CHAPTER
21
SUMMARY
Electric charge, conductors, and insulators: The fundamental quantity in electrostatics is electric
charge. There are two kinds of charge, positive and negative. Charges of the same sign repel each
other; charges of opposite sign attract. Charge is conserved; the total charge in an isolated system is
constant.
All ordinary matter is made of protons, neutrons, and electrons. The positive protons and electrically neutral neutrons in the nucleus of an atom are bound together by the nuclear force; the negative electrons surround the nucleus at distances much greater than the nuclear size. Electric
interactions are chiefly responsible for the structure of atoms, molecules, and solids.
Conductors are materials in which charge moves easily; in insulators, charge does not move easily. Most metals are good conductors; most nonmetals are insulators.
Coulomb’s law: For charges q1 and q2 separated by a distance r, the magnitude of the electric force on either
charge is proportional to the product q1 q2 and inversely
proportional to r 2. The force on each charge is along the
line joining the two charges—repulsive if q1 and q2 have
the same sign, attractive if they have opposite signs. In SI
units the unit of electric charge is the coulomb, abbreviated C. (See Examples 21.1 and 21.2.)
When two or more charges each exert a force on a
charge, the total force on that charge is the vector sum of
the forces exerted by the individual charges. (See Examples 21.3 and 21.4.)
S
Electric field: Electric field E, a vector quantity, is the
force per unit charge exerted on a test charge at any point.
The electric field produced by a point charge is directed
radially away from or toward the charge. (See Examples
21.5–21.7.)
F =
1 ƒ q1 q2 ƒ
4pP0 r 2
+
+
–
+
S
F2 on 1
(21.2)
r
q1
S
1
= 8.988 * 10 9 N # m2>C2
4pP0
F1 on 2
q2
q1 S
F2 on 1
r
S
F1 on 2
q2
S
S
F0
q0
(21.3)
1 q
Nr
Eⴝ
4pP0 r 2
(21.7)
S
Eⴝ
E
q
S
S
Superposition of electric fields: The electric field E of any combination of charges is the vector
sum of the fields caused by the individual charges. To calculate the electric field caused by a continuous distribution of charge, divide the distribution into small elements, calculate the field caused by
each element, and then carry out the vector sum, usually by integrating. Charge distributions are
described by linear charge density l, surface charge density s, and volume charge density r. (See
Examples 21.8–21.12.)
y dQ
ds r
⫽
x 2⫹
a
a2
dEx
x
a
P
a
x dEy
O
r
dE
Q
S
Electric field lines: Field lines provide a graphical representation
of electric fields. At any point on a
S
field line, the tangent to the line is in the direction of E at that point. The number
of lines per unit
S
area (perpendicular to their direction) is proportional to the magnitude of E at the point.
S
E
E
+
–
S
E
Electric dipoles: An electric dipole is a pair of electric
charges of equal magnitude q but opposite sign, separated
S
by a distance d. The electric dipole moment p has magniS
tude p = qd. The direction of p is from negative toward
S
positive charge. An electric dipole in an electric field E
S
S
experiences a torque T equal to the vector product of p
S
and E. The magnitude
of the torque depends on the angle f
S
S
between p and E. The potential energy U for an electric
dipole in an electric
field also depends on the relative oriS
S
entation of p and E. (See Examples 21.13 and 21.14.)
714
t = pE sin f
(21.15)
S
S
S
S
Tⴝp : E
S
#
S
U = -p E
S
(21.16)
d
E
1q
d sin f
f
(21.18)
S
S
F25 2qE
S
F15 qE
S
p
2q
Discussion Questions
BRIDGING PROBLEM
715
Calculating Electric Field: Half a Ring of Charge
Positive charge Q is uniformly distributed around a semicircle of
radius a as shown in Fig. 21.34. Find the magnitude and direction
of the resulting electric field at point P, the center of curvature of
the semicircle.
SOLUTION GUIDE
See MasteringPhysics® study area for a Video Tutor solution.
IDENTIFY and SET UP
1. The target variables are the components of the electric field at P.
2. Divide the semicircle into infinitesimal segments, each of which
is a short circular arc of radius a and angle du. What is the
length of such a segment? How much charge is on a segment?
21.34
Q
a
Problems
EXECUTE
4. Integrate your expressions for dE x and dE y from u ⫽ 0 to u ⫽ p.
The results will be the x-component and y-component of the
electric field at P.
5. Use your results from step 4 to find the magnitude and direction of the field at P.
EVALUATE
6. Does your result for the electric-field magnitude have the correct units?
7. Explain how you could have found the x-component of the
electric field using a symmetry argument.
8. What would be the electric field at P if the semicircle were
extended to a full circle centered at P?
y
P
3. Consider an infinitesimal segment located at an angular position u on the semicircle, measured from the lower right corner of the semicircle at x ⫽ a, y ⫽ 0. (Thus u ⫽ p> 2 at x ⫽ 0,
y ⫽ a and u = p at x = - a, y = 0.) What are the x- and
y-components of the electric field at P (dE x and dE y) produced
by just this segment?
x
For instructor-assigned homework, go to www.masteringphysics.com
. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems
requiring calculus. BIO: Biosciences problems.
DISCUSSION QUESTIONS
Q21.1 If you peel two strips of transparent tape off the same roll
and immediately let them hang near each other, they will repel
each other. If you then stick the sticky side of one to the shiny side
of the other and rip them apart, they will attract each other. Give a
plausible explanation, involving transfer of electrons between the
strips of tape, for this sequence of events.
Q21.2 Two metal spheres are hanging from nylon threads. When
you bring the spheres close to each other, they tend to attract.
Based on this information alone, discuss all the possible ways that
the spheres could be charged. Is it possible that after the spheres
touch, they will cling together? Explain.
Q21.3 The electric force between two charged particles becomes
weaker with increasing distance. Suppose instead that the electric
force were independent of distance. In this case, would a charged
comb still cause a neutral insulator to become polarized as in Fig.
21.8? Why or why not? Would the neutral insulator still be
attracted to the comb? Again, why or why not?
Q21.4 Your clothing tends to cling together after going through the
dryer. Why? Would you expect more or less clinging if all your
clothing were made of the same material (say, cotton) than if you
dried different kinds of clothing together? Again, why? (You may
want to experiment with your next load of laundry.)
Q21.5 An uncharged metal sphere hangs from a nylon thread.
When a positively charged glass rod is brought close to the metal
sphere, the sphere is drawn toward the rod. But if the sphere
touches the rod, it suddenly flies away from the rod. Explain why
the sphere is first attracted and then repelled.
Q21.6 The free electrons in a metal are gravitationally attracted
toward the earth. Why, then, don’t they all settle to the bottom of
the conductor, like sediment settling to the bottom of a river?
Q21.7 . Figure Q21.7 shows some of the Figure Q21.7
electric field lines due to three point
charges arranged along the vertical axis.
All three charges have the same magnitude. (a) What are the signs of the three
charges? Explain your reasoning. (b) At
what point(s) is the magnitude of the electric field the smallest? Explain your reasoning. Explain how the fields produced
by each individual point charge combine to
give a small net field at this point or points.
Q21.8 Good electrical conductors, such as
metals, are typically good conductors of
heat; electrical insulators, such as wood, are typically poor conductors of heat. Explain why there should be a relationship
between electrical conduction and heat conduction in these
materials.
716
CHAPTER 21 Electric Charge and Electric Field
Q21.9 . Suppose the charge shown in Fig. 21.28a is fixed in position. A small, positively charged particle is then placed at some
point in the figure and released. Will the trajectory of the particle
follow an electric field line? Why or why not? Suppose instead that
the particle is placed at some point in Fig. 21.28b and released (the
positive and negative charges shown in the figure are fixed in position). Will its trajectory follow an electric field line? Again, why or
why not? Explain any differences between your answers for the
two different situations.
Q21.10 Two identical metal objects are mounted on insulating
stands. Describe how you could place charges of opposite sign but
exactly equal magnitude on the two objects.
Q21.11 You can use plastic food wrap to cover a container by
stretching the material across the top and pressing the overhanging
material against the sides. What makes it stick? (Hint: The answer
involves the electric force.) Does the food wrap stick to itself with
equal tenacity? Why or why not? Does it work with metallic containers? Again, why or why not?
Q21.12 If you walk across a nylon rug and then touch a large metal
object such as a doorknob, you may get a spark and a shock. Why
does this tend to happen more on dry days than on humid days?
(Hint: See Fig. 21.30.) Why are you less likely to get a shock if
you touch a small metal object, such as a paper clip?
Q21.13 You have a negatively charged object. How can you use it
to place a net negative charge on an insulated metal sphere? To
place a net positive charge on the sphere?
Q21.14 When two point charges of equal mass and charge are
released on a frictionless table, each has an initial acceleration a0.
If instead you keep one fixed and release the other one, what will
be its initial acceleration: a0, 2a0, or a0 >2? Explain.
Q21.15 A point charge of mass m and charge Q and another point
charge of mass m but charge 2Q are released on a frictionless table.
If the charge Q has an initial acceleration a0, what will be the
acceleration of 2Q: a0, 2a0, 4a0, a0 >2, or a0 >4? Explain.
Q21.16 A proton is placed in a uniform electric field and then
released. Then an electron is placed at this same point and released.
Do these two particles experience the same force? The same acceleration? Do they move in the same direction when released?
Q21.17 In Example 21.1 (Section 21.3) we saw that the electric
force between two a particles is of the order of 10 35 times as
strong as the gravitational force. So why do we readily feel the
gravity of the earth but no electrical force from it?
Q21.18 What similarities do electrical forces have with gravitational forces? What are the most significant differences?
Q21.19 Two irregular objects A
Figure Q21.19
and B carry charges of opposite
sign. Figure Q21.19 shows the
electric field lines near each of
these objects. (a) Which object
S
is positive, A or B? How do you
E
B
know? (b) Where is the electric
field stronger, close to A or
close to B? How do you know?
A
Q21.20 Atomic nuclei are
made of protons and neutrons.
This shows that there must be
another kind of interaction in
addition to gravitational and electric forces. Explain.
Q21.21 Sufficiently strong electric fields can cause atoms to
become positively ionized—that is, to lose one or more electrons.
Explain how this can happen. What determines how strong the
field must be to make this happen?
Q21.22 The electric fields at point P due to Figure Q21.22
the positive charges q1 and q2 are shown in
S
S
E2
E1
Fig. Q21.22. Does the fact that they cross
each other violate the statement in Section
21.6 that electric field lines never cross?
Explain.
P
Q21.23 The air temperature and the velocity of the air have different values at differq2
ent places in the earth’s atmosphere. Is the q1
air velocity a vector field? Why or why not?
Is the air temperature a vector field? Again, why or why not?
EXERCISES
Section 21.3 Coulomb’s Law
21.1 .. Excess electrons are placed on a small lead sphere with
mass 8.00 g so that its net charge is -3.20 * 10 -9 C. (a) Find the
number of excess electrons on the sphere. (b) How many excess
electrons are there per lead atom? The atomic number of lead is 82,
and its atomic mass is 207 g>mol.
21.2 . Lightning occurs when there is a flow of electric charge
(principally electrons) between the ground and a thundercloud.
The maximum rate of charge flow in a lightning bolt is about
20,000 C>s; this lasts for 100 ms or less. How much charge flows
between the ground and the cloud in this time? How many electrons flow during this time?
21.3 .. BIO Estimate how many electrons there are in your body.
Make any assumptions you feel are necessary, but clearly state
what they are. (Hint: Most of the atoms in your body have equal
numbers of electrons, protons, and neutrons.) What is the combined charge of all these electrons?
21.4 . Particles in a Gold Ring. You have a pure (24-karat)
gold ring with mass 17.7 g. Gold has an atomic mass of
197 g>mol and an atomic number of 79. (a) How many protons
are in the ring, and what is their total positive charge? (b) If the
ring carries no net charge, how many electrons are in it?
21.5 . BIO Signal Propagation in Neurons. Neurons are components of the nervous system of the body that transmit signals as
electrical impulses travel along their length. These impulses propagate when charge suddenly rushes into and then out of a part of the
neuron called an axon. Measurements have shown that, during the
inflow part of this cycle, approximately 5.6 * 10 11 Na + (sodium
ions) per meter, each with charge + e, enter the axon. How many
coulombs of charge enter a 1.5-cm length of the axon during this
process?
21.6 . Two small spheres spaced 20.0 cm apart have equal
charge. How many excess electrons must be present on each
sphere if the magnitude of the force of repulsion between them is
4.57 * 10 -21 N?
21.7 .. An average human weighs about 650 N. If two such
generic humans each carried 1.0 coulomb of excess charge, one
positive and one negative, how far apart would they have to be for
the electric attraction between them to equal their 650-N weight?
21.8 .. Two small aluminum spheres, each having mass 0.0250 kg,
are separated by 80.0 cm. (a) How many electrons does each sphere
contain? (The atomic mass of aluminum is 26.982 g>mol, and its
atomic number is 13.) (b) How many electrons would have to be
removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00 * 10 4 N
(roughly 1 ton)? Assume that the spheres may be treated as point
charges. (c) What fraction of all the electrons in each sphere does
this represent?
717
Exercises
21.9 .. Two small plastic spheres are given positive electrical
charges. When they are 15.0 cm apart, the repulsive force between
them has magnitude 0.220 N. What is the charge on each sphere
(a) if the two charges are equal and (b) if one sphere has four times
the charge of the other?
21.10 .. What If We Were Not Neutral? A 75-kg person holds
out his arms so that his hands are 1.7 m apart. Typically, a person’s
hand makes up about 1.0% of his or her body weight. For round
numbers, we shall assume that all the weight of each hand is due to
the calcium in the bones, and we shall treat the hands as point
charges. One mole of Ca contains 40.18 g, and each atom has 20
protons and 20 electrons. Suppose that only 1.0% of the positive
charges in each hand were unbalanced by negative charge.
(a) How many Ca atoms does each hand contain? (b) How many
coulombs of unbalanced charge does each hand contain? (c) What
force would the person’s arms have to exert on his hands to prevent them from flying off? Does it seem likely that his arms are
capable of exerting such a force?
21.11 .. Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to
each of them. Disregarding all other forces, how many electrons
would you have to add to each sphere so that the two spheres will
accelerate at 25.0g when released? Which way will they accelerate?
21.12 .. Just How Strong Is the Electric Force? Suppose you
had two small boxes, each containing 1.0 g of protons. (a) If one were
placed on the moon by an astronaut and the other were left on the
earth, and if they were connected by a very light (and very long!)
string, what would be the tension in the string? Express your answer
in newtons and in pounds. Do you need to take into account the gravitational forces of the earth and moon on the protons? Why? (b) What
gravitational force would each box of protons exert on the other box?
21.13 . In an experiment in space, one proton is held fixed and
another proton is released from rest a distance of 2.50 mm away.
(a) What is the initial acceleration of the proton after it is released?
(b) Sketch qualitative (no numbers!) acceleration–time and
velocity–time graphs of the released proton’s motion.
21.14 . A negative charge of -0.550 mC exerts an upward 0.200-N
force on an unknown charge 0.300 m directly below it. (a) What is
the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on
the - 0.550-mC charge?
21.15 .. Three point charges are arranged on a line. Charge
q3 = + 5.00 nC and is at the origin. Charge q2 = - 3.00 nC and is
at x = + 4.00 cm. Charge q1 is at x = + 2.00 cm. What is q1
(magnitude and sign) if the net force on q3 is zero?
21.16 .. In Example 21.4, suppose the point charge on the y-axis
at y = - 0.30 m has negative charge - 2.0 mC, and the other
charges remain the same. Find the magnitude and direction of the
net force on Q. How does your answer differ from that in Example
21.4? Explain the differences.
21.17 .. In Example 21.3, calculate the net force on charge q1.
21.18 .. In Example 21.4, what is the net force (magnitude and
direction) on charge q1 exerted by the other two charges?
21.19 .. Three point charges are arranged along the x-axis. Charge
q1 = + 3.00 mC is at the origin, and charge q2 = - 5.00 mC is at
x = 0.200 m. Charge q3 = - 8.00 mC. Where is q3 located if the
net force on q1 is 7.00 N in the -x-direction?
21.20 .. Repeat Exercise 21.19 for q3 = + 8.00 mC.
21.21 .. Two point charges are located on the y-axis as follows:
charge q1 = - 1.50 nC at y = - 0.600 m, and charge q2 =
+ 3.20 nC at the origin 1y = 02. What is the total force (magnitude and direction) exerted by these two charges on a third charge
q3 = + 5.00 nC located at y = - 0.400 m?
21.22 .. Two point charges are placed on the x-axis as follows:
Charge q1 = + 4.00 nC is located at x = 0.200 m, and charge
q2 = + 5.00 nC is at x = - 0.300 m. What are the magnitude and
direction of the total force exerted by these two charges on a negative point charge q3 = - 6.00 nC that is placed at the origin?
21.23 .. BIO Base Pairing in DNA, I. The two sides of the
DNA double helix are connected by pairs of bases (adenine,
thymine, cytosine, and guanine). Because of the geometric shape
of these molecules, adenine bonds with thymine and cytosine
bonds with guanine. Figure E21.23 shows the thymine–adenine
bond. Each charge shown is ⫾e, and the H ¬ N distance is
0.110 nm. (a) Calculate the net force that thymine exerts on adenine.
Is it attractive or repulsive? To keep the calculations fairly simple,
yet reasonable, consider only the forces due to the O ¬ H ¬ N and
the N ¬ H ¬ N combinations, assuming that these two combinations are parallel to each other. Remember, however, that in the
O ¬ H ¬ N set, the O - exerts a force on both the H + and the N -,
and likewise along the N ¬ H ¬ N set. (b) Calculate the force on the
electron in the hydrogen atom, which is 0.0529 nm from the proton.
Then compare the strength of the bonding force of the electron in
hydrogen with the bonding force of the adenine–thymine molecules.
Figure E21.23
H
H
H
Thymine
O
2
1
C
C
H
0.280
H
nm
Adenine
H
N
1 2
N
0.300
C
C
nm
C
H
N
1
C
N
2
C
C
N
N
2
C
C
O
H
N
H
21.24 .. BIO Base Pairing in DNA, II. Refer to Exercise 21.23.
Figure E21.24 shows the bonding of the cytosine and guanine molecules. The O ¬ H and H ¬ N distances are each 0.110 nm. In this
case, assume that the bonding is due only to the forces along the
O ¬ H ¬ O, N ¬ H ¬ N, and O ¬ H ¬ N combinations, and
assume also that these three combinations are parallel to each other.
Calculate the net force that cytosine exerts on guanine due to the
preceding three combinations. Is this force attractive or repulsive?
Figure E21.24
H
H
Cytosine
H
O
2
C
C
C
N
N
2
1
C
O
2
0.290
nm
H
O
1
2
1
0.300
C
nm
H
N
1 2
0.290
C
nm
H
N
1 2
H
Guanine
N
H
C
C
C
N
N
Section 21.4 Electric Field and Electric Forces
21.25 . CP A proton is placed in a uniform electric field of
2.75 * 10 3 N>C. Calculate: (a) the magnitude of the electric force
felt by the proton; (b) the proton’s acceleration; (c) the proton’s
speed after 1.00 ms in the field, assuming it starts from rest.
718
CHAPTER 21 Electric Charge and Electric Field
21.26 . A particle has charge - 3.00 nC. (a) Find the magnitude
and direction of the electric field due to this particle at a point
0.250 m directly above it. (b) At what distance from this particle
does its electric field have a magnitude of 12.0 N>C?
21.27 . CP A proton is traveling horizontally to the right at
4.50 * 10 6 m>s. (a) Find the magnitude and direction of the
weakest electric field that can bring the proton uniformly to rest
over a distance of 3.20 cm. (b) How much time does it take the
proton to stop after entering the field? (c) What minimum field
(magnitude and direction) would be needed to stop an electron
under the conditions of part (a)?
21.28 . CP An electron is released from rest in a uniform electric
field. The electron accelerates vertically upward, traveling 4.50 m
in the first 3.00 ms after it is released. (a) What are the magnitude
and direction of the electric field? (b) Are we justified in ignoring
the effects of gravity? Justify your answer quantitatively.
21.29 .. (a) What must the charge (sign and magnitude) of a
1.45-g particle be for it to remain stationary when placed in a
downward-directed electric field of magnitude 650 N>C? (b) What
is the magnitude of an electric field in which the electric force on a
proton is equal in magnitude to its weight?
21.30 .. A point charge is placed at Figure E21.30
each corner of a square with side
1q
1q
a
length a. The charges all have the
same magnitude q. Two of the
charges are positive and two are negative, as shown in Fig. E21.30. What
a
is the direction of the net electric
field at the center of the square due to
the four charges, and what is its magnitude in terms of q and a?
2q
2q
21.31 . Two point charges are separated by 25.0 cm (Fig. E21.31). Find the net electric field these
charges produce at (a) point A and (b) point B. (c) What would be
the magnitude and direction of the electric force this combination
of charges would produce on a proton at A?
Figure E21.31
B
26.25 nC
A
10.0 cm
212.5 nC
10.0 cm
25.0 cm
21.32 .. Electric Field of the Earth. The earth has a net electric charge that causes a field at points near its surface equal to
150 N>C and directed in toward the center of the earth. (a) What
magnitude and sign of charge would a 60-kg human have to
acquire to overcome his or her weight by the force exerted by the
earth’s electric field? (b) What would be the force of repulsion
between two people each with the charge calculated in part (a) and
separated by a distance of 100 m? Is use of the earth’s electric
field a feasible means of flight? Why or why not?
21.33 .. CP An electron is projected Figure E21.33
with an initial speed v0 = 1.60 *
2.00 cm
10 6 m>s into the uniform field
v0
between the parallel plates in Fig.
S
E21.33. Assume that the field between 1.00 cm –
E
the plates is uniform and directed vertically downward, and that the field
outside the plates is zero. The electron enters the field at a point
midway between the plates. (a) If the electron just misses the upper
plate as it emerges from the field, find the magnitude of the electric
field. (b) Suppose that in Fig. E21.33 the electron is replaced by a
proton with the same initial speed v0. Would the proton hit one of
the plates? If the proton would not hit one of the plates, what
would be the magnitude and direction of its vertical displacement
as it exits the region between the plates? (c) Compare the paths
traveled by the electron and the proton and explain the differences.
(d) Discuss whether it is reasonable to ignore the effects of gravity
for each particle.
21.34 .. Point charge q1 = - 5.00 nC is at the origin and point
charge q2 = + 3.00 nC is on the x-axis at x = 3.00 cm. Point P is
S
on the
y-axis at y = 4.00 cm. (a) Calculate the electric fields E 1
S
and E 2 at point P due to the charges q1 and q2. Express your results
in terms of unit vectors (see Example 21.6). (b) Use the results of
part (a) to obtain the resultant field at P, expressed in unit vector
form.
21.35 .. CP In Exercise 21.33, what is the speed of the electron
as it emerges from the field?
21.36 . (a) Calculate the magnitude and direction (relative to the
+x-axis) of the electric field in Example 21.6. (b) A - 2.5-nC point
charge is placed at point P in Fig. 21.19. Find the magnitude and
direction of (i) the force that the - 8.0-nC charge at the origin
exerts on this charge and (ii) the force that this charge exerts on the
- 8.0-nC charge at the origin.
21.37 .. If two electrons are each 1.50 *
Figure E21.37
10 -10 m from a proton, as shown in Fig.
e
E21.37, find the magnitude and direction
of the net electric force they will exert on
the proton.
65.0°
21.38 .. CP A uniform electric field
p
exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged
plate and strikes the surface of the opposite plate, 1.60 cm distant
from the first, in a time interval of 1.50 * 10 -6 s. (a) Find the
magnitude of the electric field. (b) Find the speed of the proton
when it strikes the negatively charged plate.
21.39 . A point charge is at the origin. With this point charge as
the source point, what is the unit vector rN in the direction of (a) the
field point at x = 0, y = - 1.35 m; (b) the field point at
x = 12.0 cm, y = 12.0 cm; (c) the field point at x = - 1.10 m,
y = 2.60 m? Express your results in terms of the unit vectors nı
and n≥ .
21.40 .. A +8.75-mC point Figure E21.40
charge is glued down on a horiS
E
zontal frictionless table. It is tied
to a -6.50-mC point charge by
2.50 cm
a light, nonconducting 2.50-cm
wire. A uniform electric field of
8.75 mC
magnitude 1.85 * 10 8 N>C is 26.50 mC
directed parallel to the wire, as
shown in Fig. E21.40. (a) Find the tension in the wire. (b) What
would the tension be if both charges were negative?
21.41 .. (a) An electron is moving east in a uniform electric field
of 1.50 N>C directed to the west. At point A, the velocity of the
electron is 4.50 * 10 5 m>s toward the east. What is the speed of
the electron when it reaches point B, 0.375 m east of point A? (b) A
proton is moving in the uniform electric field of part (a). At point
A, the velocity of the proton is 1.90 * 10 4 m>s, east. What is the
speed of the proton at point B?
Exercises
Section 21.5 Electric-Field Calculations
21.42 . Two point charges Q and + q
Figure E21.42
(where q is positive) produce the net electric
Q
field shown at point P in Fig. E21.42. The
field points parallel to the line connecting
the two charges. (a) What can you conclude
S
d
E
about the sign and magnitude of Q? Explain
your reasoning. (b) If the lower charge were
P
negative instead, would it be possible for
the field to have the direction shown in the
d
figure? Explain your reasoning.
21.43 .. Two positive point charges q are
1q
placed on the x-axis, one at x = a and one
at x = - a. (a) Find the magnitude and
direction of the electric field at x = 0. (b) Derive an expression for
the electric field at points on the x-axis. Use your result to graph
the x-component of the electric field as a function of x, for values
of x between -4a and +4a.
21.44 . The two charges q1
Figure E21.44
and q2 shown in Fig. E21.44
have equal magnitudes. What
B
is the direction of the net electric field due to these two
charges at points A (midway
A
between the charges), B, and C
if (a) both charges are negative,
q1
q2
(b) both charges are positive,
(c) q1 is positive and q2 is negC
ative.
21.45 . A + 2.00-nC point
charge is at the origin, and a second - 5.00-nC point charge is on
the x-axis at x = 0.800 m. (a) Find the electric field (magnitude
and direction) at each of the following points on the x-axis: (i) x =
0.200 m; (ii) x = 1.20 m; (iii) x = - 0.200 m. (b) Find the net
electric force that the two charges would exert on an electron
placed at each point in part (a).
21.46 .. Repeat Exercise 21.44, but now let q1 = - 4.00 nC.
21.47 . Three negative point charges lie
along a line as shown in Fig. E21.47. Find Figure E21.47
the magnitude and direction of the electric
25.00 mC
field this combination of charges produces
at point P, which lies 6.00 cm from the
8.00 cm 6.00 cm
- 2.00-mC charge measured perpendicular
to the line connecting the three charges.
P
21.48 .. BIO Electric Field of Axons. A
22.00 mC
nerve signal is transmitted through a neu- 8.00 cm
ron when an excess of Na + ions suddenly
enters the axon, a long cylindrical part
25.00 mC
of the neuron. Axons are approximately
10.0 mm in diameter, and measurements
show that about 5.6 * 10 11 Na + ions per meter (each of charge
+ e) enter during this process. Although the axon is a long cylinder,
the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the
axon. Let us look at a 0.10-mm length of the axon and model it as a
point charge. (a) If the charge that enters each meter of the axon
gets distributed uniformly along it, how many coulombs of charge
enter a 0.10-mm length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at
the surface of the body if the axon is 5.00 cm below the skin? (c)
Certain sharks can respond to electric fields as weak as 1.0 mN>C .
719
How far from this segment of axon could a shark be and still detect
its electric field?
21.49 . In a rectangular coordinate system a positive point charge
q = 6.00 * 10 -9 C is placed at the point x = + 0.150 m, y = 0,
and an identical point charge is placed at x = - 0.150 m, y = 0.
Find the x- and y-components, the magnitude, and the direction of
the electric field at the following points: (a) the origin; (b)
x = 0.300 m, y = 0; (c) x = 0.150 m, y = - 0.400 m; (d) x = 0,
y = 0.200 m.
21.50 .. A point charge q1 = - 4.00 nC is at the point x =
0.600 m, y = 0.800 m, and a second point charge q2 = + 6.00 nC
is at the point x = 0.600 m, y = 0. Calculate the magnitude and
direction of the net electric field at the origin due to these two point
charges.
21.51 .. Repeat Exercise 21.49 for the case where the point
charge at x = + 0.150 m, y = 0 is positive and the other is negative, each with magnitude 6.00 * 10 -9 C.
21.52 .. A very long, straight wire has charge per unit length
1.50 * 10 -10 C>m. At what distance from the wire is the electricfield magnitude equal to 2.50 N>C?
21.53 . A ring-shaped conductor with radius a = 2.50 cm has a
total positive charge Q = + 0.125 nC uniformly distributed
around it, as shown in Fig. 21.23. The center of the ring is at the
origin of coordinates O. (a) What is the electric field (magnitude
and direction) at point P, which is on the x-axis at x = 40.0 cm?
(b) A point charge q = - 2.50 mC is placed at the point P
described in part (a). What are the magnitude and direction of the
force exerted by the charge q on the ring?
21.54 .. A straight, nonconducting plastic wire 8.50 cm long carries a charge density of +175 nC>m distributed uniformly along
its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point
6.00 cm directly above its midpoint. (b) If the wire is now bent
into a circle lying flat on the table, find the magnitude and direction
of the electric field it produces at a point 6.00 cm directly above its
center.
21.55 .. A charge of -6.50 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 cm.
(a) Find the magnitude and direction of the electric field this disk
produces at a point P on the axis of the disk a distance of
2.00 cm from its center. (b) Suppose that the charge were all
pushed away from the center and distributed uniformly on the
outer rim of the disk. Find the magnitude and direction of the
electric field at point P. (c) If the charge is all brought to the center of the disk, find the magnitude and direction of the electric
field at point P. (d) Why is the field in part (a) stronger than the
field in part (b)? Why is the field in part (c) the strongest of the
three fields?
Section 21.7 Electric Dipoles
21.56 . The ammonia molecule 1NH32 has a dipole moment of
in the gas phase are
5.0 * 10 -30 C # m. Ammonia molecules
S
placed in a uniform electric field E with magnitude 1.6 *
10 6 N>C. (a) What is the change in electric potential energy when
the dipole Smoment of a molecule changes its orientation with
respect to E from parallel to perpendicular? (b) At what absolute
temperature T is the average translational kinetic energy 32 kT of a
molecule equal to the change in potential energy calculated in part
(a)? (Note: Above this temperature, thermal agitation prevents the
dipoles from aligning with the electric field.)
720
CHAPTER 21 Electric Charge and Electric Field
21.57 . Point charges q1 = - 4.5 nC and q2 = + 4.5 nC are separated by 3.1 mm, forming an electric dipole. (a) Find the electric
dipole moment (magnitude and direction). (b) The charges are in a
uniform electric field whose direction makes an angle of 36.9°
with the line connecting the charges. What is the magnitude of
this field if the torque exerted on the dipole has magnitude
7.2 * 10 -9 N # m?
21.58 . The dipole moment of the water molecule 1H2O2 is
6.17 * 10 -30 C # m. Consider a water molecule located at the
S
origin whose dipole moment p points in the +x-direction. A chlorine ion 1C1 2, of charge - 1.60 * 10 -19 C, is located at
x = 3.00 * 10 -9 m. Find the magnitude and direction of the electric force that the water molecule exerts on the chlorine ion. Is this
force attractive or repulsive? Assume that x is much larger than the
separation d between the charges in the dipole, so that the approximate expression for the electric field along the dipole axis derived
in Example 21.14 can be used.
21.59 . Torque on a Dipole. An electric
dipole with dipole
S
S
moment p is in a uniform electric field E. (a) Find the orientations
of the dipole for which the torque on the dipole is zero. (b) Which
of the orientations in part (a) is stable, and which is unstable?
(Hint: Consider a small displacement away from the equilibrium
position and see what happens.) (c) Show that for the stable orientation in part (b), the dipole’s own electric field tends to oppose the
external field.
21.60 .. Consider the electric dipole of Example 21.14. (a)
Derive an expression for the magnitude of the electric field produced by the dipole at a point on the x-axis in Fig. 21.33. What is
the direction of this electric field? (b) How does the electric field at
points on the x-axis depend on x when x is very large?
21.61 . Three charges are at
the corners of an isosceles trian- Figure E21.61
gle as shown in Fig. E21.61.
15.00 mC
The ⫾5.00-mC charges form
a dipole. (a) Find the force
2.00 cm
(magnitude and direction) the
- 10.00-mC charge exerts on the
210.00 mC
dipole. (b) For an axis perpendi- 3.00 cm
cular to the line connecting the
2.00 cm
⫾5.00-mC charges at the midpoint of this line, find the torque
25.00 mC
(magnitude and direction) exerted
on the dipole by the -10.00-mC
charge.
21.62 . A dipole consisting of charges ⫾e, 220 nm apart, is
placed between two very large (essentially infinite) sheets carrying
equal but opposite charge densities of 125 mC>m2. (a) What is the
maximum potential energy this dipole can have due to the sheets,
and how should it be oriented relative to the sheets to attain this
value? (b) What is the maximum torque the sheets can exert on the
dipole, and how should it be oriented relative to the sheets to attain
this value? (c) What net force do the two sheets exert on the
dipole?
PROBLEMS
21.63 ... Four identical charges Q are placed at the corners of a
square of side L. (a) In a free-body diagram, show all of the
forces that act on one of the charges. (b) Find the magnitude and
direction of the total force exerted on one charge by the other
three charges.
21.64 ... Two charges, one of 2.50 mC and the other of
- 3.50 mC, are placed on the x-axis, one at the origin and the other
at x = 0.600 m, as shown in Fig. P21.64. Find the position on the
x-axis where the net force on a small charge + q would be zero.
Figure P21.64
12.50 mC
0
23.50 mC
x (m)
0.600 m
21.65 .. Three point charges are arranged along the x-axis.
Charge q1 = - 4.50 nC is located at x = 0.200 m, and charge
q2 = + 2.50 nC is at x = - 0.300 m. A positive point charge q3 is
located at the origin. (a) What must the value of q3 be for the net
force on this point charge to have magnitude 4.00 mN? (b) What is
the direction of the net force on q3? (c) Where along the x-axis can
q3 be placed and the net force on it be zero, other than the trivial
answers of x = + q and x = - q ?
21.66 .. A charge q1 = + 5.00 nC is placed at the origin of an
xy-coordinate system, and a charge q2 = - 2.00 nC is placed on
the positive x-axis at x = 4.00 cm. (a) If a third charge q3 =
+ 6.00 nC is now placed at the point x = 4.00 cm, y = 3.00 cm,
find the x- and y-components of the total force exerted on this
charge by the other two. (b) Find the magnitude and direction of
this force.
21.67 .. CP Two positive point charges Q are held fixed on the
x-axis at x = a and x = - a. A third positive point charge q, with
mass m, is placed on the x-axis away from the origin at a coordinate x such that ƒ x ƒ V a. The charge q, which is free to move
along the x-axis, is then released. (a) Find the frequency of oscillation of the charge q. (Hint: Review the definition of simple harmonic motion in Section 14.2. Use the binomial expansion
11 + z2n = 1 + nz + n1n - 12z 2 >2 + Á , valid for the case
ƒ z ƒ 6 1.) (b) Suppose instead that the charge q were placed on the
y-axis at a coordinate y such that ƒ y ƒ V a, and then released. If
this charge is free to move anywhere in the xy-plane, what will
happen to it? Explain your answer.
21.68 .. CP Two identical spheres Figure P21.68
with mass m are hung from silk
threads of length L, as shown in
Fig. P21.68. Each sphere has the same
charge, so q1 = q2 = q. The radius of
L
L
each sphere is very small compared to
the distance between the spheres, so
u u
they may be treated as point charges.
Show that if the angle u is small,
the equilibrium separation d between
the spheres is d = 1q 2L>2pP0mg21>3.
(Hint: If u is small, then tan u ⬵
mass m
mass m
charge q1
charge q2
sin u.)
...
21.69
CP Two small spheres with
mass m = 15.0 g are hung by silk threads of length L = 1.20 m
from a common point (Fig. P21.68). When the spheres are given
equal quantities of negative charge, so that q1 = q2 = q, each
thread hangs at u = 25.0° from the vertical. (a) Draw a diagram
showing the forces on each sphere. Treat the spheres as point
charges. (b) Find the magnitude of q. (c) Both threads are now
shortened to length L = 0.600 m, while the charges q1 and q2
remain unchanged. What new angle will each thread make with the
vertical? (Hint: This part of the problem can be solved numerically
Problems
3.
00
cm
21.76 ... Two point charges q1 and Figure P21.76
q2 are held in place 4.50 cm apart.
q1
Another point charge Q = - 1.75 mC
of mass 5.00 g is initially located
S
a
3.00 cm from each of these charges
(Fig. P21.76) and released from rest.
You observe that the initial accelera- Q
4.50 cm
tion of Q is 324 m>s2 upward, parallel
to the line connecting the two point
charges. Find q1 and q2.
21.77 . Three identical point charges
q2
q are placed at each of three corners of
a square of side L. Find the magnitude
and direction of the net force on a point charge -3q placed (a) at
the center of the square and (b) at the vacant corner of the square.
In each case, draw a free-body diagram showing the forces exerted
on the - 3q charge by each of the other three charges.
21.78 ... Three point charges are placed on the y-axis: a charge q
at y = a, a charge -2q at the origin, and a charge q at y = - a.
Such an arrangement is called an electric quadrupole. (a) Find the
magnitude and direction of the electric field at points on the positive x-axis. (b) Use the binomial expansion to find an approximate
expression for the electric field valid for x W a. Contrast this
behavior to that of the electric field of a point charge and that of
the electric field of a dipole.
21.79 .. CP Strength of the Electric Force. Imagine two
1.0-g bags of protons, one at the earth’s north pole and the other at
the south pole. (a) How many protons are in each bag? (b) Calculate the gravitational attraction and the electrical repulsion that
each bag exerts on the other. (c) Are the forces in part (b) large
enough for you to feel if you were holding one of the bags?
21.80 . Electric Force Within the Nucleus. Typical dimensions of atomic nuclei are of the order of 10 -15 m (1 fm2. (a) If
two protons in a nucleus are 2.0 fm apart, find the magnitude of
the electric force each one exerts on the other. Express the answer
in newtons and in pounds. Would this force be large enough for a
person to feel? (b) Since the protons repel each other so strongly,
why don’t they shoot out of the nucleus?
21.81 .. If Atoms Were Not Neutral . . . Because the charges
on the electron and proton have the same absolute value, atoms are
electrically neutral. Suppose this were not precisely true, and the
absolute value of the charge of the electron were less than the
charge of the proton by 0.00100%. (a) Estimate what the net
charge of this textbook would be under these circumstances. Make
any assumptions you feel are justified, but state clearly what they
are. (Hint: Most of the atoms in this textbook have equal numbers
of electrons, protons, and neutrons.) (b) What would be the magnitude of the electric force between two textbooks placed 5.0 m
apart? Would this force be attractive or repulsive? Estimate what
the acceleration of each book would be if the books were 5.0 m
apart and there were no non- Figure P21.82
electric forces on them. (c)
Discuss how the fact that ordinary matter is stable shows
that the absolute values of the
u
charges on the electron and
L
L
proton must be identical to a
very high level of accuracy.
21.82 ... CP Two tiny spheres of mass 6.80 mg carry
S
E
charges of equal magnitude,
00
3.
cm
by using trial values for u and adjusting the values of u until a selfconsistent answer is obtained.)
21.70 .. CP Two identical spheres are each attached to silk
threads of length L = 0.500 m and hung from a common point
(Fig. P21.68). Each sphere has mass m = 8.00 g. The radius of
each sphere is very small compared to the distance between the
spheres, so they may be treated as point charges. One sphere is
given positive charge q1, and the other a different positive charge
q2 ; this causes the spheres to separate so that when the spheres are
in equilibrium, each thread makes an angle u = 20.0° with the
vertical. (a) Draw a free-body diagram for each sphere when in
equilibrium, and label all the forces that act on each sphere.
(b) Determine the magnitude of the electrostatic force that acts on
each sphere, and determine the tension in each thread. (c) Based
on the information you have been given, what can you say about
the magnitudes of q1 and q2 ? Explain your answers. (d) A small
wire is now connected between the spheres, allowing charge to
be transferred from one sphere to the other until the two spheres
have equal charges; the wire is then removed. Each thread now
makes an angle of 30.0° with the vertical. Determine the
original charges. (Hint: The total charge on the pair of spheres is
conserved.)
21.71 .. Sodium chloride (NaCl, ordinary table salt) is made up
of positive sodium ions 1Na+2 and negative chloride ions 1C1-2.
(a) If a point charge with the same charge and mass as all the
Na+ ions in 0.100 mol of NaCl is 2.00 cm from a point charge
with the same charge and mass as all the C1- ions, what is the
magnitude of the attractive force between these two point
charges? (b) If the positive point charge in part (a) is held in
place and the negative point charge is released from rest, what is
its initial acceleration? (See Appendix D for atomic masses.)
(c) Does it seem reasonable that the ions in NaCl could be separated in this way? Why or why not? (In fact, when sodium chloride dissolves in water, it breaks up into Na+ and C1- ions.
However, in this situation there are additional electric forces
exerted by the water molecules on the ions.)
21.72 .. A -5.00-nC point charge is on the x-axis at x = 1.20 m.
A second point charge Q is on the x-axis at -0.600 m. What must
be the sign and magnitude of Q for the resultant electric field at the
origin to be (a) 45.0 N> C in the + x-direction, (b) 45.0 N> C in the
- x-direction?
21.73 .. CP A small 12.3-g plastic ball is tied Figure P21.73
to a very light 28.6-cm string that is attached to
the vertical wall of a room (Fig. P21.73). A uniform horizontal electric field exists in this
room. When the ball has been given an excess
charge of - 1.11 mC, you observe that it
remains suspended, with the string making an
17.4°
angle of 17.4° with the wall. Find the magnitude and direction of the electric field in the
room.
21.74 .. CP At t = 0 a very small object
with mass 0.400 mg and charge + 9.00 mC is traveling at 125 m> s
in the - x-direction. The charge is moving in a uniform electric field
that is in the +y-direction and that has magnitude E = 895 N>C.
The gravitational force on the particle can be neglected. How far is
the particle from the origin at t = 7.00 ms?
21.75 .. Two particles having charges q1 = 0.500 nC and
q2 = 8.00 nC are separated by a distance of 1.20 m. At what point
along the line connecting the two charges is the total electric field
due to the two charges equal to zero?
721
722
CHAPTER 21 Electric Charge and Electric Field
72.0 nC, but opposite sign. They are tied to the same ceiling hook
by light strings of length 0.530 m. When a horizontal uniform electric
field E that is directed to the left is turned on, the spheres hang at rest
with the angle u between the strings equal to 50.0 o (Fig. P21.82). (a)
Which ball (the one on the right or the one on the left) has positive
charge? (b) What is the magnitude E of the field?
21.83 .. CP Consider a model of a hydrogen atom in which an
electron is in a circular orbit of radius r = 5.29 * 10 -11 m around
a stationary proton. What is the speed of the electron in its orbit?
21.84 .. CP A small sphere with mass 9.00 mg and charge
- 4.30 mC is moving in a circular orbit around a stationary sphere
that has charge + 7.50 mC. If the speed of the small sphere is
5.90 * 10 3 m>s, what is the radius of its orbit? Treat the spheres
as point charges and ignore gravity.
21.85 .. Two small copper spheres each have radius 1.00 mm.
(a) How many atoms does each sphere contain? (b) Assume that each
copper atom contains 29 protons and 29 electrons. We know that
electrons and protons have charges of exactly the same magnitude,
but let’s explore the effect of small differences (see also Problem
21.81). If the charge of a proton is + e and the magnitude of the
charge of an electron is 0.100% smaller, what is the net charge of
each sphere and what force would one sphere exert on the other if
they were separated by 1.00 m?
21.86 ... CP Operation of an Inkjet Printer. In an inkjet
printer, letters are built up by squirting drops of ink at the paper
from a rapidly moving nozzle. The ink drops, which have a mass
of 1.4 * 10 -8 g each, leave the nozzle and travel toward the paper
at 20 m>s, passing through a charging unit that gives each drop a
positive charge q by removing some electrons from it. The drops
then pass between parallel deflecting plates 2.0 cm long where
there is a uniform vertical electric field with magnitude
8.0 * 10 4 N>C. If a drop is to be deflected 0.30 mm by the time it
reaches the end of the deflection plates, what magnitude of charge
must be given to the drop?
21.87 .. CP A proton is projected into a uniform electric field
that points vertically upward and has magnitude E. The initial
velocity of the proton has a magnitude v0 and is directed at an
angle a below the horizontal. (a) Find the maximum distance h max
that the proton descends vertically below its initial elevation. You
can ignore gravitational forces. (b) After what horizontal distance
d does the proton return to its original elevation? (c) Sketch the
trajectory of the proton. (d) Find the numerical values of h max and
d if E = 500 N>C, v0 = 4.00 * 10 5 m>s, and a = 30.0°.
21.88 . A negative point charge q1 = - 4.00 nC is on the x-axis
at x = 0.60 m. A second point charge q2 is on the x-axis at
x = - 1.20 m. What must the sign and magnitude of q2 be for the
net electric field at the origin to be (a) 50.0 N>C in the
+ x-direction and (b) 50.0 N> C in the - x-direction?
21.89 .. CALC Positive charge
Q is distributed uniformly along Figure P21.89
the x-axis from x = 0 to x = a.
y
A positive point charge q is
located on the positive x-axis at
q
x = a + r, a distance r to the
Q
right of the end of Q (Fig.
x
O
a
r
P21.89). (a) Calculate the x- and
y-components of the electric field
produced by the charge distribution Q at points on the positive x-axis
where x 7 a. (b) Calculate the force (magnitude and direction) that
the charge distribution Q exerts on q. (c) Show that if r W a, the
magnitude of the force in part (b) is approximately Qq>4pP0r 2.
Explain why this result is obtained.
21.90 .. CALC Positive charge Figure P21.90
Q is distributed uniformly along
y
the positive y-axis between
a
y = 0 and y = a. A negative
point charge - q lies on the positive x-axis, a distance x from the
Q
origin (Fig. P21.90). (a) Calculate the x- and y-components of
x
O
the electric field produced by the
2q
charge distribution Q at points on
the positive x-axis. (b) Calculate the x- and y-components of the
force that the charge distribution Q exerts on q. (c) Show that if
x W a, Fx ⬵ - Qq>4pP0x 2 and Fy ⬵ + Qqa>8pP0x 3. Explain
why this result is obtained.
21.91 .. A charged line like that shown in Fig. 21.24 extends
from y = 2.50 cm to y = - 2.50 cm. The total charge distributed
uniformly along the line is -7.00 nC. (a) Find the electric field
(magnitude and direction) on the x-axis at x = 10.0 cm. (b) Is the
magnitude of the electric field you calculated in part (a) larger or
smaller than the electric field 10.0 cm from a point charge that has
the same total charge as this finite line of charge? In terms of the
approximation used to derive E = Q>4pP0x 2 for a point charge
from Eq. (21.9), explain why this is so. (c) At what distance x does
the result for the finite line of charge differ by 1.0% from that for
the point charge?
21.92 . CP A Parallel Universe. Imagine a parallel universe in
which the electric force has the same properties as in our universe
but there is no gravity. In this parallel universe, the sun carries
charge Q, the earth carries charge -Q, and the electric attraction
between them keeps the earth in orbit. The earth in the parallel universe has the same mass, the same orbital radius, and the same
orbital period as in our universe. Calculate the value of Q. (Consult
Appendix F as needed.)
21.93 ... A uniformly charged disk like the disk in Fig. 21.25 has
radius 2.50 cm and carries a total charge of 7.0 * 10 -12 C. (a)
Find the electric field (magnitude and direction) on the x-axis at
x = 20.0 cm. (b) Show that for x W R, Eq. (21.11) becomes
E = Q>4pP0x 2, where Q is the total charge on the disk. (c) Is the
magnitude of the electric field you calculated in part (a) larger or
smaller than the electric field 20.0 cm from a point charge that has
the same total charge as this disk? In terms of the approximation
used in part (b) to derive E = Q>4pP0x 2 for a point charge from
Eq. (21.11), explain why this is so. (d) What is the percent difference between the electric fields produced by the finite disk and by
a point charge with the same charge at x = 20.0 cm and at
x = 10.0 cm?
21.94 .. BIO Electrophoresis.
Figure P21.94
Electrophoresis is a process
used by biologists to separate
different biological molecules
(such as proteins) from each
other according to their ratio of
charge to size. The materials to
be separated are in a viscous
solution that produces a drag
force FD proportional to the
size and speed of the molecule.
We can express this relationship as FD = KRy, where R is
the radius of the molecule (modeled as being spherical), v is its
speed, and K is a constant that depends on the viscosity of the
Challenge Problems
solution. The solution is placed in an external electric field E so
that the electric force on a particle of charge q is F = qE. (a) Show
that when the electric field is adjusted so that the two forces (electric and viscous drag) just balance, the ratio of q to R is Kv>E.
(b) Show that if we leave the electric field on for a time T, the distance
x that the molecule moves during that time is x = 1ET>k21q>R2.
(c) Suppose you have a sample containing three different biological molecules for which the molecular ratio q>R for material 2 is
twice that of material 1 and the ratio for material 3 is three times
that of material 1. Show that the distances migrated by these molecules after the same amount of time are x 2 = 2x 1 and x 3 = 3x 1.
In other words, material 2 travels twice as far as material 1, and
material 3 travels three times as far as material 1. Therefore, we
have separated these molecules according to their ratio of charge to
size. In practice, this process can be carried out in a special gel or
paper, along which the biological molecules migrate. (Fig. P21.94).
The process can be rather slow, requiring several hours for separations of just a centimeter or so.
21.95 . CALC Positive charge + Q is distributed uniformly along
the + x-axis from x = 0 to x = a. Negative charge -Q is distributed uniformly along the - x-axis from x = 0 to x = - a. (a) A
positive point charge q lies on the positive y-axis, a distance y from
the origin. Find the force (magnitude and direction) that the positive and negative charge distributions together exert on q. Show
that this force is proportional to y -3 for y W a. (b) Suppose
instead that the positive point charge q lies on the positive x-axis, a
distance x 7 a from the origin. Find the force (magnitude and
direction) that the charge distribution exerts on q. Show that this
force is proportional to x -3 for x W a.
21.96 .. CP A small sphere with mass m carries a positive charge
q and is attached to one end of a silk fiber of length L. The other
end of the fiber is attached to a large vertical insulating sheet that
has a positive surface charge density s. Show that when the sphere
is in equilibrium, the fiber makes an angle equal to arctan
1qs>2mgP02 with the vertical sheet.
21.97 .. CALC Negative charge - Q is distributed uniformly
around a quarter-circle of radius a that lies in the first quadrant, with
the center of curvature at the origin. Find the x- and y-components of
the net electric field at the origin.
21.98 .. CALC A semicircle
of radius a is in the first and Figure P21.98
y
second quadrants, with the
center of curvature at the origin. Positive charge +Q is dis1Q
2Q
tributed uniformly around the
a
left half of the semicircle, and
negative charge - Q is distribx
uted uniformly around the right
half of the semicircle (Fig.
P21.98). What are the magnitude and direction of the net
Figure P21.99
electric field at the origin pro1.20 m
duced by this distribution of
+ + + + + + + + +
charge?
–
21.99 .. Two 1.20-m noncon–
ducting wires meet at a right
–
angle. One segment carries
–
+2.50 mC of charge distrib- 1.20 m –
P
–
uted uniformly along its length,
–
and the other carries - 2.50 mC
–
distributed uniformly along it,
–
as shown in Fig. P21.99.
723
(a) Find the magnitude and direction of the electric field these
wires produce at point P, which is 60.0 cm from each wire. (b) If
an electron is released at P, what are the magnitude and direction
of the net force that these wires exert on it?
21.100 . Two very large parallel sheets are 5.00 cm apart. Sheet
A carries a uniform surface charge density of -9.50mC>m2, and
sheet B, which is to the right of A, carries a uniform charge density
of -11.6 mC>m2. Assume the sheets are large enough to be treated
as infinite. Find the magnitude and direction of the net electric field
these sheets produce at a point (a) 4.00 cm to the right of sheet A;
(b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B.
21.101 . Repeat Problem 21.100 for the case where sheet B is
positive.
21.102 . Two very large horizontal sheets are 4.25 cm apart and
carry equal but opposite uniform surface charge densities of magnitude s. You want to use these sheets to hold stationary in the
region between them an oil droplet of mass 324 mg that carries an
excess of five electrons. Assuming that the drop is in vacuum, (a)
which way should the electric field between the plates point, and
(b) what should s be?
21.103 .. An infinite sheet with positive charge per unit area s
lies in the xy-plane. A second infinite sheet with negative charge
per unit area -s lies in the yz-plane. Find the net electric field at
all points that do not lie in either of these planes. Express your
answer in terms of the unit vectors nı , n≥ , and kN .
21.104 .. CP A thin disk with a
circular hole at its center, called Figure P21.104
an annulus, has inner radius R1
x
and outer radius R2 (Fig.
R2
P21.104). The disk has a uniform
positive surface charge density s
R1
z
on its surface. (a) Determine the
total electric charge on the annuy
s
lus. (b) The annulus lies in the
O
yz-plane, with its center at the origin. For an arbitrary point on the
x-axis (the axis of the annulus),
S
find the magnitude and direction of the electric field E. Consider
points both above and below the annulus in Fig. P21.104. (c) Show
that at points on the x-axis that are sufficiently close to the origin,
the magnitude of the electric field is approximately proportional to
the distance between the center of the annulus and the point. How
close is “sufficiently close”? (d) A point particle with mass m and
negative charge -q is free to move along the x-axis (but cannot
move off the axis). The particle is originally placed at rest at
x = 0.01R1 and released. Find the frequency of oscillation of the
particle. (Hint: Review Section 14.2. The annulus is held stationary.)
CHALLENGE PROBLEMS
21.105 ... Three charges are
Figure P21.105
placed as shown in Fig.
q3
S
P21.105. The magnitude of q1 is
F
2.00 mC, but its sign and the
4.00 cm
3.00 cm
value of the charge q2 are not
known. Charge q3 is +4.00 mC,
S
and the net force F on q3 is
q2
5.00 cm
entirely in the negative x-direc- q1
tion. (a) Considering the different possible signs of q1, there are four
S
S
possible force diagrams representing the forces F1 and F2 that q1 and
q2 exert on q3 . Sketch these four possible force configurations.
724
CHAPTER 21 Electric Charge and Electric Field
S
(b) Using the sketches from part (a) and the direction of F, deduce
the signs of the charges q1 and q2. (c) Calculate the magnitude of
q2. (d) Determine F, the magnitude of the net force on q3.
21.106 ... Two charges are
Figure P21.106
placed as shown in Fig.
P
P21.106. The magnitude of q1 is
3.00 mC, but its sign and the
12.0 cm
value of the charge q2 are not 5.0 cm
S
E
known. The direction
of the net
S
electric field E at point P is
q2
q1
13.0 cm
entirely in the negative y-direction. (a) Considering the different possible signs of q1 and q2, there are four possible diagrams
S
S
that could represent the electric fields E 1 and E 2 produced by q1
and q2. Sketch the four possible electric-field configurations.
S
(b) Using the sketches from part (a) and the direction of
E, deduce
S
the signs of q1 and q2. (c) Determine the magnitude of E.
21.107 ... CALC Two thin rods of length L lie along the x-axis,
one between x = a>2 and x = a>2 + L and the other between
x = - a>2 and x = - a>2 - L. Each rod has positive charge Q
distributed uniformly along its length. (a) Calculate the electric
field produced by the second rod at points along the positive
x-axis. (b) Show that the magnitude of the force that one rod exerts
on the other is
F =
Q2
2
4pP0L
ln c
(a + L)2
d
a1a + 2L)
(c) Show that if a W L, the magnitude of this force reduces to
F = Q 2 >4pP0a 2. (Hint: Use the expansion ln11 + z2 = z z 2 >2 + z 3 >3 - Á , valid for ƒ z ƒ V 1. Carry all expansions to at
least order L2 >a 2.) Interpret this result.
Answers
Chapter Opening Question
?
Water molecules have a permanent electric dipole moment: One
end of the molecule has a positive charge and the other end has a
negative charge. These ends attract negative and positive ions,
respectively, holding the ions apart in solution. Water is less effective as a solvent for materials whose molecules do not ionize
(called nonionic substances), such as oils.
Test Your Understanding Questions
21.1 Answers: (a) the plastic rod weighs more, (b) the glass rod
weighs less, (c) the fur weighs less, (d) the silk weighs more The
plastic rod gets a negative charge by taking electrons from the fur,
so the rod weighs a little more and the fur weighs a little less after
the rubbing. By contrast, the glass rod gets a positive charge by giving electrons to the silk. Hence, after they are rubbed together, the
glass rod weighs a little less and the silk weighs a little more. The
weight change is very small: The number of electrons transferred is a
small fraction of a mole, and a mole of electrons has a mass of
only 16.02 * 10 23 electrons219.11 * 10 -31 kg>electron2 = 5.48 *
10 -7 kg = 0.548 milligram!
21.2 Answers: (a) (i), (b) (ii) Before the two spheres touch, the
negatively charged sphere exerts a repulsive force on the electrons
in the other sphere, causing zones of positive and negative
induced charge (see Fig. 21.7b). The positive zone is closer to the
negatively charged sphere than the negative zone, so there is a net
force of attraction that pulls the spheres together, like the comb
and insulator in Fig. 21.8b. Once the two metal spheres touch,
some of the excess electrons on the negatively charged sphere will
flow onto the other sphere (because metals are conductors). Then
both spheres will have a net negative charge and will repel each
other.
21.3 Answer: (iv) The force exerted by q1 on Q is still as in Example 21.4. The magnitude of the force exerted by q2 on Q is still
equal to F1 on Q, but the direction of the force is now toward q2 at
an angle a below the x-axis. Hence the x-components of the two
forces cancel while the (negative) y-components add together, and
the total electric force is in the negative y-direction.
S
21.4 Answers: (a) (ii), (b) (i) The electric field E produced by a
positive point charge points directly away from the charge (see
Fig. 21.18a) and has a magnitude that depends on the distance r
from the charge to the field point.
Hence
a second, negative point
S
S
charge q 6 0 will feel a force F ⴝ qE that points directly toward
the positive charge and has a magnitude that depends on the
distance r between the two charges. If the negative charge moves
directly toward the positive charge, the direction of the force
remains the same but the force magnitude increases as the distance
r decreases. If the negative charge moves in a circle around
the positive charge, the force magnitude stays the same (because
the distance r is constant) but the force direction changes.
21.5 Answer: (iv) Think of a pair of segments of length dy, one at
coordinate y 7 0 and the other at coordinate -y 6 0. The upper
S
segment has a positive charge and produces an electric
field dE at
S
P that points away from the segment, so this dE has a positive
S
x-component and a negative y-component, like the vector dE in
Fig. 21.24. The lower segment
has the same amount of negative
S
charge. It produces a dE that has the same magnitude but points
toward the lower segment, so it has a negative x-component and a
negative y-component. By symmetry, the two x-components are
equal but opposite, so they cancel. Thus the total electric field has
only a negative y-component.
S
21.6 Answer: yes If the field lines are straight, E must S
point inS the
same direction throughout the region. Hence the force F = qE on
a particle of charge q is always in the same direction. A particle
released
from rest accelerates in a straight line in the direction of
S
F, and so its trajectory is a straight line along a field line.
21.7 Answer: (ii) Equations (21.17) and (21.18) tell us thatSthe
S
potential energy for a dipole in an electric field is U = - p E =
S
-pE cos f, where
f is the angle between the directions of p and
S
S
S
E. If p and E point in opposite directions, so that f = 180°, we
have cos f = - 1 and U = + pE. This is the maximum value that
U can have. From our discussion of energy diagrams in Section
7.5, it follows that this is a situation of unstable equilibrium.
#
Bridging Problem
Answer: E = 2kQ>pa 2 in the –y-direction
21
ELECTRIC CHARGE AND ELECTRIC FIELD
21.1.
(a) IDENTIFY and SET UP: Use the charge of one electron (−1.602 × 10−19 C) to find the number of
electrons required to produce the net charge.
EXECUTE: The number of excess electrons needed to produce net charge q is
q
−3.20 × 10−9 C
=
= 2.00 × 1010 electrons.
−e −1.602 × 10−19 C/electron
(b) IDENTIFY and SET UP: Use the atomic mass of lead to find the number of lead atoms in
8.00 × 10−3 kg of lead. From this and the total number of excess electrons, find the number of excess
electrons per lead atom.
EXECUTE: The atomic mass of lead is 207 × 10−3 kg/mol, so the number of moles in 8.00 × 10−3 kg is
n=
8.00 × 1023 kg
mtot
=
= 0.03865 mol. N A (Avogadro’s number) is the number of atoms in 1 mole,
M
207 × 10−3 kg/mol
so the number of lead atoms is N = nN A = (0.03865 mol)(6.022 × 1023 atoms/mol) = 2.328 × 1022 atoms.
21.2.
2.00 × 1010 electrons
= 8.59 × 10−13.
2.328 × 1022 atoms
EVALUATE: Even this small net charge corresponds to a large number of excess electrons. But the number
of atoms in the sphere is much larger still, so the number of excess electrons per lead atom is very small.
IDENTIFY: The charge that flows is the rate of charge flow times the duration of the time interval.
SET UP: The charge of one electron has magnitude e = 1.60 × 10−19 C.
The number of excess electrons per lead atom is
EXECUTE: The rate of charge flow is 20,000 C/s and t = 100 μs = 1.00 × 10−4 s.
Q = (20,000 C/s)(1.00 × 10−4 s) = 2.00 C. The number of electrons is ne =
21.3.
Q
= 1.25 × 1019.
1.60 × 10−19 C
EVALUATE: This is a very large amount of charge and a large number of electrons.
IDENTIFY: From your mass estimate the number of protons in your body. You have an equal number of
electrons.
SET UP: Assume a body mass of 70 kg. The charge of one electron is −1.60 × 10−19 C.
EXECUTE: The mass is primarily protons and neutrons of m = 1.67 × 10−27 kg. The total number of
protons and neutrons is np and n =
70 kg
1.67 × 10−27 kg
= 4.2 × 1028. About one-half are protons, so
np = 2.1 × 1028 = ne . The number of electrons is about 2.1 × 1028. The total charge of these electrons is
Q = (−1.60 × 10−19 C/electron )(2.10 × 1028 electrons) = −3.35 × 109 C.
EVALUATE: This is a huge amount of negative charge. But your body contains an equal number of
protons and your net charge is zero. If you carry a net charge, the number of excess or missing electrons is
a very small fraction of the total number of electrons in your body.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-1
21-2
Chapter 21
21.4.
IDENTIFY: Use the mass m of the ring and the atomic mass M of gold to calculate the number of gold
atoms. Each atom has 79 protons and an equal number of electrons.
SET UP: N A = 6.02 × 1023 atoms/mol. A proton has charge +e.
EXECUTE: The mass of gold is 17.7 g and the atomic weight of gold is 197 g/mol. So the number of atoms is
⎛ 17.7 g ⎞
22
N A n = (6.02 × 1023 atoms/mol) ⎜
⎟ = 5.41 × 10 atoms. The number of protons is
⎝ 197 g/mol ⎠
np = (79 protons/atom)(5.41×1022 atoms) = 4.27 ×1024 protons.
Q = (np )(1.60 × 10−19 C/proton) = 6.83 × 105 C.
(b) The number of electrons is ne = np = 4.27 × 1024.
21.5.
EVALUATE: The total amount of positive charge in the ring is very large, but there is an equal amount of
negative charge.
IDENTIFY: Each ion carries charge as it enters the axon.
SET UP: The total charge Q is the number N of ions times the charge of each one, which is e. So Q = Ne,
where e = 1.60 × 10−19 C.
EXECUTE: The number N of ions is N = (5.6 × 1011ions/m)(1.5 × 10−2 m) = 8.4 × 109 ions. The total charge
Q carried by these ions is Q = Ne = (8.4 × 109 )(1.60 × 10−19 C) = 1.3 × 10−9 C = 1.3 nC.
21.6.
EVALUATE: The amount of charge is small, but these charges are close enough together to exert large
forces on nearby charges.
IDENTIFY: Apply Coulomb’s law and calculate the net charge q on each sphere.
SET UP: The magnitude of the charge of an electron is e = 1.60 × 10−19 C.
EXECUTE: F =
1
q2
4π ⑀0 r 2
. This gives
q = 4π ⑀0 Fr 2 = 4π ⑀ 0 (4.57 × 10−21 N)(0.200 m) 2 = 1.43 × 10−16 C. And therefore, the total
number of electrons required is n = q /e = (1.43 × 10−16 C)/(1.60 × 10−19 C/electron) = 890 electrons.
21.7.
EVALUATE: Each sphere has 890 excess electrons and each sphere has a net negative charge. The two like
charges repel.
k q1q2
IDENTIFY: Apply F =
and solve for r.
r2
SET UP: F = 650 N.
EXECUTE: r =
21.8.
k q1q2
(8.99 × 109 N ⋅ m 2 /C2 )(1.0 C) 2
=
= 3.7 × 103 m = 3.7 km
F
650 N
EVALUATE: Charged objects typically have net charges much less than 1 C.
IDENTIFY: Use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum
atoms in one sphere. Each atom has 13 electrons. Apply Coulomb’s law and calculate the magnitude of
charge q on each sphere.
SET UP: N A = 6.02 × 1023 atoms/mol. q = n′ee, where n′e is the number of electrons removed from one
sphere and added to the other.
EXECUTE: (a) The total number of electrons on each sphere equals the number of protons.
⎛
⎞
0.0250 kg
24
ne = np = (13)( N A ) ⎜
⎟ = 7.25 × 10 electrons.
0.026982
kg/mol
⎝
⎠
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
(b) For a force of 1.00 × 104 N to act between the spheres, F = 1.00 × 104 N =
1
q2
4π ⑀0 r 2
21-3
. This gives
q = 4π ⑀ 0 (1.00 × 104 N)(0.800 m) 2 = 8.43 × 10−4 C. The number of electrons removed from one sphere
and added to the other is n′e = q /e = 5.27 × 1015 electrons.
(c) n′e /n e = 7.27 × 10−10.
21.9.
EVALUATE: When ordinary objects receive a net charge the fractional change in the total number of
electrons in the object is very small.
IDENTIFY: Apply Coulomb’s law.
SET UP: Consider the force on one of the spheres.
(a) EXECUTE: q1 = q2 = q
F=
1
q1q2
4π ⑀0 r
2
=
q2
4π ⑀0r 2
so q = r
F
0.220 N
= 0.150 m
= 7.42 × 10−7 C (on each)
(1/4π ⑀0 )
8.988 × 109 N ⋅ m 2 /C2
so q1 = r
F
F
= 12 r
= 1 (7.42 × 10−7 C) = 3.71 × 10−7 C.
4(1/4π ⑀0 )
(1/4π ⑀ 0 ) 2
(b) q2 = 4q1
F=
1
q1q2
4π ⑀0 r
2
=
4q12
4π ⑀0r
2
And then q2 = 4q1 = 1.48 × 10−6 C.
21.10.
EVALUATE: The force on one sphere is the same magnitude as the force on the other sphere, whether the
spheres have equal charges or not.
IDENTIFY: We first need to determine the number of charges in each hand. Then we can use Coulomb’s
law to find the force these charges would exert on each hand.
SET UP: One mole of Ca contains N A = 6.02 × 1023 atoms. Each proton has charge e = 1.60 × 10−19 C.
The force each hand exerts on the other is F = k
q2
.
r2
EXECUTE: (a) The mass of one hand is (0.010)(75 kg) = 0.75 kg = 750 g . The number of moles of Ca is
n=
750 g
= 18.7 mol. The number of atoms is
40.18 g/mol
N = nN A = (18.7 mol)(6.02 × 1023 atoms/mol) = 1.12 × 1025 atoms.
(b) Each Ca atom contains positive charge 20e. The total positive charge in each hand is
N e = (1.12 × 1025 )(20)(1.60 × 10−19 C) = 3.58 × 107 C. If 1.0% is unbalanced by negative charge, the net
positive charge of each hand is q = (0.010)(3.58 × 107 C) = 3.6 × 105 C.
21.11.
(c) The repulsive force each hand exerts on the other would be
(3.6 × 105 C) 2
q2
F = k 2 = (8.99 × 109 N ⋅ m 2 /C2 )
= 4.0 × 1020 N. This is an immense force; our hands
(1.7 m) 2
r
would fly off.
EVALUATE: Ordinary objects contain a very large amount of charge. But negative and positive charge is
present in almost equal amounts and the net charge of a charged object is always a very small fraction of
the total magnitude of charge that the object contains.
qq
IDENTIFY: Apply F = ma, with F = k 1 22 .
r
SET UP: a = 25.0 g = 245 m/s 2 . An electron has charge − e = −1.60 × 10−19 C.
EXECUTE: F = ma = (8.55 × 10−3 kg)(245 m/s 2 ) = 2.09 N. The spheres have equal charges q, so
F =k
q2
r
2
and q = r
2.09 N
F
= (0.150 m)
= 2.29 × 10−6 C.
k
8.99 × 109 N ⋅ m 2 /C2
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-4
Chapter 21
q
2.29 × 10−6 C
=
= 1.43 × 1013 electrons. The charges on the spheres have the same sign so the
e 1.60 × 10−19 C
electrical force is repulsive and the spheres accelerate away from each other.
EVALUATE: As the spheres move apart the repulsive force they exert on each other decreases and their
acceleration decreases.
IDENTIFY: We need to determine the number of protons in each box and then use Coulomb’s law to
calculate the force each box would exert on the other.
SET UP: The mass of a proton is 1.67 × 10−27 kg and the charge of a proton is 1.60 × 10−19 C. The
N=
21.12.
distance from the earth to the moon is 3.84 × 108 m. The electrical force has magnitude Fe = k
where k = 8.99 × 109 N ⋅ m 2 /C2 . The gravitational force has magnitude Fgrav = G
m1 m2
r2
q1 q2
r2
,
, where
G = 6.67 × 10−11 N ⋅ m 2 /kg 2 .
EXECUTE: (a) The number of protons in each box is N =
1.0 × 10−3 kg
1.67 × 10−27 kg
= 5.99 × 1023. The total charge
of each box is q = Ne = (5.99 × 1023 )(1.60 × 10−19 C) = 9.58 × 104 C. The electrical force on each box is
Fe = k
q2
(9.58 × 104 C) 2
r
(3.84 × 108 m) 2
= (8.99 × 109 N ⋅ m 2 /C2 )
2
= 560 N = 130 lb. The tension in the string must equal
this repulsive electrical force. The weight of the box on earth is w = mg = 9.8 × 10−3 N and the weight of
the box on the moon is even less, since g is less on the moon. The gravitational forces exerted on the boxes
by the earth and by the moon are much less than the electrical force and can be neglected.
mm
(1.0 × 10−3 kg) 2
(b) Fgrav = G 1 2 2 = (6.67 × 10−11 N ⋅ m 2 /kg 2 )
= 4.5 × 10−34 N.
8
2
r
(3.84 × 10 m)
21.13.
EVALUATE: Both the electrical force and the gravitational force are proportional to 1/r 2 . But in SI units
the coefficient k in the electrical force is much greater than the coefficient G in the gravitational force. And
a small mass of protons contains a large amount of charge. It would be impossible to put 1.0 g of protons
into a small box, because of the very large repulsive electrical forces the protons would exert on each other.
IDENTIFY: In a space satellite, the only force accelerating the free proton is the electrical repulsion of the
other proton.
SET UP: Coulomb’s law gives the force, and Newton’s second law gives the acceleration:
a = F/m = (1/4π ⑀0 )(e 2 /r 2 )/m.
EXECUTE: (a) a = (9.00 ×109 N ⋅ m 2 /C 2 )(1.60 ×10−19 C)2 /[(0.00250 m) 2 (1.67 ×10−27 kg)] = 2.21×104 m/s 2 .
(b) The graphs are sketched in Figure 21.13.
EVALUATE: The electrical force of a single stationary proton gives the moving proton an initial
acceleration about 20,000 times as great as the acceleration caused by the gravity of the entire earth. As the
protons move farther apart, the electrical force gets weaker, so the acceleration decreases. Since the protons
continue to repel, the velocity keeps increasing, but at a decreasing rate.
Figure 21.13
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21.14.
21-5
IDENTIFY: Apply Coulomb’s law.
SET UP: Like charges repel and unlike charges attract.
EXECUTE: (a) F =
1
q1q2
4π ⑀0 r 2
. This gives 0.200 N =
1
(0.550 × 10−6 C) q2
4π ⑀ 0
(0.30 m)2
and
q2 = +3.64 × 10−6 C. The force is attractive and q1 < 0, so q2 = +3.64 × 10−6 C.
(b) F = 0.200 N. The force is attractive, so is downward.
21.15.
EVALUATE: The forces between the two charges obey Newton’s third law.
IDENTIFY: Apply Coulomb’s law. The two forces on q3 must have equal magnitudes and opposite
directions.
SET UP: Like charges repel and unlike charges attract.
G
qq
EXECUTE: The force F2 that q2 exerts on q3 has magnitude F2 = k 2 2 3 and is in the +x-direction.
r2
G
q q
q q
F1 must be in the −x-direction, so q1 must be positive. F1 = F2 gives k 1 2 3 = k 2 2 3 .
r1
r2
2
2
⎛r ⎞
⎛ 2.00 cm ⎞
q1 = q2 ⎜ 1 ⎟ = ( 3.00 nC ) ⎜
⎟ = 0.750 nC.
⎝ 4.00 cm ⎠
⎝ r2 ⎠
EVALUATE: The result for the magnitude of q1 doesn’t depend on the magnitude of q2 .
21.16.
IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on Q.
SET UP: The force that q1 exerts on Q is repulsive, as in Example 21.4, but now the force that q2 exerts is
attractive.
EXECUTE: The x-components cancel. We only need the y-components, and each charge contributes
1 (2.0 × 10−6 C)(4.0 × 10−6 C)
sin α = −0.173 N (since sinα = 0.600).
equally. F1 y = F2 y = −
4π ⑀0
(0.500 m) 2
Therefore, the total force is 2 F = 0.35 N, in the − y -direction.
EVALUATE: If q1 is −2.0 μC and q2 is +2.0 μC, then the net force is in the + y -direction.
21.17.
IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on q1.
G
G
SET UP: Like charges repel and unlike charges attract, so F2 and F3 are both in the + x -direction.
EXECUTE: F2 = k
q1q2
r122
= 6.749 × 1025 N, F3 = k
q1q3
r132
= 1.124 × 10−4 N. F = F2 + F3 = 1.8 × 10−4 N.
F = 1.8 × 10−4 N and is in the + x -direction.
G
G
EVALUATE: Comparing our results to those in Example 21.3, we see that F1 on 3 = − F3 on 1, as required
21.18.
by Newton’s third law.
IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on q2 .
G
SET UP: F2 on 1 is in the + y -direction.
EXECUTE: F2 on 1 =
(9.0 × 109 N ⋅ m 2 /C2 )(2.0 × 1026 C)(2.0 × 1026 C)
(0.60 m) 2
= 0.100 N. ( F2 on 1) x = 0 and
( F2 on 1) y = +0.100 N. FQ on 1 is equal and opposite to F1 on Q (Example 21.4), so ( FQ on 1 ) x = −0.23 N
and ( FQ on 1 ) y = 0.17 N. Fx = ( F2 on 1) x + ( FQ on 1) x = −0.23 N.
Fy = ( F2 on 1 ) y + ( FQ on 1 ) y = 0.100 N + 0.17 N = 0.27 N. The magnitude of the total force is
G
0.23
= 40°, so F is 40° counterclockwise from the +y-axis,
F = (0.23 N) 2 + (0.27 N) 2 = 0.35 N. tan −1
0.27
or 130° counterclockwise from the +x- axis.
EVALUATE: Both forces on q1 are repulsive and are directed away from the charges that exert them.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-6
21.19.
Chapter 21
IDENTIFY and SET UP: Apply Coulomb’s law to calculate the force exerted by q2 and q3 on q1. Add
these forces as vectors to get the net force. The target variable is the x-coordinate of q3 .
G
EXECUTE: F2 is in the x-direction.
F2 = k
q1q2
= 3.37 N, so F2 x = +3.37 N
r122
Fx = F2 x + F3 x and Fx = −7.00 N
F3 x = Fx − F2 x = − 7.00 N − 3.37 N = − 10.37 N
For F3 x to be negative, q3 must be on the − x -axis.
F3 = k
q1q3
x
2
, so x =
k q1q3
F3
= 0.144 m, so x = −0.144 m
EVALUATE: q2 attracts q1 in the +x-direction so q3 must attract q1 in the −x-direction, and q3 is at
21.20.
negative x.
IDENTIFY: Apply Coulomb’s law.
G
SET UP: Like charges repel and unlike charges attract. Let F21 be the force that q2 exerts on q1 and let
G
F31 be the force that q3 exerts on q1.
EXECUTE: The charge q3 must be to the right of the origin; otherwise both q2 and q3 would exert forces
in the +x-direction. Calculating the two forces:
1 q1q2 (9 × 109 N ⋅ m 2 /C2 )(3.00 × 10−6 C)(5.00 × 10−6 C)
F21 =
=
= 3.375 N, in the +x-direction.
4π ⑀0 r122
(0.200 m) 2
F31 =
(9 × 109 N ⋅ m 2 /C2 )(3.00 × 10−6 C)(8.00 × 10−6 C)
r132
We need Fx = F21 − F31 = −7.00 N, so 3.375 N −
r13 =
=
0.216 N ⋅ m 2
0.216 N ⋅ m 2
r132
r132
, in the −x-direction.
= −7.00 N.
0.216 N ⋅ m 2
= 0.144 m. q3 is at x = 0.144 m.
3.375 N + 7.00 N
EVALUATE: F31 = 10.4 N. F31 is larger than F21, because q3 is larger than q2 and also because r13 is
less than r12 .
21.21.
IDENTIFY: Apply Coulomb’s law to calculate the force each of the two charges exerts on the third charge.
Add these forces as vectors.
SET UP: The three charges are placed as shown in Figure 21.21a.
Figure 21.21a
EXECUTE: Like charges repel and unlike attract, so the free-body diagram for q3 is as shown in
Figure 21.21b.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
F1 =
F2 =
21-7
1 q1q3
4π ⑀0 r132
1
q2q3
2
4π ⑀0 r23
Figure 21.21b
F1 = (8.988 × 109 N ⋅ m 2 /C2 )
(1.50 × 10−9 C)(5.00 × 10−9 C)
F2 = (8.988 × 109 N ⋅ m 2 /C2 )
= 1.685 × 10−6 N
(0.200 m) 2
(3.20 × 10−9 C)(5.00 × 10−9 C)
G G G
The resultant force is R = F1 + F2 .
(0.400 m) 2
= 8.988 × 10−7 N
Rx = 0.
R y = −( F1 + F2 ) = −(1.685 × 10−6 N + 8.988 × 10−7 N) = −2.58 × 10−6 N.
The resultant force has magnitude 2.58 × 10−6 N and is in the − y -direction.
EVALUATE: The force between q1 and q3 is attractive and the force between q2 and q3 is replusive.
21.22.
IDENTIFY: Apply F = k
qq′
r2
to each pair of charges. The net force is the vector sum of the forces due to
q1 and q2 .
SET UP: Like charges repel and unlike charges attract. The charges and their forces on q3 are shown in
Figure 21.22.
EXECUTE: F1 = k
F2 = k
q2q3
r22
q1q3
r12
= (8.99 × 109 N ⋅ m 2 /C2 )
= (8.99 × 109 N ⋅ m 2 /C2 )
(4.00 × 10−9 C)(6.00 × 10−9 C)
(0.200 m) 2
(5.00 × 10−9 C)(6.00 × 10−9 C)
(0.300 m) 2
= 5.394 × 10−6 N.
= 2.997 × 10−6 N.
Fx = F1x + F2 x = + F1 − F2 = 2.40 × 10−6 N. The net force has magnitude 2.40 × 10−6 N and is in the
+x-direction.
EVALUATE: Each force is attractive, but the forces are in opposite directions because of the placement of the
charges. Since the forces are in opposite directions, the net force is obtained by subtracting their magnitudes.
Figure 21.22
21.23.
IDENTIFY: We use Coulomb’s law to find each electrical force and combine these forces to find the net
force.
SET UP: In the O-H-N combination the O− is 0.170 nm from the H + and 0.280 nm from the N − . In the
N-H-N combination the N − is 0.190 nm from the H + and 0.300 nm from the other N − . Like charges
repel and unlike charges attract. The net force is the vector sum of the individual forces. The force due to
qq
e2
each pair of charges is F = k 1 2 2 = k 2 .
r
r
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-8
Chapter 21
EXECUTE: (a) F = k
q1 q2
r2
=k
e2
r2
.
O-H-N:
O − - H + : F = (8.99 × 109 N ⋅ m 2 /C2 )
O − - N − : F = (8.99 × 109 N ⋅ m 2 /C2 )
(1.60 × 10−19 C)2
(0.170 × 10−9 m) 2
(1.60 × 10−19 C) 2
(0.280 × 10
−9
m)
2
= 7.96 × 10−9 N, attractive
= 2.94 × 10−9 N, repulsive
N-H-N:
N − - H + : F = (8.99 × 109 N ⋅ m 2 /C2 )
N − - N − : F = (8.99 × 109 N ⋅ m 2 /C2 )
(1.60 × 10−19 C) 2
(0.190 × 10−9 m) 2
(1.60 × 10−19 C)2
(0.300 × 10−9 m)2
= 6.38 × 10−9 N, attractive
= 2.56 × 10−9 N, repulsive
The total attractive force is 1.43 × 10−8 N and the total repulsive force is 5.50 × 10−9 N. The net force is
attractive and has magnitude 1.43 × 10−8 N − 5.50 × 10−9 N = 8.80 × 10−9 N.
21.24.
e2
(1.60 × 10−19 C)2
= 8.22 × 10−8 N.
(0.0529 × 10−9 m) 2
r
EVALUATE: The bonding force of the electron in the hydrogen atom is a factor of 10 larger than the
bonding force of the adenine-thymine molecules.
IDENTIFY: We use Coulomb’s law to find each electrical force and combine these forces to find the net
force.
SET UP: In the O-H-O combination the O− is 0.180 nm from the H + and 0.290 nm from the other O− .
(b) F = k
2
= (8.99 × 109 N ⋅ m 2 /C2 )
In the N-H-N combination the N − is 0.190 nm from the H + and 0.300 nm from the other N − . In the
O-H-N combination the O − is 0.180 nm from the H + and 0.290 nm from the other N − . Like charges
repel and unlike charges attract. The net force is the vector sum of the individual forces. The force due to
qq
e2
each pair of charges is F = k 1 2 2 = k 2 .
r
r
q1 q2
e2
EXECUTE: Using F = k 2 = k 2 , we find that the attractive forces are: O− - H + , 7.10 × 10−9 N;
r
r
N − - H + , 6.37 × 10−9 N; O− - H + , 7.10 × 10−9 N. The total attractive force is 2.06 × 10−8 N. The repulsive
forces are: O − - O − , 2.74 × 10−9 N; N − - N − , 2.56 × 10−9 N; O − - N − , 2.74 × 10−9 N. The total repulsive
21.25.
force is 8.04 × 10−9 N. The net force is attractive and has magnitude 1.26 × 10−8 N.
EVALUATE: The net force is attractive, as it should be if the molecule is to stay together.
IDENTIFY: F = q E. Since the field is uniform, the force and acceleration are constant and we can use a
constant acceleration equation to find the final speed.
SET UP: A proton has charge +e and mass 1.67 × 10−27 kg.
EXECUTE: (a) F = (1.60 × 10−19 C)(2.75 × 103 N/C) = 4.40 × 10−16 N
(b) a =
F 4.40 × 10−16 N
=
= 2.63 × 1011 m/s 2
m 1.67 × 10−27 kg
(c) vx = v0 x + axt gives v = (2.63 × 1011 m/s 2 )(1.00 × 10−6 s) = 2.63 × 105 m/s
21.26.
EVALUATE: The acceleration is very large and the gravity force on the proton can be ignored.
q
IDENTIFY: For a point charge, E = k 2 .
r
G
SET UP: E is toward a negative charge and away from a positive charge.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-9
EXECUTE: (a) The field is toward the negative charge so is downward.
3.00 × 10−9 C
E = (8.99 × 109 N ⋅ m 2 /C2 )
= 432 N/C.
(0.250 m) 2
kq
(8.99 × 109 N ⋅ m 2 /C2 )(3.00 × 1029 C)
=
= 1.50 m
E
12.0 N/C
EVALUATE: At different points the electric field has different directions, but it is always directed toward
the negative point charge.
IDENTIFY: The acceleration that stops the charge is produced by the force that the electric field exerts on it.
Since the field and the acceleration are constant, we can use the standard kinematics formulas to find
acceleration and time.
(a) SET UP: First use kinematics to find the proton’s acceleration. vx = 0 when it stops. Then find the
electric field needed to cause this acceleration using the fact that F = qE.
(b) r =
21.27.
EXECUTE: vx2 = v02x + 2ax ( x − x0 ). 0 = (4.50 × 106 m/s) 2 + 2a (0.0320 m) and a = 3.16 × 1014 m/s 2 . Now
find the electric field, with q = e. eE = ma and
E = ma/e = (1.67 × 10−27 kg)(3.16 × 1014 m/s 2 )/(1.60 × 10−19 C) = 3.30 × 106 N/C, to the left.
(b) SET UP: Kinematics gives v = v0 + at , and v = 0 when the electron stops, so t = v0 /a.
EXECUTE: t = v0 /a = (4.50 × 106 m/s)/(3.16 × 1014 m/s 2 ) = 1.42 × 10−8 s = 14.2 ns
(c) SET UP: In part (a) we saw that the electric field is proportional to m, so we can use the ratio of the
electric fields. Ee /Ep = me /mp and Ee = ( me /mp ) Ep .
EXECUTE: Ee = [(9.11 ×10 −31 kg)/(1.67 ×10 −27 kg)](3.30 ×106 N/C) = 1.80 ×103 N/C, to the right
21.28.
EVALUATE: Even a modest electric field, such as the ones in this situation, can produce enormous
accelerations for electrons and protons.
IDENTIFY: Use constant acceleration equations to calculate the upward acceleration a and then apply
G
G
F = qE to calculate the electric field.
SET UP: Let +y be upward. An electron has charge q = − e.
EXECUTE: (a) v0 y = 0 and a y = a, so y − y0 = v0 yt + 12 a yt 2 gives y − y0 = 12 at 2 . Then
a=
E=
2( y − y0 )
t
2
=
2(4.50 m)
(3.00 × 10−6 s) 2
= 1.00 × 1012 m/s 2 .
F ma (9.11 × 10−31 kg)(1.00 × 1012 m/s 2 )
=
=
= 5.69 N/C
q
q
1.60 × 10−19 C
The force is up, so the electric field must be downward since the electron has negative charge.
(b) The electron’s acceleration is ~ 1011 g , so gravity must be negligibly small compared to the electrical force.
21.29.
EVALUATE: Since the electric field is uniform, the force it exerts is constant and the electron moves with
constant acceleration.
(a) IDENTIFY: Eq. (21.4) relates the electric field, charge of the particle, and the force on the particle. If
the particle is to remain stationary the net force on it must be zero.
SET UP: The free-body diagram for the particle is sketched in Figure 21.29. The weight is mg, downward. For
the net force to be zero the force exerted by the electric field must be upward. The electric field is downward.
Since the electric field and the electric force are in opposite directions the charge of the particle is negative.
mg = q E
Figure 21.29
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-10
Chapter 21
mg (1.45 × 1023 kg)(9.80 m/s 2 )
=
= 2.19 × 10−5 C and q = −21.9 μC
E
650 N/C
(b) SET UP: The electrical force has magnitude FE = q E = eE . The weight of a proton is w = mg .
EXECUTE:
q =
FE = w so eE = mg
mg (1.673 × 10−27 kg)(9.80 m/s 2 )
=
= 1.02 × 10−7 N/C.
e
1.602 × 10−19 C
This is a very small electric field.
EVALUATE: In both cases q E = mg and E = ( m/ q ) g . In part (b) the m/ q ratio is much smaller
EXECUTE: E =
21.30.
(∼ 10−8 ) than in part (a) (∼ 102 ) so E is much smaller in (b). For subatomic particles gravity can usually
be ignored compared to electric forces.
IDENTIFY: The net electric field is the vector sum of the individual fields.
SET UP: The distance from a corner to the center of the square is r = ( a/2)2 + (a/2) 2 = a/ 2 . The
magnitude of the electric field due to each charge is the same and equal to Eq =
kq
r
2
=2
kq
a2
. All four
y-components add and the x-components cancel.
EXECUTE: Each y-component is equal to Eqy = − Eq cos 45° = −
is
21.31.
4 2kq
a2
Eq
2
=
−2kq
2a
2
=−
2kq
a2
. The resultant field
, in the − y -direction.
EVALUATE: We must add the y-components of the fields, not their magnitudes.
q
IDENTIFY: For a point charge, E = k 2 . The net field is the vector sum of the fields produced by each
r
G
G
G
charge. A charge q in an electric field E experiences a force F = qE .
SET UP: The electric field of a negative charge is directed toward the charge. Point A is 0.100 m from q2
and 0.150 m from q1. Point B is 0.100 m from q1 and 0.350 m from q2 .
EXECUTE: (a) The electric fields at point A due to the charges are shown in Figure 21.31a.
q
6.25 × 10−9 C
= 2.50 × 103 N/C
E1 = k 21 = (8.99 × 109 N ⋅ m 2 /C2 )
(0.150 m) 2
rA1
q2
rA22
q2
= (8.99 × 109 N ⋅ m 2 /C2 )
12.5 × 10−9 C
= 1.124 × 104 N/C
(0.100 m) 2
Since the two fields are in opposite directions, we subtract their magnitudes to find the net field.
E = E2 − E1 = 8.74 × 103 N/C, to the right.
(b) The electric fields at point B are shown in Figure 21.31b.
q
6.25 × 10−9 C
= 5.619 × 103 N/C
E1 = k 21 = (8.99 × 109 N ⋅ m 2 /C2 )
(0.100 m)2
rB1
E2 = k
= (8.99 × 109 N ⋅ m 2 /C2 )
12.5 × 10−9 C
= 9.17 × 102 N/C
rB22
(0.350 m)2
Since the fields are in the same direction, we add their magnitudes to find the net field.
E = E1 + E2 = 6.54 × 103 N/C, to the right.
E2 = k
(c) At A, E = 8.74 × 103 N/C, to the right. The force on a proton placed at this point would be
F = qE = (1.60 × 10−19 C)(8.74 × 103 N/C) = 1.40 × 10−15 N, to the right.
EVALUATE: A proton has positive charge so the force that an electric field exerts on it is in the same
direction as the field.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-11
Figure 21.31
21.32.
G
G
IDENTIFY: The electric force is F = qE .
SET UP: The gravity force (weight) has magnitude w = mg and is downward.
EXECUTE: (a) To balance the weight the electric force must be upward. The electric field is downward,
so for an upward force the charge q of the person must be negative. w = F gives mg = q E and
mg (60 kg)(9.80 m/s 2 )
=
= 3.9 C.
E
150 N/C
qq ′
(3.9 C) 2
(b) F = k 2 = (8.99 × 109 N ⋅ m 2 /C2 )
= 1.4 × 107 N. The repulsive force is immense and this is
r
(100 m)2
not a feasible means of flight.
EVALUATE: The net charge of charged objects is typically much less than 1 C.
IDENTIFY: Eq. (21.3) gives the force on the particle in terms of its charge and the electric field between
the plates. The force is constant and produces a constant acceleration. The motion is similar to projectile
motion; use constant acceleration equations for the horizontal and vertical components of the motion.
(a) SET UP: The motion is sketched in Figure 21.33a.
q =
21.33.
For an electron q = − e.
Figure 21.33a
G
G
G
G
G
F = qE and q negative gives that F and E are in opposite directions, so F is upward. The free-body
diagram for the electron is given in Figure 21.33b.
EXECUTE: ∑ Fy = ma y
eE = ma
Figure 21.33b
Solve the kinematics to find the acceleration of the electron: Just misses upper plate says that
x − x0 = 2.00 cm when y − y0 = +0.500 cm.
x-component
v0 x = v0 = 1.60 × 106 m/s, a x = 0, x − x0 = 0.0200 m, t = ?
x − x0 = v0 xt + 12 axt 2
x − x0
0.0200 m
=
= 1.25 × 10−8 s
v0 x
1.60 × 106 m/s
In this same time t the electron travels 0.0050 m vertically:
t=
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-12
Chapter 21
y-component
t = 1.25 × 10−8 s, v0 y = 0, y − y0 = +0.0050 m, a y = ?
y − y0 = v0 yt + 12 a yt 2
2(0.0050 m)
=
= 6.40 × 1013 m/s 2
t2
(1.25 × 10−8 s)2
(This analysis is very similar to that used in Chapter 3 for projectile motion, except that here the acceleration
is upward rather than downward.) This acceleration must be produced by the electric-field force: eE = ma
ay =
2( y − y0 )
ma (9.109 × 10−31 kg)(6.40 × 1013 m/s 2 )
=
= 364 N/C
e
1.602 × 10−19 C
Note that the acceleration produced by the electric field is much larger than g, the acceleration produced by
gravity, so it is perfectly ok to neglect the gravity force on the elctron in this problem.
eE (1.602 × 10−19 C)(364 N/C)
(b) a =
=
= 3.49 × 1010 m/s 2
mp
1.673 × 10−27 kg
E=
This is much less than the acceleration of the electron in part (a) so the vertical deflection is less and the
proton won’t hit the plates. The proton has the same initial speed, so the proton takes the same time
t = 1.25 × 10−8 s to travel horizontally the length of the plates. The force on the proton is downward (in the
G
same direction as E , since q is positive), so the acceleration is downward and a y = −3.49 × 1010 m/s 2 .
y − y0 = v0 yt + 12 a yt 2 = 12 (−3.49 × 1010 m/s 2 )(1.25 × 10−8 s)2 = −2.73 × 10−6 m. The displacement is
21.34.
2.73 × 10−6 m, downward.
(c) EVALUATE: The displacements are in opposite directions because the electron has negative charge and
the proton has positive charge. The electron and proton have the same magnitude of charge, so the force
the electric field exerts has the same magnitude for each charge. But the proton has a mass larger by a
factor of 1836 so its acceleration and its vertical displacement are smaller by this factor.
(d) In each case a g and it is reasonable to ignore the effects of gravity.
IDENTIFY: Apply Eq. (21.7) to calculate the electric field due to each charge and add the two field vectors
to find the resultant field.
G
SET UP: For q1 , rˆ = ˆj. For q2 , rˆ = cos θ iˆ + sin θ ˆj , where θ is the angle between E2 and the +x-axis.
G
EXECUTE: (a) E1 =
G
E2 =
21.35.
q2
=
2
4π ⑀0r2
9
2 2
−9
ˆj = (9.0 × 10 N ⋅ m /C )( −5.00 × 10 C) ˆj = (−2.813 × 104 N/C) ˆj.
4π ⑀0r12
(0.0400 m) 2
q1
(9.0 × 109 N ⋅ m 2 /C2 )(3.00 × 10−9 C)
2
(0.0300 m) + (0.0400 m)
2
G
= 1.080 × 104 N/C. The angle of E2 , measured from
⎛ 4.00 cm ⎞
the x -axis, is 180° − tan −1 ⎜
⎟ = 126.9° Thus
⎝ 3.00 cm ⎠
G
E2 = (1.080 × 104 N/C)(iˆ cos126.9° + ˆj sin126.9°) = (−6.485 × 103 N/C) iˆ + (8.64 × 103 N/C) ˆj
G
G
(b) The resultant field is E1 + E 2 = (−6.485 × 103 N/C) iˆ + (−2.813 × 104 N/C + 8.64 × 103 N/C) ˆj.
G
G
E1 + E2 = (−6.485 × 103 N/C) iˆ − (1.95 × 104 N/C ) ˆj.
G
G
EVALUATE: E1 is toward q1 since q1 is negative. E2 is directed away from q2 , since q2 is positive.
IDENTIFY: Apply constant acceleration equations to the motion of the electron.
SET UP: Let +x be to the right and let + y be downward. The electron moves 2.00 cm to the right and
0.50 cm downward.
EXECUTE: Use the horizontal motion to find the time when the electron emerges from the field.
x − x0 = 0.0200 m, ax = 0, v0 x = 1.60 × 106 m/s. x − x0 = v0 xt + 12 axt 2 gives t = 1.25 × 10−8 s. Since
⎛ v0 y + v y ⎞
ax = 0, vx = 1.60 × 106 m/s. y − y0 = 0.0050 m, v0y = 0, t = 1.25 × 10−8 s. y − y0 = ⎜
⎟ t gives
2
⎝
⎠
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-13
v y = 8.00 × 105 m/s. Then v = vx2 + v 2y = 1.79 × 106 m/s.
EVALUATE: v y = v0 y + a yt gives a y = 6.4 × 1013 m/s 2 . The electric field between the plates is
E=
21.36.
ma y
(9.11 × 10−31 kg)(6.4 × 1013 m/s 2 )
= 364 V/m. This is not a very large field.
1.60 × 10−19 C
G
G
IDENTIFY: Use the components of E from Example 21.6 to calculate the magnitude and direction of E .
G
G
Use F = qE to calculate the force on the −2.5 nC charge and use Newton’s third law for the force on the
−8.0 nC charge.
G
SET UP: From Example 21.6, E = ( −11 N/C) iˆ + (14 N/C) ˆj.
e
=
EXECUTE: (a) E = E x2 + E y2 = (−11 N/C) 2 + (14 N/C) 2 = 17.8 N/C.
⎛ Ey ⎞
⎟ = tan −1 (14/11) = 51.8°, so θ = 128° counterclockwise from the +x-axis.
tan −1 ⎜
⎜ Ex ⎟
⎝
⎠
G G
(b) (i) F = Eq so F = (17.8 N/C)( 2.5 × 1029 C) = 4.45 × 10−8 N, at 52° below the +x-axis.
21.37.
(ii) 4.45 × 10−8 N at 128° counterclockwise from the +x-axis.
EVALUATE: The forces in part (b) are repulsive so they are along the line connecting the two charges and
in each case the force is directed away from the charge that exerts it.
IDENTIFY: The forces the charges exert on each other are given by Coulomb’s law. The net force on the
proton is the vector sum of the forces due to the electrons.
SET UP: qe = −1.60 × 10−19 C. qp = +1.60 × 10−19 C. The net force is the vector sum of the forces exerted
by each electron. Each force has magnitude F = k
q1q2
r2
=k
e2
r2
and is attractive so is directed toward the
electron that exerts it.
EXECUTE: Each force has magnitude
qq
e2 (8.988 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C)2
F1 = F2 = k 1 2 2 = k 2 =
= 1.023 × 10−8 N. The vector force
r
r
(1.50 × 10−10 m)2
diagram is shown in Figure 21.37.
Figure 21. 37
Taking components, we get F1x = 1.023 × 10−8 N; F1 y = 0. F2 x = F2 cos65.0° = 4.32 × 10−9 N;
F2 y = F2 sin 65.0° = 9.27 × 10−9 N. Fx = F1x + F2 x = 1.46 × 10−8 N; Fy = F1 y + F2 y = 9.27 × 10−9 N.
F = Fx2 + Fy2 = 1.73 × 10−8 N. tan θ =
Fy
Fx
=
9.27 × 10−9 N
1.46 × 10−8 N
= 0.6349 which gives
θ = 32.4°. The net force is 1.73 × 10−8 N and is directed toward a point midway between the two electrons.
EVALUATE: Note that the net force is less than the algebraic sum of the individual forces.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-14
21.38.
Chapter 21
IDENTIFY: Apply constant acceleration equations to the motion of the proton. E = F/ q .
SET UP: A proton has mass mp = 1.67 × 10−27 kg and charge + e. Let +x be in the direction of motion of
the proton.
EXECUTE: (a) v0 x = 0. a =
E=
2(0.0160 m)(1.67 × 10−27 kg)
(1.60 × 10−19 C)(1.50 × 10−6 s) 2
(b) vx = v0 x + axt =
21.39.
eE
1
1 eE 2
t . Solving for E gives
. x − x0 = v0 xt + 12 axt 2 gives x − x0 = a xt 2 =
mp
2
2 mp
= 148 N/C.
eE
t = 2.13 × 104 m/s.
mp
EVALUATE: The electric field is directed from the positively charged plate toward the negatively charged
plate and the force on the proton is also in this direction.
IDENTIFY: Find the angle θ that r̂ makes with the +x-axis. Then rˆ = (cos θ ) iˆ + (sin θ ) ˆj. .
SET UP: tan θ = y/x
π
⎛ −1.35 ⎞
ˆ
EXECUTE: (a) tan −1 ⎜
⎟ = − rad. rˆ = − j.
2
⎝ 0 ⎠
2ˆ
2ˆ
⎛ 12 ⎞ π
(b) tan −1 ⎜ ⎟ = rad. rˆ =
i+
j.
2
2
⎝ 12 ⎠ 4
21.40.
⎛ 2.6 ⎞
ˆ
ˆ
(c) tan −1 ⎜
⎟ = 1.97 rad = 112.9°. rˆ = −0.39i + 0.92 j (Second quadrant).
⎝ +1.10 ⎠
EVALUATE: In each case we can verify that r̂ is a unit vector, because rˆ ⋅ rˆ = 1.
IDENTIFY: The net force on each charge must be zero.
SET UP: The force diagram for the −6.50 μC charge is given in Figure 21.40. FE is the force exerted on
the charge by the uniform electric field. The charge is negative and the field is to the right, so the force
exerted by the field is to the left. Fq is the force exerted by the other point charge. The two charges have
opposite signs, so the force is attractive. Take the +x axis to be to the right, as shown in the figure.
EXECUTE: (a) FE = q E = (6.50 × 10−6 C)(1.85 × 108 N/C) = 1.20 × 103 N
Fq = k
q1q2
r2
= (8.99 × 109 N ⋅ m 2 /C2 )
(6.50 × 1026 C)(8.75 × 1026 C)
(0.0250 m)2
= 8.18 × 102 N
∑ Fx = 0 gives T + Fq − FE = 0 and T = FE − Fq = 382 N.
(b) Now Fq is to the left, since like charges repel.
∑ Fx = 0 gives T − Fq − FE = 0 and T = FE + Fq = 2.02 × 103 N.
EVALUATE: The tension is much larger when both charges have the same sign, so the force one charge
exerts on the other is repulsive.
Figure 21.40
21.41.
G
G G
G
G
IDENTIFY and SET UP: Use E in Eq. (21.3) to calculate F , F = ma to calculate a , and a constant
acceleration equation to calculate the final velocity. Let +x be east.
(a) EXECUTE: Fx = q E = (1.602 × 10−19 C)(1.50 N/C) = 2.403 × 10−19 N
a x = Fx /m = (2.403 × 10−19 N)/(9.109 × 10−31 kg) = +2.638 × 1011 m/s 2
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-15
v0 x = +4.50 × 105 m/s, a x = +2.638 × 1011 m/s 2 , x − x0 = 0.375 m, vx = ?
vx2 = v02x + 2a x ( x − x0 ) gives vx = 6.33 × 105 m/s
G
G
EVALUATE: E is west and q is negative, so F is east and the electron speeds up.
(b) EXECUTE: Fx = − q E = −(1.602 × 10−19 C)(1.50 N/C) = −2.403 × 10−19 N
a x = Fx /m = (−2.403 × 10−19 N)/(1.673 × 10−27 kg) = −1.436 × 108 m/s 2
v0 x = +1.90 × 104 m/s, ax = 21.436 × 108 m/s 2 , x − x0 = 0.375 m, vx = ?
vx2 = v02x + 2a x ( x − x0 ) gives vx = 1.59 × 104 m/s
G
EVALUATE: q > 0 so F is west and the proton slows down.
21.42.
IDENTIFY: The net electric field is the vector sum of the fields due to the individual charges.
SET UP: The electric field points toward negative charge and away from positive charge.
Figure 21.42
G
G
G
EXECUTE: (a) Figure 21.42a shows EQ and E1q at point P. EQ must have the direction shown, to
G
produce a resultant field in the specified direction. EQ is toward Q, so Q is negative. In order for the
horizontal components of the two fields to cancel, Q and q must have the same magnitude.
(b) No. If the lower charge were negative, its field would be in the direction shown in Figure 21.42b. The
G
G
two possible directions for the field of the upper charge, when it is positive ( E+ ) or negative ( E− ), are
21.43.
shown. In neither case is the resultant field in the direction shown in the figure in the problem.
EVALUATE: When combining electric fields, it is always essential to pay attention to their directions.
IDENTIFY: Calculate the electric field due to each charge and find the vector sum of these two fields.
SET UP: At points on the x-axis only the x component of each field is nonzero. The electric field of a
point charge points away from the charge if it is positive and toward it if it is negative.
EXECUTE: (a) Halfway between the two charges, E = 0.
(b) For x < a, E x =
For x > a, E x =
⎛
⎞
q
q
4q
ax
−
.
⎜⎜
⎟=−
2
2
4π ⑀0 ⎝ (a + x)
4π ⑀0 ( x 2 − a 2 )2
(a − x) ⎟⎠
1
⎛
⎞
q
q
2q x 2 + a 2
+
=
.
⎜⎜
⎟
4π ⑀0 ⎝ ( a + x)2 (a − x)2 ⎟⎠ 4π ⑀0 ( x 2 − a 2 ) 2
For x < − a, E x =
1
⎞
−1 ⎛
q
q
2q x 2 + a 2
+
=
−
.
⎜⎜
⎟
4π ⑀0 ⎝ (a + x) 2 (a − x) 2 ⎟⎠
4π ⑀0 ( x 2 − a 2 ) 2
The graph of E x versus x is sketched in Figure 21.43.
EVALUATE: The magnitude of the field approaches infinity at the location of one of the point charges.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-16
Chapter 21
Figure 21.43
21.44.
IDENTIFY: Add the individual electric fields to obtain the net field.
SET UP: The electric field points away from positive charge and toward negative charge.
Figure 21.44
G
G
G
EXECUTE: (a) The electric fields E1 and E2 and their vector sum, the net field E , are shown for each
21.45.
point in Figure 21.44a. The electric field is toward A at points B and C and the field is zero at A.
(b) The electric fields are shown in Figure 21.44b. The electric field is away from A at B and C. The field
is zero at A.
(c) The electric fields are shown in Figure 21.44c. The field is horizontal and to the right at points A, B and C.
EVALUATE: Compare your results to the field lines shown in Figure 21.28a and b in the textbook.
IDENTIFY: Eq. (21.7) gives the electric field of each point charge. Use the principle of superposition and
add the electric field vectors. In part (b) use Eq. (21.3) to calculate the force, using the electric field
calculated in part (a).
(a) SET UP: The placement of charges is sketched in Figure 21.45a.
\
Figure 21.45a
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-17
The electric field of a point charge is directed away from the point charge if the charge is positive and
1 q
toward the point charge if the charge is negative. The magnitude of the electric field is E =
, where
4π ⑀0 r 2
r is the distance between the point where the field is calculated and the point charge.
G
G
(i) At point a the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45b.
Figure 21.45b
EXECUTE: E1 =
E2 =
q1
2.00 × 10−9 C
4π ⑀0 r1
(0.200 m) 2
1
= (8.988 × 109 N ⋅ m 2 /C2 )
2
q2
5.00 × 10−9 C
4π ⑀0 r2
(0.600 m) 2
1
= (8.988 × 109 N ⋅ m 2 /C 2 )
2
= 449.4 N/C
= 124.8 N/C
E1x = 449.4 N/C, E1 y = 0
E2 x = 124.8 N/C, E2 y = 0
E x = E1x + E2 x = +449.4 N/C + 124.8 N/C = +574.2 N/C
E y = E1 y + E2 y = 0
The resultant field at point a has magnitude 574 N/C and is in the +x-direction.
G
G
(ii) SET UP: At point b the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45c.
Figure 21.45c
EXECUTE: E1 =
E2 =
1
q1
= (8.988 × 109 N ⋅ m 2 /C2 )
2
2.00 × 10−9 C
4π ⑀0 r1
q2
5.00 × 10−9 C
4π ⑀0 r2
(0.400 m) 2
1
= (8.988 × 109 N ⋅ m 2 /C2 )
2
(1.20 m) 2
= 12.5 N/C
= 280.9 N/C
E1x = 12.5 N/C, E1 y = 0
E2 x = −280.9 N/C, E2 y = 0
E x = E1x + E2 x = +12.5 N/C − 280.9 N/C = −268.4 N/C
E y = E1 y + E2 y = 0
The resultant field at point b has magnitude 268 N/C and is in the − x -direction.
G
G
(iii) SET UP: At point c the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45d.
Figure 21.45d
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-18
Chapter 21
EXECUTE: E1 =
E2 =
1
q2
4π ⑀0 r22
q1
2.00 × 10−9 C
4π ⑀0 r1
(0.200 m) 2
1
= (8.988 × 109 N ⋅ m 2 /C2 )
2
= (8.988 × 109 N ⋅ m 2 /C2 )
5.00 × 10−9 C
(1.00 m) 2
= 449.4 N/C
= 44.9 N/C
E1x = −449.4 N/C, E1 y = 0
E2 x = +44.9 N/C, E2 y = 0
E x = E1x + E2 x = −449.4 N/C + 44.9 N/C = −404.5 N/C
E y = E1 y + E2 y = 0
The resultant field at point b has magnitude 404 N/C and is in the − x -direction.
G
(b) SET UP: Since we have calculated E at each point the simplest way to get the force is to use
G
G
F = − eE .
EXECUTE: (i) F = (1.602 × 10−19 C)(574.2 N/C) = 9.20 × 10−17 N, − x -direction
(ii) F = (1.602 × 10 −19 C)(268.4 N/C) = 4.30 × 10−17 N, + x -direction
(iii) F = (1.602 × 10−19 C)(404.5 N/C) = 6.48 × 10−17 N, + x -direction
21.46.
EVALUATE: The general rule for electric field direction is away from positive charge and toward negative
charge. Whether the field is in the +x- or −x-direction depends on where the field point is relative to the
charge that produces the field. In part (a), for (i) the field magnitudes were added because the fields were in
the same direction and in (ii) and (iii) the field magnitudes were subtracted because the two fields were in
opposite directions. In part (b) we could have used Coulomb’s law to find the forces on the electron due to
the two charges and then added these force vectors, but using the resultant electric field is much easier.
IDENTIFY: Apply Eq. (21.7) to calculate the field due to each charge and then require that the vector sum
of the two fields to be zero.
SET UP: The field of each charge is directed toward the charge if it is negative and away from the charge
if it is positive.
EXECUTE: The point where the two fields cancel each other will have to be closer to the negative charge,
because it is smaller. Also, it can’t be between the two charges, since the two fields would then act in the
same direction. We could use Coulomb’s law to calculate the actual values, but a simpler way is to note
that the 8.00 nC charge is twice as large as the −4.00 nC charge. The zero point will therefore have to be a
factor of
2 farther from the 8.00 nC charge for the two fields to have equal magnitude. Calling x the
distance from the –4.00 nC charge: 1.20 + x = 2 x and x = 2.90 m.
21.47.
EVALUATE: This point is 4.10 m from the 8.00 nC charge. The two fields at this point are in opposite
directions and have equal magnitudes.
q
IDENTIFY: E = k 2 . The net field is the vector sum of the fields due to each charge.
r
SET UP: The electric field of a negative charge is directed toward the charge. Label the charges q1, q2
and q3 , as shown in Figure 21.47a. This figure also shows additional distances and angles. The electric
fields at point P are shown in Figure 21.47b. This figure also shows the xy coordinates we will use and the
G G
G
x and y components of the fields E1 , E2 and E3.
EXECUTE: E1 = E3 = (8.99 × 109 N ⋅ m 2 /C2 )
E2 = (8.99 × 109 N ⋅ m 2 /C2 )
2.00 × 10−6 C
(0.0600 m) 2
5.00 × 10−6 C
(0.100 m) 2
= 4.49 × 106 N/C
= 4.99 × 106 N/C
E y = E1 y + E2 y + E3 y = 0 and E x = E1x + E2 x + E3 x = E2 + 2 E1 cos53.1° = 1.04 × 107 N/C
E = 1.04 × 107 N/C, toward the −2.00 μC charge.
EVALUATE: The x-components of the fields of all three charges are in the same direction.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-19
Figure 21.47
21.48.
IDENTIFY: We can model a segment of the axon as a point charge.
SET UP: If the axon segment is modeled as a point charge, its electric field is E = k
q
r2
. The electric field
of a point charge is directed away from the charge if it is positive.
EXECUTE: (a) 5.6 × 1011 Na + ions enter per meter so in a 0.10 mm = 1.0 × 10−4 m section, 5.6 × 107 Na + ions
enter. This number of ions has charge q = (5.6 × 107 )(1.60 × 10−19 C) = 9.0 × 10−12 C.
(b) E = k
(c) r =
21.49.
q
9.0 × 10−12 C
r
(5.00 × 10−2 m) 2
= (8.99 × 109 N ⋅ m 2 /C2 )
2
= 32 N/C, directed away from the axon.
kq
(8.99 × 109 N ⋅ m 2 /C2 )(9.0 × 10−12 C)
=
= 280 m.
E
1.0 × 10−6 N/C
EVALUATE: The field in (b) is considerably smaller than ordinary laboratory electric fields.
IDENTIFY: The electric field of a positive charge is directed radially outward from the charge and has
1 q
magnitude E =
. The resultant electric field is the vector sum of the fields of the individual charges.
4π ⑀0 r 2
SET UP: The placement of the charges is shown in Figure 21.49a.
Figure 21.49a
EXECUTE: (a) The directions of the two fields are shown in Figure 21.49b.
E1 = E2 =
1 q
with r = 0.150 m.
4π ⑀0 r 2
E = E2 − E1 = 0; Ex = 0, E y = 0
Figure 21. 49b
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-20
Chapter 21
(b) The two fields have the directions shown in Figure 21.49c.
E = E1 + E2 , in the + x-direction
Figure 21. 49c
E1 =
E2 =
1
q1
4π ⑀0 r12
= (8.988 × 109 N ⋅ m 2 /C2 )
6.00 × 10−9 C
(0.150 m) 2
q2
6.00 × 10−9 C
4π ⑀0 r2
(0.450 m) 2
1
= (8.988 × 109 N ⋅ m 2 /C2 )
2
= 2396.8 N/C
= 266.3 N/C
E = E1 + E2 = 2396.8 N/C + 266.3 N/C = 2660 N/C; E x = +2660 N/C, E y = 0
(c) The two fields have the directions shown in Figure 21.49d.
sin θ =
0.400 m
= 0.800
0.500 m
cos θ =
0.300 m
= 0.600
0.500 m
Figure 21. 49d
E1 =
1 q1
4π ⑀0 r12
E1 = (8.988 × 109 N ⋅ m 2 /C2 )
E2 =
6.00 × 10−9 C
(0.400 m) 2
= 337.1 N/C
1 q2
4π ⑀0 r22
E2 = (8.988 × 109 N ⋅ m 2 /C2 )
6.00 × 10−9 C
(0.500 m)2
= 215.7 N/C
E1x = 0, E1 y = − E1 = −337.1 N/C
E2 x = + E2 cosθ = + (215.7 N/C)(0.600) = +129.4 N/C
E2 y = − E2 sin θ = −(215.7 N/C)(0.800) = −172.6 N/C
E x = E1x + E2 x = +129 N/C
E y = E1 y + E2 y = −337.1 N/C − 172.6 N/C = −510 N/C
E = E x2 + E y2 = (129 N/C) 2 + (−510 N/C)2 = 526 N/C
G
E and its components are shown in Figure 21.49e.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
tan α =
21-21
Ey
Ex
−510 N/C
= −3.953
+129 N/C
α = 284°, counterclockwise from + x-axis
tan α =
Figure 21. 49e
(d) The two fields have the directions shown in Figure 21.49f.
sin θ =
0.200 m
= 0.800
0.250 m
Figure 21. 49f
The components of the two fields are shown in Figure 21.49g.
E1 = E2 =
1
q
4π ⑀0 r 2
E1 = (8.988 × 109 N ⋅ m 2 /C2 )
6.00 × 10−9 C
(0.250 m) 2
E1 = E2 = 862.8 N/C
Figure 21. 49g
E1x = − E1 cosθ , E2 x = + E2 cosθ
E x = E1x + E2 x = 0
E1 y = + E1 sin θ , E2 y = + E2 sin θ
E y = E1 y + E2 y = 2 E1 y = 2 E1 sin θ = 2(862.8 N/C)(0.800) = 1380 N/C
E = 1380 N/C, in the + y -direction.
EVALUATE: Point a is symmetrically placed between identical charges, so symmetry tells us the electric
field must be zero. Point b is to the right of both charges and both electric fields are in the +x-direction and
the resultant field is in this direction. At point c both fields have a downward component and the field of
G
q2 has a component to the right, so the net E is in the 4th quadrant. At point d both fields have an upward
21.50.
component but by symmetry they have equal and opposite x-components so the net field is in the
+y-direction. We can use this sort of reasoning to deduce the general direction of the net field before doing
any calculations.
IDENTIFY: Apply Eq. (21.7) to calculate the field due to each charge and then calculate the vector sum of
those fields.
SET UP: The fields due to q1 and to q2 are sketched in Figure 21.50.
EXECUTE:
G
E2 =
1 (6.00 × 10−9 C)
4π ⑀0
(0.6 m) 2
(− iˆ) = −150iˆ N/C.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-22
Chapter 21
⎛
⎞
1
1
1
(4.00 × 10−9 C) ⎜⎜
(0.600) iˆ +
(0.800) ˆj ⎟⎟ = (21.6 iˆ + 28.8 ˆj )N/C.
2
2
4π ⑀0
(1.00 m)
⎝ (1.00 m)
⎠
G G
G
E = E1 + E 2 = ( −128.4 N/C) iˆ + (28.8 N/C) ˆj. E = (128.4 N/C) 2 + (28.8 N/C)2 = 131.6 N/C at
G
E1 =
⎛ 28.8 ⎞
⎟ = 12.6° above the − x -axis and therefore 167.4° counterclockwise from the +x-axis.
⎝ 128.4 ⎠
G
G
EVALUATE: E1 is directed toward q1 because q1 is negative and E2 is directed away from q2 because
θ = tan −1 ⎜
q2 is positive.
Figure 21.50
21.51.
G
G
IDENTIFY: The resultant electric field is the vector sum of the field E1 of q1 and E2 of q2 .
SET UP: The placement of the charges is shown in Figure 21.51a.
Figure 21.51a
EXECUTE: (a) The directions of the two fields are shown in Figure 21.51b.
E1 = E2 =
1 q1
4π ⑀0 r12
E1 = (8.988 × 109 N ⋅ m 2 /C2 )
6.00 × 10−9 C
(0.150 m) 2
E1 = E2 = 2397 N/C
Figure 21. 51b
E1x = −2397 N/C, E1 y = 0 E2 x = −2397 N/C, E2 y = 0
E x = E1x + E2 x = 2 ( −2397 N/C) = −4790 N/C
E y = E1 y + E2 y = 0
The resultant electric field at point a in the sketch has magnitude 4790 N/C and is in the − x -direction.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-23
(b) The directions of the two fields are shown in Figure 21.51c.
Figure 21.51c
E1 =
E2 =
1
q1
4π ⑀0 r12
= (8.988 × 109 N ⋅ m 2 /C2 )
6.00 × 10−9 C
(0.150 m)2
q2
6.00 × 10−9 C
4π ⑀0 r2
(0.450 m) 2
1
= (8.988 × 109 N ⋅ m 2 /C2 )
2
= 2397 N/C
= 266 N/C
E1x = +2397 N/C, E1 y = 0 E2 x = −266 N/C, E2 y = 0
E x = E1x + E2 x = +2397 N/C − 266 N/C = +2130 N/C
E y = E1 y + E2 y = 0
The resultant electric field at point b in the sketch has magnitude 2130 N/C and is in the + x -direction.
(c) The placement of the charges is shown in Figure 21.51d.
sin θ =
0.300 m
= 0.600
0.500 m
cos θ =
0.400 m
= 0.800
0.500 m
Figure 21. 51d
The directions of the two fields are shown in Figure 21.51e.
E1 =
1
q1
4π ⑀0 r12
E1 = (8.988 × 109 N ⋅ m 2 /C2 )
6.00 × 10−9 C
(0.400 m) 2
E1 = 337.0 N/C
E2 =
1 q2
4π ⑀0 r22
E2 = (8.988 × 109 N ⋅ m 2 /C2 )
6.00 × 10−9 C
(0.500 m) 2
E2 = 215.7 N/C
Figure 21. 51e
E1x = 0, E1 y = − E1 = −337.0 N/C
E2 x = − E2 sin θ = − (215.7 N/C)(0.600) = −129.4 N/C
E2 y = + E2 cos θ = + (215.7 N/C)(0.800) = +172.6 N/C
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-24
Chapter 21
E x = E1x + E2 x = −129 N/C
E y = E1 y + E2 y = −337.0 N/C + 172.6 N/C = −164 N/C
E = E x2 + E y2 = 209 N/C
G
The field E and its components are shown in Figure 21.51f.
tan α =
tan α =
Ey
Ex
−164 N/C
= +1.271
−129 N/C
α = 232°, counterclockwise from + x -axis
Figure 21. 51f
(d) The placement of the charges is shown in Figure 21.51g.
sin θ =
0.200 m
= 0.800
0.250 m
cos θ =
0.150 m
= 0.600
0.250 m
Figure 21. 51g
The directions of the two fields are shown in Figure 21.51h.
E1 = E2 =
1
q
4π ⑀0 r 2
E1 = (8.988 × 109 N ⋅ m 2 /C2 )
6.00 × 10−9 C
(0.250 m) 2
E1 = 862.8 N/C
E2 = E1 = 862.8 N/C
Figure 21. 51h
E1x = − E1 cos θ , E2 x = − E2 cos θ
E x = E1x + E2 x = −2(862.8 N/C)(0.600) = −1040 N/C
E1 y = + E1 sin θ , E2 y = − E2 sin θ
E y = E1 y + E2 y = 0
E = 1040 N/C, in the − x-direction.
EVALUATE: The electric field produced by a charge is toward a negative charge and away from a positive
charge. As in Exercise 21.45, we can use this rule to deduce the direction of the resultant field at each point
before doing any calculations.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21.52.
IDENTIFY: For a long straight wire, E =
SET UP:
.
1
= 1.80 × 1010 N ⋅ m 2 /C2 .
2π ⑀0
1.5 × 10−10 C/m
= 1.08 m
2π ⑀0 (2.50 N/C)
EXECUTE: r =
21.53.
λ
2π ⑀0r
21-25
EVALUATE: For a point charge, E is proportional to 1/r 2 . For a long straight line of charge, E is
proportional to 1/r.
G
G
IDENTIFY: For a ring of charge, the electric field is given by Eq. (21.8). F = qE . In part (b) use
Newton’s third law to relate the force on the ring to the force exerted by the ring.
SET UP: Q = 0.125 × 10−9 C, a = 0.025 m and x = 0.400 m.
G
1
Qx
EXECUTE: (a) E =
iˆ = (7.0 N/C) iˆ.
4π ⑀0 ( x 2 + a 2 )3/2
G
G
G
(b) F
= −F
= − qE = − (−2.50 × 10−6 C)(7.0 N/C) iˆ = (1.75 × 10−5 N) iˆ
on ring
21.54.
on q
EVALUATE: Charges q and Q have opposite sign, so the force that q exerts on the ring is attractive.
(a) IDENTIFY: The field is caused by a finite uniformly charged wire.
SET UP: The field for such a wire a distance x from its midpoint is
E=
EXECUTE: E =
⎛ 1 ⎞
λ
= 2⎜
.
⎟
2
2π ⑀0 x ( x/a) 2 + 1
4
π
⑀
0 ⎠ x ( x/a ) + 1
⎝
λ
1
(18.0 × 109 N ⋅ m 2 /C2 )(175 × 10−9 C/m)
2
= 3.03 × 104 N/C, directed upward.
⎛ 6.00 cm ⎞
(0.0600 m) ⎜
⎟ +1
⎝ 4.25 cm ⎠
(b) IDENTIFY: The field is caused by a uniformly charged circular wire.
SET UP: The field for such a wire a distance x from its midpoint is E =
1
Qx
4π ⑀0 ( x + a 2 )3/2
2
. We first find
the radius of the circle using 2π r = l.
EXECUTE: Solving for r gives r = l/2π = (8.50 cm)/2π = 1.353 cm.
The charge on this circle is Q = λl = (175 nC/m)(0.0850 m) = 14.88 nC.
The electric field is
E=
21.55.
1
Qx
4π ⑀ 0 ( x + a )
2
2 3/2
=
(9.00 × 109 N ⋅ m 2 /C2 )(14.88 × 1029 C/m)(0.0600 m)
⎡(0.0600 m)2 + (0.01353 m)2 ⎤
⎣
⎦
3/ 2
E = 3.45 × 104 N/C, upward.
EVALUATE: In both cases, the fields are of the same order of magnitude, but the values are different
because the charge has been bent into different shapes.
IDENTIFY: We must use the appropriate electric field formula: a uniform disk in (a), a ring in (b) because
all the charge is along the rim of the disk, and a point-charge in (c).
(a) SET UP: First find the surface charge density (Q/A), then use the formula for the field due to a disk of
⎤
σ ⎡⎢
1
⎥.
1−
charge, E x =
2
2⑀0 ⎢
⎥
+
R
x
(
/
)
1
⎣
⎦
EXECUTE: The surface charge density is σ =
Q
Q
6.50 × 10−9 C
= 1.324 × C/m 2 .
= 2=
2
A πr
π (0.0125 m)
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-26
Chapter 21
The electric field is
⎡
⎤
⎢
⎥
−5
2
⎡
⎤
σ ⎢
1
1.324 × 10 C/m
⎢
1
⎥
⎥
1−
=
1−
Ex =
⎢
⎥
12
2
2
−
2
2
2⑀0 ⎢
C /N ⋅ m ) ⎢
( R/x) + 1 ⎥⎦ 2(8.85 × 10
⎥
⎛ 1.25 cm ⎞
⎣
⎜
⎟ +1⎥
⎢
.
2
00
cm
⎝
⎠
⎣
⎦
E x = 1.14 × 105 N/C, toward the center of the disk.
(b) SET UP: For a ring of charge, the field is E =
1
Qx
4π ⑀0 ( x + a 2 )3/2
2
.
EXECUTE: Substituting into the electric field formula gives
1
Qx
(9.00 × 109 N ⋅ m 2 /C2 )(6.50 × 10−9 C)(0.0200 m)
E=
=
2
2
3/2
3/ 2
4π ⑀0 ( x + a )
⎡(0.0200 m) 2 + (0.0125 m)2 ⎤
⎣
⎦
E = 8.92 × 104 N/C, toward the center of the disk.
(c) SET UP: For a point charge, E = (1/4π ⑀0 ) q/r 2 .
EXECUTE: E = (9.00 × 109 N ⋅ m 2 /C 2 )(6.50 × 10−9 C)/(0.0200 m) 2 = 1.46 × 105 N/C
21.56.
(d) EVALUATE: With the ring, more of the charge is farther from P than with the disk. Also with the ring
the component of the electric field parallel to the plane of the ring is greater than with the disk, and this
component cancels. With the point charge in (c), all the field vectors add with no cancellation, and all the
charge is closer to point P than in the other two cases.
(a) IDENTIFY: The potential energy is given by Eq. (21.17).
G
G G
G
SET UP: U (φ ) = − p ⋅ E = − pE cos φ , where φ is the angle between p and E .
EXECUTE: parallel: φ = 0 and U (0°) = − pE
perpendicular: φ = 90° and U (90°) = 0
ΔU = U (90°) − U (0°) = pE = (5.0 × 10−30 C ⋅ m)(1.6 × 106 N/C) = 8.0 × 10−24 J.
(b) 32 kT = ΔU so T =
21.57.
2ΔU
2(8.0 × 10−24 J)
=
= 0.39 K
3k
3(1.381 × 10−23 J/K)
EVALUATE: Only at very low temperatures are the dipoles of the molecules aligned by a field of this
strength. A much larger field would be required for alignment at room temperature.
(a) IDENTIFY and SET UP: Use Eq. (21.14) to relate the dipole moment to the charge magnitude and the
separation d of the two charges. The direction is from the negative charge toward the positive charge.
G
EXECUTE: p = qd = (4.5 × 10−9 C)(3.1 × 10−3 m) = 1.4 × 10−11 C ⋅ m; The direction of p is from q1 toward q2 .
(b) IDENTIFY and SET UP: Use Eq. (21.15) to relate the magnitudes of the torque and field.
EXECUTE: τ = pE sin φ , with φ as defined in Figure 21.57, so
E=
E=
τ
p sin φ
7.2 × 10−9 N ⋅ m
(1.4 × 10−11 C ⋅ m)sin 36.9°
= 860 N/C
Figure 21. 57
EVALUATE: Eq. (21.15) gives the torque about an axis through the center of the dipole. But the forces on
the two charges form a couple (Problem 11.21) and the torque is the same for any axis parallel to this one.
The force on each charge is q E and the maximum moment arm for an axis at the center is d/2, so the
maximum torque is 2( q E )( d/2) = 1.2 × 10−8 N ⋅ m. The torque for the orientation of the dipole in the
problem is less than this maximum.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21.58.
21-27
G
G
IDENTIFY: Calculate the electric field due to the dipole and then apply F = qE .
SET UP: From Example 21.14, Edipole ( x) =
EXECUTE: Edipole =
6.17 × 10−30 C ⋅ m
2π ⑀0 (3.0 × 10−9 m)3
p
2π ⑀ 0 x3
.
= 4.11 × 106 N/C. The electric force is
F = qE = (1.60 × 10−19 C)(4.11 × 106 N/C) = 6.58 × 10−13 N and is toward the water molecule (negative
21.59.
x-direction).
G
G
EVALUATE: Edipole is in the direction of p, so is in the +x-direction. The charge q of the ion is negative,
G
G
so F is directed opposite to E and is therefore in the −x-direction.
G G G
IDENTIFY: The torque on a dipole in an electric field is given by τ = p × E .
G
G
SET UP: τ = pE sin φ , where φ is the angle between the direction of p and the direction of E .
G
G
EXECUTE: (a) The torque is zero when p is aligned either in the same direction as E or in the opposite
direction, as shown in Figure 21.59a.
G
G
(b) The stable orientation is when p is aligned in the same direction as E . In this case a small rotation of
G
G
G
the dipole results in a torque directed so as to bring p back into alignment with E . When p is directed
G
G
G
opposite to E , a small displacement results in a torque that takes p farther from alignment with E .
(c) Field lines for Edipole in the stable orientation are sketched in Figure 21.59b.
EVALUATE: The field of the dipole is directed from the + charge toward the − charge.
Figure 21. 59
21.60.
IDENTIFY: Find the vector sum of the fields due to each charge in the dipole.
SET UP: A point on the x-axis with coordinate x is a distance r = (d/2) 2 + x 2 from each charge.
EXECUTE: (a) The magnitude of the field the due to each charge is E =
⎞
1 q
q ⎛
1
=
⎜
⎟,
4π ⑀ 0 r 2 4π ⑀0 ⎜⎝ ( d/2)2 + x 2 ⎟⎠
where d is the distance between the two charges. The x-components of the forces due to the two charges
are equal and oppositely directed and so cancel each other. The two fields have equal y-components,
⎞
2q ⎛
1
⎜
⎟ sin θ , where θ is the angle below the x-axis for both fields.
so E = 2 E y =
2
2
4π ⑀0 ⎜ ( d/2 ) + x ⎟
⎝
⎠
⎞
⎞⎛
⎛ 2q ⎞ ⎛
d/2
d/2
qd
1
⎟=
⎜
⎟⎜
. The
and Edipole = ⎜
sin θ =
⎟
2
2
2 3/2
⎜
⎟
2
2
2
2
⎝ 4π ⑀0 ⎠ ⎜⎝ ( d/2 ) + x ⎟⎠ ⎜ ( d/2 ) + x 2 ⎟ 4π ⑀0 ((d/2) + x )
(d/2) + x
⎝
⎠
field is the −y-direction.
(b) At large x, x 2 (d/2) 2 , so the expression in part (a) reduces to the approximation Edipole ≈
EVALUATE: Example 21.14 shows that at points on the +y-axis far from the dipole, Edipole ≈
qd
4π ⑀0 x3
qd
2π ⑀ 0 y 3
.
.
The expression in part (b) for points on the x-axis has a similar form.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-28
21.61.
Chapter 21
(a) IDENTIFY: Use Coulomb’s law to calculate each force and then add them as vectors to obtain the net
force. Torque is force times moment arm.
SET UP: The two forces on each charge in the dipole are shown in Figure 21.61a.
sin θ = 1.50/2.00 so θ = 48.6°
Opposite charges attract and like charges repel.
Fx = F1x + F2 x = 0
Figure 21. 61a
EXECUTE: F1 = k
qq ′
r2
=k
(5.00 × 10−6 C)(10.0 × 10−6 C)
(0.0200 m)2
= 1.124 × 103 N
F1 y = − F1 sin θ = −842.6 N
F2 y = −842.6 N so Fy = F1 y + F2 y = −1680 N (in the direction from the +5.00-μ C charge toward the
−5.00-μ C charge).
EVALUATE: The x-components cancel and the y-components add.
(b) SET UP: Refer to Figure 21.61b.
The y-components have zero moment arm
and therefore zero torque.
F1x and F2 x both produce clockwise torques.
Figure 21. 61b
EXECUTE: F1x = F1 cosθ = 743.1 N
21.62.
τ = 2( F1x )(0.0150 m) = 22.3 N ⋅ m, clockwise
EVALUATE: The electric field produced by the −10.00 μ C charge is not uniform so Eq. (21.15) does not
apply.
IDENTIFY: The plates produce a uniform electric field in the space between them. This field exerts torque
on a dipole and gives it potential energy.
SET UP: The electric field between the plates is given by E = σ /⑀0 , and the dipole moment is p = ed. The
G G
potential energy of the dipole due to the field is U = − p ⋅ E = − pE cos φ , and the torque the field exerts on
it is τ = pE sinφ.
G G
EXECUTE: (a) The potential energy, U = − p ⋅ E = − pE cos φ , is a maximum when φ = 180°. The field
between the plates is E = σ /⑀0 , giving
U max = (1.60 × 10−19 C)(220 × 10−9 m)(125 × 10−6 C/m 2 )/(8.85 × 10−12 C2 /N ⋅ m 2 ) = 4.97 × 10−19 J
The orientation is parallel to the electric field (perpendicular to the plates) with the positive charge of the
dipole toward the positive plate.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-29
(b) The torque, τ = pE sinφ , is a maximum when φ = 90° or 270°. In this case
τ max = pE = pσ /⑀0 = edσ /⑀0
τ max = (1.60 × 10−19 C )( 220 × 10−9 m )(125 × 10−6 C/m 2 )/(8.85 × 10−12 C2 /N ⋅ m 2 )
τ max = 4.97 × 10−19 N ⋅ m
21.63.
The dipole is oriented perpendicular to the electric field (parallel to the plates).
(c) F = 0.
EVALUATE: When the potential energy is a maximum, the torque is zero. In both cases, the net force on
the dipole is zero because the forces on the charges are equal but opposite (which would not be true in a
nonuniform electric field).
IDENTIFY: Apply Coulomb’s law to calculate the force exerted on one of the charges by each of the other
three and then add these forces as vectors.
(a) SET UP: The charges are placed as shown in Figure 21.63a.
q1 = q2 = q3 = q4 = Q
Figure 21.63a
Consider forces on q4 . The free-body diagram is given in Figure 21.63b. Take the y-axis to be parallel to the
G
diagonal between q2 and q4 and let + y be in the direction away from q2 . Then F2 is in the + y -direction.
EXECUTE: F3 = F1 =
F2 =
1
1 Q2
4π ⑀0 L2
Q2
4π ⑀0 2 L2
F1x = − F1 sin 45° = − F1/ 2
F1 y = + F1 cos 45° = + F1/ 2
F3 x = + F3 sin 45° = + F3/ 2
F3 y = + F3 cos 45° = + F3/ 2
F2 x = 0, F2 y = F2
Figure 21.63b
(b) Rx = F1x + F2 x + F3 x = 0
R y = F1 y + F2 y + F3 y = (2/ 2)
1 Q2
1 Q2
Q2
+
=
(1 + 2 2)
4π ⑀0 L2 4π ⑀0 2 L2 8π ⑀0 L2
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-30
Chapter 21
R=
21.64.
Q2
(1 + 2 2). Same for all four charges.
8π ⑀0 L2
EVALUATE: In general the resultant force on one of the charges is directed away from the opposite corner.
The forces are all repulsive since the charges are all the same. By symmetry the net force on one charge
can have no component perpendicular to the diagonal of the square.
k qq ′
IDENTIFY: Apply F = 2 to find the force of each charge on + q. The net force is the vector sum of
r
the individual forces.
SET UP: Let q1 = +2.50 μC and q2 = −3.50 μC. The charge + q must be to the left of q1 or to the right of
q2 in order for the two forces to be in opposite directions. But for the two forces to have equal magnitudes,
+ q must be closer to the charge q1, since this charge has the smaller magnitude. Therefore, the two forces
can combine to give zero net force only in the region to the left of q1. Let + q be a distance d to the left of
q1, so it is a distance d + 0.600 m from q2 .
EXECUTE: F1 = F2 gives
d must be positive, so d =
kq q1
2
d
=
kq q2
( d + 0.600 m)
2
. d =±
q1
(d + 0.600 m) = ±(0.8452)(d + 0.600 m).
q2
(0.8452)(0.600 m)
= 3.27 m. The net force would be zero when + q is at
1 − 0.8452
x = −3.27 m.
G
G
EVALUATE: When + q is at x = −3.27 m, F1 is in the − x direction and F2 is in the +x direction.
21.65.
IDENTIFY: The forces obey Coulomb’s law, and the net force is the vector sum of the individual forces.
qq ′
SET UP: F = k 2 . Like charges repel and unlike charges attract. Charges q1 and q2 and the forces
r
G
they exert on q3 at the origin are sketched in Figure 21.65a. For the net force on q3 to be zero, F1 and
G
F2 from q1 and q2 must be equal in magnitude and opposite in direction.
Figure 21.65
G G
G
EXECUTE: (a) Since F1 , F2 and Fnet are all in the +x-direction, F = F1 + F2 . This gives
4.00 × 10−6 N = k
q1 q3
r12
+k
q2 q3
r22
.
4.00 × 10−6 N
⎛ 4.50 × 10−9 C 2.50 × 10−9 C ⎞
= q3 ⎜
+
⎟ and
⎜ [0.200 m]2
8.99 × 10 N ⋅ m /C
[0.300 m]2 ⎟⎠
⎝
9
2
2
q3 = 3.17 × 10−9 C = 3.17 nC.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-31
G
G
G
G G
(b) Both F1 and F2 are in the +x-direction, so Fnet = F1 + F2 is in the +x-direction.
G
G
(c) The forces F1 and F2 on q3 in each of the three regions are sketched in Figure 21.65b. Only in
G
G
regions I (to the left of q2 ) and III (to the right of q1 ) are F1 and F2 in opposite directions. But since
q2 < q1 , q3 must be closer to q2 than to q1 in order for F1 = F2 , and this is the case only in region I.
Let q3 be a distance d to the left of q2 , so it is a distance d + 0.500 m from q1. F1 = F2 gives
4.50 nC
2.50 nC
=k
. 1.80d 2 = (d + 0.500 m) 2 . 1.80d = ± (d + 0.500 m). The positive solution
(d + 0.500 m) 2
d2
is d = 1.46 m. This point is at x = −0.300 m − 1.46 m = −1.76 m.
EVALUATE: At the point found in part (c) the electric field is zero. The force on any charge placed at this
point will be zero.
qq′
IDENTIFY: Apply F = k 2 for each pair of charges and find the vector sum of the forces that q1 and
r
q2 exert on q3.
k
21.66.
SET UP: Like charges repel and unlike charges attract. The three charges and the forces on q3 are shown
in Figure 21.66.
Figure 21.66
EXECUTE: (a) F1 = k
q1q3
r12
= (8.99 × 109 N ⋅ m 2 /C2 )
(5.00 × 10−9 C)(6.00 × 10−9 C)
(0.0500 m) 2
= 1.079 × 10−4 C.
θ = 36.9°. F1x = + F1 cos θ = 8.63 × 10−5 N. F1 y = + F1 sin θ = 6.48 × 10−5 N.
F2 = k
q2q3
r22
= (8.99 × 109 N ⋅ m 2 /C2 )
(2.00 × 10−9 C)(6.00 × 10−9 C)
(0.0300 m) 2
= 1.20 × 10−4 C.
F2 x = 0, F2 y = − F2 = −1.20 × 10−4 N. Fx = F1x + F2 x = 8.63 × 10−5 N.
Fy = F1 y + F2 y = 6.48 × 10−5 N + (−1.20 × 10−4 N) = −5.52 × 10−5 N.
(b) F = Fx2 + Fy2 = 1.02 × 10−4 N. tan φ =
Fy
Fx
= 0.640. φ = 32.6°, below the +x-axis.
EVALUATE: The individual forces on q3 are computed from Coulomb’s law and then added as vectors,
using components.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-32
21.67.
Chapter 21
(a) IDENTIFY: Use Coulomb’s law to calculate the force exerted by each Q on q and add these forces as
vectors to find the resultant force. Make the approximation x a and compare the net force to F = − kx
to deduce k and then f = (1/2π ) k/m .
SET UP: The placement of the charges is shown in Figure 21.67a.
Figure 21. 67a
EXECUTE: Find the net force on q.
Fx = F1x + F2 x and F1x = + F1, F2 x = − F2
Figure 21. 67b
F1 =
1
qQ
1
qQ
, F2 =
4π ⑀ 0 ( a + x )2
4π ⑀ 0 ( a − x) 2
Fx = F1 − F2 =
Fx =
qQ ⎡ 1
1 ⎤
−
⎢
⎥
4π ⑀0 ⎣ (a + x) 2 (a − x) 2 ⎦
−2
−2
⎡ ⎛
x⎞
x⎞ ⎤
⎛
+
+
−
−
1
1
⎢
⎜
⎟
⎜
⎟ ⎥
4π ⑀0 a 2 ⎢⎣ ⎝ a ⎠
⎝ a ⎠ ⎥⎦
qQ
Since x a we can use the binomial expansion for (1 − x/a ) −2 and (1 + x/a ) −2 and keep only the first two
terms: (1 + z ) n ≈ 1 + nz. For (1 − x/a ) −2 , z = − x/a and n = −2 so (1 − x/a ) −2 ≈ 1 + 2 x/a. For (1 + x/a )−2 ,
z = + x/a and n = −2 so (1 + x/a ) −2 ≈ 1 − 2 x/a. Then F ≈
qQ
⎡⎛
2 ⎢⎜
4π ⑀0a ⎣⎝
1−
⎛ qQ ⎞
2 x ⎞ ⎛ 2 x ⎞⎤
x.
⎟ − ⎜ 1 + ⎟ ⎥ = 2⎜⎜
3⎟
⎟
a ⎠ ⎝
a ⎠⎦
⎝ π ⑀0a ⎠
For simple harmonic motion F = − kx and the frequency of oscillation is f = (1/2π ) k/m . The net force
here is of this form, with k = qQ/π ⑀0 a3 . Thus f =
1
qQ
.
2π π ⑀0ma 3
(b) The forces and their components are shown in Figure 21.67c.
Figure 21.67c
The x-components of the forces exerted by the two charges cancel, the y-components add, and the net force
is in the + y -direction when y > 0 and in the − y -direction when y < 0. The charge moves away from the
origin on the y-axis and never returns.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-33
EVALUATE: The directions of the forces and of the net force depend on where q is located relative to the
other two charges. In part (a), F = 0 at x = 0 and when the charge q is displaced in the + x- or − x-direction
the net force is a restoring force, directed to return q to x = 0. The charge oscillates back and forth, similar
21.68.
to a mass on a spring.
IDENTIFY: Apply ∑ Fx = 0 and ∑ Fy = 0 to one of the spheres.
SET UP: The free-body diagram is sketched in Figure 21.68. Fe is the repulsive Coulomb force between
the spheres. For small θ , sin θ ≈ tan θ.
EXECUTE: ∑ Fx = T sin θ − Fe = 0 and ∑ Fy = T cos θ − mg = 0. So
mg sin θ
kq 2
= Fe = 2 . But
cos θ
d
1/3
tan θ ≈ sin θ =
⎛ q2L ⎞
d
2kq 2 L
, so d 3 =
and d = ⎜
⎜ 2π ⑀ mg ⎟⎟
mg
2L
0
⎝
⎠
.
EVALUATE: d increases when q increases.
Figure 21.68
21.69.
IDENTIFY: Use Coulomb’s law for the force that one sphere exerts on the other and apply the 1st
condition of equilibrium to one of the spheres.
(a) SET UP: The placement of the spheres is sketched in Figure 21.69a.
Figure 21.69a
The free-body diagrams for each sphere are given in Figure 21.69b.
Figure 21.69b
Fc is the repulsive Coulomb force exerted by one sphere on the other.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-34
Chapter 21
(b) EXECUTE: From either force diagram in part (a): ∑ Fy = ma y
T cos 25.0° − mg = 0 and T =
mg
cos 25.0°
∑ Fx = ma x
T sin 25.0° − Fc = 0 and Fc = T sin 25.0°
Use the first equation to eliminate T in the second: Fc = (mg/ cos 25.0°)(sin 25.0°) = mg tan 25.0°
Fc =
1
q1q2
4π ⑀0 r
2
=
q2
1
4π ⑀0 r
2
=
q2
1
4π ⑀0 [ 2(1.20 m)sin 25.0°]2
Combine this with Fc = mg tan 25.0° and get mg tan 25.0° =
q = ( 2.40 m) sin 25.0°
q2
1
4π ⑀ 0 [ 2(1.20 m)sin 25.0°]2
mg tan 25.0°
(1/4π ⑀ 0 )
(15.0 × 10−3 kg)(9.80 m/s 2 ) tan 25.0°
= 2.80 × 10−6 C
8.988 × 109 N ⋅ m 2 /C 2
(c) The separation between the two spheres is given by 2 L sin θ . q = 2.80μC as found in part (b).
q = (2.40 m)sin 25.0°
Fc = (1/4π ⑀ 0 ) q 2 /(2 L sin θ )2 and Fc = mg tan θ . Thus (1/4π ⑀ 0 )q 2 /(2 L sin θ ) 2 = mg tan θ .
(sin θ ) 2 tan θ =
1
q2
= (8.988 × 109 N ⋅ m 2 /C2 )
(2.80 × 10−6 C) 2
= 0.3328.
4(0.600 m) (15.0 × 10−3 kg)(9.80 m/s 2 )
Solve this equation by trial and error. This will go quicker if we can make a good estimate of the value of
θ that solves the equation. For θ small, tan θ ≈ sin θ . With this approximation the equation becomes
4π ⑀0 4 L mg
2
2
sin 3 θ = 0.3328 and sin θ = 0.6930, so θ = 43.9°. Now refine this guess:
θ
sin 2 θ tan θ
0.5000
0.3467
0.3361
0.3335
0.3309
45.0°
40.0°
39.6°
39.5°
39.4°
21.70.
so θ = 39.5°
EVALUATE: The expression in part (c) says θ → 0 as L → ∞ and θ → 90° as L → 0. When L is decreased
from the value in part (a), θ increases.
IDENTIFY: Apply ∑ Fx = 0 and ∑ Fy = 0 to each sphere.
SET UP: (a) Free body diagrams are given in Figure 21.70. Fe is the repulsive electric force that one
sphere exerts on the other.
EXECUTE: (b) T = mg/cos 20° = 0.0834 N, so Fe = T sin 20° = 0.0285 N =
kq1q2
r12
.
(Note: r1 = 2(0.500 m)sin 20° = 0.342 m.)
(c) From part (b), q1q2 = 3.71 × 10−13 C2 .
(d) The charges on the spheres are made equal by connecting them with a wire, but we still have
q +q
1 Q2
, where Q = 1 2 . But the separation r2 is known:
Fe = mg tan θ = 0.0453 N =
2
2
4π ⑀0 r
2
q1 + q2
= 4π ⑀0 Fe r22 = 1.12 × 10−6 C. This equation, along
2
with that from part (c), gives us two equations in q1 and q2 : q1 + q2 = 2.24 × 10−6 C and
r2 = 2(0.500 m)sin 30° = 0.500 m. Hence: Q =
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-35
q1q2 = 3.71 × 10−13 C2 . By elimination, substitution and after solving the resulting quadratic equation, we
find: q1 = 2.06 × 10−6 C and q2 = 1.80 × 10−7 C.
EVALUATE: After the spheres are connected by the wire, the charge on sphere 1 decreases and the charge
on sphere 2 increases. The product of the charges on the sphere increases and the thread makes a larger
angle with the vertical.
Figure 21.70
21.71.
IDENTIFY and SET UP: Use Avogadro’s number to find the number of Na + and Cl− ions and the total
G
G
positive and negative charge. Use Coulomb’s law to calculate the electric force and F = ma to calculate
the acceleration.
(a) EXECUTE: The number of Na + ions in 0.100 mol of NaCl is N = nN A . The charge of one ion is + e,
so the total charge is q1 = nN A e = (0.100 mol)(6.022 × 1023 ions/mol)(1.602 × 10−19 C/ion) = 9.647 × 103 C.
There are the same number of Cl− ions and each has charge −e, so q2 = −9.647 × 103 C.
F=
1
q1q2
4π ⑀ 0 r 2
= (8.988 × 109 N ⋅ m 2 /C2 )
(9.647 × 103 C) 2
(0.0200 m)2
= 2.09 × 1021 N
(b) a = F/m. Need the mass of 0.100 mol of Cl− ions. For Cl, M = 35.453 × 10−3 kg/mol, so
m = (0.100 mol)(35.453 × 10−3 kg/mol) = 35.45 × 10−4 kg. Then a =
F
2.09 × 1021 N
=
= 5.90 × 1023 m/s 2 .
m 35.45 × 10−4 kg
(c) EVALUATE: Is is not reasonable to have such a huge force. The net charges of objects are rarely larger
than 1 μC; a charge of 104 C is immense. A small amount of material contains huge amounts of positive
21.72.
and negative charges.
IDENTIFY: The net electric field at the origin is the vector sum of the fields due to the two charges.
G
G
q
SET UP: E = k 2 . E is toward a negative charge and away from a positive charge. At the origin, E1
r
due to the −5.00 nC charge is in the +x-direction, toward the charge.
EXECUTE: (a) E1 = (8.99 × 109 N ⋅ m 2 /C2 )
(5.00 × 10−9 C)
(1.20 m) 2
= 31.2 N/C. E1x = +31.2 N/C.
G
E x = E1x + E2 x . E x = +45.0 N/C, so E2 x = Ex − E1x = +45.0 N/C − 31.2 N/C = 13.8 N/C. E is away from Q
E2 r 2 (13.8 N/C)(0.600 m) 2
=
= 5.53 × 10−10 C.
k
r2
8.99 × 109 N ⋅ m 2 /C2
G
(b) E x = −45.0 N/C, so E2 x = Ex − E1x = −45.0 N/C − 31.2 N/C = −76.2 N/C. E is toward Q so Q is
so Q is positive. E2 = k
Q
gives Q =
E2 r 2 (76.2 N/C)(0.600 m) 2
=
= 3.05 × 10−9 C.
k
8.99 × 109 N ⋅ m 2 /C2
q
EVALUATE: Use of the equation E = k 2 gives only the magnitude of the electric field. When
r
combining fields, you still must figure out whether to add or subtract the magnitudes depending on the
direction in which the fields point.
negative. Q =
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-36
21.73.
Chapter 21
IDENTIFY: The electric field exerts a horizontal force away from the wall on the ball. When the ball hangs
at rest, the forces on it (gravity, the tension in the string, and the electric force due to the field) add to zero.
SET UP: The ball is in equilibrium, so for it ∑ Fx = 0 and ∑ Fy = 0. The force diagram for the ball is
G
G
given in Figure 21.73. FE is the force exerted by the electric field. F = qE . Since the electric field is
G
horizontal, FE is horizontal. Use the coordinates shown in the figure. The tension in the string has been
replaced by its x- and y-components.
Figure 21.73
EXECUTE: ∑ Fy = 0 gives Ty − mg = 0. T cos θ − mg = 0 and T =
mg
. ∑ Fx = 0 gives FE − Tx = 0.
cos θ
FE − T sin θ = 0. Combing the equations and solving for FE gives
21.74.
⎛ mg ⎞
−3
2
−2
FE = ⎜
⎟ sin θ = mg tan θ = (12.3 × 10 kg)(9.80 m/s )(tan17.4°) = 3.78 × 10 N. FE = q E so
⎝ cosθ ⎠
G
G
F
3.78 × 10−2 N
E= E =
= 3.41 × 104 N/C. Since q is negative and FE is to the right, E is to the left in the figure.
−6
q
1.11 × 10 C
EVALUATE: The larger the electric field E the greater the angle the string makes with the wall.
IDENTIFY: We can find the force on the charged particle due to the electric field. Then we can use
Newton’s second law to find its acceleration and the constant-acceleration kinematics formulas to find the
components of the distance it moves.
SET UP: The x-component of the electric force on a charged particle is Fx = qE x and Fx = ma x . For
1
constant acceleration in the x-direction, x − x0 = v0 xt + a xt 2 . Similar equations apply in the y-direction.
2
EXECUTE: The only nonzero acceleration is in the y-direction, so a x = 0 and
Fy = qE y = (9.00 × 10−6 C)(895 N/C) = 8.055 × 10−3 N. a y =
21.75.
Fy
m
=
8.055 × 10−3 N
0.400 × 10−6 kg
= 2.014 × 104 m/s 2 .
1
x − x0 = v0 xt + a xt 2 = (−125 m/s)(7.00 × 10−3 s) = −0.875 m.
2
1
1
y − y0 = v0 yt + a yt 2 = (2.014 × 104 m/s 2 )(7.00 × 10−3 s) 2 = 0.4934 m. r = x 2 + y 2 = 1.00 m.
2
2
EVALUATE: The 1.00 m is the distance of the particle from the origin at the end of 7.00 ms, but it is not
the distance the particle has traveled in 7.00 ms.
G
G
q
IDENTIFY: For a point charge, E = k 2 . For the net electric field to be zero, E1 and E2 must have equal
r
magnitudes and opposite directions.
G
SET UP: Let q1 = +0.500 nC and q2 = +8.00 nC. E is toward a negative charge and away from a
positive charge.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-37
EXECUTE: The two charges and the directions of their electric fields in three regions are shown in Figure 21.75.
Only in region II are the two electric fields in opposite directions. Consider a point a distance x from
0.500 nC
8.00 nC
q1 so a distance 1.20 m − x from q2 . E1 = E2 gives k
. 16 x 2 = (1.20 m − x) 2 .
=k
2
x
(1.20 m − x )2
4 x = ± (1.20 m − x ) and x = 0.24 m is the positive solution. The electric field is zero at a point between the
two charges, 0.24 m from the 0.500 nC charge and 0.96 m from the 8.00 nC charge.
EVALUATE: There is only one point along the line connecting the two charges where the net electric field
is zero. This point is closer to the charge that has the smaller magnitude.
Figure 21.75
21.76.
IDENTIFY: For the acceleration (and hence the force) on Q to be upward, as indicated, the forces due to
q1 and q2 must have equal strengths, so q1 and q2 must have equal magnitudes. Furthermore, for the
force to be upward, q1 must be positive and q2 must be negative.
SET UP: Since we know the acceleration of Q, Newton’s second law gives us the magnitude of the force
on it. We can then add the force components using F = FQq1 cosθ + FQq2 cosθ = 2 FQq1 cosθ . The electrical
force on Q is given by Coulomb’s law, FQq1 =
1 Qq1
(for q1 ) and likewise for q2 .
4π ⑀0 r 2
EXECUTE: First find the net force: F = ma = (0.00500 kg)(324 m/s 2 ) = 1.62 N. Now add the force
components, calling θ the angle between the line connecting q1 and q2 and the line connecting q1 and Q.
F
1.62 N
=
= 1.08 N. Now find the charges
2cosθ
⎛ 2.25 cm ⎞
2⎜
⎟
⎝ 3.00 cm ⎠
by solving for q1 in Coulomb’s law and use the fact that q1 and q2 have equal magnitudes but opposite
F = FQq1 cosθ + FQq2 cosθ = 2 FQq1 cosθ and FQq1 =
signs. FQq1 =
1
Q q1
4π ⑀ 0 r 2
and q1 =
r 2 FQq1
(0.0300 m)2 (1.08 N)
= 6.17 × 10−8 C.
=
9
2 2
26
1
(9
.
00
×
10
N
⋅
m
/C
)(1
.
75
×
10
C)
Q
4π ⑀0
q2 = −q1 = −6.17 × 10−8 C.
EVALUATE: Simple reasoning allows us first to conclude that q1 and q2 must have equal magnitudes but
21.77.
opposite signs, which makes the equations much easier to set up than if we had tried to solve the problem
in the general case. As Q accelerates and hence moves upward, the magnitude of the acceleration vector
will change in a complicated way.
IDENTIFY: Use Coulomb’s law to calculate the forces between pairs of charges and sum these forces as
vectors to find the net charge.
(a) SET UP: The forces are sketched in Figure 21.77a.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-38
Chapter 21
G G
G G
EXECUTE: F1 + F3 = 0, so the net force is F = F2 .
F=
1
q (3q )
4π ⑀0 ( L/ 2)
2
=
6q 2
4π ⑀0 L2
, away from the vacant corner.
Figure 21. 77a
(b) SET UP: The forces are sketched in Figure 21.77b.
EXECUTE: F2 =
F1 = F3 =
4π ⑀0 ( 2 L)
1
q (3q )
4π ⑀0
2
L
3q 2
q (3q )
1
=
=
2
4π ⑀0 (2 L2 )
3q 2
4π ⑀0 L2
The vector sum of F1 and F3 is F13 = F12 + F32 .
Figure 21. 77b
G
3 2q 2 G
F13 = 2 F1 =
; F13 and F2 are in the same direction.
2
4π ⑀0 L
3q 2 ⎛
1⎞
⎜ 2 + ⎟ , and is directed toward the center of the square.
2⎠
4π ⑀0 L2 ⎝
EVALUATE: By symmetry the net force is along the diagonal of the square. The net force is only slightly
larger when the −3q charge is at the center. Here it is closer to the charge at point 2 but the other two
F = F13 + F2 =
21.78.
forces cancel.
IDENTIFY: Use Eq. (21.7) for the electric field produced by each point charge. Apply the principle of
superposition and add the fields as vectors to find the net field.
(a) SET UP: The fields due to each charge are shown in Figure 21.78a.
cosθ =
x
2
x + a2
Figure 21.78a
EXECUTE: The components of the fields are given in Figure 21.78b.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
E1 = E2 =
E3 =
21-39
1 ⎛ q ⎞
⎜
⎟
4π ⑀0 ⎝ a 2 + x 2 ⎠
1 ⎛ 2q ⎞
⎜ ⎟
4π ⑀0 ⎝ x 2 ⎠
Figure 21.78b
E1 y = − E1 sin θ , E2 y = + E2 sin θ so E y = E1 y + E2 y = 0.
E1x = E2 x = + E1 cosθ =
⎞
1 ⎛ q ⎞⎛
x
⎟ , E3 x = − E3
⎜ 2
⎟ ⎜⎜ 2
2
4π ⑀0 ⎝ a + x ⎠ ⎝ x + a 2 ⎟⎠
⎛ 1 ⎛ q ⎞⎛
⎞⎞
x
2q
E x = E1x + E2 x + E3 x = 2 ⎜
⎜
⎟⎟ −
⎜
⎟
2
2
⎜ 4π ⑀ 0 ⎝ a + x ⎠ ⎜ x 2 + a 2 ⎟ ⎟ 4π ⑀0 x 2
⎝
⎠⎠
⎝
Ex = −
⎞
⎞
2q ⎛ 1
x
2q ⎛
1
⎜⎜ 2 − 2
⎟⎟ = −
⎜⎜ 1 −
⎟
2
3/
2
2
2
2
3/2
4π ⑀0 ⎝ x
(a + x ) ⎠
4π ⑀0 x ⎝ (1 + a /x ) ⎟⎠
Thus E =
2q
⎛
2⎜
⎜
4π ⑀0 x ⎝
1−
1
2
⎞
2 3/2 ⎟
⎟
(1 + a /x )
⎠
, in the − x-direction.
(b) x a implies a 2 /x 2 1 and (1 + a 2 /x 2 ) −3/2 ≈ 1 − 3a 2 /2 x 2 .
Thus E ≈
21.79.
⎛ ⎛ 3a 2 ⎞ ⎞
3qa 2
⎜1 − ⎜1 − 2 ⎟ ⎟ =
.
4π ⑀0 x ⎝ ⎜⎝ 2 x ⎟⎠ ⎟⎠ 4π ⑀0 x 4
2q
2⎜
EVALUATE: E ∼ 1/x 4 . For a point charge E ∼ 1/x 2 and for a dipole E ∼ 1/x3 . The total charge is zero so
at large distances the electric field should decrease faster with distance than for a point charge. By
G
symmetry E must lie along the x-axis, which is the result we found in part (a).
IDENTIFY: The small bags of protons behave like point-masses and point-charges since they are
extremely far apart.
SET UP: For point-particles, we use Newton’s formula for universal gravitation (F = Gm1m 2 /r 2 ) and
Coulomb’s law. The number of protons is the mass of protons in the bag divided by the mass of a single
proton.
EXECUTE: (a) (0.0010 kg)/(1.67 × 10−27 kg) = 6.0 × 1023 protons
(b) Using Coulomb’s law, where the separation is twice the radius of the earth, we have
Felectrical = (9.00 × 109 N ⋅ m 2 /C2 )(6.0 × 1023 × 1.60 × 10−19 C) 2 /(2 × 6.38 × 106 m) 2 = 5.1 × 105 N
Fgrav = (6.67 × 10−11 N ⋅ m 2 /kg 2 )(0.0010 kg) 2 /(2 × 6.38 × 106 m) 2 = 4.1 × 10−31 N
(c) EVALUATE: The electrical force (≈200,000 lb!) is certainly large enough to feel, but the gravitational
21.80.
force clearly is not since it is about 1036 times weaker.
IDENTIFY: We can treat the protons as point-charges and use Coulomb’s law.
SET UP: (a) Coulomb’s law is F = (1/4π ⑀0 ) q1q2 /r 2 .
EXECUTE: F = (9.00 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2 /(2.0 × 10−15 m) 2 = 58 N = 13 lb, which is
certainly large enough to feel.
(b) EVALUATE: Something must be holding the nucleus together by opposing this enormous repulsion.
This is the strong nuclear force.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-40
21.81.
Chapter 21
IDENTIFY: Estimate the number of protons in the textbook and from this find the net charge of the
textbook. Apply Coulomb’s law to find the force and use Fnet = ma to find the acceleration.
SET UP: With the mass of the book about 1.0 kg, most of which is protons and neutrons, we find that the
number of protons is 12 (1.0 kg)/(1.67 × 10−27 kg) = 3.0 × 1026.
EXECUTE: (a) The charge difference present if the electron’s charge was 99.999, of the proton’s is
Δq = (3.0 × 1026 )( 0.00001)(1.6 × 10−19 C ) = 480 C.
(b) F = k (Δq ) 2 /r 2 = k (480 C)2 /(5.0 m)2 = 8.3 × 1013 N, and is repulsive.
a = F/m = (8.3 × 1013 N)/(1 kg) = 8.3 × 1013 m/s 2 .
21.82.
EXECUTE: (c) Even the slightest charge imbalance in matter would lead to explosive repulsion!
IDENTIFY: The positive sphere will be deflected in the direction of the electric field but the negative sphere
will be deflected in the direction opposite to the electric field. Since the spheres hang at rest, they are in
equilibrium so the forces on them must balance. The external forces on each sphere are gravity, the tension in
the string, the force due to the uniform electric field and the electric force due to the other sphere.
SET UP: The electric force on one sphere due to the other is FC = k
q2
in the horizontal direction, the
r2
force on it due to the uniform electric field is FE = qE in the horizontal direction, the gravitational force is
mg vertically downward and the force due to the string is T directed along the string. For equilibrium
∑ Fx = 0 and ∑ Fy = 0.
EXECUTE: (a) The positive sphere is deflected in the same direction as the electric field, so the one that is
deflected to the left is positive.
(b) The separation between the two spheres is 2(0.530 m)sin 25o = 0.4480 m.
FC = k
q2
(8.99 × 109 N ⋅ m 2 /C2 )(72.0 × 10−9 C) 2
r
(0.4480 m)2
=
2
= 2.322 × 10−4 N. FE = qE. ∑ Fy = 0 gives
mg
. ∑ Fx = 0 gives T sin 25o + FC − FE = 0. mg tan 25o + FC = qE.
cos 25o
Combining the equations and solving for E gives
mg tan 25o + FC (6.80 × 10−6 kg)(9.8 m/s 2 ) tan 25o + 2.322 × 10−4 N
E=
=
= 3.66 × 103 N/C.
q
72.0 × 10−9 C
EVALUATE: Since the charges have opposite signs, they attract each other, which tends to reduce the
angle between the strings. Therefore if their charges were negligibly small, the angle between the strings
would be greater than 50°.
IDENTIFY: The only external force acting on the electron is the electrical attraction of the proton, and its
acceleration is toward the center of its circular path (that is, toward the proton). Newton’s second law
applies to the proton and Coulomb’s law gives the electrical force on it due to the proton.
v2
e2
v2
SET UP: Newton’s second law gives FC = m . Using the electrical force for FC gives k 2 = m .
r
r
r
T cos 25o − mg = 0 and T =
21.83.
EXECUTE: Solving for v gives v =
21.84.
ke2
(8.99 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2
=
= 2.19 × 106 m/s.
mr
(9.109 × 10−31 kg)(5.29 × 10−11 m)
EVALUATE: This speed is less than 1% the speed of light, so it is reasonably safe to use Newtonian
physics.
IDENTIFY: Since we can ignore gravity, the only external force acting on the moving sphere is the
electrical attraction of the stationary sphere, and its acceleration is toward the center of its circular path
(that is, toward the stationary sphere). Newton’s second law applies to the moving sphere and Coulomb’s
law gives the electrical force on it due to the stationary sphere.
qq
v2
v2
SET UP: Newton’s second law gives FC = m . Using the electrical force for FC gives k 1 22 = m .
r
r
r
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
EXECUTE: Solving for r gives r =
21.85.
k q1q2
=
21-41
(8.99 × 109 N ⋅ m 2 /C2 )(4.3 × 10−6 C)(7.5 × 10−6 C)
= 0.925 m.
mv 2
(9.00 × 10−9 kg)(5.9 × 103 m/s) 2
EVALUATE: We can safely ignore gravity in most cases because it is normally much weaker than the
electric force.
IDENTIFY and SET UP: Use the density of copper to calculate the number of moles and then the number of
atoms. Calculate the net charge and then use Coulomb’s law to calculate the force.
⎛4
⎞
⎛4 ⎞
EXECUTE: (a) m = ρV = ρ ⎜ π r 3 ⎟ = (8.9 × 103 kg/m3 ) ⎜ π ⎟ (1.00 × 10−3 m)3 = 3.728 × 10−5 kg
3
⎝
⎠
⎝3 ⎠
n = m/M = (3.728 × 10−5 kg)/(63.546 × 10−3 kg/mol) = 5.867 × 10−4 mol
N = nN A = 3.5 × 1020 atoms
(b) N e = (29)(3.5 × 1020 ) = 1.015 × 1022 electrons and protons
qnet = eN e − ( 0.99900)eN e = (0.100 × 10−2 )(1.602 × 10−19 C)(1.015 × 1022 ) = 1.6 C
21.86.
q2
(1.6 C)2
= 2.3 × 1010 N
(1.00 m) 2
EVALUATE: The amount of positive and negative charge in even small objects is immense. If the charge
of an electron and a proton weren’t exactly equal, objects would have large net charges.
IDENTIFY: Apply constant acceleration equations to a drop to find the acceleration. Then use F = ma to
find the force and F = q E to find q .
F =k
r
2
=k
SET UP: Let D = 2.0 cm be the horizontal distance the drop travels and d = 0.30 mm be its vertical
displacement. Let + x be horizontal and in the direction from the nozzle toward the paper and let + y be
vertical, in the direction of the deflection of the drop. a x = 0 and a y = a.
1
EXECUTE: Find the time of flight: t = D/v = (0.020 m)/(20 m/s) = 0.00100 s. d = at 2 .
2
a=
2d
2(3.00 × 10−4 m)
t
(0.001 s) 2
=
2
= 600 m/s 2 .
21.87.
(1.4 × 10−11 kg)(600 m/s 2 )
= 1.05 × 10−13 C.
8.00 × 104 N/C
EVALUATE: Since q is positive the vertical deflection is in the direction of the electric field.
IDENTIFY: Eq. (21.3) gives the force exerted by the electric field. This force is constant since the electric
field is uniform and gives the proton a constant acceleration. Apply the constant acceleration equations for
the x- and y-components of the motion, just as for projectile motion.
(a) SET UP: The electric field is upward so the electric force on the positively charged proton is upward
G
G
and has magnitude F = eE. Use coordinates where positive y is downward. Then applying ∑ F = ma to
the proton gives that a x = 0 and a y = −eE/m. In these coordinates the initial velocity has components
Then a = F/m = qE/m gives q = ma/E =
vx = +v0 cos α and v y = + v0 sin α , as shown in Figure 21.87a.
Figure 21.87a
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-42
Chapter 21
EXECUTE: Finding hmax : At y = hmax the y-component of the velocity is zero.
v y = 0, v0 y = v0 sin α , a y = −eE/m, y − y0 = hmax = ?
v 2y = v02 y + 2a y ( y − y0 )
y − y0 =
hmax =
v 2y − v02 y
2a y
−v02 sin 2 α mv02 sin 2 α
=
2( −eE/m)
2eE
(b) Use the vertical motion to find the time t: y − y0 = 0, v0 y = v0 sin α , a y = −eE/m, t = ?
y − y0 = v0 yt + 12 a yt 2
With y − y0 = 0 this gives t = −
2v0 y
ay
=−
2(v0 sin α ) 2mv0 sin α
=
eE
−eE/m
Then use the x-component motion to find d: a x = 0, v0 x = v0 cos α , t = 2mv0 sin α /eE , x − x0 = d = ?
2
2
⎛ 2mv0 sin α ⎞ mv0 2sin α cos α mv0 sin 2α
x − x0 = v0 xt + 12 a xt 2 gives d = v0 cos α ⎜
=
⎟=
eE
eE
eE
⎝
⎠
(c) The trajectory of the proton is sketched in Figure 21.87b.
Figure 21.87b
(d) Use the expression in part (a): hmax =
Use the expression in part (b): d =
[(4.00 × 105 m/s)(sin 30.0°)]2 (1.673 × 10−27 kg)
2(1.602 × 10−19 C)(500 N/C)
(1.673 × 10−27 kg)(4.00 × 105 m/s) 2 sin 60.0°
(1.602 × 10−19 C)(500 N/C)
= 0.418 m
= 2.89 m
EVALUATE: In part (a), a y = −eE/m = −4.8 × 1010 m/s 2 . This is much larger in magnitude than g, the
acceleration due to gravity, so it is reasonable to ignore gravity. The motion is just like projectile motion,
except that the acceleration is upward rather than downward and has a much different magnitude. hmax
and d increase when α or v0 increase and decrease when E increases.
21.88.
IDENTIFY: E x = E1x + E2 x . Use Eq. (21.7) for the electric field due to each point charge.
G
SET UP: E is directed away from positive charges and toward negative charges.
−9
1 q1
C
9
2 2 4.00 × 10
=
(
8
.
99
×
10
N
⋅
m
/C
)
= +99.9 N/C.
EXECUTE: (a) E x = +50.0 N/C. E1x =
2
2
4π ⑀0 r1
(0.60 m)
E x = E1x + E2 x , so E2 x = Ex − E1x = +50.0 N/C − 99.9 N/C = 249.9 N/C. Since E2 x is negative, q2 must
be negative. q2 =
E2 x r22
(1/4π ⑀0 )
=
(49.9 N/C)(1.20 m)2
9
2
8.99 × 10 N ⋅ m /C
2
= 7.99 × 10−9 C. q2 = −7.99 × 10−9 C
(b) E x = −50.0 N/C. E1x = +99.9 N/C, as in part (a). E2 x = Ex − E1x = −149.9 N/C. q2 is negative.
q2 =
E2 x r22
(1/4π ⑀0 )
=
(149.9 N/C)(1.20 m)2
8.99 × 109 N ⋅ m 2 /C2
= 2.40 × 10−8 C. q2 = −2.40 × 10−8 C.
EVALUATE: q2 would be positive if E2 x were positive.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21.89.
21-43
IDENTIFY: Divide the charge distribution into infinitesimal segments of length dx. Calculate E x and E y
due to a segment and integrate to find the total field.
SET UP: The charge dQ of a segment of length dx is dQ = (Q/a )dx. The distance between a segment at
x and the charge q is a + r − x. (1 − y ) −1 ≈ 1 + y when y 1.
EXECUTE: (a) dEx =
a + r = x, so E x =
G
G
(b) F = qE =
1 a
Qdx
1 Q⎛1
1 ⎞
1
dQ
so E x =
=
⎜ −
⎟.
4π ⑀0 ∫0 a ( a + r − x) 2 4π ⑀0 a ⎝ r a + r ⎠
4π ⑀0 (a + r − x )2
1 Q⎛ 1
1⎞
− ⎟ . E y = 0.
⎜
4π ⑀0 a ⎝ x − a x ⎠
1 qQ ⎛ 1
1 ⎞ˆ
⎜ −
⎟ i.
4π ⑀0 a ⎝ r a + r ⎠
EVALUATE: (c) For x a, F =
kqQ
kqQ
kqQ
1 qQ
((1 − a/x) −1 − 1) =
(1 + a/x + ⋅⋅⋅ − 1) ≈ 2 ≈
. (Note
π
⑀0 r 2
ax
ax
4
x
that for x a, r = x − a ≈ x.) The charge distribution looks like a point charge from far away, so the force
21.90.
takes the form of the force between a pair of point charges.
IDENTIFY: Use Eq. (21.7) to calculate the electric field due to a small slice of the line of charge and
G
integrate as in Example 21.10. Use Eq. (21.3) to calculate F .
SET UP: The electric field due to an infinitesimal segment of the line of charge is sketched in Figure 21.90.
sin θ =
y
2
x + y2
x
cosθ =
2
x + y2
Figure 21.90
Slice the charge distribution up into small pieces of length dy. The charge dQ in each slice is
dQ = Q (dy/a ). The electric field this produces at a distance x along the x-axis is dE. Calculate the
G
components of dE and then integrate over the charge distribution to find the components of the total field.
1 ⎛ dQ ⎞
Q ⎛ dy ⎞
EXECUTE: dE =
⎜⎜ 2
⎟⎟ =
⎜
⎟
2
4π ⑀0 ⎝ x + y ⎠ 4π ⑀0 a ⎜⎝ x 2 + y 2 ⎟⎠
⎞
Qx ⎛
dy
dE x = dE cosθ =
⎜⎜ 2
⎟
2
3/2
4π ⑀0 a ⎝ ( x + y ) ⎟⎠
⎞
Q ⎛
ydy
dE y = − dE sin θ = −
⎜⎜ 2
⎟
2
3/2
4π ⑀0 a ⎝ ( x + y ) ⎟⎠
Qx ⎡ 1
⎢
E x = ∫ dE x = −
=
Ñ
2
2
3/2
4π ⑀0 a 0 ( x + y )
4π ⑀0a ⎢ x 2
⎣
Qx
a
dy
a
⎤
Q
⎥ =
2
2⎥
π
⑀0 x
4
x + y ⎦0
y
1
2
x + a2
a
⎡
⎤
⎞
1
Q ⎛1
1
⎢−
⎥ =−
= −
E y = ∫ dEy = −
⎜ −
⎟
Ñ
2
2
3/2
⎟
4π ⑀ 0a 0 ( x + y )
4π ⑀0a ⎢
4π ⑀0a ⎜⎝ x
x 2 + y 2 ⎥⎦ 0
x2 + a2 ⎠
⎣
Q
a
ydy
Q
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-44
Chapter 21
G
G
(b) F = q0 E
Fx = −qE x =
⎞
1
qQ ⎛ 1
1
− qQ
; Fy = − qE y =
⎜ −
⎟
⎜
⎟
2
2
2
2
4π ⑀0 x x + a
4π ⑀0a ⎝ x
x +a ⎠
21.91.
−1/2
1 ⎛ a2 ⎞
1⎛
a2 ⎞ 1 a2
= ⎜1 + 2 ⎟
= ⎜1 − 2 ⎟ = − 3
⎜
⎟
⎜
x ⎝ 2 x ⎟⎠ x 2 x
x ⎠
x2 + a 2 x ⎝
qQ
qQ ⎛ 1 1 a 2 ⎞
qQa
Fx ≈ −
, Fy ≈
⎜ − +
⎟=
2
4π ⑀0a ⎜⎝ x x 2 x3 ⎟⎠ 8π ⑀ 0 x3
4π ⑀0 x
G
qQ
EVALUATE: For x a, Fy Fx and F ≈ Fx =
and F is in the − x-direction. For x a the
2
4π P0 x
charge distribution Q acts like a point charge.
IDENTIFY: Apply Eq. (21.9) from Example 21.10.
G
SET UP: a = 2.50 cm. Replace Q by Q . Since Q is negative, E is toward the line of charge and
1
(c) For x a,
G
Q
1
E=−
iˆ.
4π ⑀0 x x 2 + a 2
G
Q
1
1
7.00 × 10−9 C
EXECUTE: E = −
iˆ = −
iˆ = (−6110 N/C) iˆ.
4π ⑀0 x x 2 + a 2
4π ⑀0 (0.100 m) (0.100 m)2 + (0.025 m) 2
21.92.
(b) The electric field is less than that at the same distance from a point charge (6300 N/C). For large x,
⎞
1
1⎛
a2 ⎞
1 Q⎛
a2
−
+ ⋅⋅⋅ ⎟ . The first
( x + a ) −1/2 = (1 + a 2 /x 2 ) −1/2 ≈ ⎜1 − 2 ⎟ , which gives E x →∞ =
1
⎜
2
2
⎜
⎟
⎜
⎟
x
x ⎝ 2x ⎠
4π ⑀0 x ⎝ 2 x
⎠
correction term to the point charge result is negative.
(c) For a 1% difference, we need the first term in the expansion beyond the point charge result to be less
a2
≈ 0.010 ⇒ x ≈ a 1/(2(0.010)) = (0.025 m) 1/0.020 ⇒ x ≈ 0.177 m.
than 0.010:
2x2
EVALUATE: At x = 10.0 cm (part b), the exact result for the line of charge is 3.1% smaller than for a point
charge. It is sensible, therefore, that the difference is 1.0% at a somewhat larger distance, 17.7 cm.
G
kQ 2
G
IDENTIFY: The electrical force has magnitude F = 2 and is attractive. Apply ∑ F = ma to the earth.
r
SET UP: For a circular orbit, a =
v2
2π r
. The mass of the earth is mE = 5.97 × 1024 kg,
. The period T is
r
v
the orbit radius of the earth is 1.50 × 1011 m and its orbital period is 3.146 × 107 s.
EXECUTE: F = ma gives
Q=
21.93.
mE 4π 2r 3
kT
2
=
kQ
r
2
= mE
4π 2r 2
v2
, so
. v2 =
r
T2
(5.97 × 1024 kg)(4)(π 2 )(1.50 × 1011 m)3
(8.99 × 109 N ⋅ m 2 /C2 )(3.146 × 107 s)2
= 2.99 × 1017 C.
EVALUATE: A very large net charge would be required.
IDENTIFY: Apply Eq. (21.11).
SET UP: σ = Q/A = Q/π R 2 . (1 + y 2 )−1/2 ≈ 1 − y 2 /2, when y 2 1.
EXECUTE: (a) E =
E=
σ
2⑀ 0
[1 − ( R 2 /x 2 + 1) −1/2 ].
⎡
⎞
7.00 pC/π (0.025 m) 2 ⎢ ⎛ (0.025 m) 2
1− ⎜
+ 1⎟
2
⎟
⎢ ⎜⎝ (0.200 m)
2⑀0
⎠
⎣
−1/2 ⎤
⎥ = 1.56 N/C, in the + x-direction.
⎥
⎦
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
(b) For x R, E =
21.94.
21-45
σ
σ R2
σπ R 2
Q
[1 − (1 − R 2 /2 x 2 + ⋅⋅⋅)] ≈
=
=
.
2
2
2⑀0
2⑀0 2 x
4π ⑀0 x
4π ⑀0 x 2
(c) The electric field of (a) is less than that of the point charge (0.90 N/C) since the first correction term to
the point charge result is negative.
(1.58 − 1.56)
(d) For x = 0.200 m, the percent difference is
= 0.01 = 1%. For x = 0.100 m,
1.56
(6.30 − 6.00)
= 0.047 ≈ 5%.
Edisk = 6.00 N/C and Epoint = 6.30 N/C, so the percent difference is
6.30
EVALUATE: The field of a disk becomes closer to the field of a point charge as the distance from the disk
increases. At x = 10.0 cm, R/x = 25% and the percent difference between the field of the disk and the field
of a point charge is 5%.
IDENTIFY: When the forces on it balance, the acceleration of a molecule is zero and it moves with
constant velocity.
SET UP: The electrical force is FE = qE and the viscous drag force is FD = KRv.
q Kv
.
=
R E
Eq
⎛ Eq ⎞
⎛ ET ⎞ q
T =⎜
(b) The speed is constant and has magnitude v =
. Therefore x = vt = ⎜
⎟
⎟ .
KR
⎝ KR ⎠
⎝ K ⎠R
EXECUTE: (a) F = FD so qE = KRv and
ET
⎛ ET ⎞ q
⎛q⎞
⎛q⎞
⎛q⎞
⎛q⎞
(c) x = ⎜
is constant. ⎜ ⎟ = 2 ⎜ ⎟ and ⎜ ⎟ = 3⎜ ⎟ .
⎟ , where
K
⎝ K ⎠R
⎝ R ⎠2
⎝ R ⎠1
⎝ R ⎠3
⎝ R ⎠1
21.95.
⎛ ET ⎞⎛ q ⎞
⎛ ET ⎞⎛ q ⎞
⎛ ET ⎞⎛ q ⎞
⎛ ET ⎞⎛ q ⎞
x2 = ⎜
⎟⎜ ⎟ = 2 ⎜
⎟⎜ ⎟ = 2 x1; x3 = ⎜
⎟⎜ ⎟ = 3⎜
⎟⎜ ⎟ = 3x1.
K
R
K
R
K
R
⎝
⎠⎝ ⎠2
⎝
⎠⎝ ⎠1
⎝
⎠⎝ ⎠3
⎝ K ⎠⎝ R ⎠1
EVALUATE: The distance a particle moves is not proportional to its charge, but rather is proportional to
the ratio of its charge to its radius (size).
G
G
IDENTIFY: Find the resultant electric field due to the two point charges. Then use F = qE to calculate the
force on the point charge.
SET UP: Use the results of Problems 21.90 and 21.89.
EXECUTE: (a) The y-components of the electric field cancel, and the x-component from both charges, as
⎞
1 −2Q ⎛ 1
1
given in Problem 21.90, is E x =
⎜⎜ − 2
⎟ . Therefore,
2
1/2
4π ⑀0 a ⎝ y ( y + a ) ⎟⎠
G
G
⎞ˆ
1 −2Qq ⎛ 1
1
1 −2Qq
1 Qqa ˆ
F=
i.
(1 − (1 − a 2 /2 y 2 + ⋅⋅⋅)) iˆ = −
⎜⎜ − 2
⎟⎟ i . If y a, F ≈
2
1/2
4π ⑀0 a ⎝ y ( y + a ) ⎠
4π ⑀ 0 ay
4π ⑀0 y 3
(b) If the point charge is now on the x-axis the two halves of the charge distribution provide different
G
G
1 Qq ⎛ 1
1⎞
forces, though still along the x-axis, as given in Problem 21.89: F+ = qE+ =
− ⎟ iˆ
⎜
4π ⑀0 a ⎝ x − a x ⎠
G G
G
G
G
1 Qq ⎛ 1
1 ⎞ˆ
1 Qq ⎛ 1
2
1 ⎞ˆ
and F− = qE− = −
− +
⎜ −
⎟ i . Therefore, F = F+ + F− =
⎜
⎟ i . For
4π ⑀0 a ⎝ x x + a ⎠
4π ⑀0 a ⎝ x − a x x + a ⎠
G
x a, F ≈
⎞
⎛ a a2
⎞⎞
Qq ⎛ ⎛ a a 2
1 2Qqa ˆ
⎜ ⎜1 + + 2 + . . . ⎟ − 2 + ⎜ 1 − + 2 − . . . ⎟ ⎟ iˆ =
i.
⎜
⎟
⎜
⎟
⎟
4π ⑀0 ax ⎜⎝ ⎝
4
π
⑀0 x3
x x
x
x
⎠
⎝
⎠⎠
1
EVALUATE: If the charge distributed along the x-axis were all positive or all negative, the force would be
proportional to 1/y 2 in part (a) and to 1/x 2 in part (b), when y or x is very large.
21.96.
IDENTIFY: Apply ∑ Fx = 0 and ∑ Fy = 0 to the sphere, with x horizontal and y vertical.
G
SET UP: The free-body diagram for the sphere is given in Figure 21.96. The electric field E of the sheet
is directed away from the sheet and has magnitude E =
σ
2⑀0
(Eq. 21.12).
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-46
Chapter 21
EXECUTE: ∑ Fy = 0 gives T cos α = mg and T =
T=
mg
qσ
. ∑ Fx = 0 gives T sin α =
and
cos α
2⑀0
qσ
mg
qσ
qσ
. Combining these two equations we have
=
and tan α =
. Therefore,
2⑀0 sin α
cos α 2⑀0 sin α
2⑀0mg
⎛ qσ ⎞
⎟.
⎝ 2⑀0mg ⎠
α = arctan ⎜
EVALUATE: The electric field of the sheet, and hence the force it exerts on the sphere, is independent of
the distance of the sphere from the sheet.
Figure 21.96
21.97.
IDENTIFY: Divide the charge distribution into small segments, use the point charge formula for the
electric field due to each small segment and integrate over the charge distribution to find the x and y
components of the total field.
SET UP: Consider the small segment shown in Figure 21.97a.
EXECUTE: A small segment that subtends
angle dθ has length a dθ and contains charge
⎛ adθ ⎞
2Q
dQ = ⎜ 1
dθ . ( 12 π a is the total
⎟Q =
⎜ πa ⎟
π
⎝2
⎠
length of the charge distribution.)
Figure 21.97a
The charge is negative, so the field at the origin is directed toward the small segment. The small segment is
located at angle θ as shown in the sketch. The electric field due to dQ is shown in Figure 21.97b, along
with its components.
dE =
dE =
1 dQ
4π ⑀0 a 2
Q
2
2π ⑀0a 2
dθ
Figure 21.97b
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
21-47
dE x = dE cosθ = (Q/2π 2⑀0a 2 )cosθ dθ
E x = ∫ dE x =
Q
Ñ
π /2
2π 2⑀0a 2 0
cosθ dθ =
Q
(sin θ
2π 2⑀0a 2
π /2
0
)=
Q
2π 2⑀0a 2
dE y = dE sin θ = (Q/2π 2⑀0a 2 )sin θ dθ
E y = ∫ dE y =
Q
2π
2
Ñ
π /2
⑀0a 2 0
sin θ dθ =
Q
2π 2⑀0a 2
(− cosθ
π /2
Q
)= 2 2
0
2π ⑀ 0a
EVALUATE: Note that Ex = E y , as expected from symmetry.
21.98.
IDENTIFY: We must add the electric field components of the positive half and the negative half.
SET UP: From Problem 21.97, the electric field due to the quarter-circle section of positive charge has
Q
Q
components E x = + 2 2 , E y = − 2 2 . The field due to the quarter-circle section of negative
2π ⑀0a
2π ⑀0a
Q
Q
, Ey = + 2 2 .
2π 2⑀0a 2
2π ⑀0a
EXECUTE: The components of the resultant field is the sum of the x- and y-components of the fields due
Q
to each half of the semicircle. The y-components cancel, but the x-components add, giving E x = + 2 2 ,
π ⑀0a
in the + x-direction.
EVALUATE: Even though the net charge on the semicircle is zero, the field it produces is not zero because
of the way the charge is arranged.
IDENTIFY: Each wire produces an electric field at P due to a finite wire. These fields add by vector addition.
1
Q
. The field due to the negative wire points to the left,
SET UP: Each field has magnitude
4π ⑀0 x x 2 + a 2
charge has components E x = +
21.99.
while the field due to the positive wire points downward, making the two fields perpendicular to each other
and of equal magnitude. The net field is the vector sum of these two, which is
1
Q
cos 45°. In part (b), the electrical force on an electron at P is eE.
Enet = 2E1 cos 45° = 2
4π ⑀0 x x 2 + a 2
EXECUTE: (a) The net field is Enet = 2
Enet =
1
Q
cos 45°.
4π ⑀0 x x 2 + a 2
2(9.00 × 109 N ⋅ m 2 /C2 )(2.50 × 10−6 C)cos 45°
2
(0.600 m) (0.600 m) + (0.600 m)
2
= 6.25 × 104 N/C.
The direction is 225° counterclockwise from an axis pointing to the right at point P.
(b) F = eE = (1.60 × 10−19 C)(6.25 × 104 N/C) = 1.00 × 10−14 N, opposite to the direction of the electric
21.100.
field, since the electron has negative charge.
EVALUATE: Since the electric fields due to the two wires have equal magnitudes and are perpendicular to
each other, we only have to calculate one of them in the solution.
IDENTIFY: Each sheet produces an electric field that is independent of the distance from the sheet. The net
field is the vector sum of the two fields.
SET UP: The formula for each field is E = σ /2⑀0 , and the net field is the vector sum of these,
Enet =
σB σ A σB ±σ A
, where we use the + or − sign depending on whether the fields are in the
±
=
2⑀0 2⑀0
2⑀ 0
same or opposite directions and σ B and σ A are the magnitudes of the surface charges.
EXECUTE: (a) The two fields oppose and the field of B is stronger than that of A, so
Enet =
σ B σ A σ B − σ A 11.6 μ C/m 2 − 9.50 μ C/m 2
−
=
=
= 1.19 × 105 N/C, to the right.
−12 2
2
2⑀0 2⑀0
2⑀0
2(8.85 × 10 C /N ⋅ m )
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-48
Chapter 21
(b) The fields are now in the same direction, so their magnitudes add.
Enet = (11.6 μ C/m 2 + 9.50 μ C/m 2 )/2⑀0 = 1.19 × 106 N/C, to the right
(c) The fields add but now point to the left, so Enet = 1.19 × 106 N/C, to the left.
21.101.
EVALUATE: We can simplify the calculations by sketching the fields and doing an algebraic solution first.
IDENTIFY: Each sheet produces an electric field that is independent of the distance from the sheet. The net
field is the vector sum of the two fields.
SET UP: The formula for each field is E = σ /2⑀0 , and the net field is the vector sum of these,
Enet =
σB
2⑀0
±
σA
2⑀0
=
σB ±σ A
, where we use the + or − sign depending on whether the fields are in the
2⑀ 0
same or opposite directions and σ B and σ A are the magnitudes of the surface charges.
EXECUTE: (a) The fields add and point to the left, giving Enet = 1.19 × 106 N/C.
(b) The fields oppose and point to the left, so Enet = 1.19 × 105 N/C.
(c) The fields oppose but now point to the right, giving Enet = 1.19 × 105 N/C.
21.102.
EVALUATE: We can simplify the calculations by sketching the fields and doing an algebraic solution first.
IDENTIFY: The sheets produce an electric field in the region between them which is the vector sum of the
fields from the two sheets.
SET UP: The force on the negative oil droplet must be upward to balance gravity. The net electric field
between the sheets is E = σ /⑀0 , and the electrical force on the droplet must balance gravity, so qE = mg .
EXECUTE: (a) The electrical force on the drop must be upward, so the field should point downward since
the drop is negative.
(b) The charge of the drop is 5e, so qE = mg . (5e)(σ /⑀0 ) = mg and
σ=
21.103.
mg⑀0 (324 × 10−9 kg)(9.80 m/s 2 )(8.85 × 10−12 C2 /N ⋅ m 2 )
=
= 35.1 C/m 2
5e
5(1.60 × 10−19 C)
EVALUATE: Balancing oil droplets between plates was the basis of the Milliken Oil-Drop Experiment
which produced the first measurement of the mass of an electron.
IDENTIFY and SET UP: Example 21.11 gives the electric field due to one infinite sheet. Add the two fields
as vectors.
G
EXECUTE: The electric field due to the first sheet, which is in the xy-plane, is E1 = (σ /2⑀0 ) kˆ for z > 0 and
G
G
E1 = − (σ /2⑀ 0 )kˆ for z < 0. We can write this as E1 = (σ /2⑀0 )( z/ z )kˆ , since z/ z = +1 for z > 0 and
z/ z = − z/z = −1 for z < 0. Similarly, we can write the electric field due to the second sheet as
G
E2 = − (σ /2⑀ 0 )( x/ x )iˆ, since its charge density is −σ . The net field is
G G
G
E = E1 + E2 = (σ /2⑀0 )( −( x/ x )iˆ + ( z/ z )kˆ ).
EVALUATE: The electric field is independent of the y-component of the field point since displacement in the
± y -direction is parallel to both planes. The field depends on which side of each plane the field is located.
21.104.
IDENTIFY: Apply Eq. (21.11) for the electric field of a disk. The hole can be described by adding a disk of
charge density −σ and radius R1 to a solid disk of charge density +σ and radius R2 .
SET UP: The area of the annulus is π ( R22 − R12 )σ . The electric field of a disk, Eq. (21.11) is
E=
σ ⎡
1 − 1/ ( R/x) 2 + 1 ⎤ .
⎥⎦
2⑀ 0 ⎢⎣
EXECUTE: (a) Q = Aσ = π ( R22 − R12 )σ
G
σ ⎛⎡
2
2
⎤ ⎡
⎤⎞ x
(b) E ( x) =
⎜ ⎢1 − 1/ ( R2 /x) + 1 ⎥ − ⎢1 − 1/ ( R1/x) + 1 ⎥ ⎟ iˆ.
⎦ ⎣
⎦⎠ x
2⑀0 ⎝ ⎣
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
(
21-49
)
G
x ˆ
σ
1/ ( R1/x) 2 + 1 − 1/ ( R2 /x) 2 + 1
E ( x) =
i . The electric field is in the + x-direction at points above
2⑀0
x
the disk and in the −x-direction at points below the disk, and the factor
(c) Note that 1/ ( R1/x ) 2 + 1 =
x ˆ
i specifies these directions.
x
x
x
(1 + ( x/R1 ) 2 ) −1/2 ≈ . This gives
R1
R1
2
G
σ ⎛1 1 ⎞x ˆ σ ⎛1 1 ⎞ ˆ
2
E ( x) =
i=
⎜ −
⎟
⎜ −
⎟ x i . Sufficiently close means that ( x/R1) 1.
2⑀0 ⎝ R1 R2 ⎠ x
2⑀0 ⎝ R1 R2 ⎠
(d) Fx = − qEx = 2
k=
qσ ⎛ 1
1 ⎞
⎜ −
⎟ x . The force is in the form of Hooke’s law: Fx = −kx, with
2⑀0 ⎝ R1 R2 ⎠
qσ ⎛ 1
1 ⎞
⎜ −
⎟.
2⑀0 ⎝ R1 R2 ⎠
f =
1
2π
k
1
=
m 2π
qσ ⎛ 1
1 ⎞
⎜ −
⎟.
2⑀0m ⎝ R1 R2 ⎠
EVALUATE: The frequency is independent of the initial position of the particle, so long as this position is
sufficiently close to the center of the annulus for ( x/R1 ) 2 to be small.
21.105.
IDENTIFY: Apply Coulomb’s law to calculate the forces that q1 and q2 exert on q3 , and add these force
vectors to get the net force.
SET UP: Like charges repel and unlike charges attract. Let + x be to the right and + y be toward the top of
the page.
EXECUTE: (a) The four possible force diagrams are sketched in Figure 21.105a.
Only the last picture can result in a net force in the − x -direction.
(b) q1 = −2.00 μ C, q3 = +4.00 μ C, and q2 > 0.
G
G
(c) The forces F1 and F2 and their components are sketched in Figure 21.105b.
Fy = 0 = −
q2 =
q1 q3
q2 q3
1
1
sin θ1 +
sin θ 2 . This gives
4π ⑀0 (0.0400 m) 2
4π ⑀0 (0.0300 m) 2
9
sin θ1 9
3/5 27
q1
=
q1
=
q1 = 0.843 μ C.
16 sin θ 2 16
4/5 64
(d) Fx = F1x + F2 x and Fy = 0, so F = q3
q1
q2
1 ⎛
4
3⎞
+
⎜⎜
⎟ = 56.2 N.
2
2
4π ⑀0 ⎝ (0.0400 m) 5 (0.0300 m) 5 ⎟⎠
G
EVALUATE: The net force F on q3 is in the same direction as the resultant electric field at the location of
q3 due to q1 and q2 .
Figure 21.105
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21-50
21.106.
Chapter 21
IDENTIFY: Calculate the electric field at P due to each charge and add these field vectors to get the net
field.
SET UP: The electric field of a point charge is directed away from a positive charge and toward a negative
charge. Let + x be to the right and let + y be toward the top of the page.
EXECUTE: (a) The four possible diagrams are sketched in Figure 21.106a.
The first diagram is the only one in which the electric field must point in the negative y-direction.
(b) q1 = −3.00 μ C, and q2 < 0.
G
G
5
(c) The electric fields E1 and E2 and their components are sketched in Figure 24.106b. cos θ1 = ,
13
k q1
k q2
5
12
12
12
5
. This gives
+
sin θ1 = , cos θ 2 = and sin θ 2 = . E x = 0 = −
13
13
13
(0.050 m) 2 13 (0.120 m) 2 13
k q2
(0.120 m)
Ey = −
2
=
k q1
5
(0.050 m)
2 12
k q1
12
(0.050 m)
2 13
−
. Solving for q2 gives q2 = 7.2 μ C, so q2 = −7.2 μ C. Then
5
kq2
(0.120 m)
2 13
= −1.17 × 107 N/C. E = 1.17 × 107 N/C.
G
EVALUATE: With q1 known, specifying the direction of E determines both q2 and E.
Figure 21.106
21.107.
IDENTIFY: To find the electric field due to the second rod, divide that rod into infinitesimal segments of
length dx, calculate the field dE due to each segment and integrate over the length of the rod to find the
G
G
total field due to the rod. Use dF = dq E to find the force the electric field of the second rod exerts on
each infinitesimal segment of the first rod.
SET UP: An infinitesimal segment of the second rod is sketched in Figure 21.107. dQ = (Q/L )dx′.
EXECUTE: (a) dE =
L
E x = Ñ dE x =
0
k dQ
( x + a/2 + L − x′) 2
=
kQ
dx′
.
L ( x + a/2 + L − x′)2
L
kQ L
dx′
kQ ⎡
1
kQ ⎛ 1
1
⎤
⎞
=
=
−
⎜
⎟.
L Ñ0 ( x + a/2 + L − x′) 2
L ⎢⎣ x + a/2 + L − x′ ⎥⎦ 0
L ⎝ x + a/2 x + a/2 + L ⎠
2kQ ⎛ 1
1
⎞
−
⎜
⎟.
L ⎝ 2 x + a 2L + 2 x + a ⎠
(b) Now consider the force that the field of the second rod exerts on an infinitesimal segment dq of the first
rod. This force is in the + x -direction. dF = dq E.
Ex =
F = ∫ E dq = Ñ
L + a/2 EQ
a/ 2
L
dx =
(
2kQ 2
2
L
L + a/2 ⎛
Ña/2
1
1
⎞
−
⎜
⎟ dx.
2
2
2
+
+
+
x
a
L
x
a
⎝
⎠
)
2kQ 2 1
kQ 2 ⎛ a + 2 L + a ⎞⎛ 2 L + 2a ⎞ ⎞
F= 2
[ln (a + 2 x)]aL/2+ a/2 − [ln(2 L + 2 x + a)]aL/2+ a/2 = 2 1n ⎜ ⎜⎛
⎟⎜
⎟ ⎟.
2a
L 2
L
⎠⎝ 4 L + 2a ⎠ ⎠
⎝⎝
F=
⎛ (a + L)2 ⎞
1n
⎜⎜
⎟⎟ .
L2
⎝ a (a + 2 L) ⎠
kQ 2
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
(c) For a L, F =
21-51
⎛ a 2 (1 + L/a ) 2 ⎞ kQ 2
1n
⎜⎜ 2
⎟⎟ = 2 (2 1n (1 + L/a ) − ln(1 + 2 L/a )).
L2
⎝ a (1 + 2 L/a ) ⎠ L
kQ 2
z2
. Therefore, for a L,
2
⎞ ⎛ 2 L 2 L2
⎞ ⎞ kQ 2
kQ 2 ⎛ ⎛ L L2
− 2 + ⋅⋅⋅ ⎟ ⎟ ≈ 2 .
F ≈ 2 ⎜ 2 ⎜ − 2 + ⋅⋅⋅ ⎟ − ⎜
⎟ ⎜ a
⎟⎟ a
L ⎜⎝ ⎜⎝ a 2a
a
⎠ ⎝
⎠⎠
EVALUATE: The distance between adjacent ends of the rods is a. When a L the distance between the
rods is much greater than their lengths and they interact as point charges.
For small z, ln(1 + z ) ≈ z −
Figure 21.107
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.