q44.33
The frequency increases linearly with the magnetic field strength.
Q44.34
The decay of a radioactive nucleus at one particular moment instead of at another instant cannot be
predicted and has no cause. Natural events are not just like a perfect clockworks. In history, the idea
of a determinate mechanical Universe arose temporarily from an unwarranted wild extrapolation of
Isaac Newton’s account of planetary motion. Before Newton’s time [really you can blame Pierre
Simon de Laplace] and again now, no one thought of natural events as just like a perfect row of
falling dominos. We can and do use the word “cause” more loosely to describe antecedent enabling
events. One gear turning another is intelligible. So is the process of a hot dog getting toasted over a
campfire, even though random molecular motion is at the essence of that process. In summary, we
say that the future is not determinate. All natural events have causes in the ordinary sense of the
word, but not necessarily in the contrived sense of a cause operating infallibly and predictably in a
way that can be calculated. We have better reason now than ever before to think of the Universe as
intelligible. First describing natural events, and second determining their causes form the basis of
science, including physics but also scientific medicine and scientific bread-baking. The evidence
alone of the past hundred years of discoveries in physics, finding causes of natural events from the
photoelectric effect to x-rays and jets emitted by black holes, suggests that human intelligence is a
good tool for figuring out how things go. Even without organized science, humans have always
been searching for the causes of natural events, with explanations ranging from “the will of the
gods” to Schrödinger’s equation. We depend on the principle that things are intelligible as we make
significant strides towards understanding the Universe. To hope that our search is not futile is the
best part of human nature.
Chapter 44
571
SOLUTIONS TO PROBLEMS
Section 44.1
P44.1
Some Properties of Nuclei
An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water
molecule there are eight neutrons and ten protons.
So protons and neutrons are nearly equally numerous in your body, each contributing mass
(say) 35 kg:
35 kg
P44.2
F 1 nucleon I
GH 1.67 × 10 kg JK
−27
~ 10 28 protons
and
~ 10 28 neutrons .
The electron number is precisely equal to the proton number,
~ 10 28 electrons .
1
mv 2 = q∆V
2
and
2m∆V = qr 2 B 2 :
mv 2
= qvB
r
2 m∆ V
r=
qB
2
=
e
r = 5.59 × 10 11
(a)
b
2 1 000 V
g
2
m
−27
kg
j
−27
kg
j
e1.60 × 10 Cja0.200 Tf
m kg j m
−19
e
For 12 C , m = 12 u and r = 5.59 × 10 11 m
kg
j 12e1.66 × 10
r = 0.078 9 m = 7.89 cm .
For 13 C :
e
r = 5.59 × 10 11 m
kg
j 13e1.66 × 10
r = 0.082 1 m = 8.21 cm .
(b)
With
the ratio gives
r1 =
r1
=
r2
2 m 1 ∆V
qB
2
and r2 =
m1
m2
r1 7.89 cm
=
= 0.961
r2 8.21 cm
and
so they do agree.
m1
12 u
=
= 0.961
13 u
m2
2 m 2 ∆V
qB 2
572
P44.3
P44.4
Nuclear Structure
Q 1Q 2
a2fa6fe1.60 × 10 Cj
C j
e1.00 × 10 mj
−19
e
9
2
(a)
F = ke
(b)
a=
(c)
QQ
U = k e 1 2 = 8.99 × 10 9 N ⋅ m 2 C 2
r
r2
= 8.99 × 10 N ⋅ m
2
e
= 27.6 N
2
−14
27.6 N
F
=
= 4.17 × 10 27 m s 2
m 6.64 × 10 −27 kg
2
away from the nucleus.
a2fa6f 1.60 × 10 C
j e1.e00 × 10 mj j
−19
2
= 2.76 × 10 −13 J = 1.73 MeV
−14
Eα = 7.70 MeV
4k e Ze 2
e
ja fe
−19
9
2 k e Ze 2 2 8.99 × 10 79 1.60 × 10
=
=
Eα
7.70 1.60 × 10 −13
(a)
d min =
(b)
The de Broglie wavelength of the α is
λ=
(c)
mv 2
h
=
mα vα
h
2mα Eα
=
e
j
j
2
= 29.5 × 10 −15 m = 29.5 fm
6.626 × 10 −34
e
2 6.64 × 10
−27
j7.70e1.60 × 10 j
−13
= 5.18 × 10 −15 m = 5.18 fm .
Since λ is much less than the distance of closest approach , the α may be considered a
particle.
P44.5
(a)
The initial kinetic energy of the alpha particle must equal the electrostatic potential energy
at the distance of closest approach.
Ki = U f =
rmin
(b)
e
ja fa fe
8.99 × 10 9 N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C
k qQ
= e
=
Ki
0.500 MeV 1.60 × 10 −13 J MeV
Since K i =
vi =
k e qQ
rmin
a
fe
j
j
2
= 4.55 × 10 −13 m
k qQ
1
mα vi2 = e
,
2
rmin
2 k e qQ
=
mα rmin
e
ja fa fe
kg uje3.00 × 10
2 8.99 × 10 9 N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C
a4.00 ufe1.66 × 10
−27
−13
m
j
j
2
= 6.04 × 10 6 m s .
Chapter 44
P44.6
It must start with kinetic energy equal to K i = U f =
the 24 He and
197
79 Au
13
k e qQ
. Here r f stands for the sum of the radii of
rf
nuclei, computed as
e
13
je
j
r f = r0 A1 + r0 A 2 = 1.20 × 10 −15 m 41 3 + 197 1 3 = 8.89 × 10 −15 m .
Thus, K i = U f
P44.7
P44.8
e8.99 × 10
=
e
ja f
e
ja f
13
r = r0 A 1 3 = 1.20 × 10 −15 m 4
(b)
r = r0 A 1 3 = 1.20 × 10 −15 m 238
2
= 4.09 × 10 −12 J = 25.6 MeV .
= 1.90 × 10 −15 m
13
= 7.44 × 10 −15 m
a f
From r = r0 A 1 3 , the radius of uranium is rU = r0 238
a f
1
1
rU then r0 A 1 3 = r0 238
2
2
13
.
13
A = 30 .
from which
The number of nucleons in a star of two solar masses is
A=
Therefore
P44.10
j
8.89 × 10 −15 m
(a)
Thus, if r =
P44.9
ja fa fe
N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C
9
V=
e
2 1.99 × 10 30 kg
1.67 × 10
−27
j
kg nucleon
e
= 2.38 × 10 57 nucleons .
je
r = r0 A 1 3 = 1.20 × 10 −15 m 2.38 × 10 57
b
g
4 4 4
π r = π 0.021 5 m
3
3
3
j
13
= 16.0 km .
= 4.16 × 10 −5 m3
We take the nuclear density from Example 44.2
e
je
j
m = ρV = 2.3 × 10 17 kg m3 4.16 × 10 −5 m3 = 9.57 × 10 12 kg
and
F =G
m1 m 2
r2
9.57 × 10 kg
j e a1.00 mf j
12
e
= 6.67 × 10 −11 N ⋅ m 2 kg 2
2
2
F = 6.11 × 10 15 N toward the other ball.
P44.11
The stable nuclei that correspond to magic numbers are:
Z magic:
2 He
8O
20 Ca
28 Ni
50 Sn
82 Pb
An artificially produced nucleus with Z = 126 might be more stable than other nuclei with lower
values for Z, since this number of protons is magic.
N magic:
3
1T ,
4
2 He,
15
7 N,
16
8O,
37
17 Cl ,
39
19 K ,
40
20 Ca ,
51
23 V ,
89
39 Y ,
90
40 Zr ,
136
54 Xe ,
138
56 Ba ,
139
57 La,
140
58 Ce ,
141
59 Pr ,
142
208
60 Nd , 82 Pb,
210
84 Po
573
52
24 Cr ,
88
38 Sr ,
209
83 Bi ,
574
*P44.12
Nuclear Structure
(a)
For even Z, even N, even A, the list begins 24 He, 126 C , and ends
194
196
202
208
78 Pt , 78 Pt , 80 Hg , 82 Pb,
containing 48 isotopes.
(b)
195
The whole even Z, odd N, odd A list is 94 Be , 129
54 Xe , 78 Pt , with 3 entries.
(c)
The odd Z, even N, odd A list has 46 entries, represented as 11 H , 73 Li , …,
203
205
81Tl , 81Tl ,
209
83 Bi .
(d)
The odd Z, odd N, even A list has 1 entry, 147 N . Do not be misled into thinking that nature
favors nuclei with even numbers of neutrons. The form of the question here forces a count
with essentially equal numbers of odd-Z and even-Z nuclei. If we counted all of the stable
nuclei we would find many even-even isotopes but also lots of even-Z odd-N nuclei and
odd-Z even-N nuclei; we would find roughly equal numbers of these two kinds of odd-A
nuclei. A nucleus with one odd neutron is no more likely to be unstable than a nucleus with
one odd proton.
With the arbitrary 25% abundance standard, we can note that most elements have a
single predominant isotope. Ni, Cu, Zn, Ga, Ge, Pd, Ag, Os, Ir, and Pt form a compact patch
on the periodic table and have two common isotopes, as do some others. Tungsten is the
only element with three isotopes over 25% abundance.
*P44.13
(a)
Z1 = 8Z 2
N1 = 5 N 2
b
g
N 1 + Z1 = 6 N 1 + Z 2 and N 1 = Z1 + 4
Thus:
N 1 + Z1 = 6
FG 1 N
H5
1
+
1
Z1
8
IJ
K
5
Z1
4
5
∴ Z1 + 4 = Z1
4
∴ Z1 = 16
∴ N1 =
N 1 = Z1 + 4 = 20 , A1 = Z1 + N 1 = 36
N2 =
Hence:
(b)
6
2 He
36
16 S
1
1
N 1 = 4, Z 2 = Z1 = 2 , A 2 = Z 2 + N 2 = 6
5
8
and 62 He.
is unstable. Two neutrons must be removed to make it stable
e Hej .
4
2
Chapter 44
Section 44.2
P44.14
Nuclear Binding Energy
Using atomic masses as given in Table A.3,
b
g b
g
−2.014 102 + 1 1.008 665 + 1 1.007 825
For 21 H:
(a)
2
Eb
931.5 MeV
= 0.001 194 u
= 1.11 MeV nucleon .
u
A
gFGH
b
b
g b
IJ
K
g
2 1.008 665 + 2 1.007 825 − 4.002 603
(b)
For 24 He:
(c)
For
56
26 Fe :
30 1.008 665 + 26 1.007 825 − 55.934 942 = 0.528 u
Eb 0.528
=
= 0.009 44 uc 2 = 8.79 MeV nucleon .
56
A
(d)
For
238
92 U :
146 1.008 665 + 92 1.007 825 − 238.050 783 = 1.934 2 u
Eb 1.934 2
=
= 0.008 13 uc 2 = 7.57 MeV nucleon .
A
238
4
Eb
2
= 0.007 59 uc = 7.07 MeV nucleon .
A
b
g b
b
g b
g
g
a
∆M = Zm H + Nm n − M
BE ∆M 931.5
=
A
A
Nuclei
Z
N
M in u
∆M in u
55
25
26
27
30
30
32
54.938 050
55.934 942
58.933 200
0.517 5
0.528 46
0.555 35
P44.15
Mn
Fe
59
Co
56
∴56 Fe has a greater
P44.16
575
f
BE
in MeV
A
8.765
8.790
8.768
BE
than its neighbors. This tells us finer detail than is shown in Figure 44.5.
A
Use Equation 44.2.
The
23
11 Na ,
and for
23
12 Mg ,
Eb
= 8.11 MeV nucleon
A
Eb
= 7.90 MeV nucleon .
A
The binding energy per nucleon is greater for
23
11 Na
by 0.210 MeV . (There is less proton repulsion
23
in Na .)
P44.17
(a)
(b)
(c)
The neutron-to-proton ratio
139
A−Z
is greatest for
Z
139
55 Cs
and is equal to 1.53.
La has the largest binding energy per nucleon of 8.378 MeV.
139
Cs with a mass of 138.913 u. We locate the nuclei carefully on Figure 44.4, the
neutron–proton plot of stable nuclei. Cesium appears to be farther from the center of the
zone of stability. Its instability means extra energy and extra mass.
576
P44.18
Nuclear Structure
(a)
40
The radius of the
ja f
e
R = r0 A 1 3 = 1.20 × 10 −15 m 40
Ca nucleus is:
13
= 4.10 × 10 −15 m .
The energy required to overcome electrostatic repulsion is
e
j e
−19
2
2
9
C
3 k Q 2 3 8.99 × 10 N ⋅ m C 20 1.60 × 10
=
U= e
5R
5 4.10 × 10 −15 m
(b)
e
The binding energy of
b
40
20 Ca
j
j
2
= 1.35 × 10 −11 J = 84.1 MeV .
is
g b
g
b
g
Eb = 20 1.007 825 u + 20 1.008 665 u − 39.962 591 u 931.5 MeV u = 342 MeV .
(c)
P44.19
The nuclear force is so strong that the binding energy greatly exceeds the minimum energy
needed to overcome electrostatic repulsion.
f a f
e Xj b931.494 MeV ug .
= 8b1.007 825 ug + 7b1.008 665 ug − 15.003 065 u b931.494 MeV ug = 111.96 MeV .
= 7b1.007 825 ug + 8b1.008 665 ug − 15.000 109 u b931.494 MeV ug = 115.49 MeV .
For 158 O :
Eb
For 157 N:
Eb
Therefore, the binding energy of
P44.20
15
7N
Sn
A
Z
is larger by 3.54 MeV .
Removal of a neutron from 43
20 Ca would result in the residual nucleus,
separation energy is Sn , the overall process can be described by
42
20 Ca .
j + massanf
= b 41.958 618 + 1.008 665 − 42.958 767 g u = b0.008 516 ugb931.5 MeV ug =
mass
Section 44.3
P44.21
a
Eb MeV = ZM H + Nmn − M
The binding energy of a nucleus is
e
43
20 Ca
j+S
n
= mass
e
If the required
42
20 Ca
7.93 MeV .
Nuclear Models
∆Eb = Ebf − Ebi
Eb
= 7.4 MeV
A
For
A = 200,
so
Ebi = 200 7.4 MeV = 1 480 MeV .
For
A ≈ 100 ,
so
Ebf = 2 100 8.4 MeV = 1 680 MeV .
a
f
Eb
= 8.4 MeV
A
a fa
f
∆Eb = Ebf − Ebi : Eb = 1 680 MeV − 1 480 MeV = 200 MeV
FIG. P44.21
Chapter 44
P44.22
P44.23
(a)
The first term overstates the importance of volume and the second term subtracts this
overstatement.
(b)
For spherical volume
(a)
“Volume” term:
b4 3gπ R
3
R
R3
R
.
For
cubical
volume
.
=
2
2
3
6
4π R
6R
The maximum binding energy or lowest state of energy is achieved by building “nearly”
spherical nuclei.
=
a
fa f
E1 = C1 A = 15.7 MeV 56 = 879 MeV .
a
fa f = −260 MeV .
ZaZ − 1f
a26fa25f = −121 MeV .
= −C
= −a0.71 MeV f
A
a56f
a A − 2Zf = −a23.6 MeVf a56 − 52f = −6.74 MeV .
=C
56
A
“Surface” term:
E2 = −C 2 A
“Coulomb” term:
E3
“Asymmetry” term:
E4
23
3
= − 17.8 MeV 56
23
13
13
2
4
Eb = 491 MeV
(b)
Section 44.4
E1
E
E
E
= 179%; 2 = −53.0%, 3 = −24.6% ; 4 = −1.37%
Eb
Eb
Eb
Eb
Radioactivity
a
f
FG
H
IJ e
K
P44.24
R = R0 e − λ t = 6. 40 mCi e
P44.25
dN
= − λN
dt
so
λ=
T1 2
P44.26
b
ga
− ln 2 8.04 d 40. 2 d
f = a6.40 mCifee − ln 2 j5 = a6.40 mCifFG
1I
H 2 JK =
je
0.200 mCi
j
a
λ
f
F ln 2 I F 1.00 g I e6.02 × 10 j
GH 5.27 yr JK GH 59.93 g mol JK
F 1 yr IJ = 4.18 × 10
R = e1.32 × 10 decays yr jG
H 3.16 × 10 s K
23
R = λN =
(a)
5
1
dN
−
= 1.00 × 10 −15 6.00 × 10 11 = 6.00 × 10 −4 s −1
N
dt
ln 2
=
= 1.16 × 10 3 s = 19.3 min
21
P44.27
2
7
13
Bq
R = R0 e − λ t ,
R
1
1
10.0
ln
λ = ln 0 =
= 5.58 × 10 −2 h −1 = 1.55 × 10 −5 s −1
4.00 h
8.00
t
R
ln 2
T1 2 =
= 12.4 h
From
FG IJ FG
H K H
IJ FG IJ
K H K
λ
R0
F
GH
I
JK
N0 =
(c)
R = R0 e − λ t = 10.0 mCi exp −5.58 × 10 −2 × 30.0 = 1.88 mCi
λ
=
10.0 × 10 −3 Ci 3.70 × 10 10 s
= 2.39 × 10 13 atoms
1 Ci
1.55 × 10 −5 s
(b)
a
f e
j
577
578
P44.28
Nuclear Structure
where
λ=
R
= 0.100 = e − λ t
R0
so
ln 0.100 = − λ t
2.30 =
P44.29
ln 2
= 0.026 6 h −1
26.0 h
R = R0 e − λ t
a
FG 0.026 6 IJ t
H h K
f
t = 86.4 h
e
j
The number of nuclei which decay during the interval will be N 1 − N 2 = N 0 e − λ t1 − e − λ t 2 .
ln 2 0.693
=
= 0.010 7 h −1 = 2.97 × 10 −6 s −1
T1 2 64.8 h
First we find l :
λ=
and
N0 =
Substituting these values,
N 1 − N 2 = 4.98 × 10 11 e
R0
λ
b40.0 µCige3.70 × 10
=
2.97 × 10
jLMN
e
−6
4
s
s −1 µCi
−1
ja
e
− 0.010 7 h −1 10.0 h
j = 4.98 × 10
11
nuclei .
f − e −e0.010 7 h ja12.0 hf O .
−1
PQ
Hence, the number of nuclei decaying during the interval is N1 − N 2 = 9.47 × 10 9 nuclei .
P44.30
*P44.31
e
j
The number of nuclei which decay during the interval will be N 1 − N 2 = N 0 e − λ t1 − e − λ t 2 .
ln 2
T1 2
First we find λ:
λ=
so
e −λ t = e
and
N0 =
Substituting in these values
N1 − N 2 =
e
ln 2 − t T1 2
R0
λ
=
j = 2 −t T
R0 T1 2
ln 2
R0 T1 2
af
ln 2
12
.
ee
af
We have all this information: N x 0 = 2.50 N y 0
a f
N a0 fe
a f
N x 3d = 4.20 N y 3d
x
−λ x 3d
af
= 4.20 N y 0 e
−λ y 3d
= 4.20
2.5 3 dλ y
e
4.2
2.5
+ 3 dλ y
3dλ x = ln
4.2
0.693
2.5
0.693
= ln
+ 3d
= 0.781
3d
T1 2 x
4.2
1.60 d
e 3 dλ x =
FG IJ
H K
FG IJ
H K
T1 2 x = 2.66 d
af
N x 0 −λ y 3d
e
2.50
− λ t1
j
− e − λ t2 =
R0 T1 2
ln 2
e2
− t1 T1 2
−2
− t 2 T1 2
j.
Chapter 44
*P44.32
(a)
579
dN 2
= rate of change of N 2
dt
= rate of production of N 2 – rate of decay of N 2
= rate of decay of N 1 – rate of decay of N 2
= λ 1 N1 − λ 2 N 2
(b)
From the trial solution
af
N2 t =
∴
∴
e
j
dN 2
N λ
= 10 1 − λ 2 e − λ 2 t + λ 1 e − λ 1t
λ1 − λ 2
dt
e
j
(1)
dN 2
N λ
+ λ 2 N 2 = 10 1 − λ 2 e − λ 2 t + λ 1 e − λ 1t + λ 2 e − λ 2 t − λ 2 e − λ 1t
λ1 − λ 2
dt
N λ
= 10 1 λ 1 − λ 2 e − λ 1t
λ1 − λ 2
= λ 1 N1
e
b
So
(c)
N10 λ 1 − λ 2 t
− e − λ 1t
e
λ1 − λ 2
j
g
dN 2
= λ 1 N 1 − λ 2 N 2 as required.
dt
The functions to be plotted are
e
af
L
N at f = 1 130.8 M e e
N
N 1 t = 1 000 e
Decay of
j
j
2
−e
Po and
214
Pb
1 200
− 0. 223 6 min −1 t
− 0 . 223 6 min −1 t
218
e
jO
− 0 .025 9 min −1 t
PQ
From the graph: t m ≈ 10.9 min
1 000
Po
Pb
800
600
400
200
0
0
10
20
time (min)
30
FIG. P44.32(c)
(d)
From (1),
b
ln λ 1 λ 2
λ
dN 2
= 0 if λ 2 e − λ 2 t = λ 1 e − λ 1t . ∴ e b λ 1 − λ 2 gt = 1 . Thus, t = t m =
λ1 − λ 2
λ2
dt
g
.
With λ 1 = 0.223 6 min −1 , λ 2 = 0.025 9 min −1 , this formula gives t m = 10.9 min , in
agreement with the result of part (c).
Section 44.5
P44.33
The Decay Processes
b
gb
g
Q = b 238.050 783 − 234.043 596 − 4.002 603 g ub931.5 MeV ug =
Q = M U- 238 − M Th- 234 − M He- 4 931.5 MeV u
4.27 MeV
40
580
P44.34
P44.35
Nuclear Structure
(a)
A gamma ray has zero charge and it contains no protons or neutrons. So for a gamma ray
Z = 0 and A = 0. Keeping the total values of Z and A for the system conserved then requires
Z = 28 and A = 65 for X. With this atomic number it must be nickel, and the nucleus must be
*
in an exited state, so it is 65
28 Ni .
(b)
α = 24 He has Z = 2
so for X
and
A=4
we require
Z = 84 − 2 = 82
for Pb
and
A = 215 − 4 = 211, X= 211
82 Pb.
(c)
A positron e + = 01 e has charge the same as a nucleus with Z = 1 . A neutrino 00 ν has no charge.
Neither contains any protons or neutrons. So X must have by conservation Z = 26 + 1 = 27 . It
is Co. And A = 55 + 0 = 55 . It is 55
27 Co.
Similar reasoning about balancing the sums of Z and A across the reaction reveals:
(d)
0
−1 e
(e)
1
1H
(or p). Note that this process is a nuclear reaction, rather than radioactive decay. We can
solve it from the same principles, which are fundamentally conservation of charge and
conservation of baryon number.
NC =
F 0.021 0 g I e6.02 × 10
GH 12.0 g mol JK
23
molecules mol
j
eN = 1.05 × 10 carbon atomsj of which 1 in 7.70 × 10 is a C atom
ln 2
= 1.21 × 10 yr = 3.83 × 10
s
bN g = 1.37 × 10 , λ = 5 730
yr
21
C
0 C-14
11
9
−4
C-14
−1
−12
R = λN = λN 0 e − λ t
e
837
= 951 decays week .
0.88
R
−1
R
= −λ t
ln
so
t=
ln
R0
λ
R0
Taking logarithms,
t=
N = N0 e −λ t
R
R0
L 400 sg OP = 3.17 × 10
jMM 7b186week
PQ
N
3
decays week .
FG
H
IJ
K
FG IJ
H K
951
−1
ln
= 9.96 × 10 3 yr .
−4
−1
3
1.21 × 10 yr
3.17 × 10
dN
= R = − λN 0 e − λ t = R0 e − λ t
dt
FG R IJ = ln 2 t
H RK T
lnb0.25 0.13g
If R = 0.13 Bq , t = 5 730 yr
= 5 406 yr .
0.693
lnb0.25 0.11g
= 6 787 yr .
If R = 0.11 Bq , t = 5 730 yr
e −λ t =
−1
R=
At time t,
*P44.36
je
R0 = λN 0 = 3.83 × 10 −12 s −1 1.37 × 10 9
At t = 0 ,
14
eλ t =
R0
R
λ t = ln
0
12
t = T1 2
b
ln R0 R
g
ln 2
0.693
The range is most clearly written as between 5 400 yr and 6 800 yr , without understatement.
Chapter 44
P44.37
3
1H
nucleus→ 32 He nucleus + e − + ν
3
1H
becomes
nucleus + e − → 32 He nucleus + 2e − + ν .
Ignoring the slight difference in ionization energies,
3
1H
we have
atom→ 32 He atom + ν
3.016 049 u = 3.016 029 u + 0 +
Q
c2
Q = 3.016 049 u − 3.016 029 u 931.5 MeV u = 0.018 6 MeV = 18.6 keV
b
P44.38
(a)
gb
g
For e + decay,
b
g
b
gb
Q = M X − M Y − 2m e c 2 = 39.962 591 u − 39.963 999 u − 2 0.000 549 u 931.5 MeV u
Q = −2.33 MeV
Since Q < 0 , the decay cannot occur spontaneously.
(b)
For alpha decay,
b
g
b
Q = M X − Mα − M Y c 2 = 91.905 287 u − 4.002 603 u − 93.905 088 u 931.5 MeV u
g
Q = −2.24 MeV
Since Q < 0 , the decay cannot occur spontaneously.
(c)
For alpha decay,
b
g
b
Q = M X − Mα − M Y c 2 = 143.910 083 u − 4.002 603 u − 139.905 434 u 931.5 MeV u
Q = 1.91 MeV
Since Q > 0 , the decay can occur spontaneously.
P44.39
(a)
(b)
e− + p → n + ν
For nuclei,
15
O + e − → 15 N + ν .
15
8O
Add seven electrons to both sides to obtain
(c)
From Table A.3,
m
e Oj = me Nj + cQ
15
15
2
∆m = 15.003 065 u − 15.000 109 u = 0.002 956 u
b
gb
g
Q = 931.5 MeV u 0.002 956 u = 2.75 MeV
atom→ 157 N atom + ν .
g
g
581
582
Nuclear Structure
Section 44.6
P44.40
(a)
Natural Radioactivity
Let N be the number of
238
U nuclei and N ′ be
206
a
Pb nuclei.
f
Then N = N 0 e − λ t and N 0 = N + N ′ so N = N + N ′ e − λ t or e λ t = 1 +
FG
H
Taking logarithms,
λ t = ln 1 +
Thus,
t=
If
N
= 1.164 for the
N′
238
From above, e λ t = 1 +
IJ
K
where λ =
ln 2
.
T1 2
F T I lnFG 1 + N ′ IJ .
GH ln 2 JK H N K
12
U → 206 Pb chain with T1 2 = 4.47 × 10 9 yr , the age is:
t=
(b)
N′
N
N′
.
N
F 4.47 × 10 yr I lnFG 1 + 1 IJ =
GH ln 2 JK H 1.164 K
9
4.00 × 10 9 yr .
N
N
e −λ t
N′
. Solving for
=
gives
.
N′
N ′ 1 − e−λ t
N
With t = 4.00 × 10 9 yr and T1 2 = 7.04 × 10 8 yr for the
235
U → 207 Pb chain,
F ln 2 I t = aln 2fe4.00 × 10 yrj = 3.938 and
GH T JK 7.04 × 10 yr
9
λt=
8
12
With t = 4.00 × 10 9 yr and T1 2 = 1.41 × 10 10 yr for the
Th→ 208 Pb chain,
aln 2fe4.00 × 10 yrj = 0.196 6 and
9
λt=
232
1.41 × 10 10 yr
P44.41
FIG. P44.41
N
= 0.019 9 .
N′
N
= 4.60 .
N′
Chapter 44
P44.42
(a)
(b)
(c)
F 4.00 × 10 Ci I F 3.70 × 10 Bq I F 1.00 × 10 L I = 148 Bq m
GH 1 L JK GH 1 Ci JK GH 1 m JK
F T I = e148 Bq m jFG 3.82 d IJ FG 86 400 s IJ = 7.05 × 10 atoms m
R
N = = RG
H ln2 K H 1 d K
λ
H ln 2 JK
F 1 mol IJ FG 222 g IJ = 2.60 × 10 g m
mass = e7.05 × 10 atoms m jG
H 6.02 × 10 atoms K H 1 mol K
10
−12
4.00 pCi L =
3
3
3
12
7
3
7
3
583
3
−14
23
3
Since air has a density of 1.20 kg m3 , the fraction consisting of radon is
fraction =
*P44.43
(a)
2.60 × 10 −14 g m 3
1 200 g m
3
= 2.17 × 10 −17
.
Let x, y denote the half-lives of the nuclei X, Y.
R X R0 e − λ X t
− a 0.685 h fa ln 2 fb1 x −1 y g
=
=e
= 1.04 , which gives
RY R0 e − λ Y t
1 1
− = −0.082 603 69 h −1 .
x y
(1)
From the data: x − y = 77.2 h .
(2)
1
1
−
= −0.082 603 69 h −1 .
x x − 77.2 h
Substitute (2) into (1):
This reduces to the quadratic equation
x 2 − 77.2 x − 934.6 = 0
x = 87.84 h or −10.64 h .
which has solutions:
Thus: x = T1 2 , X = 87.84 h = 3.66 days is the only physical root.
From (2): y = T1 2 , Y = 87.84 h − 77.2 h = 10.6 h .
P44.44
(b)
From Table A.3, X is
(c)
From Figure 44.18,
224
224
212
Ra and Y is
Ra decays to
212
Pb .
Pb by three successive alpha-decays.
a f
− ln 2 t T1 2
Number remaining:
N = N0 e
Fraction remaining:
N
− a ln 2 ft T1 2
= e−λ t = e
.
N0
.
(a)
With T1 2 = 3.82 d and t = 7.00 d ,
N
= e − aln 2 fa7.00 f
N0
(b)
When t = 1.00 yr = 365.25 d ,
N
= e − aln 2 fa365. 25 f
N0
(c)
a3.82 f =
a3.82 f =
0. 281 .
1.65 × 10 −29 .
Radon is continuously created as one daughter in the series of decays starting from the
long-lived isotope
238
U.
584
Nuclear Structure
Section 44.7
P44.45
Nuclear Reactions
Q = M 27 Al + Mα − M 30 P − mn c 2
b
g
Q = 26.981 539 + 4.002 603 − 29.978 314 − 1.008 665 u 931.5 MeV u = −2.64 MeV
P44.46
(a)
For X,
A = 24 + 1 − 4 = 21
and
Z = 12 + 0 − 2 = 10 ,
21
10 Ne
.
so X is
144
54 Xe
.
A = 235 + 1 − 90 − 2 = 144
(b)
and
Z = 92 + 0 − 38 − 0 = 54 ,
A=2−2=0
(c)
and
Z = 2 − 1 = +1,
so X must be a positron.
X = 01 e +
As it is ejected, so is a neutrino:
P44.47
so X is
(a)
*
197
1
198
198
79 Au + 0 n → 79 Au → 80 Hg
(b)
Consider adding 79 electrons:
197
79 Au
+
and X ′= 00 ν .
0
−1 e + ν
atom+ 01 n→ 198
80 Hg atom + ν + Q
Q = M 197 Au + mn − M 198 Hg c 2
b
g
Q = 196.966 552 + 1.008 665 − 197.966 752 u 931.5 MeV u = 7.89 MeV
P44.48
Neglect recoil of product nucleus, (i.e., do not require momentum conservation for the system of
colliding particles). The energy balance gives K emerging = K incident + Q . To find Q:
b
g b
g
Q = b1.007 825 + 26.981 539g − b 26.986 705 + 1.008 665 g ub931.5 MeV ug = −5.59 MeV
Q = M H + M Al − M Si + mn c 2
Thus,
P44.49
9
4 Be + 1.665
K emerging = 6.61 MeV − 5.59 MeV = 1.02 MeV .
MeV → 84 Be+ 01 n , so M 8 Be = M 9 Be −
M 8 Be = 9.012 182 u −
4
4
Q
− mn
c2
a−1.665 MeVf − 1.008 665 u =
931.5 MeV u
9
1
10
4 Be+ 0 n→ 4 Be + 6.812
8.005 3 u
MeV , so M 10 Be = M 9 Be + mn −
M 10 Be = 9.012 182 u + 1.008 665 u −
4
4
4
4
Q
c2
6.812 MeV
= 10.013 5 u
931.5 MeV u
Chapter 44
P44.50
10
5B
(a)
+ 24 He→ 136 C + 11 H
The product nucleus is
13
6C
(b)
236
92 U
→
.
10
5B
.
+ 11 H → 105 B + 24 He
The product nucleus is
P44.51
13
6C
90
143
37 Rb + 55 Cs
+ 3 01 n ,
so Q = M 236 U − M 90 Rb − M 143 Cs − 3mn c 2
92
37
55
From Table A.3,
b
g b
g
Q = 236.045 562 − 89.914 809 − 142.927 330 − 3 1.008 665 u 931.5 MeV u = 165 MeV .
Section 44.8
Nuclear Magnetic Resonance and Magnetic Resonance Imagining
P44.52
FIG. P44.52
P44.53
2 µB
b
ja
ge
(a)
fn =
(b)
fp =
(c)
In the Earth’s magnetic field,
fp =
h
b
=
2 1.913 5 5.05 × 10 −27 J T 1.00 T
6.626 × 10 −34 J ⋅ s
ja
ge
2 2.792 8 5.05 × 10 −27 J T 1.00 T
6.626 × 10
b
−34
ge
J⋅s
je
6.626 × 10
29.2 MHz
f=
42.6 MHz
j=
2.13 kHz .
2 2.792 8 5.05 × 10 −27 50.0 × 10 −6
−34
f=
585
586
Nuclear Structure
Additional Problems
*P44.54
(a)
With mn and vn as the mass and speed of the neutrons, Eq. 9.23 of the text becomes, after
making appropriate notational changes, for the two collisions v1 =
FG 2m IJ v
Hm +m K
∴ bm + m g v = bm + m g v
∴ m bv − v g = m v − m v
n
v2 =
n
2
n
∴ mn =
P44.55
n
n
n
1
n
2
2
n
FG 2m IJ v ,
Hm +m K
2
1
n
1
1 1
1
= 2 m n vn
2 2
m 1 v1 − m 2 v 2
v 2 − v1
a1 ufe3.30 × 10
j a fe
7
m s − 14 u 4.70 × 10 6 m s
(b)
mn =
(a)
Q = M 9 Be + M 4 He − M 12 C − mn c 2
6
7
4.70 × 10 m s − 3.30 × 10 m s
j=
1.16 u
b
g
Q = 9.012 182 u + 4.002 603 u − 12.000 000 u − 1.008 665 u 931.5 MeV u = 5.70 MeV
(b)
Q = 2 M 2 H − M 3 He − mn
b
g
b
a
g
Q = 2 2.014 102 − 3.016 029 − 1.008 665 u 931.5 MeV u = 3.27 MeV exothermic
P44.56
(a)
f
At threshold, the particles have no kinetic energy relative to each other. That is, they move
like two particles that have suffered a perfectly inelastic collision. Therefore, in order to
calculate the reaction threshold energy, we can use the results of a perfectly inelastic
collision. Initially, the projectile M a moves with velocity v a while the target M X is at rest.
We have from momentum conservation for the projectile-target system:
Ma va = M a + M X vc .
1
The initial energy is: Ei = M a v a2 .
2
The final kinetic energy is:
b
Ef =
g
b
g
b
1
1
M a + M X v c2 = M a + M X
2
2
gLMN MM+ vM OPQ = LMN M M+ M OPQE .
2
a a
a
a
X
a
i
X
From this, we see that E f is always less than Ei and the change in energy, E f − Ei , is given by
E f − Ei =
LM M
NM + M
a
a
X
OP
Q
− 1 Ei = −
LM M OPE .
NM + M Q
X
a
X
i
This loss of kinetic energy in the isolated system corresponds to an increase in mass-energy
during the reaction. Thus, the absolute value of this kinetic energy change is equal to –Q
(remember that Q is negative in an endothermic reaction). The initial kinetic energy Ei is the
threshold energy Eth .
Therefore,
LM M OPE
NM + M Q
L M + M OP =
= −Q M
N M Q
X
−Q =
a
or
continued on next page
Eth
X
X
th
a
X
LM
N
−Q 1 +
Ma
MX
OP
Q
.
Chapter 44
(b)
First, calculate the Q-value for the reaction:
587
Q = M N-14 + M He- 4 − M O-17 − M H-1 c 2
b
OP = −a−1.19 MeVfLM1 + 4.002 603 OP =
Q
N 14.003 074 Q
g
Q = 14.003 074 + 4.002 603 − 16.999 132 − 1.007 825 u 931.5 MeV u = −1.19 MeV .
Eth = −Q
Then,
P44.57
1
1H
LM M + M
N M
X
a
X
1.53 MeV .
+ 73 Li → 74 Be + 01 n
b
g b
g b931.5 MeV ug
Q = b1.007 825 u + 7.016 004 ug − b7.016 929 u + 1.008 665 ug b931.5 MeV ug
Q = e −1.765 × 10 ujb931.5 MeV ug = −1.644 MeV
F m
I Q = F 1 + 1.007 825 I a1.644 MeVf = 1.88 MeV
= G1 +
Thus, KE
JK GH 7.016 004 JK
H m
Q = M H + M Li − M Be + M n
−3
incident projectile
min
P44.58
.
target nucleus
(a)
N0 =
(b)
λ=
1.00 kg
mass
=
= 2.52 × 10 24
mass per atom
239.05 u 1.66 × 10 −27 kg u
a
fe
j
ln 2
ln 2
=
= 9.106 × 10 −13 s −1
4
T1 2
2.412 × 10 yr 3.156 × 10 7 s yr
e
je
e
j
je
j
R0 = λN 0 = 9.106 × 10 −13 s −1 2.52 × 10 24 = 2.29 × 10 12 Bq
(c)
t=
P44.59
(a)
FG R IJ = 1 lnFG R IJ
λ HR K λ H R K
F 2.29 × 10 Bq I = 3.38 × 10 sFG 1 yr IJ =
lnG
H 3.156 × 10 s K
s
H 0.100 Bq JK
R = R0 e − λ t , so t =
−1
0
ln
0
12
1
9.106 × 10 −13
13
−1
7
1.07 × 10 6 yr
57
57
0
0
27 Co → 26 Fe + +1 e + 0 ν
The Q-value for this positron emission is Q = M 57 Co − M 57 Fe − 2m e c 2 .
b
g b
g
Q = 56.936 296 − 56.935 399 − 2 0.000 549 u 931.5 MeV u = −0.187 MeV
Since Q < 0 , this reaction cannot spontaneously occur .
(b)
14
6C
→
14
7N
+
0
0
−1 e + 0 ν
The Q-value for this e − decay is Q = M 14 C − M 14 N c 2 .
b
g
Q = 14.003 242 − 14.003 074 u 931.5 MeV u = 0.156 MeV = 156 keV
Since Q > 0 , the decay can spontaneously occur .
(c)
The energy released in the reaction of (b) is shared by the electron and neutrino. Thus,
K e can range from zero to 156 keV .
588
P44.60
Nuclear Structure
(a)
r = r0 A 1 3 = 1.20 × 10 −15 A 1 3 m.
r = 2.75 × 10 −15 m .
When A = 12 ,
(b)
(c)
(d)
P44.61
(a)
F=
a
f
ke Z − 1 e2
r2
When Z = 6
e8.99 × 10
=
9
ja
r2
r = 2.75 × 10 −15 m,
and
a
f
fe
N ⋅ m 2 C 2 Z − 1 1.60 × 10 −19 C
ja
e
fe
j
2
F = 152 N .
8.99 × 10 9 Z − 1 1.6 × 10 −19
k e q1 q 2 k e Z − 1 e 2
=
=
U=
r
r
r
j
2
When Z = 6
and
r = 2.75 × 10 −15 m,
U = 4.19 × 10 −13 J = 2.62 MeV .
A = 238;
Z = 92,
r = 7.44 × 10 −15 m
F = 379 N
and
U = 2.82 × 10 −12 J = 17.6 MeV .
Because the reaction p → n + e + + ν would violate the law of conservation of energy
m p = 1.007 276 u
me + = 5.49 × 10 −4 u .
mn = 1.008 665 u
mn + m e+ > m p .
Note that
(b)
The required energy can come from the electrostatic repulsion of protons in the nucleus.
(c)
Add seven electrons to both sides of the reaction for nuclei
to obtain the reaction for neutral atoms
13
7N
atom →
13
6C
13
13
7 N → 6C
+ e+ + ν
atom + e + + e − + ν
e N j − me C j − m − m − m
Q = b931.5 MeV ug 13.005 739 − 13.003 355 − 2e5.49 × 10 j − 0 u
Q = b931.5 MeV uge1.286 × 10 uj = 1.20 MeV
Q = c2 m
13
13
e+
ν
e−
−4
−3
P44.62
(a)
A least-square fit to the graph yields:
e
j
λ = − slope = − −0.250 h −1 = 0.250 h −1
and
b g = intercept = 8.30 .
F 1 h IJ = 4.17 × 10
λ = 0.250 h G
H 60.0 min K
ln cpm
(b)
t= 0
−1
T1 2 =
ln 2
=
min −1
ln 2
λ
4.17 × 10 −3 min −1
= 166 min = 2.77 h
continued on next page
−3
FIG. P44.62
589
Chapter 44
(c)
0
= 8.30 .
bcpmg = e counts min = 4.02 × 10 counts min .
4.02 × 10 counts min
1 bcpmg
=
=
= 9.65 × 10 atoms
λ Eff
e4.17 × 10 min ja0.100f
8.30
Thus,
P44.63
b g
intercept = ln cpm
From (a),
3
0
R0
3
0
6
(d)
N0 =
(a)
The reaction is 145
61 Pm →
(b)
Q = M Pm − Mα − M Pr 931.5 = 144.912 744 − 4.002 603 − 140.907 648 931.5 = 2.32 MeV
(c)
The alpha and daughter have equal and opposite momenta
λ
−3
b
Eα =
141
59 Pr + α
g
pα2
2 mα
−1
b
Ed =
g
pα = p d
p d2
2m d
pα2 2mα
1 2 mα
Eα
Eα
md
141
=
= 2
=
=
=
= 97. 2% or
2
+
+4
Etot Eα + Ed
m
m
141
+
m
m
1
2
1
2
pα 2mα + pα 2m d
α
d
α
d
e
j e
j b
g b
g
2.26 MeV.
This is carried away by the alpha.
P44.64
(a)
If ∆E is the energy difference between the excited and ground states of the nucleus of mass
M, and hf is the energy of the emitted photon, conservation of energy for the nucleusphoton system gives
∆E = hf + Er .
(1)
Where Er is the recoil energy of the nucleus, which can be expressed as
Er =
a f
2
Mv
Mv 2
=
.
2
2M
(2)
Since system momentum must also be conserved, we have
Mv =
(b)
hf
.
c
2
hf
b g
Hence, Er can be expresses as
Er =
When
hf << Mc 2
we can make the approximation that
hf ≈ ∆E
so
Er ≈
Er =
a ∆E f
2 Mc 2
a ∆E f
(3)
.
2
.
2 Mc 2
2
2 Mc 2
and
where
∆E = 0.014 4 MeV
a fb
g
Mc 2 = 57 u 931.5 MeV u = 5.31 × 10 4 MeV .
e1.44 × 10 MeVj =
=
2e5.31 × 10 MeV j
2
−2
Therefore,
Er
4
1.94 × 10 −3 eV .
590
P44.65
Nuclear Structure
(a)
One liter of milk contains this many
23
−17
(a)
−1
λ=
with
FG
H
FG IJ
H K
18
nuclei
−1
18
For the iodine, R = R0 e − λ t
t=
−17
7
9
12
P44.66
K nuclei:
mol I F 0.011 7 I
gFGH 6.02 ×3910.1 gnuclei
JK GH 100 JK = 3.60 × 10
mol
ln 2
ln 2
FG 1 yr IJ = 1.72 × 10 s
λ=
=
T
1.28 × 10 yr H 3.156 × 10 s K
R = λN = e1.72 × 10
s je3.60 × 10 j = 61.8 Bq
b
N = 2.00 g
(b)
40
IJ
K
ln 2
8.04 d
2 000
R
1
8.04 d
ln 0 =
ln
= 40.3 d
λ
R
ln 2
61.8
For cobalt–56,
λ=
F
GH
I
JK
ln 2
ln 2 365.25 d
=
= 3.28 yr −1 .
T1 2 77.1 d
1 yr
The elapsed time from July 1054 to July 2003 is 949 yr.
R = R0 e − λ t
R
− e 3. 28 yr −1 jb 949 yr g
= e−λ t = e
= e −3 116 = e − aln 10 f1 353 = ~ 10 −1 353 .
R0
implies
(b)
For carbon–14,
λ=
ln 2
= 1.21 × 10 −4 yr −1
5 730 yr
R
− e1.21 ×10 −4 yr −1 jb 949 yr g
= e−λ t = e
= e −0.115 = 0.892
R0
P44.67
We have
N 235 = N 0 , 235 e − λ 235 t
and
N 238 = N 0 , 238 e − λ 238 t
N 235
e −aln 2 ft Th, 235 +aln 2 ft Th , 238 j .
= 0.007 25 = e
N 238
Taking logarithms,
F
GH 0.704ln× 102
F
1
−4.93 = G −
H 0.704 × 10
−4.93 = −
or
t=
9
9
−4.93
e−1.20 × 10
−9
yr
−1
yr
yr
+
+
j ln 2
I
JK
I aln 2ft
J
yr K
ln 2
t
4.47 × 10 9 yr
1
4.47 × 10 9
= 5.94 × 10 9 yr .
Chapter 44
P44.68
(a)
Add two electrons to both sides of the reaction to have it in energy terms:
411 H atom → 24 He atom + Q
b
Q = ∆mc 2 = 4M 1 H − M 4 He c 2
1
g
b
Q = 4 1.007 825 u − 4.002 603 u 931.5 MeV u
1.99 × 10 30 kg
10
JI
gFGH 1.601 ×MeV
JK =
−13
4.28 × 10 −12 J
= 1.19 × 10 57 atoms = 1.19 × 10 57 protons
(b)
N=
(c)
The energy that could be created by this many protons in this reaction is:
1.67 × 10 −27 kg atom
e1.19 × 10
P=
P44.69
2
57
protons
E
∆t
× 10
JI
jFGH 4.428protons
JK = 1.27 × 10
−12
∆t =
so
E
P
=
45
J
1.27 × 10 45 J
= 3.38 × 10 18 s = 107 billion years .
3.77 × 10 26 W
E = − µ ⋅ B so the energies are E1 = + µB and E2 = − µB
µ = 2.792 8 µ n and µ n = 5.05 × 10 −27 J T
b
ja
ge
f
∆E = 2 µB = 2 2.792 8 5.05 × 10 −27 J T 12.5 T = 3.53 × 10 −25 J = 2.20 × 10 −6 eV .
P44.70
(a)
λ=
FG
H
IJ
K
1 yr
ln 2
ln 2
=
= 4.17 × 10 −9 s −1
T1 2 5.27 yr 3.156 × 10 7 s
b
t = 30.0 months = 2.50 yr
gFGH 3.1561 ×yr10 s IJK = 7.89 × 10
7
b g
F R I L a10.0 Cife3.70 × 10
= G Je = M
H λ K MN 4.17 × 10 s
7
s
R = R0 e − λ t = λN 0 e − λ t
10
so N 0
λt
−9
N 0 = 1.23 × 10 20 nuclei
e
Mass = 1.23 × 10 20 atoms
(b)
Bq Ci
−1
j OPee
PQ
je
4.17 × 10 −9 s −1 7.89 × 10 7 s
I = 1.23 × 10
mol
jFGH 6.02 ×5910.93 gatoms
mol JK
23
−2
j
g = 12.3 mg
We suppose that each decaying nucleus promptly puts out both a beta particle and two
gamma rays, for
a
f
Q = 0.310 + 1.17 + 1.33 MeV = 2.81 MeV
a
fe
je
j
P = QR = 2.81 MeV 1.6 × 10 −13 J MeV 3.70 × 10 11 s −1 = 0.166 W
591
592
P44.71
Nuclear Structure
For an electric charge density ρ =
Ze
b g
4 3 π R3
.
Using Gauss’s Law inside the sphere,
E ⋅ 4π r 2 =
b4 3gπ r
3
Ze
b4 3gπ R
∈0
3
:
=
P44.72
(a)
z FGH 4π1∈ IJK
R
0
0
2
Z 2 e2r 2
R
6
1 Zer
4π ∈0 R 3
ar ≤ R f
E=
1 Ze
4π ∈0 r 2
ar ≥ R f
U=
We now find the electrostatic energy:
1
U = ∈0
2
E=
1
4π r dr + ∈0
2
2
z
∞
1
∈0 E 2 4π r 2 dr
2
r =0
z FGH 4π1∈ IJK
∞
R
2
Z2 e2
0
r
4
4π r 2 dr =
LM
N
1
Z 2 e 2 R5
+
6
8π ∈0 5 R
R
OP
Q
2 2
3 Z e
20 π ∈0 R
For the electron capture,
93
0
93
43 Tc + −1 e → 42 Mo + γ
.
The disintegration energy is Q = M 93 Tc − M 93 Mo c 2 .
b
g
Q = 92.910 2 − 92.906 8 u 931.5 MeV u = 3.17 MeV > 2.44 MeV
Electron capture is allowed
For positron emission,
to all specified excited states in
93
93
0
43 Tc → 42 Mo + +1 e + γ
93
42 Mo.
.
The disintegration energy is Q ′ = M 93 Tc − M 93 Mo − 2m e c 2 .
b
g b
g
Q ′ = 92.910 2 − 92.906 8 − 2 0.000 549 u 931.5 MeV u = 2.14 MeV
Positron emission can reach
the 1.35, 1.48, and 2.03 MeV states
but there is insufficient energy to
reach the 2.44 MeV state.
(b)
The daughter nucleus in both forms of
decay is
93
42 Mo
.
FIG. P44.72
Chapter 44
P44.73
K=
1
mv 2 ,
2
so
v=
2K
=
m
b
ge
2 0.040 0 eV 1.60 × 10 −19 J eV
1.67 × 10
The time for the trip is t =
−27
kg
j = 2.77 × 10
3
m s.
1.00 × 10 4 m
x
=
= 3.61 s .
v 2.77 × 10 3 m s
The number of neutrons finishing the trip is given by N = N 0 e − λ t .
N
− a ln 2 ft T1 2
− a ln 2 fb 3 .61 s 624 s g
=1−e
=1−e
= 0.004 00 = 0.400% .
N0
The fraction decaying is 1 −
P44.74
(a)
If we assume all the
then
N = N0 e
yields
t=
P44.75
λ
ln
Sr came from
87
Rb,
−λ t
FG N IJ = T lnFG N IJ
H N K ln 2 H N K
12
0
0
where
N = N Rb-87
and
N 0 = N Sr-87 + N Rb-87
t=
(b)
−1
87
e4.75 × 10
10
ln 2
yr
j lnF 1.82 × 10 + 1.07 × 10 I =
GH 1.82 × 10
JK
10
9
10
It could be no older . The rock could be younger if some
b g
R = R0 exp − λ t lets us write
ln R = ln R0 − λ t
which is the equation of a straight line with
slope = λ .
3.91 × 10 9 yr .
87
Sr were originally present.
The logarithmic plot shown in Figure P44.75 is fitted by
ln R = 8.44 − 0.262 t .
If t is measured in minutes, then decay constant λ is 0.262 per
minute. The half–life is
T1 2 =
ln 2
λ
=
ln 2
= 2.64 min .
0.262 min
The reported half–life of 137 Ba is 2.55 min. The difference
reflects experimental uncertainties.
FIG. P44.75
ANSWERS TO EVEN PROBLEMS
P44.2
(a) 7.89 cm and 8.21 cm ;
(b) see the solution
P44.4
(a) 29.5 fm; (b) 5.18 fm; (c) see the solution
P44.6
25.6 MeV
30
Si with A = 30
P44.8
a nucleus such as
P44.10
6.11 PN toward the other ball
P44.12
(a) 48; (b) 3; (c) 46; (d) 1
593
594
P44.14
P44.16
P44.18
Nuclear Structure
(a) 1.11 MeV nucleon;
(b) 7.07 MeV nucleon ;
(c) 8.79 MeV nucleon ;
(d) 7.57 MeV nucleon
0.210 MeV greater for
less proton repulsion
23
Na because it has
(a) 84.1 MeV ; (b) 342 MeV ; (c) The nuclear
force of attraction dominates over
electrical repulsion
P44.20
7.93 MeV
P44.22
(a) see the solution;
R
R
and ; see the solution
(b)
3
6
P44.24
0.200 mCi
P44.26
41.8 TBq
P44.28
86.4 h
P44.30
P44.32
R0 T1 2
ln 2
e2
− t1 T1 2
(a)
(e)
*
65
28 Ni ;
1
1H
(b)
j
g
211
82 Pb ;
(c)
55
27 Co;
(d)
0
−1 e;
P44.36
between 5 400 yr and 6 800 yr
P44.38
(a) cannot occur ; (b) cannot occur ;
(c) can occur
P44.40
(a) 4.00 Gyr ; (b) 0.019 9 and 4.60
P44.42
(a) 148 Bq m3 ; (b) 7.05 × 10 7 atoms m 3 ;
(c) 2.17 × 10 −17
(a) 0.281 ; (b) 1.65 × 10 −29 ;
(c) see the solution
P44.46
(a)
P44.48
1.02 MeV
P44.50
(a)
P44.52
see the solution
P44.54
(a) see the solution; (b) 1.16 u
P44.56
(a) see the solution; (b) 1.53 MeV
P44.58
(a) 2.52 × 10 24 ; (b) 2.29 TBq ; (c) 1.07 Myr
P44.60
(a) 2.75 fm; (b) 152 N ; (c) 2.62 MeV ;
(d) 7.44 fm , 379 N , 17.6 MeV
P44.62
(a) see the solution;
(b) 4.17 × 10 −3 min −1 ; 2.77 h ;
(c) 4.02 × 10 3 counts min ;
21
10 Ne;
13
6C ;
0 +
0
(b) 144
54 Xe; (c) 1 e and 0 ν
(b)
10
5B
(d) 9.65 × 10 6 atoms
(a) see the solution; (b) see the solution;
(c) see the solution; 10.9 min ;
ln λ 1 λ 2
; yes
(d) t m =
λ1 − λ 2
b
P44.34
−2
− t 2 T1 2
P44.44
P44.64
(a) see the solution; (b) 1.94 meV
P44.66
(a) ~ 10 −1 353 ; (b) 0.892
P44.68
(a) 4.28 pJ; (b) 1.19 × 10 57 atoms ;
(c) 107 Gyr
P44.70
(a) 12.3 mg ; (b) 0.166 W
P44.72
(a) electron capture to all; positron
emission to the 1.35 MeV , 1.48 MeV , and
2.03 MeV states ; (b) 93
42 Mo; see the solution
P44.74
(a) 3.91 Gyr ; (b) No older; it could be
younger if some 87 Sr were originally
present, contrary to our assumption.
45
Applications of Nuclear Physics
CHAPTER OUTLINE
45.1
45.2
45.3
45.4
45.5
45.6
45.7
Interactions Involving
Neutrons
Nuclear Fission
Nuclear Reactors
Nuclear Fusion
Radiation Damage
Radiation Detectors
Uses of Radiation
ANSWERS TO QUESTIONS
Q45.1
A moderator is used to slow down neutrons released in the
fission of one nucleus, so that they are likely to be absorbed by
another nucleus to make it fission.
Q45.2
The hydrogen nuclei in water molecules have mass similar to
that of a neutron, so that they can efficiently rob a fast-moving
neutron of kinetic energy as they scatter it. Once the neutron is
slowed down, a hydrogen nucleus can absorb it in the reaction
n + 11 H → 21 H .
Q45.3
The excitation energy comes from the binding energy of the
extra nucleon.
Q45.4
The advantage of a fission reaction is that it can generate much more electrical energy per gram of
fuel compared to fossil fuels. Also, fission reactors do not emit greenhouse gasses as combustion
byproducts like fossil fuels—the only necessary environmental discharge is heat. The cost involved
in producing fissile material is comparable to the cost of pumping, transporting and refining fossil
fuel.
The disadvantage is that some of the products of a fission reaction are radioactive—and some
of those have long half-lives. The other problem is that there will be a point at which enough fuel is
spent that the fuel rods do not supply power economically and need to be replaced. The fuel rods
are still radioactive after removal. Both the waste and the “spent” fuel rods present serious health
and environmental hazards that can last for tens of thousands of years. Accidents and sabotage
involving nuclear reactors can be very serious, as can accidents and sabotage involving fossil fuels.
Q45.5
The products of fusion reactors are generally not themselves unstable, while fission reactions result
in a chain of reactions which almost all have some unstable products.
Q45.6
For the deuterium nuclei to fuse, they must be close enough to each other for the nuclear forces to
overcome the Coulomb repulsion of the protons—this is why the ion density is a factor. The more
time that the nuclei in a sample spend in close proximity, the more nuclei will fuse—hence the
confinement time is a factor.
595
596
Applications of Nuclear Physics
Q45.7
In a fusion reaction, the main idea is to get the nuclear forces, which act over very short distances, to
overcome the Coulomb repulsion of the protons. Tritium has one more neutron in the nucleus, and
thus increases the nuclear force, decreasing the necessary kinetic energy to obtain D–T fusion as
compared to D–D fusion.
Q45.8
The biggest obstacle is power loss due to radiation. Remember that a high temperature must be
maintained to keep the fuel in a reactive plasma state. If this kinetic energy is lost due to
bremsstrahlung radiation, then the probability of nuclear fusion will decrease significantly.
Additionally, each of the confinement techniques requires power input, thus raising the bar for
sustaining a reaction in which the power output is greater than the power input.
Q45.9
Fusion of light nuclei to a heavier nucleus releases energy. Fission of a heavy nucleus to lighter
nuclei releases energy. Both processes are steps towards greater stability on the curve of binding
energy, Figure 44.5. The energy release per nucleon is typically greater for fusion, and this process is
harder to control.
Q45.10
Advantages of fusion: high energy yield, no emission of greenhouse gases, fuel very easy to obtain,
reactor can not go supercritical like a fission reactor, low amounts of radioactive waste.
Disadvantages: requires high energy input to sustain reaction, lithium and helium are scarce,
neutrons released by reaction cause structural damage to reactor housing.
Q45.11
The fusion fuel must be heated to a very high temperature. It must be contained at a sufficiently
high density for a sufficiently long time to achieve a reasonable energy output.
Q45.12
The first method uses magnetic fields to contain the plasma, reducing its contact with the walls of
the container. This way, there is a reduction in heat loss to the environment, so that the reaction may
be sustained over seconds.
The second method involves striking the fuel with high intensity, focused lasers from multiple
directions, effectively imploding the fuel. This increases the internal pressure and temperature of
the fuel to the point of ignition.
Q45.13
No. What is critical in radiation safety is the type of radiation encountered. The curie is a measure of
the rate of decay, not the products of the decay or of their energies.
Q45.14
X-ray radiation can cause genetic damage in the developing fetus. If the damaged cells survive the
radiation and reproduce, then the genetic errors will be replicated, potentially causing severe birth
defects or death of the child.
Q45.15
For each additional dynode, a larger applied voltage is needed, and hence a larger output from a
power supply—“infinite” amplification would not be practical. Nor would it be desirable: the goal is
to connect the tube output to a simple counter, so a massive pulse amplitude is not needed. If you
made the detector sensitive to weaker and weaker signals, you would make it more and more
sensitive to background noise.
Chapter 45
Q45.16
597
Sometimes the references are oblique indeed. Some must serve for more than one form of energy or
mode of transfer. Here is one list:
kinetic: ocean currents
rotational kinetic: Earth turning
gravitational: water lifted up
elastic: Elastic energy is necessary for sound, listed below.
internal: by contrast to a chilly night; or in forging a chain
chemical: flames
sound: thunder
electrical transmission: lightning
electromagnetic radiation: heavens blazing; lightning
atomic electronic: In the blazing heavens, stars have different colors because of different
predominant energy losses by atoms at their surfaces.
nuclear: The blaze of the heavens is produced by nuclear reactions in the cores of stars.
Remarkably, the word “energy” in this translation is an anachronism. Goethe wrote the song a few
years before Thomas Young coined the term.
SOLUTIONS TO PROBLEMS
Section 45.1
Interactions Involving Neutrons
Section 45.2
Nuclear Fission
*P45.1
The energy is
FG 1 eV IJ FG 1 U - 235 nucleus IJ FG
235 g
IJ FG M IJ =
H 1.60 × 10 J K H 208 MeV K H 6.02 × 10 nucleus K H 10 K
∆m = bm + M g − b M + M + 3m g
∆m = b1.008 665 u + 235.043 923 ug − d97.912 7 u + 134.916 5 u + 3b1.008 665 ugi
3.30 × 10 10 J
P45.2
−19
n
U
Zr
23
Te
6
0.387 g of U - 235 .
n
∆m = 0.197 39 u = 3.28 × 10 −28 kg
Q = ∆mc 2 = 2.95 × 10 −11 J = 184 MeV
so
Three different fission reactions are possible:
1
235
0 n + 92 U
→
90
144
38 Sr + 54 Xe +
2 01 n
144
54 Xe
1
235
0 n + 92 U
→
90
143
38 Sr + 54 Xe +
1
235
0 n + 92 U
→
90
142
38 Sr + 54 Xe +
4 01 n
142
54 Xe
P45.4
1
238
0 n + 92 U
→
239
92 U
P45.5
−
1
232
233
233
0 n + 90Th → 90Th → 91 Pa + e
P45.6
(a)
Q = ∆m c 2 = mn + M U235 − M Ba141 − M Kr92 − 3mn c 2
(b)
f=
P45.3
→
3 01 n
−
239
93 Np + e
143
54 Xe
+ν
+ν
−
239
239
93 Np → 94 Pu + e
233
233
91 Pa → 92 U
+ν
+ e− + ν
a f
∆m = b1.008 665 + 235.043 923 g − b140.914 4 + 91.926 2 + 3 × 1.008 665 g u = 0.185 993 u
Q = b0.185 993 ugb931.5 MeV ug = 173 MeV
∆m 0.185993 u
=
= 7.88 × 10 −4 = 0.078 8%
236.05 u
mi
598
*P45.7
Applications of Nuclear Physics
(a)
(b)
P45.8
The initial mass is 1.007 825 u + 11.009 306 u = 12.017 131 u . The final mass is
3 4.002 603 u = 12.007 809 u
. The rest mass annihilated is ∆m = 0.009 322 u. The energy
931.5 MeV
= 8.68 MeV .
created is Q = ∆mc 2 = 0.009 322 u
1u
b
g
FG
H
IJ
K
The proton and the boron nucleus have positive charges. The colliding particles must have
enough kinetic energy to approach very closely in spite of their electric repulsion.
If the electrical power output of 1 000 MW is 40.0% of the power derived from fission reactions, the
power output of the fission process is
1 000 MW
= 2.50 × 10 9 J s 8.64 × 10 4 s d = 2.16 × 10 14 J d .
0.400
e
je
j
e
The number of fissions per day is 2.16 × 10 14 J d
This also is the number of
e6.74 × 10
24
nuclei d
235
IJ FG 1 eV IJ = 6.74 × 10
jFGH 2001 fission
× 10 eV K H 1.60 × 10
JK
−19
6
235
U nuclei used, so the mass of
F
I = 2.63 × 10
g mol
jGH 6.02 × 235
J
10 nuclei mol K
23
3
24
d −1 .
U used per day is
g d = 2.63 kg d .
In contrast, a coal-burning steam plant producing the same electrical power uses more than
6 × 10 6 kg d of coal.
P45.9
The available energy to do work is 0.200 times the energy content of the fuel.
b1.00 kg fuelge0.034 0 U fueljFGH 1 1000kgg IJK FGH 1235molg IJK e6.02 × 10
e2.90 × 10 Jja0.200f = 5.80 × 10 J = e1.00 × 10 Nj∆r
235
12
11
F a208fe1.60 × 10 JjI
moljG
GH fission JJK = 2.90 × 10
−13
23
12
J
5
∆r = 5.80 × 10 6 m = 5.80 Mm
Section 45.3
P45.10
Nuclear Reactors
4 3
πr
3
13
For a sphere:
V=
(b)
For a cube:
V=
(c)
For a parallelepiped: V = 2 a 3
(d)
Therefore, the sphere has the least leakage and the
3
and
and
and
r=
FG 3V IJ
H 4π K
(a)
= V1 3
a=
FG V IJ
H 2K
so
4π r 2
A
=
= 4.84V −1 3 .
V
4 3 π r3
so
A 6 2
= 3 = 6V −1 3 .
V
so
2a 2 + 8a 2
A
=
= 6.30V −1 3 .
V
2a3
13
parallelepiped has the greatest leakage for a given volume.
b g
e
j
Chapter 45
P45.11
235
mass of
a
fe
U available ≈ 0.007 10 9 metric tons
number of nuclei ≈
F 7 × 10 g I 6.02 × 10
GH 235 g mol JK e
12
23
F 10 g I = 7 × 10
jGH 1 metric
ton JK
6
12
g
j
nuclei mol = 1.8 × 10 34 nuclei
The energy available from fission (at 208 MeV/event) is
jb
e
ge
j
E ≈ 1.8 × 10 34 events 208 MeV event 1.60 × 10 −13 J MeV = 6.0 × 10 23 J .
This would last for a time interval of
∆t =
P45.12
E
P
≈
jFGH
e
In one minute there are
60.0 s
= 5.00 × 10 4 fissions .
1.20 ms
b
So the rate increases by a factor of 1.000 25
P45.13
IJ
K
1 yr
6.0 × 10 23 J
= 8.6 × 10 10 s
≈ 3 000 yr .
7.0 × 10 12 J s
3.16 × 10 7 s
g
50 000
= 2.68 × 10 5 .
P = 10.0 MW = 1.00 × 10 7 J s
If each decay delivers 1.00 MeV = 1.60 × 10 −13 J , then the number of decays/s = 6.25 × 10 19 Bq .
Section 45.4
P45.14
(a)
Nuclear Fusion
The Q value for the D-T reaction is 17.59 MeV.
Specific energy content in fuel for D-T reaction:
a17.59 MeVfe1.60 × 10 J MeVj = 3.39 × 10 J kg
a5 ufe1.66 × 10 kg uj
e3.00 × 10 J sjb3 600 s hrg = 31.9 g h burning of D and T
r =
e3.39 × 10 J kg je10 kg g j
−13
14
−27
9
DT
(b)
−3
14
Specific energy content in fuel for D-D reaction: Q =
two Q values
a
f
1
3.27 + 4.03 = 3.65 MeV average of
2
a3.65 MeVfe1.60 × 10 J MeV j = 8.80 × 10 J kg
a4 ufe1.66 × 10 kg uj
e3.00 × 10 J sjb3 600 s hrg = 122 g h burning of D
r =
e8.80 × 10 J kg je10 kg g j
−13
13
−27
9
DD
13
−3
.
.
599
600
P45.15
Applications of Nuclear Physics
(a)
At closest approach, the electrostatic potential energy equals the total energy E.
Uf =
b gb g = E :
k e Z1 e Z 2 e
rmin
e8.99 × 10
E=
(b)
9
je
j
2
N ⋅ m 2 C 2 1.6 × 10 −19 C Z1 Z 2
1.00 × 10 −14 m
e2.30 × 10 JjZ Z
−14
=
1
.
2
For both the D-D and the D-T reactions, Z1 = Z 2 = 1 . Thus, the minimum energy required in
both cases is
IJ =
jFGH 1.601 ×MeV
10
JK
e
E = 2.30 × 10 −14 J
−13
144 keV .
Section 45.4 in the text gives more accurate values for the critical ignition temperatures, of
about 52 keV for D-D fusion and 6 keV for D-T fusion. The nuclei can fuse by tunneling. A
triton moves more slowly than a deuteron at a given temperature. Then D-T collisions last
longer than D-D collisions and have much greater tunneling probabilities.
P45.16
e
j a f + a3 f
e
je
13
13
(a)
r f = rD + rT = 1.20 × 10 −15 m 2
= 3.24 × 10 −15 m
(b)
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
k e2
Uf = e =
rf
3.24 × 10 −15 m
(c)
Conserving momentum,
j
2
= 7.10 × 10 −14 J = 444 keV
FG m IJ v = 2 v
Hm +m K 5
F m IJ v
1
1
K + 0 = bm + m gv + U = bm + m gG
2
2
Hm +m K
F m IJ FG 1 m v IJ + U = FG m IJ K + U
K +0=G
H m + m KH 2 K
Hm +m K
F m + m IJ = 5 a444 keVf = 740 keV
K =U G
H m K 3
b
g
D
m D vi = m D + m T v f , or v f =
D
T
i
i
2
(d)
K i + Ui = K f + U f :
D
i
D
D
P45.17
T
i
f
(e)
Possibly by tunneling.
(a)
Average KE per particle is
Therefore, v rms =
(b)
t=
T
D
i
f
T
D
D
f
2
D i
D
i
FG 1 − m IJ K = U :
H m +m K
2
f
f
D
T
D
D
D
T
T
i
T
T
3
1
k B T = mv 2 .
2
2
3 k BT
=
m
x
0.1 m
= ~ 10 −7 s
~ 6
v 10 m s
e
je
3 1.38 × 10 −23 J K 4.00 × 10 8 K
e
2 1.67 × 10
−27
kg
j
j=
2.23 × 10 6 m s .
f
2
i
+U f
Chapter 45
P45.18
(a)
mI
J = 1.32 × 10 m
e
jFGH 1 609
1 mi K
= ρV = e10 kg m je1.32 × 10 m j = 1.32 × 10 kg
F M I m = FG 2.016 IJ 1.32 × 10 kg = 1.48 × 10 kg
=G
j
H 18.015 K e
H M JK
= b0.030 0%gm = e0.030 0 × 10 je1.48 × 10 kg j = 4.43 × 10
3
V = 317 × 10 6 mi 3
3
m water
H2
mH 2
H 2O
18
3
18
3
3
21
21
H 2O
m Deuterium
20
−2
H2
601
20
16
kg
The number of deuterium nuclei in this mass is
4.43 × 10 16 kg
m Deuterium
=
= 1.33 × 10 43 .
−27
m Deuteron
2.014 u 1.66 × 10
kg u
N=
a
fe
j
Since two deuterium nuclei are used per fusion, 21 H + 21 H → 24 He + Q , the number of events
N
is
= 6.63 × 10 42 .
2
The energy released per event is
b
g
b
g
Q = M 2 H + M 2 H − M 4 He c 2 = 2 2.014 102 − 4.002 603 u 931.5 MeV u = 23.8 MeV .
The total energy available is then
E=
(b)
FG N IJ Q = e6.63 × 10 ja23.8 MeVfFG 1.60 × 10 J IJ =
H 2K
H 1 MeV K
(a)
2.53 × 10 31 J .
The time this energy could possibly meet world requirements is
∆t =
P45.19
−13
42
E
P
=
2.53 × 10 31 J
e
100 7.00 × 10
12
e
J sj
= 3.61 × 10 16 s
jFGH 3.161×yr10 s IJK =
7
1.14 × 10 9 yr ~ 1 billion years .
Including both ions and electrons, the number of particles in the plasma is N = 2nV where n
is the ion density and V is the volume of the container. Application of Equation 21.6 gives
the total energy as
E=
3
Nk BT = 3nVk B T = 3 2.0 × 10 13 cm −3
2
e
L
jMMe50 m jFGH 101 mcm
N
3
6
3
3
I OPe1.38 × 10
JK PQ
−23
je
J K 4.0 × 10 8 K
j
E = 1.7 × 10 7 J
(b)
From Table 20.2, the heat of vaporization of water is L v = 2.26 × 10 6 J kg . The mass of water
that could be boiled away is
m=
1.7 × 10 7 J
E
=
= 7.3 kg .
L v 2.26 × 10 6 J kg
602
P45.20
Applications of Nuclear Physics
(a)
Lawson’s criterion for the D-T reaction is nτ ≥ 10 14 s cm 3 . For a confinement time of
τ = 1.00 s, this requires a minimum ion density of n = 10 14 cm −3 .
(b)
At the ignition temperature of T = 4.5 × 10 7 K and the ion density found above, the plasma
pressure is
LM
MNe
P = 2nk B T = 2 10 14 cm −3
(c)
O
jFGH 101 mcm IJK PPe1.38 × 10
Q
6
3
3
je
−23
j
J K 4.5 × 10 7 K = 1.24 × 10 5 J m 3 .
The required magnetic energy density is then
uB =
B2
≥ 10 P = 10 1.24 × 10 5 J m3 = 1.24 × 10 6 J m 3 ,
2µ 0
e
j
e
je
j
B ≥ 2 4π × 10 −7 N A 2 1.24 × 10 6 J m 3 = 1.77 T .
P45.21
Let the number of 6 Li atoms, each having mass 6.015 u, be N 6 while the number of 7 Li atoms, each
with mass 7.016 u, is N 7 .
F 0.925 I N .
GH 0.075 0 JK
total mass = N a6.015 uf + N a7.016 uf e1.66 × 10
kg uj = 2.00 kg ,
L
F 0.925 I a7.016 ufOPe1.66 × 10 kg uj = 2.00 kg .
N Ma6.015 uf + G
H 0.075 0 JK
MN
PQ
b
Also,
6
or
6
−27
7
−27
6
N 6 = 1.30 × 10 25 as the number of 6 Li atoms and
This yields
N7 =
P45.22
g
N 6 = 7.50% of N total = 0.075 0 N 6 + N 7 , or N 7 =
Then,
F 0.925 I e1.30 × 10 j =
GH 0.075 0 JK
25
1.61 × 10 26
as the number of 7 Li atoms.
The number of nuclei in 1.00 metric ton of trash is
b
N = 1 000 kg 1 000 g kg
23
mol
= 1.08 × 10
g 6.02 ×5610.0 gnuclei
mol
e
28
nuclei .
ja fe
j
At an average charge of 26.0 e/nucleus,
q = 1.08 × 10 28 26.0 1.60 × 10 −19 = 4.47 × 10 10 C .
Therefore
t=
q 4. 47 × 10 10
=
= 4.47 × 10 4 s = 12.4 h .
I 1.00 × 10 6
Chapter 45
Section 45.5
P45.23
N0 =
603
Radiation Damage
mass present
5.00 kg
=
= 3.35 × 10 25 nuclei
−
27
mass of nucleus
89.907 7 u 1.66 × 10
kg u
b
λ=
ge
j
ln 2
ln 2
=
= 2.38 × 10 −2 yr −1 = 4.52 × 10 −8 min −1
T1 2 29.1 yr
e
je
j
R0 = λN 0 = 4.52 × 10 −8 min −1 3.35 × 10 25 = 1.52 × 10 18 counts min
10.0 counts min
R
= e −λ t =
= 6.60 × 10 −18
R0
1.52 × 10 18 counts min
P45.24
e
j
and
λ t = − ln 6.60 × 10 −18 = 39.6
giving
t=
39.6
λ
=
39.6
= 1.66 × 10 3 yr .
2.38 × 10 −2 yr −1
Source: 100 mrad of 2-MeV γ -rays/h at a 1.00-m distance.
(a)
For γ -rays, dose in rem = dose in rad.
Thus a person would have to stand 10.0 hours to receive 1.00 rem from a 100-mrad/h
source.
(b)
1
.
r2
If the γ -radiation is emitted isotropically, the dosage rate falls off as
Thus a dosage 10.0 mrad/h would be received at a distance r = 10.0 m = 3.16 m .
P45.25
(a)
The number of x-rays taken per year is
b
gb
gb
g
n = 8 x - ray d 5 d wk 50 wk yr = 2.0 × 10 3 x - ray yr .
The average dose per photograph is
(b)
5.0 rem yr
2.0 × 10 3 x - ray yr
= 2.5 × 10 −3 rem x - ray .
The technician receives low-level background radiation at a rate of 0.13 rem yr . The dose of
5.0 rem yr received as a result of the job is
5.0 rem yr
= 38 times background levels .
0.13 rem yr
P45.26
(a)
I = I 0 e − µ x , so
With µ = 1.59 cm −1 , the thickness when I =
(b)
When
I0
= 1.00 × 10 4 ,
I
I0
is
2
FG IJ
H K
x=
I
1
ln 0
µ
I
x=
1
ln 2 = 0.436 cm .
1.59 cm −1
x=
1
ln 1.00 × 10 4 = 5.79 cm .
−1
1.59 cm
af
e
j
604
P45.27
Applications of Nuclear Physics
1 rad = 10 −2 J kg
∆t =
mc∆T
=
P
P ∆ t = mc∆T
Q = mc∆T
b
a10fe10
ga
m 4 186 J kg ⋅° C 50.0° C
−2
ja f
J kg ⋅ s m
f=
2.09 × 10 6 s ≈ 24 days!
Note that power is the product of dose rate and mass.
P45.28
10 −2 J kg
Q absorbed energy
=
= 1 000 rad
= 10.0 J kg
m
unit mass
1 rad
b
g
The rise in body temperature is calculated from Q = mc∆T where c = 4 186 J kg for water and the
human body
∆T =
P45.29
b
g
Q
1
= 10.0 J kg
= 2.39 × 10 −3 ° C (Negligible).
mc
4 186 J kg ⋅° C
If half of the 0.140-MeV gamma rays are absorbed by the patient, the total energy absorbed is
a0.140 MeVf LMF 1.00 × 10 g I F 6.02 × 10 nuclei I OP = 4.26 × 10 MeV
2
MNGH 98.9 g mol JK GH 1 mol JK PQ
J MeV j = 0.682 J
E = e 4.26 × 10 MeV je1.60 × 10
0.682 J F 1 rad I
= 1.14 rad .
Thus, the dose received is
Dose =
60.0 kg GH 10 J kg JK
F 6.02 × 10 nuclei mol I = 6.70 × 10
The nuclei initially absorbed are
N = e1.00 × 10 g jG
H 89.9 g mol JK
a f
∆N = N − N = N e1 − e j = N e1 − e
The number of decays in time t is
j.
−8
E=
23
12
−13
12
−2
P45.30
0
0
At the end of 1 year,
The energy deposited is
Thus, the dose received is
P45.31
T1 2
=
−λ t
0
12
.
− ln 2 t T1 2
0
1.00 yr
= 0.034 4
29.1 yr
e
je
j
E = e1.58 × 10 ja1.10 MeV fe1.60 × 10
J MeV j = 0.027 7 J .
F 0.027 7 J I = 3.96 × 10 J kg = 0.039 6 rad.
Dose = G
H 70.0 kg JK
∆N = N 0 − N = 6.70 × 10 12 1 − e −0.023 8 = 1.58 × 10 11 .
and
Section 45.6
t
23
−9
−13
11
−4
Radiation Detectors
b g a f b ge
a
fe
a f e
je
je
j
j
2
(a)
2
1 2 5.00 × 10 −12 F 1.00 × 10 3 V
1 2 C ∆V
E
=
=
= 3.12 × 10 7
Eβ
0.500 MeV
0.500 MeV 1.60 × 10 −13 J MeV
(b)
N=
j
5.00 × 10 −12 F 1.00 × 10 3 V
Q C ∆V
=
=
= 3.12 × 10 10 electrons
e
e
1.60 × 10 −19 C
Chapter 45
P45.32
(a)
(b)
605
EI = 10.0 eV is the energy required to liberate an electron from a dynode. Let ni be the
number of electrons incident upon a dynode, each having gained energy e ∆V as it was
accelerated to this dynode. The number of electrons that will be freed from this dynode is
∆V
:
N i = ni e
EI
a f
a1fea100 V f =
10 1 electrons .
At the first dynode, ni = 1
and
N1 =
For the second dynode, ni = N 1 = 10 1 ,
so
N2 =
(10 1 ) e 100 V
= 10 2 .
10.0 eV
At the third dynode, ni = N 2 = 10 2
and
N3 =
(10 2 ) e 100 V
= 10 3 .
10.0 eV
10.0 eV
a
f
a
f
Observing the developing pattern, we see that the number of electrons incident on the
seventh and last dynode is n 7 = N 6 = 10 6 .
(c)
The number of electrons incident on the last dynode is n 7 = 10 6 . The total energy these
electrons deliver to that dynode is given by
a f
a
f
E = ni e ∆V = 10 6 e 700 V − 600 V = 10 8 eV .
P45.33
(a)
The average time between slams is 60 min 38 = 1.6 min . Sometimes, the actual interval is
nearly zero. Perhaps about equally as often, it is 2 × 1.6 min. Perhaps about half as often, it is
4 × 1.6 min . Somewhere around 5 × 1.6 min = 8.0 min , the chances of randomness
producing so long a wait get slim, so such a long wait might likely be due to mischief.
(b)
The midpoints of the time intervals are separated by 5.00 minutes. We use R = R0 e − λ t .
Subtracting the background counts,
a f
a f e ja
F 262 IJ = ln a0.882f = − 3.47 min T
ln G
H 297 K
337 − 5 15 = 372 − 5 15 e
or
(c)
− ln 2 T1
2
5 .00 min
1 2
f
which yields T1 2 = 27.6 min .
As in the random events in part (a), we imagine a ±5 count counting uncertainty. The
smallest likely value for the half-life is then given by
ln
FG 262 − 5 IJ = − 3.47 min T
H 297 + 5 K
1 2
e j
, or T1
2
min
= 21.1 min .
The largest credible value is found from
FG 262 + 5 IJ = − 3.47 min T , yielding eT j = 38.8 min .
H 297 − 5 K
F 38.8 + 21.1 IJ ± FG 38.8 − 21.1 IJ min = a30 ± 9f min = 30 min ± 30% .
T =G
H 2 K H 2 K
ln
Thus,
1 2
1 2
1 2
max
606
Applications of Nuclear Physics
Section 45.7
P45.34
Uses of Radiation
The initial specific activity of
b R mg
0
=
59
Fe in the steel,
F
I
GH
JK
R F RI
e
=G J e
= e3.70 × 10 Bq kg je
K
H
m
m
F 800 Bq literIJ (6.50 liters) = 86.7 Bq .
R =G
H 60.0
K
20.0 µCi 100 µCi 3.70 × 10 4 Bq
= 3.70 × 10 6 Bq kg .
=
0.200 kg
kg
1 µCi
−λ t
After 1 000 h,
jb
− 6. 40 ×10 −4 h −1 1 000 h
6
g = 1.95 × 10 6
Bq kg .
0
The activity of the oil,
P45.35
oil
86.7 Bq
Roil
=
= 4.45 × 10 −5 kg .
Rm
1.95 × 10 6 Bq kg
Therefore,
m in oil =
So that wear rate is
4. 45 × 10 −5 kg
= 4.45 × 10 −8 kg h .
1 000 h
The half-life of
The
14
14
b g
ln 2
ln 2
=
= 0.009 82 s −1 .
T1 2 70.6 s
O is 70.6 s, so the decay constant is λ =
e j
O nuclei remaining after five min is N = N 0 e − λ t = 10 10 e
ja
e
− 0.009 82 s −1 300 s
f = 5.26 × 10 8 .
The number of these in one cubic centimeter of blood is
F 1.00 cm I = e5.26 × 10 jF 1.00 cm I = 2 .63 × 10
GH total vol. of blood JK
GH 2000 cm JK
R = λ N ′ = e0.009 82 s je 2 .63 × 10 j = 2 .58 × 10 Bq ~ 10
and their activity is
3
N′ = N
3
8
−1
P45.36
5
3
5
3
3
Bq .
10 4 MeV
= 9.62 × 10 3 . Since only 50% of the photons are
1.04 MeV
detected, the number of 65 Cu nuclei decaying is twice this value, or 1.92 × 10 4 . In two half3
lives, three-fourths of the original nuclei decay, so N 0 = 1.92 × 10 4 and N 0 = 2.56 × 10 4 . This
4
is 1% of the 65 Cu , so the number of 65 Cu is 2.56 × 10 6 ~ 10 6 .
(a)
The number of photons is
(b)
Natural copper is 69.17% 63 Cu and 30.83% 65 Cu . Thus, if the sample contains N Cu copper
atoms, the number of atoms of each isotope is
N 63 = 0.691 7 NCu and N 65 = 0.308 3 N Cu .
FG
H
IJ
K
FG
H
IJ e
K
N 63 0.691 7
0.6917
0.6917
=
2 .56 × 10 6 = 5.75 × 10 6 .
N 65 =
or N 63 =
0.3083
0.3083
N 65 0.308 3
j
The total mass of copper present is then
m = b62 .93 ugN + a64.93 ufN
m = b62 .93ge5.75 × 10 j + a64.93fe 2 .56 × 10 j u e1.66 × 10
g uj
Therefore,
Cu
Cu
6
= 8.77 × 10 −16 g ~ 10 −15 g
6
63
−24
65 :
Chapter 45
P45.37
(a)
Starting with N = 0 radioactive atoms at t = 0 , the rate of increase is (production – decay)
dN
= R−λN
so
dN = R − λ N dt .
dt
The variables are separable.
b
z
N
0
(b)
607
FG R − λ N IJ = t
λ H R K
FG R − λN IJ = e .
H R K
z
t
dN
= dt :
R − λN 0
−
FG R − λN IJ = −λ t
H R K
so
ln
Therefore,
1−
λ
R
g
1
ln
−λ t
and
N = e−λ t
N=
R
λ
e1 − e j .
−λ t
R
The maximum number of radioactive nuclei would be
λ
.
Additional Problems
P45.38
(a)
Suppose each 235 U fission releases 208 MeV of energy. Then, the number of nuclei that
must have undergone fission is
total release
5 × 10 13 J
N=
=
= 1.5 × 10 24 nuclei .
−13
energy per nuclei
208 MeV 1.60 × 10
J MeV
a
P45.39
fe
j
F 1.5 × 10 nuclei I b235 g molg ≈
GH 6.02 × 10 nuclei mol JK
24
(b)
mass =
(a)
At 6 × 10 8 K , the average kinetic energy of a carbon atom is
3
k B T = 1.5 8.62 × 10 −5 eV K 6 × 10 8 K = 8 × 10 4 eV
2
Note that 6 × 10 8 K is about 6 2 = 36 times larger than 1.5 × 10 7 K , the core temperature of
the Sun. This factor corresponds to the higher potential-energy barrier to carbon fusion
compared to hydrogen fusion. It could be misleading to compare it to the temperature
~ 10 8 K required for fusion in a low-density plasma in a fusion reactor.
23
a fe
(b)
je
0.6 kg
j
The energy released is
e j e
j e j
E = 2m C 12 − m Ne 20 − m He 4 c 2
b
ga
f
E = 24.000 000 − 19.992 440 − 4.002 603 931.5 MeV = 4.62 MeV
In the second reaction,
j a931.5f MeV u
E = b 24.000 000 − 23.985 042 ga931.5 f MeV =
e j e
E = 2m C 12 − m Mg 24
continued on next page
13.9 MeV
608
Applications of Nuclear Physics
(c)
The energy released is the energy of reaction of the number of carbon nuclei in a 2.00-kg
sample, which corresponds to
F
mol I F 4.62 MeV fusion event I F
1 kWh
I
G
e
jGH 6.02 ×1210.0 gatoms
G
J
J
mol
K H 2 nuclei fusion event K H 2.25 × 10 MeV JK
e1.00 × 10 ja4.62f kWh = 1.03 × 10 kWh
∆E =
2e 2.25 × 10 j
23
∆E = 2.00 × 10 3 g
19
26
7
19
P45.40
To conserve momentum, the two fragments must move in opposite directions with speeds v1 and
v 2 such that
m1 v1 = m 2 v 2
v2 =
or
FG m IJ v .
Hm K
1
1
2
The kinetic energies after the break-up are then
K 1 = 1 m1 v12
2
*P45.41
FG IJ
H K
m1
K 2 = 1 m 2 v 22 = 1 m 2
2
2
m2
and
2
v12 =
FG m IJ K .
Hm K
1
2
1
The fraction of the total kinetic energy carried off by m1 is
K1
K1
m2
=
=
K 1 + K 2 K 1 + m1 m 2 K 1
m1 + m 2
and the fraction carried off by m 2 is
1−
(a)
b
g
m2
m1
=
.
m1 + m 2
m1 + m 2
Q = 236.045 562u c 2 − 86.920 711u c 2 − 148.934 370 u c 2 = 0.190 481u c 2 = 177 MeV
Immediately after fission, this Q-value is the total kinetic energy of the fission products.
(b)
FG m IJ Q , from Problem 45.40.
Hm +m K
F 149 u IJ a177.4 MeVf = 112 MeV
=G
H 87 u + 149 u K
K Br =
La
Br
La
K La = Q − K Br = 177.4 MeV − 112.0 MeV = 65.4 MeV
(c)
v Br =
2K Br
=
m Br
v La =
2K La
=
m La
e
je
2 112 × 10 6 eV 1.6 × 10 −19 J eV
j=
7
1.58 × 10 m s
a87 ufe1.66 × 10 kg uj
J eV j
2e65.4 × 10 eV je1.6 × 10
a149 ufe1.66 × 10 kg uj = 9.20 × 10 m s
−27
−19
6
6
−27
Chapter 45
P45.42
For a typical
235
609
U , Q = 208 MeV ; and the initial mass is 235 u. Thus, the fractional energy loss is
Q
208 MeV
=
= 9.50 × 10 − 4 = 0.095 0% .
235 u 931.5 MeV u
mc 2
a
fb
g
For the D-T fusion reaction,
Q = 17.6 MeV .
The initial mass is
m = 2.014 u + 3.016 u = 5.03 u .
The fractional loss in this reaction is
Q
17.6 MeV
=
= 3.75 × 10 − 3 = 0.375%
2
5.03 u 931.5 MeV u
mc
a
f a
a
f
fb
g
0.375%
= 3.95 or the fractional loss in D - T is about 4 times that in
0.0950%
P45.43
The decay constant is λ =
235
U fission .
ln 2
ln 2
=
= 1.78 × 10 −9 s −1 .
7
T1 2
12.3 yr 3.16 × 10 s yr
b
ge
j
The tritium in the plasma decays at a rate of
e
L
O
j MMFGH 2.00cm× 10 IJK FGH 101 mcm IJK e50.0 m jPP
N
Q
F 1 Ci I = 482 Ci
Bq = e1.78 × 10 Bq jG
H 3.70 × 10 Bq JK
14
R = λ N = 1.78 × 10 − 9 s −1
R = 1.78 × 10 13
3
3
3
13
The fission inventory is
P45.44
6
3
10
4 × 10 10 Ci
~ 10 8 times greater than this amount.
482 Ci
Momentum conservation: 0 = m Li v Li + mα v α , or, m Li v Li = mα vα .
K Li
Thus,
g = bm v g = F m I v
GH 2m JK
2m
F b4.002 6 ug I F
I
F
I
=G
GH 2 b7.016 0 ug JJK H 9.25 × 10 m sK = a1.14 ufH 9.25 × 10 m sK
= 1.14 e1.66 × 10
kg jFH 9.25 × 10 m sIK = 1.62 × 10
J = 1.01 MeV
b
1
1 m Li v Li
2
= m Li v Li
=
2
2
m Li
2
K Li
K Li
2
α α
2
α
Li
2
P45.45
.
Li
2
α
2
6
− 27
2
6
2
6
−13
The complete fissioning of 1.00 gram of U 235 releases
Q=
b1.00 g g e6.02 × 10
235 grams mol
23
jb
ge
j
atoms mol 200 MeV fission 1.60 × 10 −13 J MeV = 8.20 × 10 10 J .
If all this energy could be utilized to convert m kilograms of 20.0°C water to 400°C steam (see
Chapter 20 of text for values),
then
Q = mc w ∆T + mL v + mc s ∆T
b
ga
f
b
ga
f
Q = m 4186 J kg ° C 80.0 ° C + 2.26 × 10 6 J kg + 2010 J kg ° C 300 ° C .
Therefore
m=
10
8.20 × 10 J
= 2.56 × 10 4 kg .
3.20 × 10 6 J kg
.
610
P45.46
Applications of Nuclear Physics
When mass m of 235 U undergoes complete fission, releasing 200 MeV per fission event, the total
energy released is:
Q=
F m I N a200 MeVf where N
GH 235 g mol JK
A
A
is Avogadro’s number.
If all this energy could be utilized to convert a mass m w of liquid water at Tc into steam at Th , then,
b
g
b
Q = m w c w 100° C − Tc + L v + c s Th − 100° C
g
where c w is the specific heat of liquid water, L v is the latent heat of vaporization, and c s is the
specific heat of steam. Solving for the mass of water converted gives
mw =
P45.47
(a)
b
Q
g
b
c w 100° C − Tc + L v + c s Th − 100° C
a
mN A 200 MeV
=
g b235 g molg c b100° C − T g + L
w
c
f
v
b
+ c s Th − 100° C
g
.
The number of molecules in 1.00 liter of water (mass = 1 000 g) is
N=
F 1.00 × 10 g I 6.02 × 10
GH 18.0 g mol JK e
3
23
j
molecules mol = 3.34 × 10 25 molecules .
The number of deuterium nuclei contained in these molecules is
e
N ′ = 3.34 × 10 25 molecules
1 deuteron I
J = 1.01 × 10
j FGH 3300
molecules K
22
deuterons .
Since 2 deuterons are consumed per fusion event, the number of events possible is
N′
= 5.07 × 10 21 reactions, and the energy released is
2
e
= e1.66 × 10
jb
g
Efusion = 5.07 × 10 21 reactions 3.27 MeV reaction = 1.66 × 10 22 MeV
Efusion
(b)
22
je
j
MeV 1.60 × 10 −13 J MeV = 2.65 × 10 9 J .
In comparison to burning 1.00 liter of gasoline, the energy from the fusion of deuterium is
Efusion
2.65 × 10 9 J
=
== 78.0 times larger .
Egasoline 3.40 × 10 7 J
P45.48
j a0.05 mf = 1.23 × 10
e
2
(a)
∆V = 4π r 2 ∆ r = 4π 14.0 × 10 3 m
(b)
The force on the next layer is determined by atmospheric pressure.
e
je
8
m 3 ~ 10 8 m3
j
W = P∆V = 1.013 × 10 5 N / m 2 1.23 × 10 8 m3 = 1.25 × 10 13 J ~ 10 13 J
b
g
1
yield , so yield = 1.25 × 10 14 J ~ 10 14 J
10
(c)
1.25 × 10 13 J =
(d)
1.25 × 10 14 J
= 2.97 × 10 4 ton TNT ~ 10 4 ton TNT
4.2 × 10 9 J ton TNT
or ~ 10 kilotons
P45.49
a
The thermal power transferred to the water is Pw = 0.970 waste heat
(a)
b
g
f
Chapter 45
611
Pw = 0.970 3 065 − 1 000 MW = 2 .00 × 10 9 J s
rw is the mass of water heated per hour:
rw =
The volume used per hour is
(b)
P45.50
The
235
N0 =
λ=
210
F 155 g I e6.02 × 10
GH 209.98 g mol JK
210
The half-life of
4.91 × 10 8 kg h
1.00 × 10 3 kg m3
U fuel is consumed at a rate r f =
The number of nuclei in 0.155 kg of
23
jb
ga
e
a f b
g
f
2 .00 × 10 9 J s 3600 s h
Pw
=
= 4.91 × 10 8 kg h .
c ∆T
4186 J kg ⋅° C 3.50 ° C
= 4.91 × 10 5 m3 h .
F 3 065 × 10
GH 7.80 × 10
6
10
I F 1 kg I FG 3 600 s IJ =
J
JG
J g K H 1 000 g K H 1 h K
Js
0.141 kg h .
Po is
j
nuclei mol = 4.44 × 10 23 nuclei .
Po is 138.38 days, so the decay constant is given by
ln 2
ln 2
=
= 5.80 × 10 −8 s −1 .
T1 2
138.38 d 8.64 × 10 4 s d
a
fe
j
The initial activity is
e
je
j
R0 = λN 0 = 5.80 × 10 −8 s −1 4.44 × 10 23 nuclei = 2.58 × 10 16 Bq .
The energy released in each
210
206
4
84 Po → 82 Pb + 2 He
reaction is
Q = M 210 Po − M 206 Pb − M 4 He c 2 :
84
82
2
b
g
Q = 209.982 857 − 205.974 449 − 4.002 603 u 931.5 MeV u = 5.41 MeV .
Thus, assuming a conversion efficiency of 1.00%, the initial power output of the battery is
b
g
b
ge
F mI
=G J
H ρK
jb
ge
j
P = 0.010 0 R0Q = 0.010 0 2.58 × 10 16 decays s 5.41 MeV decay 1.60 × 10 −13 J MeV = 223 W .
P45.51
3
=
m
1 3
F 70.0 kg I
=G
H 18.7 × 10 kg m JK
13
= 0.155 m
(a)
V=
(b)
Add 92 electrons to both sides of the given nuclear reaction. Then it becomes
238
4
206
92 U atom → 8 2 He atom + 82 Pb atom + Q net .
ρ
, so
3
3
b
g
b
Q net = M 238 U − 8 M 4 He − M 206 Pb c 2 = 238.050 783 − 8 4.002 603 − 205.974 449 u 931.5 MeV u
92
2
82
Q net = 51.7 MeV
(c)
If there is a single step of decay, the number of decays per time is the decay rate R and the
energy released in each decay is Q. Then the energy released per time is P = QR . If there
is a series of decays in steady state, the equation is still true, with Q representing the net
decay energy.
continued on next page
g
612
Applications of Nuclear Physics
(d)
The decay rate for all steps in the radioactive series in steady state is set by the parent
uranium:
N=
F 7.00 × 10 g I 6.02 × 10
GH 238 g mol JK e
λ=
ln 2
ln 2
1
=
= 1.55 × 10 −10
9
T1 2 4.47 × 10 yr
yr
4
F
GH
j
nuclei mol = 1.77 × 10 26 nuclei
Ie
JK
F
P = QR = a51.7 MeV fG 2 .75 × 10
H
R = λ N = 1.55 × 10 − 10
so
23
1
1.77 × 10 26 nuclei = 2 .75 × 10 16 decays yr ,
yr
j
Ie
JK
1
1.60 × 10 −13 J MeV = 2.27 × 10 5 J yr .
yr
16
dose in rem = dose in rad × RBE
(e)
b
g
j
b
g
5.00 rem yr = dose in rad yr 1.10 , giving dose in rad yr = 4.55 rad yr
b
gb
The allowed whole-body dose is then 70.0 kg 4.55 rad yr
P45.52
a
f
ET ≡ E thermal =
ET =
FG 1 IJ
H 2K
n
3
k BT = 0.039 eV
2
E where n ≡ number of collisions, and 0.039 =
gFGH 10 1 radJ kg IJK =
−2
3.18 J yr .
FG 1 IJ e2.0 × 10 j .
H 2K
n
6
Therefore, n = 25.6 = 26 collisions .
P45.53
Conservation of linear momentum and energy can be applied to find the kinetic energy of the
neutron. We first suppose the particles are moving nonrelativistically.
The momentum of the alpha particle and that of the neutron must add to zero, so their
velocities must be in opposite directions with magnitudes related by
m n v n + mα v α = 0
b1.008 7 ugv = b4.002 6 ugv
or
n
α.
At the same time, their kinetic energies must add to 17.6 MeV
E=
b
g
b
g
1
1
1
1
mn vn2 + mα vα2 = 1.008 7 u vn2 + 4.002 6 vα2 = 17.6 MeV .
2
2
2
2
Substitute vα = 0.252 0 vn :
b
g b
g
E = 0.504 35 u vn2 + 0.127 10 u vn2 = 17.6 MeV
vn =
F
I
u
GH 931.4941 MeV
J
c K
2
0.018 9 c 2
= 0.173 c = 5.19 × 10 7 m s.
0.631 45
Since this speed is not too much greater than 0.1c, we can get a reasonable estimate of the kinetic
energy of the neutron from the classical equation,
K=
continued on next page
b
ga
1
1
mv 2 = 1.008 7 u 0.173 c
2
2
f FGH 931.494 uMeV c IJK = 14.1 MeV .
2
2
Chapter 45
For a more accurate calculation of the kinetic energy, we should use relativistic expressions.
Conservation of momentum gives
γ n m n v n + γ α mα v α = 0
vα2
yielding
P45.54
vn
1.008 7
c
=
2
1−
vn2
2
c
vn2
15.746 c − 14.746 vn2
.
bγ
and
vn = 0.171 c , implying that γ n − 1 mn c 2 = 14.0 MeV .
From Table A.3, the half-life of
32
λ=
b
1 − vα2 c 2
Then
n
g
vα
= 4.002 6
2
g
− 1 mn c 2 + γ α − 1 mα c 2 = 17.6 MeV
b
g
P is 14.26 d. Thus, the decay constant is
ln 2
ln 2
=
= 0.048 6 d −1 = 5.63 × 10 −7 s −1 .
T1 2 14. 26 d
N0 =
R0
λ
=
5.22 × 10 6 decay s
5.63 × 10 −7 s −1
= 9.28 × 10 12 nuclei
At t = 10.0 days , the number remaining is
e
j
N = N 0 e − λ t = 9.28 × 10 12 nuclei e
jb
e
− 0 .048 6 d −1 10 . 0 d
g = 5.71 × 10 12 nuclei
so the number of decays has been N 0 − N = 3.57 × 10 12 and the energy released is
ja
e
fe
j
E = 3.57 × 10 12 700 keV 1.60 × 10 −16 J keV = 0.400 J .
If this energy is absorbed by 100 g of tissue, the absorbed dose is
Dose =
P45.55
(a)
F 0.400 J I F 1 rad I =
GH 0.100 kg JK GH 10 J kg JK
−2
400 rad .
The number of Pu nuclei in 1.00 kg =
6.02 × 10 23 nuclei mol
1 000 g .
239.05 g mol
b
ja
e
g
f
The total energy = 25.2 × 10 23 nuclei 200 MeV = 5.04 × 10 26 MeV
e
je
j
E = 5.04 × 10 26 MeV 4.44 × 10 −20 kWh MeV = 2.24 × 10 7 kWh
or
(b)
22 million kWh.
b
gb
E = ∆m c 2 = 3.016 049 u + 2.014 102 u − 4.002 603 u − 1.008 665 u 931.5 MeV u
E = 17.6 MeV for each D - T fusion
(c)
a
fa fe
En = Total number of D nuclei 17.6 4.44 × 10 −20
e
En = 6.02 × 10 23
continued on next page
IJ a17.6fe4.44 × 10 j =
jFGH 12.000
014 K
−20
j
2.34 × 10 8 kWh
g
613
614
Applications of Nuclear Physics
En = the number of C atoms in 1.00 kg × 4.20 eV
(d)
En =
(e)
P45.56
F 6.02 × 10 I e4.20 × 10
GH 12 JK
26
−6
je
j
MeV 4.44 × 10 −20 = 9.36 kWh
Coal is cheap at this moment in human history. We hope that safety and waste disposal
problems can be solved so that nuclear energy can be affordable before scarcity drives up
the price of fossil fuels.
Add two electrons to both sides of the given reaction.
Then 4 11 H atom→ 24 He atom + Q
a f b
Q = a 26.7 MeV fe1.60 × 10
or
g
b
g
Q = ∆m c 2 = 4 1.007 825 − 4.002603 u 931.5 MeV u = 26.7 MeV
where
−13
j
J MeV = 4.28 × 10 −12 J .
The proton fusion rate is then
rate =
P45.57
(a)
power output
3.77 × 10 26 J s
=
= 3.53 × 10 38 protons s .
energy per proton
4. 28 × 10 −12 J 4 protons
jb
e
QI = M A + M B − M C − M E c 2 , and
g
QII = M C + M D − M F − M G c 2
Q net = QI + QII = M A + M B − M C − M E + M C + M D − M F − M G c 2
Q net = Q I + QII = M A + M B + M D − M E − M F − M G c 2
Thus, reactions may be added. Any product like C used in a subsequent reaction does not
contribute to the energy balance.
(b)
Adding all five reactions gives
1
1H
or
+ 11 H+ −01 e+ 11 H+ 11 H+
4 11 H + 2
0
4
−1 e → 2 He +
2ν + Q net
2ν + Q net .
Adding two electrons to each side
Thus,
0
4
−1 e → 2 He +
4 11 H atom → 24 He atom + Q net .
b
g
b
g
Q net = 4M 1 H − M 4 He c 2 = 4 1.007 825 − 4.002 603 u 931.5 MeV u = 26.7 MeV .
1
2
Chapter 45
P45.58
(a)
L 4π F 1.50 × 10
The mass of the pellet is m = ρV = e0.200 g cm jM G
2
MN 3 H
3
−2
cm
I
JK
3
OP
PQ = 3.53 × 10
−7
615
g.
The pellet consists of equal numbers of 2 H and 3 H atoms, so the average molar mass is 2.50
and the total number of atoms is
N=
F 3.53 × 10 g I 6.02 × 10
GH 2 .50 g mol JK e
−7
23
j
atoms mol = 8.51 × 10 16 atoms .
When the pellet is vaporized, the plasma will consist of 2Nparticles (N nuclei and N
electrons). The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ. The
temperature of the plasma is found from E = 2 N 3 k B T as
2
a fe
T=
(b)
P45.59
(a)
2 .00 × 10 3 J
E
= 5.68 × 10 8 K .
=
3 Nk B 3 8.51 × 10 16 1.38 × 10 − 23 J K
e
je
j
Each fusion event uses 2 nuclei, so
E=
j
FG N IJ Q = FG 8.51 × 10
H 2K H 2
16
N
events will occur. The energy released will be
2
I a17.59 MeVfe1.60 × 10
JK
−13
j
J MeV = 1.20 × 10 5 J = 120 kJ .
The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome the
k e 2e
. The
Coulomb-repulsion barrier to 11 H + 32 He → 24 He + e + + ν , estimated as e
r
k e 7e
7
, larger by times, so
Coulomb barrier to Bethe’s fifth and eight reactions is like e
2
r
7
6
7
the required temperature can be estimated as 15 × 10 K ≈ 5 × 10 K .
2
a fa f
a fa f
e
(b)
For 12 C + 1 H →
13
j
N + Q,
b
f
ga
Q1 = 12.000 000 + 1.007 825 − 13.005 739 931.5 MeV = 1.94 MeV
For the second step, add seven electrons to both sides to
have: 13 N atom → 13 C atom + e + + e − + Q
ga
b
f
Q 2 = 13.005 739 − 13.003 355 − 2 0.000 549 931.5 MeV = 1.20 MeV
b
ga
f
Q3 = Q7 = 2 0.000 549 931.5 MeV = 1.02 MeV
a
f
= 14.003 074 + 1.007 825 − 15.003 065 a931.5 MeV f = 7.30 MeV
= 15.003 065 − 15.000 109 − 2b0.000 549 g a931.5 MeV f = 1.73 MeV
= 15.000 109 + 1.007 825 − 12 − 4.002 603 a931.5 MeV f = 4.97 MeV
Q 4 = 13.003 355 + 1.007 825 − 14.003 074 931.5 MeV = 7.55 MeV
Q5
Q6
Q8
The sum is 26.7 MeV , the same as for the proton-proton cycle.
(c)
Not all of the energy released appears as internal energy in the star. When a neutrino is
created, it will likely fly directly out of the star without interacting with any other particle.
616
P45.60
Applications of Nuclear Physics
(a)
I 2 I0 e− µ 2x
=
= e −b µ 2 − µ 1 gx
I1 I 0 e − µ 1 x
(b)
I 50
= e − a5. 40 − 41.0 fa0.100 f = e 3.56 = 35.2
I 100
(c)
I 50
= e − a5. 40 − 41.0 fa1.00 f = e 35.6 = 2.89 × 10 15
I 100
Thus, a 1.00-cm aluminum plate has essentially removed the long-wavelength x-rays from
the beam.
*P45.61
(a)
The number of fissions ocurring in the zeroth, first, second, … nth generation is
N0 , N0 K , N0 K 2 , … , N0 K n .
The total number of fissions that have ocurred up to and including the nth generation is
e
j
− 1 = a a + 1fa a − 1f , can be
N = N0 + N0K + N0K 2 +…+ N0K n = N0 1 + K + K 2 +…+ K n .
2
Note that the factoring of the difference of two squares, a
generalized to a difference of two quantities to any power,
e
ja f
a3 − 1 = a2 + a + 1 a − 1
e
ja f
a n +1 − 1 = a n + a n − 1 + … + a 2 + a + 1 a − 1 .
K n + K n −1 + … + K 2 + K + 1 =
Thus
N = N0
and
(b)
K n +1 − 1
.
K −1
The number of U-235 nuclei is
N = 5.50 kg
K n +1 − 1
K −1
FG 1 atomIJ FG 1 u
H 235 u K H 1.66 × 10
−27
I = 1.41 × 10
J
kg K
25
nuclei .
We solve the equation from part (a) for n, the number of generations:
a
f
N
K − 1 = K n +1 − 1
N0
a
f
af
F N aK − 1f N + 1 IJ = lnFG NaK − 1f + 1IJ − ln K
n ln K = lnG
K
H
K H N
K
lne1.41 × 10 a0.1f 10 + 1j
lnc N aK − 1f N + 1h
−1=
− 1 = 99.2
n=
N
K − 1 + 1 = Kn K
N0
0
0
25
0
20
ln 1.1
ln K
Therefore time must be alotted for 100 generations:
e
j
∆tb = 100 10 × 10 −9 s = 1.00 × 10 −6 s .
continued on next page
617
Chapter 45
(c)
v=
(d)
V=
B
ρ
=
18.7 × 10
3
kg m
3
= 2.83 × 10 3 m s
4 3 m
πr =
ρ
3
F 3m IJ
r =G
H 4πρ K
∆t d =
(e)
150 × 10 9 N m 2
13
F 3b5.5 kg g I
=G
GH 4π e18.7 × 10 kg m j JJK
3
13
= 4.13 × 10 −2 m
3
r
4.13 × 10 −2 m
=
= 1.46 × 10 −5 s
v 2.83 × 10 3 m s
14.6 µs is greater than 1 µs , so the entire bomb can fission. The destructive energy released
is
1.41 × 10 25 nuclei
F 200 × 10 eV I F 1.6 × 10 J I = 4.51 × 10
GH fissioning nucleus JK GH 1 eV JK
−19
6
14
FG 1 ton TNT IJ
H 4.2 × 10 J K
J = 4.51 × 10 14 J
9
= 1.07 × 10 5 ton TNT
= 107 kilotons of TNT
What if? If the bomb did not have an “initiator” to inject 10 20 neutrons at the moment when
the critical mass is assembled, the number of generations would be
n=
e
a f
ln 1. 41 × 10 25 0.1 1 + 1
ln 1.1
j − 1 = 582 requiring 583e10 × 10 sj = 5.83 µs .
−9
This time is not very short compared with 14.6 µs , so this bomb would likely release much
less energy.
ANSWERS TO EVEN PROBLEMS
P45.2
184 MeV
P45.18
(a) 2.53 × 10 31 J ; (b) 1.14 × 10 9 yr
P45.4
see the solution
P45.20
P45.6
(a) 173 MeV ; (b) 0.078 8%
(a) 10 14 cm −3 ; (b) 1.24 × 10 5 J m3 ;
(c) 1.77 T
P45.8
2.63 kg d
P45.22
12.4 h
P45.10
(a) 4.84V −1 3 ; (b) 6V −1 3 ; (c) 6.30V −1 3 ;
(d) the sphere has minimum loss and the
parallelepiped maximum
P45.24
(a) 10.0 h; (b) 3.16 m
P45.26
(a) 0.436 cm; (b) 5.79 cm
P45.28
2.39 × 10 −3 ° C
P45.30
3.96 × 10 −4 J kg
P45.32
(a) 10; (b) 10 6 ; (c) 10 8 eV
P45.34
4.45 × 10 −8 kg h
5
P45.12
2.68 × 10
P45.14
(a) 31.9 g h ; (b) 122 g h
P45.16
2
(a) 3.24 fm ; (b) 444 keV ; (c) vi ;
5
(d) 740 keV ; (e) possibly by tunneling
618
Applications of Nuclear Physics
P45.36
(a) ~ 10 6 ; (b) ~ 10 −15 g
P45.38
(a) 1.5 × 10 24 ; (b) 0.6 kg
P45.40
see the solution
P45.42
The fractional loss in D - T is about 4 times
that in
P45.44
P45.46
235
U fission
1.01 MeV
a
mN A 200 MeV
f
°C − T g + L O
P
b235 g molgLMM+c c b100
N bT − 100° Cg PQ
w
c
s
h
v
P45.48
(a) ~ 10 8 m 3 ; (b) ~ 10 13 J ; (c) ~ 10 14 J ;
(d) ~ 10 kilotons
P45.50
223 W
P45.52
26 collisions
P45.54
400 rad
P45.56
3.53 × 10 38 protons s
P45.58
(a) 5.68 × 10 8 K ; (b) 120 kJ
P45.60
(a) see the solution; (b) 35.2 ; (c) 2.89 × 10 15
46
Particle Physics and Cosmology
CHAPTER OUTLINE
46.1
46.2
46.3
46.4
46.5
46.6
46.7
46.8
46.9
46.10
46.11
46.12
46.13
The Fundamental Forces in
Nature
Positrons and Other
Antiparticles
Mesons and the Beginning
of Particle Physics
Classification of Particles
Conservation Laws
Strange Particles and
Strangeness
Making Elementary Particles
and Measuring Their
Properties
Finding Patterns in the
Particles
Quarks
Multicolored Quarks
The Standard Model
The Cosmic Connection
Problems and Perspectives
ANSWERS TO QUESTIONS
Q46.1
Strong Force—Mediated by gluons.
Electromagnetic Force—Mediated by photons.
Weak Force—Mediated by W + , W − , and Z 0 bosons.
Gravitational Force—Mediated by gravitons.
Q46.2
The production of a single gamma ray could not satisfy the law
of conservation of momentum, which must hold true in
this—and every—interaction.
Q46.3
In the quark model, all hadrons are composed of smaller units
called quarks. Quarks have a fractional electric charge and a
1
baryon number of . There are 6 types of quarks: up, down,
3
strange, charmed, top, and bottom. Further, all baryons contain
3 quarks, and all mesons contain one quark and one anti-quark.
Leptons are thought to be fundamental particles.
Q46.4
Hadrons are massive particles with structure and size. There are two classes of hadron: mesons and
baryons. Hadrons are composed of quarks. Hadrons interact via the strong force.
Leptons are light particles with no structure or size. It is believed that leptons are fundamental
particles. Leptons interact via the weak force.
Q46.5
Baryons are heavy hadrons with spin
Q46.6
Resonances are hadrons. They decay into strongly interacting particles such as protons, neutrons,
and pions, all of which are hadrons.
Q46.7
The baryon number of a proton or neutron is one. Since baryon number is conserved, the baryon
number of the kaon must be zero.
Q46.8
Decays by the weak interaction typically take 10 −10 s or longer to occur. This is slow in particle
physics.
1
3
or , are composed of three quarks, and have long
2
2
lifetimes. Mesons are light hadrons with spin 0 or 1, are composed of a quark and an antiquark, and
have short lifetimes.
619
620
Particle Physics and Cosmology
Q46.9
The decays of the muon, tau, charged pion, kaons, neutron, lambda, charged sigmas, xis, and omega
occur by the weak interaction. All have lifetimes longer than 10 −13 s. Several produce neutrinos;
none produce photons. Several violate strangeness conservation.
Q46.10
The decays of the neutral pion, eta, and neutral sigma occur by the electromagnetic interaction.
These are three of the shortest lifetimes in Table 46.2. All produce photons, which are the quanta of
the electromagnetic force. All conserve strangeness.
Q46.11
Yes, protons interact via the weak interaction; but the strong interaction predominates.
Q46.12
You can think of a conservation law as a superficial regularity which we happen to notice, as a
person who does not know the rules of chess might observe that one player’s two bishops are
always on squares of opposite colors. Alternatively, you can think of a conservation law as
identifying some stuff of which the universe is made. In classical physics one can think of both
matter and energy as fundamental constituents of the world. We buy and sell both of them. In
classical physics you can also think of linear momentum, angular momentum, and electric charge as
basic stuffs of which the universe is made. In relativity we learn that matter and energy are not
conserved separately, but are both aspects of the conserved quantity relativistic total energy.
Discovered more recently, four conservation laws appear equally general and thus equally
fundamental: Conservation of baryon number, conservation of electron-lepton number,
conservation of tau-lepton number, and conservation of muon-lepton number. Processes involving
the strong force and the electromagnetic force follow conservation of strangeness, charm,
bottomness, and topness, while the weak interaction can alter the total S, C, B and T quantum
numbers of an isolated system.
Q46.13
No. Antibaryons have baryon number –1, mesons have baryon number 0, and baryons have baryon
number +1. The reaction cannot occur because it would not conserve baryon number, unless so
much energy is available that a baryon-antibaryon pair is produced.
Q46.14
The Standard Model consists of quantum chromodynamics (to describe the strong interaction) and
the electroweak theory (to describe the electromagnetic and weak interactions). The Standard Model
is our most comprehensive description of nature. It fails to unify the two theories it includes, and
fails to include the gravitational force. It pictures matter as made of six quarks and six leptons,
interacting by exchanging gluons, photons, and W and Z bosons.
Q46.15
All baryons and antibaryons consist of three quarks. All mesons and antimesons consist of two
1
quarks. Since quarks have spin quantum number and can be spin-up or spin-down, it follows that
2
the three-quark baryons must have a half-integer spin, while the two-quark mesons must have spin
0 or 1.
Q46.16
Each flavor of quark can have colors, designated as red, green and blue. Antiquarks are colored
antired, antigreen, and antiblue. A baryon consists of three quarks, each having a different color. By
analogy to additive color mixing we call it colorless. A meson consists of a quark of one color and
antiquark with the corresponding anticolor, making it colorless as a whole.
Q46.17
In 1961 Gell-Mann predicted the omega-minus particle, with quark composition sss. Its discovery in
1964 confirmed the quark theory.
Chapter 46
Q46.18
621
1
, B = 1, L e = L µ = Lτ = 0, and strangeness –2.
2
All of these are described by its quark composition dss (Table 46.5). The properties of the quarks
1
1
1
1 1 1 1
from Table 46.3 let us add up charge: − e − e − e = − e; spin + − + = , supposing one of the
3
3
3
2 2 2 2
1 1 1
quarks is spin-down relative to the other two; baryon number + + = 1 ; lepton numbers, charm,
3 3 3
bottomness, and topness zero; and strangeness 0 − 1 − 1 = −2 .
The Ξ − particle has, from Table 46.2, charge –e, spin
Q46.19
The electroweak theory of Glashow, Salam, and Weinberg predicted the W + , W − , and Z particles.
Their discovery in 1983 confirmed the electroweak theory.
Q46.20
Hubble determined experimentally that all galaxies outside the Local Group are moving away from
us, with speed directly proportional to the distance of the galaxy from us.
Q46.21
Before that time, the Universe was too hot for the electrons to remain in any sort of stable orbit
around protons. The thermal motion of both protons and electrons was too rapid for them to be in
close enough proximity for the Coulomb force to dominate.
Q46.22
The Universe is vast and could on its own terms get along very well without us. But as the cosmos is
immense, life appears to be immensely scarce, and therefore precious. We must do our work,
growing corn to feed the hungry while preserving our planet for future generations. One person has
singular abilities and opportunities for effort, faithfulness, generosity, honor, curiosity,
understanding, and wonder. His or her place is to use those abilities and opportunities, unique in all
the Universe.
SOLUTIONS TO PROBLEMS
Section 46.1
The Fundamental Forces in Nature
Section 46.2
Positrons and Other Antiparticles
P46.1
Assuming that the proton and antiproton are left nearly at rest after they are produced, the energy E
of the photon must be
a
f
E = 2E0 = 2 938.3 MeV = 1 876.6 MeV = 3.00 × 10 −10 J .
Thus,
E = hf = 3.00 × 10 −10 J
f=
3.00 × 10 −10 J
= 4.53 × 10 23 Hz
6.626 × 10 −34 J ⋅ s
λ=
c 3.00 × 10 8 m s
=
= 6.62 × 10 −16 m .
f 4.53 × 10 23 Hz
622
P46.2
Particle Physics and Cosmology
The minimum energy is released, and hence the minimum frequency photons are produced, when
the proton and antiproton are at rest when they annihilate.
That is, E = E0 and K = 0 . To conserve momentum, each photon must carry away one-half the
energy.
Thus,
Emin =
2 E0
= E0 = 938.3 MeV = hf min .
2
Thus,
fmin =
a938.3 MeVfe1.60 × 10 J MeVj =
e6.626 × 10 J ⋅ sj
−13
λ=
P46.3
c
fmin
−34
=
3.00 × 10 8 m s
2.27 × 10 23 Hz
2.27 × 10 23 Hz
= 1.32 × 10 −15 m .
In γ → p + + p − ,
we start with energy
2.09 GeV
we end with energy
938.3 MeV + 938.3 MeV + 95.0 MeV + K 2
where K 2 is the kinetic energy of the second proton.
Conservation of energy for the creation process gives
Section 46.3
P46.4
K 2 = 118 MeV .
Mesons and the Beginning of Particle Physics
µ + + e− → ν + ν
The reaction is
muon-lepton number before reaction:
electron-lepton number before reaction:
a−1f + a0f = −1
a0f + a1f = 1 .
Therefore, after the reaction, the muon-lepton number must be –1. Thus, one of the neutrinos must
be the anti-neutrino associated with muons, and one of the neutrinos must be the neutrino
associated with electrons:
νµ
νe .
µ + + e− → ν µ + ν e .
Then
P46.5
and
The creation of a virtual Z 0 boson is an energy fluctuation ∆E = 93 × 10 9 eV . It can last no longer
=
and move no farther than
than ∆t =
2 ∆E
a f
c ∆t =
e
je
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
4π ∆E
4π 93 × 10 9 eV
e
j
j F 1 eV I = 1.06 × 10
GH 1.60 × 10 J JK
−19
−18
m = ~ 10 −18 m .
623
Chapter 46
P46.6
A proton has rest energy 938.3 MeV. The time interval during which a virtual proton could exist is at
=
most ∆t in ∆E∆t = . The distance it could move is at most
2
c∆ t =
e
je
j
1.055 × 10 −34 J ⋅ s 3 × 10 8 m s
=c
=
~ 10 −16 m .
2 ∆E
2 938.3 1.6 × 10 −13 J
a
fe
j
According to Yukawa’s line of reasoning, this distance is the range of a force that could be associated
with the exchange of virtual protons between high-energy particles.
P46.7
By Table 46.2,
Mπ 0 = 135 MeV c 2 .
Therefore,
Eγ = 67.5 MeV for each photon
p=
f=
and
P46.8
c
Eγ
h
= 67.5 MeV c
= 1.63 × 10 22 Hz .
The time interval for a particle traveling with the speed of light to travel a distance of 3 × 10 −15 m is
∆t =
P46.9
Eγ
(a)
d 3 × 10 −15 m
=
= ~ 10 −23 s .
v 3 × 10 8 m s
e
j
∆E = m n − m p − m e c 2
From Table A-3,
(b)
b
f
ga
∆E = 1.008 665 − 1.007 825 931.5 = 0.782 MeV .
Assuming the neutron at rest, momentum conservation for the decay process implies
p p = p e . Relativistic energy for the system is conserved
em c j
a938.3f
p
Since p p = p e ,
Solving the algebra,
If p e c = γ m e v e c = 1.19 MeV , then
Solving,
2 2
2
em c j + p c = m c .
+ a0.511f + ( p c ) = 939.6 MeV .
+ p p2 c 2 +
+ ( p c) 2
e
2 2
2 2
e
2
n
2
2
pc = 1.19 MeV .
γ v e 1.19 MeV
v
x
=
=
= 2.33 where x = e .
2
0.511 MeV
c
c
1− x
v
x 2 = 1 − x 2 5. 43
and x = e = 0.919
c
e
j
v e = 0.919 c .
Then m p v p = γ e m e v e :
a
fe
−13
J MeV
γ e m e v e c 1.19 MeV 1.60 × 10
=
vp =
−
27
8
mp c
1.67 × 10
3.00 × 10 m s
e
je
v p = 3.80 × 10 5 m s = 380 km s .
(c)
The electron is relativistic, the proton is not.
j
j
624
Particle Physics and Cosmology
Section 46.4
P46.10
Classification of Particles
In ? + p + → n + µ + , charge conservation requires the unknown particle to be neutral. Baryon
number conservation requires baryon number = 0. The muon-lepton number of ? must be –1.
So the unknown particle must be ν µ .
P46.11
Ω + → Λ0 + K +
Λ0 → p + π +
Section 46.5
P46.12
P46.13
K S0 → π + + π −
(or π 0 + π 0 )
n → p + e+ + ν e
Conservation Laws
p + p → µ + + e−
Le
0+0→ 0+1
and
Lµ
0 + 0 → −1 + 0
(b)
π− +p→ p+π+
charge
−1 + 1 → +1 + 1
(c)
p+p→ p+π+
baryon number :
1+1→1+0
(d)
p+p→p+p+n
baryon number :
1+1→1+1+1
(e)
γ +p→ n+π0
charge
0+1→ 0+0
(a)
Baryon number and charge are conserved, with values of
0 +1 = 0 +1
and
1 + 1 = 1 + 1 in both reactions.
(a)
(b)
Strangeness is not conserved in the second reaction.
P46.14
Baryon number conservation allows the first and forbids the second .
P46.15
(a)
π − → µ− + νµ
Lµ :
0→1−1
(b)
K+ → µ+ + ν µ
Lµ :
0 → −1 + 1
(c)
ν e + p + → n + e+
Le :
−1 + 0 → 0 − 1
(d)
ν e + n → p + + e−
Le :
1+0→ 0+1
(e)
ν µ + n → p+ + µ −
Lµ :
1+0→ 0+1
(f)
µ − → e− + ν e + ν µ
Lµ :
1→ 0+0+1
and
Le :
0→1−1+0
Chapter 46
P46.16
Momentum conservation for the decay requires the pions to have equal speeds.
The total energy of each is
497.7 MeV
2
e j gives
a248.8 MeVf = bpcg + a139.6 MeVf .
mc
FG v IJ
pc = 206 MeV = γ mvc =
H cK
1 − bv cg
pc
206 MeV
1
FG v IJ = 1.48
=
=
H cK
139.6 MeV
mc
1 − bv cg
E 2 = p 2 c 2 + mc 2
so
2
2
2
2
2
Solving,
2
2
2
FG IJ
H K
FG v IJ = 2.18LM1 − FG v IJ OP = 2.18 − 2.18FG v IJ
H cK
H cK
MN H c K PQ
F vI
3.18G J = 2.18
H cK
v
v
= 1.48 1 −
c
c
2
2
and
2
2
2
v
=
c
so
v = 0.828 c .
and
P46.17
2.18
= 0.828
3.18
(a)
p+ → π + + π 0
(b)
p+ + p+ → p+ + p+ + π 0
(c)
p+ + p+ → p+ + π +
(d)
π + → µ+ +ν µ
This reaction can occur .
(e)
n0 → p+ + e− + ν e
This reaction can occur .
(f)
π + → µ+ + n
Violates baryon number :
0→ 0+1
Violates muon - lepton number :
0 → −1 + 0
Baryon number :
1→0+0
This reaction can occur .
Baryon number is violated:
1+1→1+0
625
626
P46.18
Particle Physics and Cosmology
(a)
p → e+ + γ
+1 → 0 + 0
Baryon number:
∆B ≠ 0 , so baryon number conservation is violated.
(b)
(c)
P46.19
From conservation of momentum for the decay:
p e = pγ .
Then, for the positron,
Ee2 = p e c
2
2
0, e
becomes
Ee2
2
2
0, e
From conservation of energy for the system:
E0, p = Ee + Eγ
or
Ee = E0, p − Eγ
so
Ee2 = E02, p − 2E0 , p Eγ + Eγ2 .
Equating this to the result from above gives
Eγ2 + E02, e = E02, p − 2E0 , p Eγ + Eγ2
E02, p − E02, e
b g +E
= d p ci + E
γ
a938.3 MeVf − a0.511 MeVf
2a938.3 MeV f
2
2
or
Eγ =
Thus,
Ee = E0 , p − Eγ = 938.3 MeV − 469 MeV = 469 MeV .
Also,
pγ =
and
p e = pγ =
2 E0 , p
Eγ
c
=
= Eγ2 + E02, e .
469 MeV
c
=
469 MeV
.
c
The total energy of the positron is
Ee = 469 MeV .
But,
E e = γ E0 , e =
FG v IJ
H cK
2
=
E0 , e
b g
1− v c
E0 , e
so
1−
which yields:
v = 0.999 999 4c .
The relevant conservation laws are:
= 469 MeV .
Ee
=
2
0.511 MeV
= 1.09 × 10 −3
469 MeV
∆L e = 0
∆L µ = 0
∆Lτ = 0 .
and
(a)
π + → π 0 + e+ + ?
Le : 0 → 0 − 1 + Le
implies L e = 1
and we have a ν e
(b)
?+ p → µ − + p + π +
L µ : L µ + 0 → +1 + 0 + 0
implies L µ = 1
and we have a ν µ
continued on next page
Chapter 46
(c)
Λ0 → ρ + µ − + ?
Lµ : 0 → 0 + 1 + Lµ
implies L µ = −1
and we have a ν µ
(d)
τ + → µ + + ?+ ?
L µ : 0 → −1 + L µ
implies L µ = 1
and we have a ν µ
Lτ : −1 → 0 + Lτ
implies L τ = −1
and we have a ν τ
L µ = 1 for one particle, and Lτ = −1 for the other particle.
Conclusion for (d):
Section 46.6
P46.20
627
We have
νµ
and
ντ .
Strange Particles and Strangeness
The ρ 0 → π + + π − decay must occur via the strong interaction.
The K S0 → π + + π − decay must occur via the weak interaction.
P46.21
P46.22
(a)
Λ0 → p + π −
Strangeness: −1 → 0 + 0
(strangeness is not conserved )
(b)
π − + p → Λ0 + K 0
Strangeness: 0 + 0 → −1 + 1
(0 = 0 and strangeness is conserved )
(c)
p + p → Λ0 + Λ0
Strangeness: 0 + 0 → +1 − 1
(0 = 0 and strangeness is conserved )
(d)
π − + p → π − + Σ+
Strangeness: 0 + 0 → 0 − 1
(0 ≠ −1 : strangeness is not conserved )
(e)
Ξ − → Λ0 + π −
Strangeness: −2 → −1 + 0
(−2 ≠ −1 so strangeness is not conserved )
(f)
Ξ0 → p + π −
Strangeness: −2 → 0 + 0
(−2 ≠ 0 so strangeness is not conserved )
(a)
µ − → e− + γ
Le :
0 → 1 + 0,
and
Lµ :
1→0
(b)
n → p + e− + ν e
Le :
0→ 0+1+1
(c)
Λ0 → p + π 0
Strangeness:
−1 → 0 + 0 ,
and
charge:
0 → +1 + 0
(d)
p → e+ + π 0
Baryon number:
+1 → 0 + 0
(e)
Ξ0 → n + π 0
Strangeness:
−2 → 0 + 0
628
P46.23
Particle Physics and Cosmology
(a)
π − + p → 2η violates conservation of baryon number as 0 + 1 → 0 , not allowed .
(b)
K − + n → Λ0 + π −
Baryon number,
0+1→1+0
Charge,
−1 + 0 → 0 − 1
Strangeness,
−1 + 0 → −1 + 0
Lepton number,
0→0
The interaction may occur via the strong interaction since all are conserved.
(c)
K− → π − + π 0
Strangeness,
−1 → 0 + 0
Baryon number,
0→0
Lepton number,
0→0
Charge,
−1 → −1 + 0
Strangeness is violated by one unit, but everything else is conserved. Thus, the reaction can
occur via the weak interaction , but not the strong or electromagnetic interaction.
(d)
Ω− → Ξ− + π 0
Baryon number,
1→1+0
Lepton number,
0→0
Charge,
−1 → −1 + 0
Strangeness,
−3 → −2 + 0
May occur by weak interaction , but not by strong or electromagnetic.
(e)
η → 2γ
Baryon number,
0→0
Lepton number,
0→0
Charge,
0→0
Strangeness,
0→0
No conservation laws are violated, but photons are the mediators of the electromagnetic
interaction. Also, the lifetime of the η is consistent with the electromagnetic interaction .
Chapter 46
P46.24
(a)
Ξ − → Λ0 + µ − + ν µ
Baryon number:
+1 → +1 + 0 + 0
Charge:
−1 → 0 − 1 + 0
Le :
0→0+0+0
Lµ :
0→ 0+1+1
Lτ :
0→0+0+0
Strangeness:
−2 → −1 + 0 + 0
B , charge, L e , and Lτ
Conserved quantities are:
(b)
K S0 → 2π 0
Baryon number:
0→0
Charge:
0→0
Le :
0→0
Lµ :
0→0
Lτ :
0→0
Strangeness:
+1 → 0
Conserved quantities are:
(c)
B , charge, L e , L µ , and Lτ
K − + p → Σ0 + n
Baryon number:
0+1→1+1
Charge:
−1 + 1 → 0 + 0
Le :
0+0→0+0
Lµ :
0+0→0+0
Lτ :
0+0→0+0
Strangeness:
−1 + 0 → −1 + 0
S , charge, L e , L µ , and Lτ
Conserved quantities are:
(d)
Σ 0 + Λ0 + γ
Baryon number:
+1 → 1 + 0
Charge:
0→0
Le :
0→0+0
Lµ :
0→0+0
Lτ :
0→0+0
Strangeness:
−1 → −1 + 0
B , S , charge, L e , L µ , and Lτ
Conserved quantities are:
(e)
e+ + e− → µ + + µ −
Baryon number:
0+0→0+0
Charge:
+1 − 1 → +1 − 1
Le :
−1 + 1 → 0 + 0
Lµ :
0 + 0 → +1 − 1
Lτ :
0+0→0+0
Strangeness:
0+0→0+0
Conserved quantities are:
(f)
629
B , S , charge, L e , L µ , and Lτ
p + n → Λ0 + Σ −
Baryon number:
−1 + 1 → −1 + 1
Charge:
−1 + 0 → 0 − 1
Le :
0+0→0+0
Lµ :
0+0→0+0
Lτ :
0+0→0+0
Strangeness:
0 + 0 → +1 − 1
Conserved quantities are:
B , S , charge, L e , L µ , and Lτ
630
P46.25
Particle Physics and Cosmology
(a)
K+ + p → ? + p
The strong interaction conserves everything.
Baryon number,
0+1→B+1
so
B=0
Charge,
+1 + 1 → Q + 1
so
Q = +1
Lepton numbers,
0+0→L+0
so
L e = L µ = Lτ = 0
Strangeness,
+1 + 0 → S + 0
so
S=1
The conclusion is that the particle must be positively charged, a non-baryon, with
strangeness of +1. Of particles in Table 46.2, it can only be the K + . Thus, this is an elastic
scattering process.
The weak interaction conserves all but strangeness, and ∆S = ±1.
(b)
Ω− → ? + π −
Baryon number,
+1 → B + 0
so
B=1
Charge,
−1 → Q − 1
so
Q=0
Lepton numbers,
0→L+0
so
L e = L µ = Lτ = 0
Strangeness,
−3 → S + 0
so
∆S = 1: S = −2
The particle must be a neutral baryon with strangeness of –2. Thus, it is the Ξ 0 .
(c)
K+ → ? + µ+ +ν µ
Baryon number,
0→B+0+0
so
B=0
Charge,
+1 → Q + 1 + 0
so
Q=0
Lepton numbers,
Le , 0 → Le + 0 + 0
so
Le = 0
Lµ , 0 → Lµ − 1 + 1
so
Lµ = 0
Lτ , 0 → Lτ + 0 + 0
so
Lτ = 0
1→S+0+0
so
∆S = ±1
Strangeness,
(for weak interaction): S = 0
The particle must be a neutral meson with strangeness = 0 ⇒ π 0 .
631
Chapter 46
Section 46.7
*P46.26
(a)
Making Elementary Particles and Measuring Their Properties
p Σ + = eBrΣ +
e1.602 177 × 10
=
5.344 288 × 10
(b)
−22
e1.602 177 × 10
pπ + = eBrπ + =
ja
fa f = 686 MeV
c
bkg ⋅ m sg bMeV cg
C ja1.15 Tfa0.580 mf
200 MeV
=
c
bkg ⋅ m sg bMeV cg
−19
−19
5.344 288 × 10 −22
C 1.15 T 1.99 m
Let ϕ be the angle made by the neutron’s path with the path of the Σ + at the moment of
decay. By conservation of momentum:
b
g
pn cos ϕ + 199.961 581 MeV c cos 64.5° = 686.075 081 MeV c
∴ pn cos ϕ = 599.989 401 MeV c
b
(1)
g
b599.989 401 MeV cg + b180.482 380 MeV cg
pn sin ϕ = 199.961 581 MeV c sin 64.5° = 180.482 380 MeV c
(c)
Eπ + =
En =
e p c j + em c j
2 2
2
π+
π+
b p c g + em c j
2
n
2
pn =
From (1) and (2):
n
2 2
=
=
b199.961 581 MeVg + a139.6 MeVf
2
b626.547 022 MeVg + a939.6 MeVf
2
2
2
(2)
2
= 627 MeV c
= 244 MeV
= 1 130 MeV
EΣ + = Eπ + + En = 243.870 445 MeV + 1 129.340 219 MeV = 1 370 MeV
(d)
d i
m Σ + c 2 = EΣ2 + − p Σ + c
2
b1 373.210 664 MeVg − b686.075 081 MeVg
2
=
∴ m Σ + = 1 190 MeV c 2
F
GH
EΣ + = γ m Σ + c 2 , where γ = 1 −
v2
c2
I
JK
−1 2
=
1 373.210 664 MeV
= 1.154 4
1 189.541 303 MeV
Solving for v, v = 0.500 c .
P46.27
Time-dilated lifetime:
T = γ T0 =
0.900 × 10 −10 s
1−v
e
2
c
2
=
0.900 × 10 −10 s
1 − ( 0.960)
je
2
= 3.214 × 10 −10 s
j
distance = 0.960 3.00 × 10 8 m s 3.214 × 10 −10 s = 9.26 cm .
2
= 1 190 MeV
632
*P46.28
Particle Physics and Cosmology
(a)
Let Emin be the minimum total energy of the bombarding particle that is needed to induce
the reaction. At this energy the product particles all move with the same velocity. The
product particles are then equivalent to a single particle having mass equal to the total mass
of the product particles, moving with the same velocity as each product particle. By
conservation of energy:
em c j + b p c g
Emin + m 2 c 2 =
2 2
3
2
.
3
p 3 = p1
By conservation of momentum:
b g = bp cg
2
e
2
1
2
= Emin
− m1 c 2
Substitute (2) in (1):
Emin + m 2 c 2 =
∴ p3 c
(1)
j.
2
(2)
em c j
3
2 2
e
2
+ Emin
− m1 c 2
j
2
.
Square both sides:
e
2
+ 2Emin m 2 c 2 + m 2 c 2
Emin
∴ Emin
em
=
2
3
j
j = em c j
2
2 2
3
e
2
+ Emin
− m1 c 2
j
2
− m12 − m 22 c 2
2m 2
∴ K min = Emin − m1 c
2
em
=
2
3
j
− m12 − m 22 − 2m1 m 2 c 2
2m 2
=
b
m 32 − m1 + m 2
g
2
2m 2
Refer to Table 46.2 for the particle masses.
(b)
K min =
a
4 938.3
f
2
e
2 938.3 MeV c 2
b497.7 + 1 115.6g
=
(c)
K min
(d)
K min =
a
MeV 2 c 2 − 2 938.3
a
f
2 938.3 + 135
2
2
K min
2
MeV 2 c 2
a
MeV 2 c 2 − 139.6 + 938.3
a
f
f
2
= 5.63 GeV
MeV 2 c 2
2 938.3 MeV c 2
a
MeV 2 c 2 − 2 938.3
a
f
f
2
MeV 2 c 2
2 938.3 MeV c 2
LMe91.2 × 10 j − a938.3 + 938.3f
=N
2a938.3f MeV c
3 2
(e)
j
f
2
2
MeV 2 c 2
OP
Q=
= 768 MeV
= 280 MeV
4.43 TeV
c2
633
Chapter 46
Section 46.8
Finding Patterns in the Particles
Section 46.9
Quarks
Section 46.10
Multicolored Quarks
Section 46.11
The Standard Model
P46.29
(a)
The number of protons
F 6.02 × 10 molecules I FG 10 protons IJ = 3.34 × 10
JK H molecule K
GH
18.0 g
F 6.02 × 10 molecules I FG 8 neutrons IJ = 2.68 × 10
= 1 000 g G
JK H molecule K
18.0 g
H
23
N p = 1 000 g
and there are
Nn
23
(b)
26
neutrons .
e
j
2 e 2.68 × 10 j + 3.34 × 10
2 3.34 × 10 26 + 2.68 × 10 26 = 9.36 × 10 26 up quarks
26
and there are
protons
3.34 × 10 26 electrons .
So there are for electric neutrality
The up quarks have number
26
26
= 8.70 × 10 26 down quarks .
Model yourself as 65 kg of water. Then you contain:
e
j ~ 10
65 e9.36 × 10 j ~ 10
65 e8.70 × 10 j ~ 10
65 3.34 × 10 26
28
electrons
26
29
up quarks
26
29
down quarks .
Only these fundamental particles form your body. You have no strangeness, charm, topness
or bottomness.
P46.30
(a)
strangeness
baryon number
charge
proton
0
1
e
u
0
1/3
2e/3
u
0
1/3
2e/3
d
0
1/3
–e/3
total
0
1
e
strangeness
baryon number
charge
neutron
0
1
0
u
0
1/3
2e/3
d
0
1/3
–e/3
d
0
1/3
–e/3
total
0
1
0
(b)
P46.31
Quark composition of proton = uud and of neutron = udd.
Thus, if we neglect binding energies, we may write
mp = 2 m u + md
and
m n = m u + 2 md .
(2)
Solving simultaneously,
we find
mu =
1
1
2 m p − m n = 2 938 MeV c 2 − 939.6 meV c 2 = 312 MeV c 2
3
3
e
(1)
j
e
and from either (1) or (2), md = 314 MeV c 2 .
j
634
P46.32
Particle Physics and Cosmology
d
0
1/3
–e/3
1
–1/3
e/3
total
1
0
0
strangeness
baryon number
charge
Λ0
–1
1
0
u
0
1/3
2e/3
d
0
1/3
–e/3
s
–1
1/3
–e/3
(b)
P46.33
(a)
(c)
(d)
total
–1
1
0
π − + p → K 0 + Λ0
In terms of constituent quarks:
(b)
s
strangeness
baryon number
charge
K0
1
0
0
(a)
ud + uud → ds + uds
up quarks:
−1 + 2 → 0 + 1,
or
1→1
down quarks:
1 + 1 → 1 + 1,
or
2→2
strange quarks:
0 + 0 → −1 + 1 ,
or
0→0
π + + p → K+ + Σ+
du + uud → us + uus
up quarks:
1 + 2 → 1 + 2,
or
3→3
down quarks:
−1 + 1 → 0 + 0 ,
or
0→0
strange quarks:
0 + 0 → −1 + 1 ,
or
0→0
K − + p → K + + K 0 + Ω−
us + uud → us + ds + sss
up quarks:
−1 + 2 → 1 + 0 + 0 ,
or
1→1
down quarks:
0 +1→ 0 +1+0,
or
1→1
strange quarks:
1 + 0 → −1 − 1 + 3 ,
or
1→1
p + p → K0 + p + π + + ?
uud + uud → ds + uud + ud + ?
The quark combination of ? must be such as to balance the last equation for up, down, and
strange quarks.
up quarks:
2+ 2 = 0+ 2+1+?
(has 1 u quark)
down quarks:
1+1 =1+1−1+?
(has 1 d quark)
strange quarks:
0 + 0 = −1 + 0 + 0 + ?
(has 1 s quark)
quark composition = uds = Λ0 or Σ 0
P46.34
In the first reaction, π − + p → K 0 + Λ0 , the quarks in the particles are: ud + uud → ds + uds . There is a
net of 1 up quark both before and after the reaction, a net of 2 down quarks both before and after,
and a net of zero strange quarks both before and after. Thus, the reaction conserves the net number
of each type of quark.
In the second reaction, π − + p → K 0 + n , the quarks in the particles are: ud + uud → ds + uds .
In this case, there is a net of 1 up and 2 down quarks before the reaction but a net of 1 up, 3 down,
and 1 anti-strange quark after the reaction. Thus, the reaction does not conserve the net number of
each type of quark.
Chapter 46
P46.35
Σ0 + p → Σ+ + γ + X
dds + uud → uds + 0 + ?
The left side has a net 3d, 2u and 1s. The right-hand side has 1d, 1u, and 1s leaving 2d and 1u
missing.
The unknown particle is a neutron, udd.
Baryon and strangeness numbers are conserved.
P46.36
P46.37
Compare the given quark states to the entries in Tables 46.4 and 46.5:
(a)
suu = Σ +
(b)
ud = π −
(c)
sd = K 0
(d)
ssd = Ξ −
(a)
uud :
(b)
udd :
Section 46.12
P46.38
FG 2 eIJ + FG − 2 eIJ + FG 1 eIJ = − e . This is the antiproton .
H 3 K H 3 K H3 K
F 2 I F1 I F1 I
charge = G − eJ + G eJ + G eJ = 0 . This is the antineutron .
H 3 K H3 K H3 K
charge = −
The Cosmic Connection
Section 39.4 says
fobserver = fsource
1 + va c
.
1 − va c
The velocity of approach, v a , is the negative of the velocity of mutual recession: v a = − v .
c
Then,
P46.39
(a)
λ′
c
λ
λ′ = λ
1+ v c
1− v c
1.18 2 =
1+v c
= 1.381
1−v c
1+
v = HR :
1−v c
1+v c
and
510 nm = 434 nm
v
v
= 1.381 − 1.381
c
c
v
= 0.160
c
(b)
=
or
2.38
λ′ = λ
1+ v c
.
1− v c
1+v c
1−v c
v
= 0.381
c
v = 0.160 c = 4.80 × 10 7 m s
R=
4.80 × 10 7 m s
v
=
= 2.82 × 10 9 ly
H 17 × 10 −3 m s ⋅ ly
635
636
P46.40
Particle Physics and Cosmology
(a)
λ n′ = λ n
1+
v=
(b)
P46.41
P46.42
a
1+v c
= Z+1
1−v c
f
f − FGH vc IJK aZ + 1f
F Z + 2Z I
cG
H Z + 2Z + 2 JK
a
v
= Z+1
c
2
a
FG v IJ eZ
H cK
2
f
2
j
2
+ 2 Z + 2 = Z2 + 2 Z
2
2
F
GH
v
c
Z 2 + 2Z
=
H
H Z 2 + 2Z + 2
I
JK
e1.7 × 10
H=
v = HR
−2
ms
j
ly
1+ v c
= 590 1.000 113 3 = 590.07 nm
1− v c
b
g
e
j
λ′ = λ
e
j
λ ′ = 590
1 + 0.011 33
= 597 nm
1 − 0.011 33
e
j
λ ′ = 590
1 + 0.113 3
= 661 nm
1 − 0.113 3
(a)
v 2.00 × 10 6 ly = 3.4 × 10 4 m s
(b)
v 2.00 × 10 8 ly = 3.4 × 10 6 m s
(c)
v 2.00 × 10 9 ly = 3.4 × 10 7 m s
(a)
Wien’s law:
λ maxT = 2.898 × 10 −3 m ⋅ K .
Thus,
λ max =
(b)
*P46.43
R=
1+ v c
= Z + 1 λn
1− v c
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
=
= 1.06 × 10 −3 m = 1.06 mm
T
2.73 K
.
This is a microwave .
We suppose that the fireball of the Big Bang is a black body.
ja
e
I = eσT 4 = (1) 5.67 × 10 −8 W m 2 ⋅ K 4 2.73 K
f
4
= 3.15 × 10 −6 W m 2
As a bonus, we can find the current power of direct radiation from the Big Bang in the section of the
universe observable to us. If it is fifteen billion years old, the fireball is a perfect sphere of radius
fifteen billion light years, centered at the point halfway between your eyes:
e
ja fe
P = IA = I ( 4π r 2 ) = 3.15 × 10 −6 W m 2 4π 15 × 10 9 ly
P = 7.98 × 10 47 W .
F
I
j GH 3 ×110ly yrm s JK e3.156 × 10
2
8
2
7
s yr
j
2
Chapter 46
P46.44
637
The density of the Universe is
ρ = 1.20 ρ c = 1.20
F 3H I .
GH 8π G JK
2
Consider a remote galaxy at distance r. The mass interior to the sphere below it is
M=ρ
FG 4 π r IJ = 1.20FG 3H IJ FG 4 π r IJ = 0.600 H r
H 3 K H 8π G K H 3 K
G
2
3
2 3
3
both now and in the future when it has slowed to rest from its current speed v = H r . The energy of
this galaxy-sphere system is constant as the galaxy moves to apogee distance R:
1
GmM
GmM
=0−
mv 2 −
2
r
R
−0.100 = −0.600
r
R
I
JK
F
GH
F
GH
so
1
Gm 0.600 H 2 r 3
Gm 0.600 H 2 r 3
=0−
mH 2 r 2 −
2
r
G
R
G
so
R = 6.00r .
I
JK
The Universe will expand by a factor of 6.00 from its current dimensions.
P46.45
(a)
k B T ≈ 2m p c 2
so
(b)
(a)
T≈
kB
=
a
f F 1.60 × 10 J I
G
J
J K j H 1 MeV K
~ 10 13 K
f F 1.60 × 10 J I
j GH 1 MeV JK
~ 10 10 K
−13
2 938.3 MeV
e1.38 × 10
−23
a
−13
2 0.511 MeV
2m e c 2
=
kB
1.38 × 10 −23 J K
e
The Hubble constant is defined in v = HR . The distance R between any two far-separated
objects opens at constant speed according to R = v t . Then the time t since the Big Bang is
found from
v = H vt
(b)
2m p c 2
k B T ≈ 2m e c 2
so
*P46.46
T≈
1= Ht
F
GH
t=
I
JK
1
.
H
3 × 10 8 m s
1
1
=
= 1.76 × 10 10 yr = 17.6 billion years
H 17 × 10 −3 m s ⋅ ly
1 ly yr
638
*P46.47
Particle Physics and Cosmology
(a)
Consider a sphere around us of radius R large compared to the size of galaxy clusters. If the
matter M inside the sphere has the critical density, then a galaxy of mass m at the surface of
the sphere is moving just at escape speed v according to
1
GMm
= 0.
mv 2 −
2
R
K +Ug = 0
The energy of the galaxy-sphere system is conserved, so this equation is true throughout the
dR
. Then
history of the Universe after the Big Bang, where v =
dt
F dR I
GH d t JK
R3 / 2
32
2
=
2 GM
R
R
= 2 GM t
0
0
*P46.48
2 R3 2
2
=
3 2 GM 3
z
T
0
dt
R
2GM R
.
2R
.
3 v
so
T=
Now Hubble’s law says
v = HR .
So
T=
e
R dR = 2 GM
2GM
=v
R
From above,
T=
R
0
2 3/2
= 2 GM T
R
3
T
T=
(b)
z
dR
= R −1 / 2 2 GM
dt
2
3 17 × 10 −3
2 R
2
=
.
3 HR 3 H
F 3 × 10 m s I =
G
J
m s ⋅ ly j H 1 ly yr K
8
1.18 × 10 10 yr = 11.8 billion years
In our frame of reference, Hubble’s law is exemplified by v 1 = HR 1 and v 2 = HR 2 . From these we
may form the equations − v 1 = − HR 1 and v 2 − v 1 = H R 2 − R 1 . These equations express Hubble’s
b
g
b g
law as seen by the observer in the first galaxy cluster, as she looks at us to find − v 1 = H − R 1 and as
she looks at cluster two to find v 2 − v 1 = H R 2 − R 1 .
b
Section 46.13
P46.49
g
Problems and Perspectives
=G
=
c3
e1.055 × 10
− 34
(a)
L=
(b)
This time is given as T =
je
J ⋅ s 6.67 × 10 −11 N ⋅ m 2 kg 2
e3.00 × 10
8
ms
j
3
Yes.
1.61 × 10 −35 m
L 1.61 × 10 − 35 m
=
= 5.38 × 10 −44 s ,
c 3.00 × 10 8 m s
which is approximately equal to the ultra-hot epoch.
(c)
j=
Chapter 46
Additional Problems
P46.50
We find the number N of neutrinos:
a
f e
10 46 J = N 6 MeV = N 6 × 1.60 × 10 −13 J
j
N = 1.0 × 10 58 neutrinos
The intensity at our location is
N
N
1.0 × 10 58
=
=
A 4π r 2 4π 1.7 × 10 5 ly
e
F
G
j GH e3.00 × 10
2
1 ly
je
8
m s 3.16 × 10 7
I
J
sj JK
2
= 3.1 × 10 14 m −2 .
The number passing through a body presenting 5 000 cm 2 = 0.50 m 2
FG 3.1 × 10
H
is then
1
m2
IJ e0.50 m j = 1.5 × 10
K
2
14
~ 10 14 .
or
*P46.51
14
A photon travels the distance from the Large Magellanic Cloud to us in 170 000 years. The
hypothetical massive neutrino travels the same distance in 170 000 years plus 10 seconds:
b
g b
c 170 000 yr = v 170 000 yr + 10 s
170 000 yr
v
=
=
c 170 000 yr + 10 s 1 + 10 s
{
g
1
e1.7 × 10 yrje3.156 × 10
5
7
s yr
=
j}
1
1 + 1.86 × 10 −12
For the neutrino we want to evaluate mc 2 in E = γ mc 2 :
v2
1
mc = = E 1 − 2 = 10 MeV 1 −
γ
c
1 + 1.86 × 10 −12
2
e
2
mc ≈ 10 MeV
e
2 1.86 × 10 −12
1
e1 + 1.86 × 10 j − 1
e1 + 1.86 × 10 j
−12 2
E
j
2
= 10 MeV
j = 10 MeVe1.93 × 10 j = 19 eV .
−6
Then the upper limit on the mass is
m=
m=
P46.52
19 eV
c2
19 eV
c
2
F
GH 931.5 × 10u
6
eV c
2
I = 2.1 × 10
JK
−8
u.
(a)
π − + p → Σ+ + π 0
is forbidden by charge conservation .
(b)
µ− → π − +νe
is forbidden by energy conservation .
(c)
p→π+ +π+ +π−
is forbidden by baryon number conservation .
−12 2
639
640
P46.53
Particle Physics and Cosmology
The total energy in neutrinos emitted per second by the Sun is:
a0.4fLMN4π e1.5 × 10
j OPQW = 1.1 × 10
11 2
23
W.
Over 10 9 years, the Sun emits 3.6 × 10 39 J in neutrinos. This represents an annihilated mass
m c 2 = 3.6 × 10 39 J
m = 4.0 × 10 22 kg .
About 1 part in 50 000 000 of the Sun’s mass, over 10 9 years, has been lost to neutrinos.
P46.54
p+p→ p+π+ + X
We suppose the protons each have 70.4 MeV of kinetic energy. From conservation of momentum for
the collision, particle X has zero momentum and thus zero kinetic energy. Conservation of system
energy then requires
e
j e
M p c 2 + Mπ c 2 + M X c 2 = M p c 2 + K p + M p c 2 + K p
a
j
f
M X c 2 = M p c 2 + 2K p − Mπ c 2 = 938.3 MeV + 2 70.4 MeV − 139.6 MeV = 939.5 MeV
X must be a neutral baryon of rest energy 939.5 MeV. Thus X is a neutron .
*P46.55
(a)
If 2N particles are annihilated, the energy released is 2 Nmc 2 . The resulting photon
E 2 Nmc 2
= 2 Nmc . Since the momentum of the system is conserved, the
momentum is p = =
c
c
rocket will have momentum 2Nmc directed opposite the photon momentum.
p = 2 Nmc
(b)
Consider a particle that is annihilated and gives up its rest energy mc 2 to another particle
which also has initial rest energy mc 2 (but no momentum initially).
e j
Thus e 2mc j = p c + emc j .
E 2 = p 2 c 2 + mc 2
2 2
2
2 2
2 2
Where p is the momentum the second particle acquires as a result of the annihilation of the
e j
first particle. Thus 4 mc 2
2
e j ,p
= p 2 c 2 + mc 2
2
2
e j . So p =
= 3 mc 2
2
3 mc .
N
N
protons and
antiprotons). Thus the total
2
2
momentum acquired by the ejected particles is 3Nmc , and this momentum is imparted to
the rocket.
This process is repeated N times (annihilate
p = 3 Nmc
(c)
Method (a) produces greater speed since 2 Nmc > 3 Nmc .
641
Chapter 46
P46.56
(a)
∆E∆t ≈ = , and
∆t =
r 1.4 × 10 −15 m
=
= 4.7 × 10 −24 s
c 3 × 10 8 m s
∆E ≈
= 1.055 × 10 −34 J ⋅ s
1 MeV
=
= 2.3 × 10 −11 J
= 1.4 × 10 2 MeV
∆t
4.7 × 10 −24 s
1.60 × 10 −13 J
m=
(b)
P46.57
IJ
K
jFGH
e
∆E
≈ 1.4 × 10 2 MeV c 2 ~ 10 2 MeV c 2
c2
From Table 46.2, mπ c 2 = 139.6 MeV a pi - meson .
m Λ c 2 = 1115.6 MeV
Λ0 → p + π −
m p c 2 = 938.3 MeV
mπ c 2 = 139.6 MeV
The difference between starting rest energy and final rest energy is the kinetic energy of the
products.
K p + K π = 37.7 MeV
and
p p = pπ = p
Applying conservation of relativistic energy to the decay process, we have
LM a938.3f
N
2
OP LM a139.6f
Q N
+ p 2 c 2 − 938.3 +
2
OP
Q
+ p 2 c 2 − 139.6 = 37.7 MeV .
Solving the algebra yields
pπ c = p p c = 100.4 MeV .
Then,
Kp =
Kπ =
P46.58
em c j + a100.4f
a139.6f + a100.4f
p
2 2
2
− m p c 2 = 5.35 MeV
2
2
− 139.6 = 32.3 MeV .
By relativistic energy conservation in the reaction,
By relativistic momentum conservation for the system,
Eγ
c
=
X=
Subtracting (2) from (1),
me c 2 =
3 − 3X
1− X
2
and X =
4
so Eγ = 4m e c 2 = 2.04 MeV .
5
1 − v2 c2
3m e v
1 − v2 c2
Eγ
Dividing (2) by (1),
Solving, 1 =
3m e c 2
Eγ + m e c 2 =
Eγ + m e c
2
=
3m e c 2
1− X2
.
(1)
.
(2)
v
.
c
−
3m e c 2 X
1− X2
.
642
P46.59
Particle Physics and Cosmology
jb
e
ga
f
Momentum of proton is
qBr = 1.60 × 10 −19 C 0.250 kg C ⋅ s 1.33 m
p p = 5.32 × 10 −20 kg ⋅ m s
c p p = 1.60 × 10 −11 kg ⋅ m 2 s 2 = 1.60 × 10 −11 J = 99.8 MeV .
Therefore,
p p = 99.8 MeV c .
The total energy of the proton is
Ep = E02 + cp
b g
2
=
a983.3f + a99.8f
2
2
= 944 MeV .
For pion, the momentum qBr is the same (as it must be from conservation of momentum in a 2particle decay).
pπ = 99.8 MeV c
E0π = 139.6 MeV
b g
Eπ = E02 + cp
2
=
a139.6f + a99.8f
2
2
= 172 MeV
Etotal after = Etotal before = Rest energy .
Thus,
Rest Energy of unknown particle = 944 MeV + 172 MeV = 1 116 MeV
(This is a Λ0 particle!)
Mass = 1 116 MeV c 2 .
P46.60
Σ 0 → Λ0 + γ
From Table 46.2, m Σ = 1 192.5 MeV c 2
m Λ = 1115.6 MeV c 2 .
and
Conservation of energy in the decay requires
e
j
E0, Σ = Eo , Λ + K Λ + Eγ
F
GH
1 192.5 MeV = 1 115.6 MeV +
or
I
JK
pΛ 2
+ Eγ .
2m Λ
System momentum conservation gives p Λ = pγ , so the last result may be written as
F
p I
1 192.5 MeV = G 1 115.6 MeV +
J +E
2m K
H
F
p c I
1 192.5 MeV = G 1 115.6 MeV +
J +E .
2m c K
H
γ
2
γ
Λ
γ
or
2 2
Λ
m Λ c 2 = 1 115.6 MeV
Recognizing that
P46.61
1 192.5 MeV = 1 115.6 MeV +
Solving this quadratic equation,
Eγ = 74.4 MeV .
γ
pγ c = Eγ
and
we now have
2
b
Eγ 2
2 1 115.6 MeV
g +E .
γ
p+p→ p+ n+π+
The total momentum is zero before the reaction. Thus, all three particles present after the reaction
may be at rest and still conserve system momentum. This will be the case when the incident protons
have minimum kinetic energy. Under these conditions, conservation of energy for the reaction gives
e
j
2 m p c 2 + K p = m p c 2 + mn c 2 + mπ c 2
so the kinetic energy of each of the incident protons is
Kp =
mn c 2 + mπ c 2 − m p c 2
2
=
a939.6 + 139.6 − 938.3f MeV =
2
70.4 MeV
.
Chapter 46
P46.62
π − → µ− +νµ :
From the conservation laws for the decay,
mπ c 2 = 139.6 MeV = Eµ + Eν
P46.63
[1]
d i + a105.7 MeVf = bp cg + a105.7 MeVf
− E = a105.7 MeV f .
2
2
and p µ = pν , Eν = pν c :
Eµ2 = p µ c
or
Eµ2
Since
Eµ + Eν = 139.6 MeV
and
dE
µ
2
ν
2
2
2
ν
i a
id
a105.7 MeVf
=
Eµ − Eν
Subtracting [3] from [1],
2Eν = 59.6 MeV
[2]
[1]
+ Eν Eµ − Eν = 105.7 MeV
then
f
2
[2]
2
139.6 MeV
= 80.0 .
and
[3]
Eν = 29.8 MeV .
The expression e − E k BT dE gives the fraction of the photons that have energy between E and E + dE .
The fraction that have energy between E and infinity is
z
z
∞
E
∞
z
z
∞
e − E k BT dE
=
e − E k BT dE
0
E
∞
b
e − E k BT − dE k B T
b
e − E k BT − dE k B T
g
g
=
e − E k BT
e
∞
E
− E k BT ∞
= e − E k BT .
0
0
We require T when this fraction has a value of 0.0100 (i.e., 1.00%)
and
E = 1.00 eV = 1.60 × 10 −19 J .
Thus,
0.010 0 = e
b
g
or ln 0.010 0 = −
P46.64
643
(a)
e
− 1.60 × 10 −19 J
1.60 × 10 −19 J
e1.38 × 10
−23
j
j e1.38 ×10
JKT
=−
−23
j
JK T
1.16 × 10 4 K
giving T = 2.52 × 10 3 K .
T
This diagram represents the annihilation of an
electron and an antielectron. From charge and
lepton-number conservation at either vertex, the
γ
γ
exchanged particle must be an electron, e − .
(b)
This is the tough one. A neutrino collides with a
neutron, changing it into a proton with release of a
muon. This is a weak interaction. The exchanged
particle has charge +e and is a W + .
(a)
(b)
FIG. P46.64
644
P46.65
Particle Physics and Cosmology
(a)
The mediator of this weak interaction is a
Z 0 boson .
(b)
The Feynman diagram shows a down quark and
its antiparticle annihilating each other. They can
produce a particle carrying energy, momentum,
and angular momentum, but zero charge, zero
baryon number, and, it may be, no color charge. In
this case the product particle is a photon .
FIG. P46.65
For conservation of both energy and momentum in the collision we would expect two
photons; but momentum need not be strictly conserved, according to the uncertainty
principle, if the photon travels a sufficiently short distance before producing another matterantimatter pair of particles, as shown in Figure P46.65. Depending on the color charges of
the d and d quarks, the ephemeral particle could also be a gluon , as suggested in the
discussion of Figure 46.14(b).
*P46.66
(a)
At threshold, we consider a photon and a proton colliding head-on to produce a proton and
a pion at rest, according to p + γ → p + π 0 . Energy conservation gives
mp c 2
2
1−u c
2
+ Eγ = m p c 2 + mπ c 2 .
m pu
Momentum conservation gives
1−u
2
c
2
−
Eγ
c
= 0.
Combining the equations, we have
mp c 2
1−u
2
c
2
+
b
mp c 2
u
= m p c 2 + mπ c 2
c
1−u c
2
938.3 MeV 1 + u c
b1 − u cgb1 + u cg
(b)
so
u
= 0.134
c
and
Eγ = 127 MeV .
g = 938.3 MeV + 135.0 MeV
λ maxT = 2.898 mm ⋅ K
λ max =
(c)
2
2.898 mm ⋅ K
= 1.06 mm
2.73 K
Eγ = hf =
hc
λ
=
continued on next page
1 240 eV ⋅ 10 −9 m
1.06 × 10 -3 m
= 1.17 × 10 −3 eV
Chapter 46
(d)
645
u′
= 0.134 and the
c
photon is moving to the left with hf ′ = 1.27 × 10 8 eV . In the unprimed frame,
In the primed reference frame, the proton is moving to the right at
hf = 1.17 × 10 −3 eV . Using the Doppler effect equation from Section 39.4, we have for the
speed of the primed frame
1+ν c
1.17 × 10 −3
1 −ν c
1.27 × 10 8 =
v
= 1 − 1.71 × 10 −22
c
Then the speed of the proton is given by
0.134 + 1 − 1.71 × 10 −22
u u′ c + ν c
=
=
= 1 − 1.30 × 10 −22 .
2
−22
c 1 + u ′ν c
1 + 0.134 1 − 1.71 × 10
e
j
And the energy of the proton is
mp c 2
1−u
2
c
2
=
938.3 MeV
e
1 − 1 − 1.30 × 10
j
−22 2
= 6.19 × 10 10 × 938.3 × 10 6 eV = 5.81 × 10 19 eV .
ANSWERS TO EVEN PROBLEMS
P46.2
2.27 × 10 23 Hz ; 1.32 fm
P46.4
ν µ and ν e
P46.6
~ 10 −16 m
P46.8
~ 10 −23 s
P46.10
νµ
P46.12
(a) electron lepton number and muon
lepton number; (b) charge;
(c) baryon number; (d) baryon number;
(e) charge
P46.14
the second violates conservation of baryon
number
P46.16
0.828 c
P46.18
(a) see the solution;
469 MeV
for both;
(b) 469 MeV ;
c
(c) 0.999 999 4c
P46.20
see the solution
P46.22
(a) electron lepton number and muon
lepton number;
(b) electron lepton number;
(c) strangeness and charge;
(d) baryon number; (e) strangeness
P46.24
see the solution
P46.26
686 MeV
200 MeV
and
;
c
c
(b) 627 MeV c ;
(c) 244 MeV , 1 130 MeV , 1 370 MeV ;
(a)
(d) 1 190 MeV c 2 , 0.500 c
P46.28
(a) see the solution; (b) 5.63 GeV ;
(c) 768 MeV ; (d) 280 MeV ; (e) 4.43 TeV
P46.30
see the solution
P46.32
see the solution
P46.34
see the solution
P46.36
(a) Σ + ; (b) π − ; (c) K 0 ; (d) Ξ −
P46.38
see the solution
646
P46.40
Particle Physics and Cosmology
(a) v = c
F Z + 2Z I ; (b) c F Z + 2Z I
GH Z + 2Z + 2 JK H GH Z + 2Z + 2 JK
2
2
2
P46.56
(a) ~ 10 2 MeV c 2 ; (b) a pi - meson
P46.58
2.04 MeV
P46.60
74.4 MeV
P46.62
29.8 MeV
P46.64
(a) electron-position annihilation; e − ;
(b) a neutrino collides with a neutron,
producing a proton and a muon; W +
P46.66
(a) 127 MeV ; (b) 1.06 mm; (c) 1.17 meV ;
(d) 5.81 × 10 19 eV
2
P46.42
(a) 1.06 mm; (b) microwave
P46.44
6.00
P46.46
(a) see the solution; (b) 17.6 Gyr
P46.48
see the solution
P46.50
~ 10 14
P46.52
(a) charge; (b) energy; (c) baryon number
P46.54
neutron