I NSTRUCTOR ’ S S OLUTIONS M ANUAL
FOR
S ERWAY
AND
V UILLE ’ S
C O L L E G E P H YS I C S
N INTH E DITION , V OLUME 2
Charles Teague
Emeritus, Eastern Kentucky University
Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States
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1 2 3 4 5 6 7 14 13 12 11 10
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TABLE OF CONTENTS
Acknowledgements
Preface
v
vii
Part 4 – Electricity and Magnetism
Chapter 15 – Electric Forces and Electric Fields
Chapter 16 – Electrical Energy and Capacitance
Chapter 17 – Current and Resistance
Chapter 18 – Direct-Current Circuits
Chapter 19 – Magnetism
Chapter 20 – Induced Voltages and Inductance
Chapter 21 – Alternating Current Circuits and Electromagnetic Waves
1
30
62
83
120
148
171
Part 5 – Light and Optics
Chapter 22 – Reflection and Refraction of Light
Chapter 23 – Mirrors and Lenses
Chapter 24 – Wave Optics
Chapter 25 – Optical Instruments
198
223
256
282
Part 6 – Modern Physics
Chapter 26 – Relativity
Chapter 27 – Quantum Physics
Chapter 28 – Atomic Physics
Chapter 29 – Nuclear Physics
Chapter 30 – Nuclear Physics and Elementary Particles
305
324
341
360
377
iii
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ACKNOWLEDGEMENTS
The author would like to thank everyone who has contributed to this work.
In particular, thanks go to the support staff at Cengage Learning for their excellent guidance
and support in all phases of this project. Special mention goes to Physics Publisher, Charles Hartford;
Development Editor, Ed Dodd; Associate Content Project Manager, Holly Schaff ; Associate Development
Editor, Brandi Kirksey; and Editorial Assistant, Brendan Killion.
Susan English of Durham Technical Community College served as accuracy reviewer for this manual.
Her contributions are deeply appreciated. Any remaining errors in this work are the responsibility of the
author alone.
I would like to acknowledge the staff of MPS Limited, a Macmillan Company for their excellent work
in assembling and typing this manual and preparing diagrams and page layouts.
Finally, the author would like to thank his wife, Carol, for her patience, understanding, and great
support during this effort.
v
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PREFACE
This manual is written to accompany College Physics, Ninth Edition, by Raymond A. Serway and Chris
Vuille. For each chapter in that text, the manual includes solutions to all end-of-chapter problems, more
detailed answers to Quick Quizzes and Multiple Choice Questions than available in the main text, and
answers to the even-numbered Conceptual Questions.
Considerable effort has been made to ensure that the solutions and answers given in this manual
comply with the rules on significant figures and rounding given in the Chapter 1 of the textbook. This
means that intermediate answers are rounded to the proper number of significant figures when written,
and that rounded value is used in all subsequent calculations. Users should not be concerned if their
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You are encouraged to keep this manual out of the hands of students as instructors in many colleges throughout the country use this textbook, and many of them use graded problem assignments as
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Attempting to keep the manual of manageable size, and recognizing that the primary users will
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questions have been shortened by not offering detailed arguments that lead to the answer. Problem
solutions often omit commentary, intermediate steps, as well as initial steps that could be necessary for
clear understanding by students. On occasions where selected problem solutions are to be shared with
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We welcome your comments on the accuracy of the solutions as presented here, as well as suggestions for alternative approaches.
Charles Teague
vii
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15
Electric Forces and Electric Fields
QUICK QUIZZES
1.
Choice (b). Object A must have a net charge because two neutral objects do not attract each other.
Since object A is attracted to positively-charged object B, the net charge on A must be negative.
2.
Choice (b). By Newton’s third law, the two objects will exert forces having equal magnitudes but
opposite directions on each other.
3.
Choice (c). The electric field at point P is due to charges other than the test charge. Thus, it is
unchanged when the test charge is altered. However, the direction of the force this field exerts on
the test change is reversed when the sign of the test charge is changed.
4.
Choice (a). If a test charge is at the center of the ring, the force exerted on the test charge by
charge on any small segment of the ring will be balanced by the force exerted by charge on the
diametrically opposite segment of the ring. The net force on the test charge, and hence the electric
field at this location, must then be zero.
5.
Choices (c) and (d). The electron and the proton have equal magnitude charges of opposite signs.
The forces exerted on these particles by the electric field have equal magnitude and opposite
directions. The electron experiences an acceleration of greater magnitude than does the proton
because the electron’s mass is much smaller than that of the proton.
6.
Choice (a). The field is greatest at point A because this is where the field lines are closest together.
The absence of lines at point C indicates that the electric field there is zero.
7.
Choice (c). When a plane area A is in a uniform electric field E, the flux through that
area is Φ E = EA cosq , where q is the angle the electric field makes with the line normal to the plane of A. If A lies in the xy-plane and E is in the z-direction, then q = 0° and
Φ E = EA = ( 5.00 N C ) ( 4.00 m 2 ) = 20.0 N ⋅ m 2 C.
8.
Choice (b). If q = 60° in Quick Quiz 15.7 above, then Φ E = EA cosq which yields
Φ E = ( 5.00 N C ) ( 4.00 m 2 ) cos ( 60° ) = 10.0 N ⋅ m 2 C.
9.
Choice (d). Gauss’s law states that the electric flux through any closed surface is equal to the net
enclosed charge divided by the permittivity of free space. For the surface shown in Figure 15.28,
the net enclosed charge is Q = −6 C, which gives Φ E = Q ∈0 = − ( 6 C ) ∈0.
10.
Choices (b) and (d). Since the net flux through the surface is zero, Gauss’s law says that the net
change enclosed by that surface must be zero as stated in (b). Statement (d) must be true because
there would be a net flux through the surface if more lines entered the surface than left it (or
vise-versa).
1
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2
Chapter 15
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
To balance the weight of the ball, the magnitude of the upward electric force must equal the magnitude of the downward gravitation force, or qE = mg, which gives
E=
−3
2
mg ( 5.0 × 10 kg ) ( 9.80 m s )
=
= 1.2 × 10 4 N C
q
4.0 × 10 −6 C
and the correct choice is (b).
2.
The magnitude of the electric field at distance r from a point charge q is E = ke q r 2 , so
E=
(8.99 × 10
9
N ⋅ m 2 C2 ) (1.60 × 10 −19 C )
(5.29 × 10
−11
m)
2
= 5.14 × 1011 N C ∼ 1012 N C
making (e) the best choice for this question.
3.
The magnitude of the electric force between two protons separated by distance r is F = ke e 2 r 2 ,
so the distance of separation must be
r=
ke e 2
=
F
(8.99 × 10
9
N ⋅ m 2 C2 ) (1.60 × 10 −19 C )
2.3 × 10 −26 N
2
= 0.10 m
and (a) is the correct choice.
4.
The ball is made of a metal, so free charges within the ball will very quickly rearrange themselves
to produce electrostatic equilibrium at all points within the ball. As soon as electrostatic equilibrium exists inside the ball, the electric field is zero at all points within the ball. Thus, the correct
choice is (c).
5.
Choosing the surface of the box as the closed surface of interest and applying Gauss’s law, the net
electric flux through the surface of the box is found to be
ΦE =
Qinside ( 3.0 − 2.0 − 7.0 + 1.0 ) × 10 −9 C
=
= −5.6 × 10 2 N ⋅ m 2 C
∈0
8.85 × 10 −12 C2 N ⋅ m 2
meaning that (b) is the correct choice.
6.
From Newton’s second law, the acceleration of the electron will be
ax =
−19
3
Fx qEx ( −1.60 × 10 C ) (1.00 × 10 N C )
=
=
= −1.76 × 1014 m s 2
m
m
9.11 × 10 −31 kg
The kinematics equation vx2 = v02 x + 2ax ( Δx ), with vx = 0, gives the stopping distance as
− ( 3.00 × 10 6 m s )
−v 2
Δx = 0 x =
= 2.56 × 10 −2 m = 2.56 cm
2ax
2 ( −1.76 × 1014 m s 2 )
2
so (a) is the correct response for this question.
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Electric Forces and Electric Fields
7.
3
The displacement from the −4.00 nC charge at point ( 0, 1.00 ) m to the point ( 4.00, − 2.00 ) m
(
(
)
)
has components rx = x f − xi = +4.00 m and ry = y f − yi = −3.00 m, so the magnitude of this
(
)
displacement is r = r + r = 5.00 m and its direction is q = tan −1 ry rx = −36.9°. The
2
x
2
y
x-component of the electric field at point ( 4.00, − 2.00 ) m is then
Ex = E cosq =
(8.99 × 109 N ⋅ m 2 C2 ) ( −4.00 × 10−9 C) cos ( −36.9°) = −1.15 N C
ke q
cosq
=
r2
( 5.00 m )2
and the correct response is (d).
8.
The magnitude of the electric force between charges Q1i and Q2 i, separated by distance ri , is
Fi = ke Q1i Q2 i ri 2 . If changes are made so Q1 f = Q1i , Q2 f = Q2 i 3, and rf = 2ri , the magnitude of the
new force will be
Ff =
ke Q1 f Q2 f
rf2
=
ke Q1i (Q2 i 3)
( 2r )
2
i
=
1 ⎛ ke Q1i Q2 i ⎞ 1
=
Fi
2
3 ( 2 ) ⎜⎝ ri 2 ⎟⎠ 12
so choice (a) is the correct answer for this question.
9.
Each of the situations described in choices (a) through (d) displays a high degree of symmetry,
and as such, readily lends itself to the use of Gauss’s law to determine the electric fields generated. Thus, the best answer for this question is choice (e), stating that Gauss’s law can be readily
applied to find the electric field in all of these contexts.
10.
When a charged insulator is brought near a metallic object, free charges within the metal move
around, causing the metallic object to become polarized. Within the metallic object, the center of
charge for the type of charge opposite to that on the insulator will be located closer to the charged
insulator than will the center of charge for the same type as that on the insulator. This causes the
attractive force between the charged insulator and the opposite type of charge in the metal to
exceed the magnitude of the repulsive force between the insulator and the same type of charge
in the metal. Thus, the net electric force between the insulator and the metallic object is one of
attraction, and choice (b) is the correct answer.
11.
The outer regions of the atoms in your body and the atoms making up the ground both contain
negatively charged electrons. When your body is in close proximity to the ground, these negatively
charged regions exert repulsive forces on each other. Since the atoms in the solid ground are rigidly locked in position and cannot move away from your body, this repulsive force prevents your
body from penetrating the ground. The best response for this question is choice (e).
12.
The positive charge +2Q makes a contribution to the electric field at
the upper right corner that is directed away from this charge in the
direction of the arrow labeled (a). The magnitude of this contribution is E+ = ke (2Q) / 2s 2, where s is the length of a side of the square.
Each of the negative charges makes a contribution of magnitude
E− Q = ke Q s 2 directed back toward that charge. The vector sum of
these two contributions due to negative charges has magnitude
E− = 2E− Q cos 45° = 2ke Q s 2
FIGURE MCQ15.12
and is directed along the diagonal of the square in the direction of the arrow labeled (d). Since
E− > E+ , the resultant electric field at the upper right corner of the square is in the direction of
arrow (d) and has magnitude E = E− − E+ = ( 2 − 1)ke Q s 2 . The correct answer to the question is
choice (d).
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4
Chapter 15
13.
If the positive charge +2Q at the lower left corner of the square in the above figure were
removed, the field contribution E+ discussed above would be eliminated. This would leave only
E− = 2ke Q s 2 as the resultant field at the upper right corner. This has a larger magnitude than
the resultant field E found above, making choice (a) the correct answer.
14.
Metal objects normally contain equal amounts of positive and negative charge and are electrically neutral. The positive charges in both metals and nonmetals are bound up in the nuclei of the
atoms and cannot move about or be easily removed. However, in metals, some of the negative
charges (the outer or valence electrons in the atoms) are quite loosely bound, can move about
rather freely, and are easily removed from the metal. When a metal object is given a positive
charge, this is accomplished by removing loosely bound electrons from the metal rather than by
adding positive charge to it. Taking away the electrons to leave a net positive charge behind very
slightly decreases the mass of the coin. Thus, choice (d) is the best choice for this question.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
Electrons are more mobile than protons and are more easily freed from atoms than the protons
which are tightly bound within the nuclei of the atoms.
4.
Conducting shoes are worn to avoid the build up of a static charge on them as the wearer walks.
Rubber-soled shoes acquire a charge by friction with the floor and could discharge with a spark,
possibly causing an explosive burning situation, where the burning is enhanced by the oxygen.
6.
No. Object A might have a charge opposite in sign to that
of B, but it also might be neutral. In this latter case, object
B causes object A to be polarized, pulling charge of the sign
opposite the change on B toward the near face of A and
pushing an equal amount of charge of the same sign as that
on B toward the far face. Then, due to difference in distances,
the force of attraction exerted by B on the induced charge
of opposite sign is slightly larger than the repulsive force
exerted by B on the induced charge of like sign. Therefore,
the net force on A is toward B.
8.
(a)
Yes. The positive charges create electric fields that extend in all directions from those
charges. The total field at point A is the vector sum of the individual fields produced by the
charges at that point.
(b)
No, because there are no field lines emanating from or converging on point A.
(c)
No. There must be a charged object present to experience a force.
10.
Electric field lines start on positive charges and end on negative charges. Thus, if the fair-weather
field is directed into the ground, the ground must have a negative charge.
12.
To some extent, a television antenna will act as a lightning rod on the house. If the antenna is
connected to the Earth by a heavy wire, a lightning discharge striking the house may pass through
the metal support rod and be safely carried to the Earth by the ground wire.
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Electric Forces and Electric Fields
14.
16.
(a)
If the charge is tripled, the flux through the surface is also tripled because the net flux is
proportional to the charge inside the surface.
(b)
The flux remains constant when the volume changes because the surface surrounds the
same amount of charge, regardless of its volume.
(c)
The flux does not change when the shape of the closed surface changes.
(d)
The flux through the closed surface remains unchanged as the charge inside the surface is
moved to another location inside that surface.
(e)
The flux is zero because the charge inside the surface is zero. All of these conclusions are
arrived at through an understanding of Gauss’s law.
5
All of the constituents of air are nonpolar except for water. The polar water molecules in the
air quite readily “steal” charge from a charged object, as any physics teacher trying to perform
electrostatics demonstrations in the summer well knows. As a result—it is difficult to accumulate
large amounts of excess charge on an object in a humid climate. During a North American winter,
the cold, dry air allows accumulation of significant excess charge, giving the possibility for
shocks caused by static electricity sparks.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
1.57 N directed to the left
4.
(a)
6.
2.25 × 10 −9 N m
8.
4.33ke q 2 a 2 to the right and 45° above the horizontal
0.115 N
(b)
1.25 cm
10.
F6 m C = 46.7 N to the left; F1.5 m C = 157 N to the right; F−2 m C = 111 N to the left
12.
5.15 × 10 3 N m
14.
16.7 mC
16.
(a)
0
(b)
30.0 N
(c)
21.6 N
(d)
17.3 N
(e)
–13.0 N
(f)
17.3 N
(g)
17.0 N
(h)
FR = 24.3 N at 44.5° above the + x-axis
(b)
40.0 N to the left
5.27 × 10 5 m s
18.
(a) 2.00 × 10 7 N C to the right
20.
(a)
5.27 × 1013 m s 2
(b)
22.
(a)
See Solution.
(b) 1.4 × 10 3 N C
24.
(a) 2.19 × 10 5 N C at 85.2° below the + x-axis
(c)
7.5 × 10 −2 N
(b) The electric field would be unchanged, but the force would double in magnitude.
68719_15_ch15_p001-029.indd 5
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6
Chapter 15
26.
4
(a) 1.80 × 10 N C to the right
28.
1.88 × 10 3 N C at 4.40° below the + x-axis
30.
(a)
32.
(a), (b), and (c)
34.
(a)
q1 q2 = −1 3
(b)
(b)
9.00 × 10 −5 N to the left
q1 < 0, q2 > 0
See Solution for Sketches.
(b)
E = 0 at the center of the triangle.
(c) 1.73ke q a 2
(b)
positive y-direction
36.
(a)
(b) 1.8 × 10 6 N m
38.
r ∼ 10 −6 m
40.
1.38 × 10 3 N ⋅ m 2 C
42.
(a)
44.
(a) 6.55 × 10 5 N ⋅ m 2 C
(b) 1.64 × 10 5 N ⋅ m 2 C
46.
(a) 1.92 × 10 7 N ⋅ m 2 C
(b) 3.20 × 10 6 N ⋅ m 2 C
See Solution.
zero
3.38 nC
(c) 1.1 × 10 5 N m
(b) The sphere contains a net positive charge in a
spherically symmetric distribution.
(c) The answer to (b) would change because of the loss of symmetry; however, the answer to
(a) would be unchanged since Qinside is unaltered.
48.
The electric field is everywhere parallel to the z-axis with the following magnitudes:
(a)
50.
Ez = +s 2 ∈0
(b)
Ez = +3s 2 ∈0
(c)
Ez = −s 2 ∈0
(a) 2.94 × 10 5 N C
(b) The magnitude of the field is independent of distance from the sheet, provided that distance
is small compared to the dimensions of the sheet.
52.
+5.25 mC
54.
E = K ed in the direction of the electron’s motion.
56.
(a)
0.307 s
58.
(a)
For 0 < x < 1.00 m, the field contributions made by the two charges are in the same direction and cannot add to zero.
(b) Yes, neglecting gravity causes a 2.28% error.
(b) Any point for x < 0 is nearer the larger charge, so the field contributions made by the two
charges cannot have equal magnitudes and add to zero.
68719_15_ch15_p001-029.indd 6
(c)
Points for x > 1.00 m are nearer the smaller charge and the field contributions by the two
charges are in opposite directions. Thus, it is possible for them to add to zero.
(d)
E = 0 at x = +9.47 m
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7
Electric Forces and Electric Fields
60.
0.951 m
62.
1.98 mC
64.
(a)
(b) 1.67 × 10 −7 s for q = 37.0°; 2.21 × 10 −7 s for q = 53.0°
37.0° or 53.0°
PROBLEM SOLUTIONS
15.1
(a)
From Coulomb’s law, F = ke
Q1Q2
, we have
r2
F = (8.99 × 10 9 N ⋅ m 2 C2 )
(b)
15.2
( 7.50 × 10
−9
C ) ( 4.20 × 10 −9 C )
(1.80 m )2
= 8.74 × 10 −8 N
Since these are like charges (both positive), the force is repulsive .
Particle A exerts a force toward the right on particle B. By Newton’s third law, particle B will then
exert a of equal magnitude force toward the left on particle A. The ratio of the final magnitude of
the force to the original magnitude of the force is
Ff
Fi
=
⎛r ⎞
=⎜ i ⎟
2
ke q1 q2 ri
⎝ rf ⎠
ke q1 q2 rf2
2
2
so
2
⎛r ⎞
⎛ 13.7 mm ⎞
Ff = Fi ⎜ i ⎟ = ( 2.62 N ) ⎜
= 1.57 N
⎟
⎝ 17.7 mm ⎠
⎝ rf ⎠
The final vector force that B exerts on A is 1.57 N directed to the left .
15.3
(a)
When the balls are an equilibrium distance apart, the tension in the string equals the magnitude of the repulsive electric force between the balls. Thus,
F=
or
15.4
ke q ( 2q )
= 2.50 N ⇒
r2
q=
( 2.50 N )( 2.00 m )2
q2 =
2 (8.99 × 10 9 N ⋅ m 2 C2 )
( 2.50 N ) r 2
2ke
= 2.36 × 10 −5 C = 23.6 mC
(b)
The charges induce opposite charges in the bulkheads, but the induced charge in the bulkhead near ball B is greater, due to B’s greater charge. Therefore, the system
moves slowly towards the bulkhead closer to ball B .
(a)
The gravitational force exerted on the upper sphere by the lower one is negligible in comparison to the gravitational force exerted by the Earth and the
downward electrical force exerted by the lower sphere. Therefore,
ΣFy = 0 ⇒ T − mg − Fe = 0
or
T = mg +
ke q1 q2
d2
T = ( 7.50 × 10
−3
T
m, q1
d
mg
Fe
m, q2
9
2
2
−9
−9
m ⎞ (8.99 × 10 N ⋅ m C ) ( 32.0 × 10 C ) ( 58.0 × 10 C )
⎛
kg ) ⎜ 9.80 2 ⎟ +
2
⎝
s ⎠
( 2.00 × 10 −2 m )
giving T = 0.115 N
continued on next page
68719_15_ch15_p001-029.indd 7
1/7/11 2:28:47 PM
8
Chapter 15
ΣFy = 0 → Fe =
(b)
ke q1 q2
= T − mg,
d2
ke q1 q2
T − mg
d=
and
Thus, if T = 0.180 N,
d=
15.5
(8.99 × 10
N ⋅ m 2 C2 ) ( 32.0 × 10 −9 C ) ( 58.0 × 10 −9 C )
= 1.25 × 10 −2 m = 1.25 cm
m
⎛
⎞
0.180 N − ( 7.50 × 10 −3 kg ) ⎜ 9.80 2 ⎟
⎝
s ⎠
2
2
(8.99 × 109 N ⋅ m 2 C2 ) ⎡⎣ 4 (1.60 × 10 −19 C) ⎤⎦
ke ( 2e )
F=
=
= 36.8 N
2
r2
(5.00 × 10 −15 m )
(a)
The mass of an alpha particle is m = 4.002 6 u, where 1 u = 1.66 × 10 −27 kg is the unified
mass unit. The acceleration of either alpha particle is then
(b)
a=
15.6
9
F
36.8 N
=
= 5.54 × 10 27 m s 2
m 4.002 6 (1.66 × 10 −27 kg )
The attractive force between the charged ends tends to compress the molecule. Its magnitude is
−19
2
⎛
k (1e )
N ⋅ m 2 ⎞ (1.60 × 10 C )
F= e 2
= ⎜ 8.99 × 10 9
= 4.89 × 10 −17 N
2
r
C2 ⎟⎠ ( 2.17 × 10 −6 m )
⎝
2
The compression of the “spring” is
x = ( 0.010 0 ) r = ( 0.010 0 ) ( 2.17 × 10 −6 m ) = 2.17 × 10 −8 m ,
so the spring constant is k =
15.7
F 4.89 × 10 −17 N
=
= 2.25 × 10 −9 N m .
x 2.17 × 10 −8 m
In the new equilibrium position,
ΣFy = Fs − Fe = kd − Fe = 0.
Fs = kd
Fs
Thus,
k q q r 2 ke q1 q2
k=
= e 1 2
=
d
d
d ⋅ r2
or
k=
(8.99 × 10
9
q1
N ⋅ m 2 C2 ) ( 0.800 × 10 −6 C ) ( −0.600 × 10 −6 C )
(3.50 × 10
−2
m ) ( 5.00 × 10
−2
m)
r
Fe
2
q2
k = 49.3 N m
15.8
See the sketch at the right. The magnitudes of the forces are
F1 = F2 = ke
q ( 2q )
q ( 3q )
q2
q2
=
2k
and
F
=
k
=
1.50k
e
3
e
e
2
a2
a2
a2
a 2
(
)
The components of the resultant force on charge q are
Fx = F1 + F3 cos 45° = ( 2 + 1.50 cos 45° ) ke
q2
q2
= 3.06ke 2
2
a
a
continued on next page
68719_15_ch15_p001-029.indd 8
1/7/11 2:28:49 PM
Electric Forces and Electric Fields
and
Fy = F1 + F3 sin 45° = ( 2 + 1.50 sin 45° ) ke
9
q2
q2
= 3.06ke 2
2
a
a
⎛
q2 ⎞
q2
The magnitude of the resultant force is Fe = Fx2 + Fy2 = 2 ⎜ 3.06ke 2 ⎟ = 4.33ke 2
a ⎠
a
⎝
⎛ Fy ⎞
and it is directed at q = tan −1 ⎜ ⎟ = tan −1 (1.00 ) = 45° above the horizontal .
⎝ Fx ⎠
15.9
(a)
The spherically symmetric charge distributions behave as if all charge was located at the
centers of the spheres. Therefore, the magnitude of the attractive force is
F=
(b)
When the spheres are connected by a conducting wire, the net charge
qnet = q1 + q2 = −6.0 × 10 −9 C will divide equally between the two identical spheres.
Thus, the force is now
F=
or
15.10
12 × 10 −9 C ) (18 × 10 −9 C )
ke q1 q2
9
2
2 (
=
8.99
×
10
N
⋅
m
C
= 2.2 × 10 −5 N
(
)
r2
( 0.30 m )2
ke ( qnet 2 )
r2
2
−9
⎛
N ⋅ m 2 ⎞ ( −6.0 × 10 C )
= ⎜ 8.99 × 10 9
2
C2 ⎟⎠ 4 ( 0.30 m )
⎝
2
F = 9.0 × 10 −7 N (repulsion)
The forces are as shown in the sketch below.
F1 =
F2 =
F3 =
2
6.00 × 10 −6 C ) (1.50 × 10 −6 C )
ke q1 q2 ⎛
9 N⋅m ⎞ (
=
8.99
×
10
= 89.9 N
2
⎜⎝
r122
C2 ⎟⎠
(3.00 × 10 −2 m )
ke q1 q3
r132
ke q2 q3
r232
−6
−6
⎛
N ⋅ m 2 ⎞ ( 6.00 × 10 C ) ( 2.00 × 10 C )
= ⎜ 8.99 × 10 9
= 43.2 N
2
C2 ⎟⎠
⎝
(5.00 × 10 −2 m )
−6
−6
⎛
N ⋅ m 2 ⎞ (1.50 × 10 C ) ( 2.00 × 10 C )
= ⎜ 8.99 × 10 9
= 67.4 N
2
C2 ⎟⎠
⎝
( 2.00 × 10−2 m )
The net force on the 6 mC charge is F6 m C = F1 − F2 = 46.7 N to the left .
The net force on the 1.5 mC charge is F1.5 m C = F1 + F3 = 157 N to the right .
The net force on the −2 mC charge is F−2 m C = F2 + F3 = 111 N to the left .
68719_15_ch15_p001-029.indd 9
1/7/11 2:28:51 PM
10
15.11
Chapter 15
In the sketch at the right, FR is the resultant of the
forces F6 and F3 that are exerted on the charge at
the origin by the 6.00 nC and the –3.00 nC charges,
respectively.
−9
−9
⎛
N ⋅ m 2 ⎞ ( 6.00 × 10 C ) ( 5.00 × 10 C )
F6 = ⎜ 8.99 × 10 9
= 3.00 × 10 −6 N
2
2
⎟
C ⎠
⎝
( 0.300 m )
−9
−9
⎛
N ⋅ m 2 ⎞ ( 3.00 × 10 C ) ( 5.00 × 10 C )
F3 = ⎜ 8.99 × 10 9
= 1.35 × 10 −5 N
C2 ⎟⎠
⎝
( 0.100 m )2
The resultant is FR =
or
15.12
(F ) + (F )
2
6
3
2
⎛F ⎞
= 1.38 × 10 −5 N at q = tan −1 ⎜ 3 ⎟ = 77.5°
⎝ F6 ⎠
FR = 1.38 × 10 −5 N at 77.5° below the − x-axis
At equilibrium, ΣFx = Fe − Fs = 0
kx =
Thus,
or
q1
Fs = Fe
Fs = kx
ke q1 q2
d2
Fe
q2
d
and the force constant of the spring is
k=
9
2
2
−6
−6
ke q1 q2 (8.99 × 10 N ⋅ m C ) ( 2.70 × 10 C ) (8.60 × 10 C )
=
2
xd 2
(5.00 × 10 −3 m ) (9.00 × 10 −2 m )
k = 5.15 × 10 3 N m
15.13
Please see the sketch at the right.
F1 =
(8.99 × 10
or
F1 = 0.288 N
F2 =
(8.99 × 10
or
F2 = 0.503 N
9
N ⋅ m 2 C2 ) ( 2.00 × 10 −6 C ) ( 4.00 × 10 −6 C )
( 0.500 m )2
9
N ⋅ m 2 C2 ) ( 2.00 × 10 −6 C ) ( 7.00 × 10 −6 C )
( 0.500 m )2
The components of the resultant force acting on the 2.00 mC charge are:
Fx = F1 − F2 cos 60.0° = 0.288 N − ( 0.503 N ) cos 60.0° = 3.65 × 10 −2 N
and Fy = −F2 sin 60.0° = − ( 0.503 N ) sin 60.0° = − 0.436 N
The magnitude and direction of this resultant force are
continued on next page
68719_15_ch15_p001-029.indd 10
1/7/11 2:28:53 PM
Electric Forces and Electric Fields
F = Fx2 + Fy2 =
at
15.14
( 0.036 5 N ) + ( 0.436 N )
2
2
11
= 0.438 N
⎛ Fy ⎞
⎛ −0.436 N ⎞
= −85.2° or 85.2° below the + x-axis
q = tan −1 ⎜ ⎟ = tan −1 ⎜
⎝ 0.036 5 N ⎟⎠
⎝ Fx ⎠
At equilibrium, Fe = Fs or
Q
Fe
ke Q 2
= kx = k ( L − Li )
L2
Fs
Q
L
Thus,
Q=
k ( L − Li ) L2
=
ke
(100
N m ) ( 0.500 m − 0.400 m ) ( 0.500 m )
8.99 × 10 9 N ⋅ m 2 C2
2
Q = 1.67 × 10 −5 C = 16.7 mC
15.15
Consider the free-body diagram of one of the spheres given
at the right. Here, T is the tension in the string and Fe is the
repulsive electrical force exerted by the other sphere.
ΣFy = 0 ⇒ T cos 5.0° = mg,
or T =
mg
cos 5.0°
ΣFx = 0 ⇒ Fe = T sin 5.0° = mg tan 5.0°
At equilibrium, the distance separating the two spheres is r = 2 L sin 5.0°.
Thus, Fe = mg tan 5.0° becomes
q = ( 2 L sin 5.0° )
mg tan 5.0°
ke
= [ 2 ( 0.300 m ) sin 5.0° ]
15.16
ke q 2
= mg tan 5.0° and yields
( 2 L sin 5.0°)2
( 0.20 × 10
−3
kg ) ( 9.80 m s 2 ) tan 5.0°
8.99 × 10 9 N ⋅ m 2 C2
In the sketch at the right,
rB C =
and
FAC
( 4.00 m ) + ( 3.00 m ) = 5.00 m
2
2
(0, 3.00 m) C
⎛ 3.00 m ⎞
q = tan −1 ⎜
= 36.9°
⎝ 4.00 m ⎟⎠
(a)
(F )
(b)
(F )
AC
AC
x
= 0
y
= ke
= 7.2 nC
1.00 × 10−4 C
q
FBC
(0, 0) A
3.00 × 10−4 C
qA qC
2
rAC
= (8.99 × 10 9 N ⋅ m 2 C2 )
(3.00 × 10
−4
rBC
q
−6.00 × 10−4 C
B
(4.00 m, 0)
C ) (1.00 × 10 −4 C )
( 3.00 m )2
= 30.0 N
continued on next page
68719_15_ch15_p001-029.indd 11
1/7/11 2:28:55 PM
12
15.17
Chapter 15
qB qC
= (8.99 × 10 9 N ⋅ m 2 C2 )
(6.00 × 10
−4
(c)
FB C = ke
(d)
(F )
x
= FB C cosq = ( 21.6 N ) cos ( 36.9° ) = 17.3 N
(e)
(F )
y
= − FB C sinq = − ( 21.6 N ) sin ( 36.9° ) = −13.0 N
(f)
(F ) = (F ) + (F )
x
= 0 + 17.3 N = 17.3 N
(g)
(F ) = (F ) + (F )
y
= 30.0 − 13.0 N = 17.0 N
(h)
FR =
BC
BC
R
R
rB2C
AC
x
AC
y
BC
x
BC
y
(F ) + (F )
2
R
x
R
2
y
=
C ) (1.00 × 10 −4 C )
( 5.00 m )2
= 21.6 N
(17.3 N )2 + (17.0 N )2 = 24.3 N
and
⎡ ( FR ) y
j = tan −1 ⎢
⎢⎣ ( FR )x
or
FR = 24.3 N at 44.5° above the + x-axis
⎤
⎛ 17.0 N ⎞
⎥ = tan −1 ⎜
= 44.5°
⎝ 17.3 N ⎟⎠
⎥⎦
In order to suspend the object in the electric field, the electric force exerted on the object by the
field must be directed upward and have a magnitude equal to the weight of the object. Thus,
Fe = qE = mg, and the magnitude of the electric field must be
(
)
3.80 g ( 9.80 m s 2 ) ⎛ 1 kg ⎞
mg
3
E=
=
⎜ 3 ⎟ = 2.07 × 10 N C
q
18 × 10 −6 C
⎝ 10 g ⎠
The electric force on a negatively charged object is in the direction opposite to that of the electric field.
Since the electric force must be directed upward, the electric field must be directed downward.
15.18
(a)
Taking to the right as
positive, the resultant
electric field at point P
is given by
ER = E1 + E3 − E2
=
ke q1 ke q3 ke q2
+ 2 − 2
r12
r3
r2
⎛
N ⋅ m 2 ⎞ ⎡ 6.00 × 10 −6 C 2.00 × 10 −6 C 1.50 × 10 −6 C ⎤
= ⎜ 8.99 × 10 9
+
−
⎢
⎥
2
C2 ⎟⎠ ⎢⎣ ( 0.020 0 m )
⎝
( 0.030 0 m )2 ( 0.010 0 m )2 ⎥⎦
This gives ER = + 2.00 × 10 7 N C
or
(b)
F = qE R = ( −2.00 × 10 −6 C ) ( 2.00 × 10 7 N C ) = −40.0 N
or
68719_15_ch15_p001-029.indd 12
E R = 2.00 × 10 7 N C to the right
F = 40.0 N to the left
1/7/11 2:28:58 PM
13
Electric Forces and Electric Fields
15.19
The force on a negative charge is opposite to the direction of the electric field and has magnitude
F = q E. Thus,
F = −6.00 × 10 −6 C ( 5.25 × 10 5 N C ) = 3.15 N
15.20
and
F = 3.15 N due north
(a)
The magnitude of the force on the electron is F = q E = eE, and the acceleration is
a=
(b)
15.21
(a)
−19
F
eE (1.60 × 10 C ) ( 300 N C )
=
=
= 5.27 × 1013 m s 2
me me
9.11 × 10 −31 kg
v = v0 + at = 0 + ( 5.27 × 1013 m s 2 ) (1.00 × 10 −8 s ) = 5.27 × 10 5 m s
y
The electrical force must be directed up the incline and have
a magnitude equal to the tangential component of the gravitational force.
n
Q,
Fe
x
m
ΣFx = 0 ⇒ Fe − mg sinq = 0
(b)
q
or
Fe = Q E = mg sinq
E=
−3
2
mg sinq ( 5.40 × 10 kg ) ( 9.80 m s ) sin 25.0°
=
= 3.19 × 10 3 N C
Q
7.00 × 10 −6 C
and
E = mg sinq Q
q
mg
Since Fe must be directed up the incline and the electrical force on a negative charge is
directed opposite to the field, it is necessary to have the electric field directed down the
incline.
E = 3.19 × 10 3 N C down the incline
Thus,
15.22
(a)
Fe
q
T
mg
(b)
ΣF
y
=0
⇒
Fe sinq − mg = 0
or
Fe =
mg
sinq
Since Fe = qE, this gives
E=
(c)
ΣF
x
and
68719_15_ch15_p001-029.indd 13
(5.8 × 10−3 kg) (9.8 m s2 ) = 1.4 × 103 N C
mg
=
q sinq
(68 × 10−6 C) sin 37°
=0
T=
⇒
Fe cosq − T = 0
(5.8 × 10
−3
kg ) ( 9.8 m s 2 )
tan 37°
or
mg
⎛ mg ⎞
T = Fe cosq = ⎜
cosq =
⎝ sinq ⎟⎠
tanq
= 7.5 × 10 −2 N
1/7/11 2:29:00 PM
14
15.23
15.24
Chapter 15
(a)
a=
(b)
t=
−19
F qE (1.60 × 10 C ) ( 640 N C )
=
=
= 6.12 × 1010 m s 2
1.673 × 10 −27 kg
m mp
Δv
1.20 × 10 6 m s
=
= 1.96 × 10 −5 s = 19.6 ms
a
6.12 × 1010 m s 2
v 2f − v02
(1.20 × 10 m s) − 0 =
=
2 ( 6.12 × 10 m s )
2
6
(c)
Δx =
(d)
KE f =
(a)
Please refer to the solution of Problem 15.13 earlier in this chapter. There it is shown
that the resultant electric force experienced by the 2.00 mC located at the origin is
F = 0.438 N at 85.2° below the + x-axis. Since the electric field at a location is defined as
the force per unit charge experienced by a test charge placed in that location, the electric
field at the origin in the charge configuration of Figure P15.13 is
2a
2
11.8 m
2
1
1
m p v 2f = (1.673 × 10 −27 kg )(1.20 × 10 6 m s ) = 1.20 × 10 −15 J
2
2
E=
(b)
10
F
0.438 N
=
at − 85.2° = 2.19 × 10 5 N C at 85.2° below the + x-axis
q0 2.00 × 10 −6 C
The electric force experienced by the charge at the origin is directly proportional to the
magnitude of that charge. Thus, doubling the magnitude of this charge would
double the magnitude of the electric force. However, the electric field is the force per unit
charge and the field would be unchanged if the charge was doubled. This is easily seen in
the calculation of part (a) above. Doubling the magnitude of the charge at the origin would
double both the numerator and the denominator of the ratio F q0 , but the value of the ratio
(i.e., the electric field) would be unchanged.
15.25
r=
(
E y = E1 sin 45.0° + 0 =
(a)
+Q
d
d
d
=
=
2 cos 45.0° 2 2 2
2
E2
r
45.0°
+Q
2d
ke Q
ke Q ke Q ⎛ 2 ⎞ ke Q
kQ ⎛ 2⎞
= 2 − 2e
− 2 ⎜
2
⎟
⎜⎝ 2 ⎟⎠ = 1 − 2 d 2
d
r ⎝ 2 ⎠
d
d
2
( )
kQ ⎛ 2⎞
ke Q ⎛ 2 ⎞
= 2e
=
2 ⎜
⎟
r ⎝ 2 ⎠ ( d 2 ) ⎜⎝ 2 ⎟⎠
Observe the figure at the right:
kQ
E1 = E2 = e 2
r
45.0°
d
P
45.0°
)
Ex = E2 − E1 cos 45.0° =
15.26
E1
From the figure at the right, observe that
2r cos 45.0° + d = 2d. Thus,
k q
and E3 = e 2
r
E y = E1 sin 30.0° − E2 sin 30.0° = 0
Ex = E1 cos 30.0° + E2 cos 30.0° − E3
(
2
)
ke Q
d2
Q = +3.00 nC
r=
4.0
0c
y
m
E3
q = −2.00 nC
r
P
E1
30.0°
30.0° x
E2
Q = +3.00 nC
continued on next page
68719_15_ch15_p001-029.indd 14
1/7/11 2:29:02 PM
Electric Forces and Electric Fields
Ex =
k q k
2ke Q
cos 30.0° − e 2 = 2e ( 2Q cos 30.0° − q )
2
r
r
r
(8.99 × 10 N ⋅ m C ) ⎡ 2 3.00 × 10
(
( 4.00 × 10 m ) ⎣
9
=
15
2
2
−9
2
−2
C ) cos 30.0° − 2.00 × 10 −9 C ⎤⎦
Ex = +1.80 × 10 4 N C
Then,
Fe = q ′E = ( −5.00 × 10 −9 C ) (1.80 × 10 4 N C ) = −9.00 × 10 −5 N
(b)
or
15.27
E = E x2 + E y2 = 1.80 × 10 4 N C and E = 1.80 × 10 4 N C to the right
Fe = 9.00 × 10 −5 N to the left
If the resultant field is zero, the
contributions from the two charges must
be in opposite directions and also have
equal magnitudes. Choose the line
connecting the charges as the x-axis, with
the origin at the –2.5 µC charge. Then, the
two contributions will have opposite
directions only in the regions x < 0 and
x > 1.0 m. For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the
point of zero resultant field is on the x-axis at x < 0.
Requiring equal magnitudes gives
k q
ke q1
= e 22
2
r1
r2
Thus,
(1.0 m + d )
6.0 mC
2.5 mC
=
2
d
(1.0 m + d )2
or
2.5
=d
6.0
Solving for d yields
d = 1.8 m,
15.28
or
1.8 m to the left of the − 2.5 mC charge
The altitude of the triangle is
h = ( 0.500 m ) sin 60.0° = 0.433 m
and the magnitudes of the fields due to
each of the charges are
9
2
2
−9
ke q1 (8.99 × 10 N ⋅ m C ) ( 3.00 × 10 C )
E1 = 2 =
h
( 0.433 m )2
= 144 N C
E2 =
ke q2
(8.99 × 109 N ⋅ m 2 C2 ) (8.00 × 10 −9 C) = 1.15 × 103 N C
=
r22
( 0.250 m )2
continued on next page
68719_15_ch15_p001-029.indd 15
1/7/11 2:29:04 PM
16
Chapter 15
E3 =
and
ke q3
r32
=
(8.99 × 10
9
N ⋅ m 2 C2 ) ( 5.00 × 10 −9 C )
( 0.250 m )2
Thus, ΣEx = E2 + E3 = 1.87 × 10 3 N C
giving
( ΣE ) + ( ΣE )
2
ER =
and
(
2
y
ΣE y = −E1 = −144 N C
= 1.88 × 10 3 N C
)
q = tan −1 ΣE y ΣEx = tan −1 ( −0.0769 ) = −4.40°
Hence
15.29
x
and
= 719 N C
E R = 1.88 × 10 3 N C at 4.40° below the +x-axis
From the symmetry of the charge distribution, students
should recognize that the resultant electric field at the
center is
ER = 0
If one does not recognize this intuitively, consider:
E R = E1 + E 2 + E 3
15.30
Ex = E1 x − E2 x =
and
E y = E1 y + E2 y − E3 =
Thus,
ER = Ex2 + E y2 = 0
(b)
68719_15_ch15_p001-029.indd 16
ke q
k q
k q
sin 30° + e 2 sin 30° − e 2 = 0
r
r
r2
The magnitude of q2 is three times the magnitude of q1 because 3 times as many lines emerge
from q2 as enter q1. q2 = 3 q1
(a)
15.31
ke q
k q
cos 30° − e 2 cos 30° = 0
2
r
r
so
Then,
q1 q2 = −1 3
q2 > 0 because lines emerge from it, and q1 < 0 because lines terminate on it.
Note in the sketches at the right that
electric field lines originate on
positive charges and terminate on
negative charges. The density of
lines is twice as great for the −2 q
charge in (b) as it is for the 1q
charge in (a).
1/7/11 2:29:06 PM
Electric Forces and Electric Fields
15.32
Rough sketches for these charge configurations are shown below.
15.33
(a)
The sketch for (a) is shown at
the right. Note that four times
as many lines should leave q1
as emerge from q2 although, for
clarity, this is not shown in this
sketch.
(b)
The field pattern looks the same
here as that shown for (a) with
the exception that the arrows
are reversed on the field lines.
(a)
The electric field in the plane of the charges has the
general appearance shown at the right:
(b)
It is zero at the center of the triangle , where (by symmetry)
one can see that the three charges individually produce
fields that cancel out. In addition to the center of the
triangle, the electric field lines in the second figure to the
right indicate three other points near the middle of each
leg of the triangle where E = 0, but they are more difficult
to find mathematically.
(c)
Using the sketch at the right, observe that
E1 = E2 = ke q a 2 and the resultant field at
point P is E = E1 + E 2. Thus,
15.34
17
Ex = E1 cos 60.0° − E2 cos 60.0° = 0
and
E y = E1 sin 60.0° + E2 sin 60.0° =
2ke q
sin 60.0°
a2
The magnitude of the resultant field is then
E = Ex2 + E y2 = 0 + E y2 = E y =
1.73ke q
a2
continued on next page
68719_15_ch15_p001-029.indd 17
1/7/11 2:29:08 PM
18
Chapter 15
Since Ex = 0 and E y > 0 , the resultant field at point P is directed vertically upward in the
positive y-direction.
(d)
15.35
15.36
(a)
Zero net charge on each surface of the sphere.
(b)
The negative charge lowered into the sphere repels − 5mC on the outside surface, and
leaves + 5 mC on the inside surface of the sphere.
(c)
The negative charge lowered inside the sphere neutralizes the inner surface, leaving
zero charge on the inside . This leaves − 5mC on the outside surface of the sphere.
(d)
When the object is removed, the sphere is left with − 5.00 mC on the outside surface and
zero charge on the inside .
(a)
The dome is a closed conducting surface. Therefore, the electric field is zero everywhere
inside it.
At the surface and outside of this spherically symmetric charge distribution, the field is as if
all the charge were concentrated at the center of the sphere.
(b)
At the surface,
E=
Outside the spherical dome, E = ke q r 2. Thus, at r = 4.0 m,
(c)
E=
15.37
9
2
2
−4
ke q (8.99 × 10 N ⋅ m C ) ( 2.0 × 10 C )
=
= 1.8 × 10 6 N C
R2
(1.0 m )2
(8.99 × 10
9
N ⋅ m 2 C2 ) ( 2.0 × 10 −4 C )
( 4.0 m )2
= 1.1 × 10 5 N C
For a uniformly charged sphere, the field is strongest at the surface.
Thus,
Em ax =
ke qm ax
,
R2
6
R 2 Em ax ( 2.0 m ) ( 3.0 × 10 N C )
=
= 1.3 × 10 −3 C
ke
8.99 × 10 9 N ⋅ m 2 C2
2
or
15.38
qm ax =
If the weight of the drop is balanced by the electric force, then mg = q E = eE, so the mass of the
drop must be
m=
−19
4
eE (1.6 × 10 C ) ( 3 × 10 N C )
=
≈ 5 × 10 −16 kg
g
9.8 m s 2
But m = rV = r ( 4p r 3 3) and the radius of the drop is r = [ 3m 4pr ] , which gives
13
⎡ 3 ( 5 × 10 −16 kg ) ⎤
r=⎢
⎥
3
⎢⎣ 4p (858 kg m ) ⎥⎦
13
68719_15_ch15_p001-029.indd 18
= 5.2 × 10 −7 m
or
r ∼ 1 mm
1/7/11 2:29:09 PM
Electric Forces and Electric Fields
15.39
(a)
(b)
Fe = ma = (1.67 × 10 −27 kg ) (1.52 × 1012 m s 2 ) = 2.54 × 10 −15 N in the direction of the
acceleration, or radially outward.
The direction of the field is the direction of the force on a positive charge (such as the proton). Thus, the field is directed radially outward . The magnitude of the field is
E=
15.40
19
Fe 2.54 × 10 −15 N
=
= 1.59 × 10 4 N C
q 1.60 × 10 −19 C
normal line
When an electric field of magnitude E is incident
on a surface of area A, the flux through the
surface is
q
Φ E = EA cosq
E
65.0°
where q is the angle between E and the line
normal to the surface. Thus, in this case,
q = 90.0° − 65.0° = 25.0°
and
15.41
Φ E = ( 435 N C ) ( 3.50 m 2 ) cos 25.0° = 1.38 × 10 3 N ⋅ m 2 C
The area of the rectangular plane is A = ( 0.350 m ) ( 0.700 m ) = 0.245 m 2 .
(a)
When the plane is parallel to the yz-plane, its normal line is parallel to the x-axis and makes
an angle q = 0° with the direction of the field. The flux is then
Φ E = EA cosq = ( 3.50 × 10 3 N C ) ( 0.245 m 2 ) cos 0° = 858 N ⋅ m 2 C
15.42
(b)
When the plane is parallel to the x-axis, q = 90° and Φ E = 0 .
(c)
Φ E = EA cosq = ( 3.50 × 10 3 N C ) ( 0.245 m 2 ) cos 40.0° = 657 N ⋅ m 2 C
(a)
Gauss’s law states that the electric flux through any closed surface equals the net charge
enclosed divided by ∈0 . We choose to consider a closed surface in the form of a sphere,
centered on the center of the charged sphere and having a radius infinitesimally larger than
that of the charged sphere. The electric field then has a uniform magnitude and is perpendicular to our surface at all points on that surface. The flux through the chosen closed
surface is therefore Φ E = EA = E ( 4p r 2 ) , and Gauss’s law gives
Qinside = ∈0 Φ E = 4p ∈0 Er 2
= 4p (8.85 × 10 −12 C2 N ⋅ m 2 ) ( 575 N C ) ( 0.230 m ) = 3.38 × 10 −9 C = 3.38 nC
2
(b)
68719_15_ch15_p001-029.indd 19
Since the electric field displays spherical symmetry, you can conclude that the charge distribution generating that field is spherically symmetric . Also, since the electric field lines are
directed outward away from the sphere, the sphere must contain a net positive charge .
1/7/11 2:29:12 PM
20
15.43
15.44
Chapter 15
From Gauss’s law, the electric flux through any closed
surface is equal to the net charge enclosed divided by ∈0 .
Thus, the flux through each surface (with a positive flux
coming outward from the enclosed interior and a negative
flux going inward toward that interior) is
For S1:
Φ E = Qnet ∈0 = ( +Q − 2Q ) ∈0 = −Q ∈0
For S2:
Φ E = Qnet ∈0 = ( +Q − Q ) ∈0 = 0
For S3:
Φ E = Qnet ∈0 = ( −2Q + Q − Q ) ∈0 = −2Q ∈0
For S4:
Φ E = Qnet ∈0 = ( 0 ) ∈0 = 0
(a)
From Gauss’s law, the total flux through this closed surface is
ΦE =
(b)
Qenclosed
+5.80 × 10 −6 C
=
= 6.55 × 10 5 N ⋅ m 2 C
∈0
8.85 × 10 −12 C2 N ⋅ m 2
When the charge is located in the center of this tetrahedron, we have total symmetry and
the flux is the same through each of the four surfaces. Thus, the flux through any one face is
Φ1 face =
15.45
Φ E 6.55 × 10 5 N ⋅ m 2 C
=
= 1.64 × 10 5 N ⋅ m 2 C
4
4
We choose a spherical Gaussian surface, concentric with the charged spherical shell and of
radius r. Then, ΣEA cosq = E ( 4p r 2 ) cos 0° = 4p r 2 E .
(a)
For r > a (that is, outside the shell), the total charge enclosed by the Gaussian surface is
Q = +q − q = 0. Thus, Gauss’s law gives 4p r 2 E = 0, or E = 0 .
(b)
Inside the shell, r < a, and the enclosed charge is Q = +q.
Therefore, from Gauss’s law, 4p r 2 E =
kq
q
q
, or E =
= e2
2
r
∈0
4p ∈0 r
The field for r < a is E = ke q r 2 directed radially outward .
15.46
(a)
The surface of the cube is a closed surface which surrounds a total charge of Q = 1.70 × 10 2 mC.
Thus, by Gauss’s law, the electric flux through the full surface of the cube is
ΦE =
(b)
Qinside
1.70 × 10 −4 C
=
= 1.92 × 10 7 N ⋅ m 2 C
∈0
8.85 × 10 −12 C2 N ⋅ m 2
Since the charge is located at the center of the cube, the six faces of the cube are symmetrically positioned around the location of the charge. Thus, one-sixth of the flux passes
through each of the faces, or
Φ face =
(c)
68719_15_ch15_p001-029.indd 20
Φ E 1.92 × 10 7 N ⋅ m 2 C
=
= 3.20 × 10 6 N ⋅ m 2 C
6
6
The answer to part (b) would change because the charge could now be at different distances
from each face of the cube, but the answer to part (a) would be unchanged because the flux
through the entire closed surface depends only on the total charge inside the surface.
1/7/11 2:29:14 PM
Electric Forces and Electric Fields
15.47
21
Note that with the point charge −2.00 nC positioned at the center of the spherical shell, we have
complete spherical symmetry in this situation. Thus, we can expect the distribution of charge on
the shell, as well as the electric fields both inside and outside of the shell, to also be spherically
symmetric.
(a)
We choose a spherical Gaussian surface, centered on
the center of the conducting shell, with radius
r = 1.50 m < a as shown at the right. Gauss’s law gives
Φ E = EA = E ( 4p r 2 ) =
so
E=
(8.99 × 10
9
Qinside
Qinside
kQ
or E =
= e center
∈0
4p ∈0 r 2
r2
N ⋅ m 2 C2 ) ( −2.00 × 10 −9 C )
(1.50 m )2
and E = −7.99 N C . The negative sign means that the field is radial inward .
(b)
All points at r = 2.20 m are in the range a < r < b, and hence are located within the conducting material making up the shell. Under conditions of electrostatic equilibrium, the
field is E = 0 at all points inside a conducting material.
(c)
If the radius of our Gaussian surface is r = 2.50 m > b, Gauss’s law (with total spheriQinside
kQ
cal symmetry) leads to E =
= e inside
just as in part (a). However, now
2
4p ∈0 r
r2
Qinside = Qshell + Qcenter = +3.00 nC − 2.00 nC = +1.00 nC. Thus, we have
(8.99 × 10
E=
9
N ⋅ m 2 C2 ) ( +1.00 × 10 −9 C )
( 2.50 m )2
= +1.44 N C
with the positive sign telling us that the field is radial outward at this location.
(d)
Under conditions of electrostatic equilibrium, all excess charge on a conductor resides
entirely on its surface. Thus, the sum of the charge on the inner surface of the shell and that
on the outer surface of the shell is Qshell = +3.00 nC. To see how much of this is on the inner
surface, consider our Gaussian surface to have a radius r that is infinitesimally larger than
a. Then, all points on the Gaussian surface lie within the conducting material, meaning that
E = 0 at all points and the total flux through the surface is Φ E = 0. Gauss’s law then states
that Qinside = Qinner + Qcenter = 0, or
surface
Qinner = − Qcenter = − ( −2.00 nC ) = +2.00 nC
surface
The charge on the outer surface must be
Qouter = Qshell − Qinner = 3.00 nC − 2.00 nC = +1.00 nC
surface
15.48
surface
Please review Example 15.8 in your textbook. There it is shown that the electric field due to a
nonconducting plane sheet of charge parallel to the xy-plane has a constant magnitude given by
Ez = s sheet 2 ∈0, where s sheet is the uniform charge per unit area on the sheet. This field is everywhere perpendicular to the xy-plane, is directed away from the sheet if it has a positive charge
density, and is directed toward the sheet if it has a negative charge density.
continued on next page
68719_15_ch15_p001-029.indd 21
1/7/11 2:29:16 PM
22
Chapter 15
In this problem, we have two plane sheets of charge, both parallel to the xy-plane and separated
by a distance of 2.00 cm. The upper sheet has charge density s sheet = −2s , while the lower sheet
has s sheet = +s . Taking upward as the positive z-direction, the fields due to each of the sheets in
the three regions of interest are:
Region
z<0
0 < z < 2.00 cm
z > 2.00 cm
Lower sheet (at z = 0)
Upper sheet (at z = 2.00 cm)
Electric Field
Electric Field
Ez = −
+s
s
=−
2 ∈0
2 ∈0
Ez = +
−2s
s
=+
∈0
2 ∈0
Ez = +
+s
s
=+
2 ∈0
2 ∈0
Ez = +
−2s
s
=+
∈0
2 ∈0
Ez = +
+s
s
=+
2 ∈0
2 ∈0
Ez = −
−2s
s
=−
∈0
2 ∈0
When both plane sheets of charge are present, the resultant electric field in each region is the
vector sum of the fields due to the individual sheets for that region.
15.49
(a)
For z < 0:
Ez = Ez ,lower + Ez ,upper = −
s
s
s
+
= +
2 ∈0 ∈0
2 ∈0
(b)
For 0 < z < 2.00 cm:
Ez = Ez ,lower + Ez ,upper = +
s
s
3s
+
= +
2 ∈0 ∈0
2 ∈0
(c)
For z > 2.00 cm:
Ez = Ez ,lower + Ez ,upper = +
s
s
s
−
= −
2 ∈0 ∈0
2 ∈0
The radius of each sphere is small in comparison to the distance to the nearest neighboring charge
(the other sphere). Thus, we shall assume that the charge is uniformly distributed over the surface
of each sphere and, in its interaction with the other charge, treat it as though it were a point
charge. In this model, we then have two identical point charges, of magnitude 35.0 mC, separated
by a total distance of 310 m (the length of the cord plus the radius of each sphere). Each of these
charges repels the other with a force of magnitude
(35.0 × 10 −3 C) = 115 N
Q2
Fe = ke 2 = (8.99 × 10 9 N ⋅ m 2 C2 )
2
r
(3.10 × 10 2 m )
2
Thus, to counterbalance this repulsion and hold each sphere in equilibrium, the cord must have a
tension of 115 N , so it will exert a 115 N on that sphere, directed toward the other sphere.
15.50
(a)
As shown in Example 15.8 in the textbook, the electric field due to a nonconducting plane
sheet of charge has a constant magnitude of E = s 2 ∈0 , where s is the uniform charge per
unit area on the sheet. The direction of the field at all locations is perpendicular to the plane
sheet and directed away from the sheet if s is positive, and toward the sheet if s is negative. Thus, if s = +5.20 mC m 2, the magnitude of the electric field at all distances greater
than zero from the plane (including the distance of 8.70 cm) is
E=
s
+5.20 × 10 −6 C m 2
=
= 2.94 × 10 5 N C
2 ∈0 2 (8.85 × 10 −12 C2 N ⋅ m 2 )
continued on next page
68719_15_ch15_p001-029.indd 22
1/7/11 2:29:19 PM
Electric Forces and Electric Fields
(b)
15.51
23
The field does not vary with distance as long as the distance is small compared with the
dimensions of the sheet.
The three contributions to the resultant
electric field at the point of interest are
shown in the sketch at the right.
The magnitude of the resultant field is
ER = −E1 + E2 + E3
ER = −
ke q1 ke q2 ke q3
+ 2 + 2 = ke
r12
r2
r3
⎡ q1
q3 ⎤
q2
⎢− 2 + 2 + 2 ⎥
r2
r3 ⎥⎦
⎢⎣ r1
⎛
N ⋅ m 2 ⎞ ⎡ 4.0 × 10 −9 C 5.0 × 10 −9 C 3.0 × 10 −9 C ⎤
ER = ⎜ 8.99 × 10 9
+
+
⎢−
⎥
2
C2 ⎟⎠ ⎢⎣ ( 2.5 m )
⎝
(1.2 m )2 ⎥⎦
( 2.0 m )2
ER = + 24 N C or E R = 24 N C in the +x-direction
15.52
Consider the force diagram shown at the right.
ΣFy = 0 ⇒ T cosq = mg or T =
mg
cosq
⎛ mg ⎞
ΣFx = 0 ⇒ Fe = T sinq = ⎜
sinq = mg tanq
⎝ cosq ⎟⎠
Since Fe = qE, we have
qE = mg tanq , or q = mg tanq E
( 2.00 × 10
q=
15.53
(a)
−3
kg ) ( 9.80 m s 2 ) tan15.0°
1.00 × 10 3 N C
= +5.25 × 10 −6 C = + 5.25 mC
At a point on the x-axis, the contributions by the two
charges to the resultant field have equal magnitudes
given by E 1 = E2 = ke q r 2.
The components of the resultant field are
⎛ k q⎞
⎛ k q⎞
E y = E1 y − E2 y = ⎜ e2 ⎟ sinq − ⎜ e2 ⎟ sinq = 0
⎝ r ⎠
⎝ r ⎠
⎡ k ( 2q ) ⎤
⎛ k q⎞
⎛ k q⎞
Ex = E1 x + E2 x = ⎜ e2 ⎟ cosq + ⎜ e2 ⎟ cosq = ⎢ e 2 ⎥ cosq
⎝ r ⎠
⎝ r ⎠
⎣ r
⎦
and
Since
cosq b r b
b
= 2 = 3 = 2
, the resultant field is
3 2
2
r
r
r
( a + b2 )
ER =
ke ( 2q ) b
(a
2
+ b2 )
3 2
in the +x-direction
continued on next page
68719_15_ch15_p001-029.indd 23
1/7/11 2:29:22 PM
24
Chapter 15
Note that the result of part (a) may be written as ER =
(b)
ke (Q ) b
(a
2
total charge in the charge distribution generating the field.
+ b2 )
3 2
, where Q = 2q is the
In the case of a uniformly charged circular ring, consider the ring to consist of a very large
number of pairs of charges uniformly spaced around the ring. Each pair consists of two
identical charges located diametrically opposite each other on the ring. The total charge of
pair number i is Qi . At a point on the axis of the ring, this pair of charges generates an elec3 2
tric field contribution that is parallel to the axis and has magnitude Ei = ke b Qi ( a 2 + b 2 ) .
The resultant electric field of the ring is the summation of the contributions by all pairs of
charges, or
⎡
kb
ER = ΣEi = ⎢ 2 e 2 3 2
⎢ (a + b )
⎣
⎤
ke b Q
⎥ ΣQi =
3 2
2
⎥
(a + b2 )
⎦
where Q = ΣQi is the total charge on the ring.
ER =
15.54
(a
ke Q b
2
+ b2 )
3 2
in the +x-direction
It is desired that the electric field exert a retarding force on the electrons, slowing them
down and bringing them to rest. For the retarding force to have maximum effect, it should
be anti-parallel to the direction of the electron’s motion. Since the force an electric field exerts
on negatively charged particles (such as electrons) is in the direction opposite to the field,
the electric field should be in the direction of the electron’s motion.
The work a retarding force of magnitude Fe = q E = eE does on the electrons as they move distance d is W = Fe d cos180° = −Fe d = −eEd. The work-energy theorem (W = ΔKE ) then gives
−eEd = KE f − KEi = 0 − K
and the magnitude of the electric field required to stop the electrons in distance d is
E=
15.55
−K
−ed
E = K ed
or
F2
Consider the sketch at the right and observe:
⎛d⎞
q = tan ⎜ ⎟ = 45.0°
⎝d⎠
+Q
−1
Fx = F1 x + F2 x = +
68719_15_ch15_p001-029.indd 24
r
d
+2Q
Thus,
or
F1
r = d2 + d2 = d 2
and
and
q
d
−Q
ke −Q Q
ke Q 2
k Q2 ⎛ 2 ⎞
2 ⎛ ke Q 2 ⎞
=
+
0.354
=
+
sin 45.0° + 0 = + e 2 ⎜
2
2
4 ⎜⎝ d ⎟⎠
r
2d ⎝ 2 ⎟⎠
d2
Fy = F1 y + F2 y = −
ke −Q Q
ke (Q ) ( 2Q )
ke Q 2 ⎛ 2 ⎞ 2ke Q 2
+
cos
45.0°
+
=
−
r2
d2
2d 2 ⎜⎝ 2 ⎟⎠
d2
⎛
ke Q 2
2 ⎞ ke Q 2
Fy = ⎜ 2 −
=
+1.65
2
4 ⎟⎠ d
d2
⎝
1/7/11 2:29:24 PM
Electric Forces and Electric Fields
15.56
(a)
25
The downward electrical force acting on the ball is
Fe = qE = ( 2.00 × 10 −6 C ) (1.00 × 10 5 N C ) = 0.200 N
The total downward force acting on the ball is then
F = Fe + mg = 0.200 N + (1.00 × 10 −3 kg ) ( 9.80 m s 2 ) = 0.210 N
Thus, the ball will behave as if it were in a modified gravitational field where the effective
free-fall acceleration is
F
0.210 N
=
= 210 m s 2
m 1.00 × 10 −3 kg
" g" =
The period of the pendulum will be
L
0.500 m
= 2p
= 0.307 s
"g"
210 m s 2
T = 2p
(b)
Yes . The force of gravity is a significant portion of the total downward force acting on
the ball. Without gravity, the effective acceleration would be
"g" =
Fe
0.200 N
=
= 200 m s 2
m 1.00 × 10 −3 kg
0.500 m
= 0.314 s
200 m s 2
giving T = 2p
a 2.28% difference from the correct value with gravity included.
15.57
The sketch at the right gives a force diagram of the positively
2
charged sphere. Here, F1 = ke q r 2 is the attractive force
exerted by the negatively charged sphere, and F2 = qE is exerted by
the electric field.
ΣFy = 0 ⇒ T cos10° = mg or T =
mg
cos10°
2
ΣFx = 0 ⇒ F2 = F1 + T sin10° or qE =
ke q
+ mg tan10°
r2
At equilibrium, the distance between the two spheres is r = 2 ( L sin10° ). Thus,
E=
=
ke q
mg tan10°
+
2
q
4 ( L sin10° )
(8.99 × 10
9
N ⋅ m 2 C2 ) ( 5.0 × 10 −8 C )
4 [( 0.100 m ) sin10° ]
or the needed electric field strength is
68719_15_ch15_p001-029.indd 25
2
+
( 2.0 × 10
−3
kg ) ( 9.80 m s 2 ) tan10°
(5.0 × 10
−8
C)
E = 4.4 × 10 5 N C
1/7/11 2:29:26 PM
26
15.58
Chapter 15
(a)
At any point on the x-axis in the range 0 < x < 1.00 m, the contributions made to the resultant
electric field by the two charges are both in the positive x-direction. Thus, it is not possible for
these contributions to cancel each other and yield a zero field.
(b)
Any point on the x-axis in the range x < 0 is located closer to the larger magnitude charge
( q = 5.00 mC) than the smaller magnitude charge ( q = 4.00 mC). Thus, the contribution to
the resultant electric field by the larger charge will always have a greater magnitude than
the contribution made by the smaller charge. It is not possible for these contributions to
cancel to give a zero resultant field.
(c)
If a point is on the x-axis in the region x > 1.00 m, the contributions made by the two charges
are in opposite directions. Also, a point in this region is closer to the smaller magnitude charge
than it is to the larger charge. Thus, there is a location in this region where the contributions of
these charges to the total field will have equal magnitudes and cancel each other.
(d)
When the contributions by the two charges cancel each other, their magnitudes must be
equal. That is,
ke
(5.00 mC) = k ( 4.00 mC )
x2
e
( x − 1.00 m )2
Thus, the resultant field is zero at
15.59
or x − 1.00 m = +x 4 5
x=
1.00 m
= + 9.47 m
1− 4 5
The two spheres have charges q1 and q2 = 2q1, so the
repulsive force that one exerts on the other has magnitude
Fe = ke q1 q2 r 2 = 2ke q12 r 2 .
.
From Figure P15.59 in the textbook, observe that the distance
separating the two spheres is
r = 3.00 cm + 2 [( 5.00 cm ) sin10.0° ] = 4.74 cm = 0.047 4 m
From the force diagram of one sphere given on the right, observe that
ΣFy = 0 ⇒ T cos10.0° = mg or T = mg cos10.0°
⎛ mg ⎞
and ΣFx = 0 ⇒ Fe = T sin10.0° = ⎜
sin10.0° = mg tan10.0°
⎝ cos10.0° ⎟⎠
Thus, 2ke q12 r 2 = mg tan10.0°
or
q1 =
giving
and
68719_15_ch15_p001-029.indd 26
mgr 2 tan10.0°
=
2ke
( 0.0150 kg) (9.80
m s 2 ) ( 0.0474 m ) tan10.0°
2
2 (8.99 × 10 9 N ⋅ m 2 C2 )
q1 = 5.69 × 10 −8 C as the charge on one sphere,
q2 = 2q1 = 1.14 × 10 −7 C as the charge on the other sphere.
1/7/11 2:29:28 PM
Electric Forces and Electric Fields
15.60
d = 1.50 m
Consider the sketch at the right. The electrical
forces acting on the third charge, Q, have
magnitudes
q1 = 3q
k q Q 3k qQ
F1 = e 21 = e 2
x
x
F2 =
and
27
F2 Q
F1
x
d−x
q2 = q
ke q2 Q
k qQ
= e
2
( d − x ) ( d − x )2
The bead with charge Q is in equilibrium when F1 = F2, or
3 ke qQ
x
2
15.61
ke qQ
( d − x )2
3(d − x ) = x 2
2
giving
and
=
x=
3d
=
1+ 3
3 (d − x ) = x
or
3 (1.50 m )
= 0.951 m
1+ 3
Because of the spherical symmetry of the
charge distribution, any electric field
present will be radial in direction. If a field
does exist at distance R from the center, it is
the same as if the net charge located within
r ≤ R were concentrated as a point charge
at the center of the inner sphere. Charge
located at r > R does not contribute to the
field at r = R.
(a)
At r = 1.00 cm, E = 0 since static
electric fields cannot exist within
conducting materials.
(b)
The net charge located at r ≤ 3.00 cm is Q = +8.00 mC.
Thus, at r = 3.00 cm,
E=
=
ke Q
r2
(8.99 × 10
9
N ⋅ m 2 C2 ) (8.00 × 10 −6 C )
(3.00 × 10
−2
m)
2
= 7.99 × 10 7 N C ( outward )
(c)
At r = 4.50 cm, E = 0 , since this is located within conducting materials.
(d)
The net charge located at r ≤ 7.00 cm is Q = + 4.00 mC.
Thus, at r = 7.00 cm,
continued on next page
68719_15_ch15_p001-029.indd 27
1/7/11 2:29:30 PM
28
Chapter 15
ke Q
r2
E=
(8.99 × 10
=
15.62
9
N ⋅ m 2 C2 ) ( 4.00 × 10 −6 C )
( 7.00 × 10 −2 m )
2
= 7.34 × 10 6 N C ( outward )
Consider the free-body diagram of the rightmost charge given at the right.
ΣFy = 0 ⇒ T cosq = mg
T = mg cosq
or
ΣFx = 0 ⇒ Fe = T sinq = ( mg cosq ) sinq = mg tanq
and
Fe =
But,
ke q 2 ke q 2
ke q 2
ke q 2
5ke q 2
+
=
+
=
2
2
r12
r22
( L sinq ) ( 2L sinq ) 4L2 sin 2 q
5ke q 2
= mg tanq
4L2 sin 2 q
Thus,
q=
or
4L2 mg sin 2 q tanq
5ke
If q = 45.0°, m = 0.100 kg, and L = 0.300 m, then
4 ( 0.300 m ) ( 0.100 kg ) ( 9.80 m s 2 ) sin 2 ( 45.0° ) tan ( 45.0° )
2
q=
or
15.63
(a)
5 (8.99 × 10 9 N ⋅ m 2 C2 )
q = 1.98 × 10 −6 C = 1.98 mC
When an electron (negative charge) moves distance Δx in the direction of an electric field,
the work done on it is
W = Fe ( Δx ) cosq = eE ( Δx ) cos180° = −eE ( Δx )
(
)
From the work-energy theorem Wnet = KE f − KEi with KE f = 0, we have
−eE ( Δx ) = −KEi ,
(b)
or
KEi
1.60 × 10 −17 J
=
= 1.00 × 10 3 N C
e ( Δx ) (1.60 × 10 −19 C ) ( 0.100 m )
v − v0 0 − 2 ( KEi ) m
=
=
− eE m
a
t=
2m ( KEi )
eE
2 ( 9.11 × 10 −31 kg ) (1.60 × 10 −17 J )
(1.60 × 10
−19
C ) (1.00 × 10 3 N C )
= 3.37 × 10 −8 s = 33.7 ns
After bringing the electron to rest, the electric force continues to act on it, causing the
electron to accelerate in the direction opposite to the field at a rate of
a=
68719_15_ch15_p001-029.indd 28
E=
The magnitude of the retarding force acting on the electron is Fe = eE, and Newton’s second
law gives the acceleration as a = − Fe m = − eE m. Thus, the time required to bring the
electron to rest is
t=
(c)
or
−19
3
eE (1.60 × 10 C ) (1.00 × 10 N C )
=
= 1.76 × 1014 m s 2
m
9.11× 10 −31 kg
1/7/11 2:29:32 PM
Electric Forces and Electric Fields
15.64
(a)
29
The acceleration of the protons is downward (in the direction of the field) and
ay =
−19
Fe eE (1.60 × 10 C ) ( 720 N C )
=
=
= 6.90 × 1010 m s 2
m
m
1.67 × 10 −27 kg
The time of flight for the proton is twice the time required to reach the peak of the arc, or
⎛ v ⎞ 2v sinq
0y
t = 2tpeak = 2 ⎜
⎟= 0
⎜⎝ a y ⎟⎠
ay
The horizontal distance traveled in this time is
⎛ 2v sinq ⎞ v 2 sin 2q
R = v0 x t = ( v0 cosq ) ⎜ 0
⎟= 0
⎜⎝
⎟⎠
ay
ay
Thus, if R = 1.27 × 10 −3 m, we must have
sin 2q =
ay R
v02
(6.90 × 10
10
=
m s 2 ) (1.27 × 10 −3 m )
(9 550
m s)
2
= 0.961
giving 2q = 73.9° or 2q = 180° − 73.9° = 106°.
Hence, q = 37.0° or 53.0°
(b)
68719_15_ch15_p001-029.indd 29
The time of flight for each possible angle of projection is:
For q = 37.0°:
t=
2v0 sinq
For q = 53.0°:
t=
2v0 sinq
ay
ay
=
2 ( 9 550 m s ) sin 37.0°
= 1.67 × 10 −7 s
6.90 × 1010 m s 2
=
2 ( 9 550 m s ) sin 53.0°
= 2.21× 10 −7 s
6.90 × 1010 m s 2
1/7/11 2:29:35 PM
16
Electrical Energy and Capacitance
QUICK QUIZZES
1.
Choice (b). The field exerts a force on the electron, causing it to accelerate in the direction
opposite to that of the field. In this process, electrical potential energy is converted into kinetic
energy of the electron. Note that the electron moves to a region of higher potential, but because
the electron has negative charge this corresponds to a decrease in the potential energy of the
electron.
2.
Choice (a). The electron, a negatively charged particle, will move toward the region of higher
electric potential. Because of the electron’s negative charge, this corresponds to a decrease in
electrical potential energy.
3.
Choice (b). Charged particles always tend to move toward positions of lower potential energy.
The electrical potential energy of a charged particle is PE = qV and, for positively-charged
particles, this decreases as V decreases. Thus, a positively-charged particle located at x = A
would move toward the left.
4.
Choice (d). For a negatively-charged particle, the potential energy (PE = qV ) decreases as V
increases. A negatively charged particle would oscillate around x = B which is a position of
minimum potential energy for negative charges.
5.
Choice (d). If the potential is zero at a point located a finite distance from charges, negative
charges must be present in the region to make negative contributions to the potential and cancel
positive contributions made by positive charges in the region.
6.
Choice (c). Both the electric potential and the magnitude of the electric field decrease as the
distance from the charged particle increases. However, the electric flux through the balloon does
not change because it is proportional to the total charge enclosed by the balloon, which does not
change as the balloon increases in size.
7.
Choice (a). From the conservation of energy, the final kinetic energy of either particle will be
given by
(
)
(
)
KE f = KEi + PEi − PE f = 0 + qVi − qV f = −q V f − Vi = −q ( ΔV )
For the electron, q = −e and ΔV = +1 V giving KE f = − ( −e ) ( +1 V ) = +1 eV.
For the proton, q = +e and ΔV = −1 V, so KE f = − ( e ) ( −1 V ) = +1 eV, the same as that of the
electron.
8.
Choice (c). The battery moves negative charge from one plate and puts it on the other. The first
plate is left with excess positive charge whose magnitude equals that of the negative charge
moved to the other plate.
9.
(a) C decreases.
(d) ΔV increases.
(b) Q stays the same.
(c)
(e) The energy stored increases.
E stays the same.
30
68719_16_ch16_p030-061.indd 30
1/7/11 2:32:36 PM
Electrical Energy and Capacitance
31
Because the capacitor is removed from the battery, charges on the plates have nowhere to go.
Thus, the charge on the capacitor plates remains the same as the plates are pulled apart. Because
E = s ∈0 = (Q A) ∈0 , the electric field is constant as the plates are separated. Because ΔV = Ed
and E does not change, ΔV increases as d increases. Because the same charge is stored at a higher
potential difference, the capacitance (C = Q ΔV ) has decreased. Because energy stored = Q 2 2C
and Q stays the same while C decreases, the energy stored increases. The extra energy must have
been transferred from somewhere, so work was done. This is consistent with the fact that the
plates attract one another, and work must be done to pull them apart.
10.
(a) C increases.
(d) ΔV remains the same.
(b) Q increases.
(c)
(e) The energy stored increases.
E stays the same.
The presence of a dielectric between the plates increases the capacitance by a factor equal to the
dielectric constant. Since the battery holds the potential difference constant while the capacitance
increases, the charge stored (Q = CΔV ) will increase. Because the potential difference and the distance between the plates are both constant, the electric field (E = ΔV d) will stay the same. The
battery maintains a constant potential difference. With ΔV constant while capacitance increases,
the stored energy [energy stored = 12 C(ΔV )2 ] will increase.
11.
Choice (a). Increased random motions associated with an increase in temperature make it more
difficult to maintain a high degree of polarization of the dielectric material. This has the effect of
decreasing the dielectric constant of the material, and in turn, decreasing the capacitance of the
capacitor.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The change in the potential energy of the proton is equal to the negative of the work done on it by
the electric field. Thus,
(
)
ΔPE = −W = −qE x ( Δx ) = − +1.6 × 10 −19 C (850 N C ) ( 2.5 m − 0 ) = −3.4 × 10 −16 J
and (b) is the correct choice for this question.
2.
Because electric forces are conservative, the kinetic energy gained is equal to the decrease in
electrical potential energy, or
(
)
KE = −PE = −q ( ΔV ) = − ( −1 e ) +1.00 × 10 4 V = +1.00 × 10 4 eV
so the correct choice is (a).
3.
In a uniform electric field, the change in electric potential is ΔV = −E x ( Δx ), giving
Ex = −
(
(
)
)
V f − Vi
ΔV
(190 V − 120 V )
=−
=−
= −35 V m = −35 N C
Δx
( 5.0 m − 3.0 m )
x f − xi
and it is seen that the correct choice is (d).
68719_16_ch16_p030-061.indd 31
1/7/11 2:32:37 PM
32
4.
Chapter 16
From conservation of energy, KE f + PE f = KEi + PEi , or
speed of the nucleus is
vB = vA2 +
(
=
1
2
mvB2 = 1 2 mvA2 + qVA − qVB , the final
2q(VA − VB )
m
(
)
2 ⎡⎣ 2 1.60 × 10 −19 C (1.50 − 4.00 ) × 10 3 V ⎤⎦
6.20 × 10 m s +
= 3.78 × 10 5 m s
6.63 × 10 −27 kg
5
)
2
Thus, the correct answer is choice (b).
5.
In a series combination of capacitors, the equivalent capacitance is always less than any individual
capacitance in the combination, meaning that choice (a) is false. Also, for a series combination
of capacitors, the magnitude of the charge is the same on all plates of capacitors in the combination, making both choices (d) and (e) false. The potential difference across the capacitance Ci is
ΔVi = Q Ci , where Q is the common charge on each capacitor in the combination. Thus, the largest potential difference (voltage) appears across the capacitor with the least capacitance, making
choice (b) the correct answer.
6.
The total potential at a point due to a set of point charges qi is
V = ∑ kqi ri
i
where ri is the distance from the point of interest to the location of the charge qi. Note that in this
case, the point at the center of the circle is equidistant from the 4 point charges located on the rim
of the circle. Note also that q2 + q3 + q4 = ( +1.5 − 1.0 − 0.5) mC = 0, so we have
ke q1 ke q2 ke q3 ke q4 ke
k
kq
+
+
+
= ( q1 + q2 + q3 + q4 ) = e ( q1 + 0 ) = e 1 = V1
r
r
r
r
r
r
r
4
= 4.5 × 10 V
or the total potential at the center of the circle is just that due to the first charge alone, and the
correct answer is choice (b).
Vcenter =
7.
With the given specifications, the capacitance of this parallel-plate capacitor will be
C=
(
)(
)(
)
2
−12
C2 N ⋅ m 2 1.00 cm 2 ⎛ 1 m 2 ⎞
k ∈0 A 1.00 × 10 8.85 × 10
=
⎜⎝ 10 4 cm 2 ⎟⎠
1.00 × 10 −3 m
d
= 8.85 × 10 −11 F = 88.5 × 10 −12 F = 88.5 pF
and the correct choice is (a).
8.
Keeping the capacitor connected to the battery means that the potential difference between the
plates is kept at a constant value equal to the voltage of the battery. Since the capacitance of a
parallel-plate capacitor is C = k ∈0 A d, doubling the plate separation d, while holding other
characteristics of the capacitor constant, means the capacitance will be decreased by a factor
of 2. The energy stored in a capacitor may be expressed as U = 12 C(ΔV )2, so when the potential
difference ΔV is held constant while the capacitance is decreased by a factor of 2, the stored
energy decreases by a factor of 2, making (c) the correct choice for this question.
9.
When the battery is disconnected, there is no longer a path for charges to use in moving onto or
off of the plates of the capacitor. This means that the charge Q is constant. The capacitance of a
68719_16_ch16_p030-061.indd 32
1/7/11 2:32:39 PM
Electrical Energy and Capacitance
33
parallel-plate capacitor is C = k ∈0 A d and the dielectric constant is k ≈ 1 when the capacitor
is air filled. When a dielectric with dielectric constant k = 2 is inserted between the plates, the
capacitance is doubled C f = 2Ci . Thus, with Q constant, the potential difference between the
plates, ΔV = Q C, is decreased by a factor of 2, meaning that choice (a) is a true statement. The
electric field between the plates of a parallel-plate capacitor is E = ΔV d and decreases when ΔV
decreases, making choice (e) false and leaving (a) as the only correct choice for this question.
(
)
10.
Once the capacitor is disconnected from the battery, there is no path for charges to move onto
or off of the plates, so the charges on the plates are constant, and choice (e) can be eliminated.
The capacitance of a parallel-plate capacitor is C = k ∈0 A d, so the capacitance decreases when
the plate separation d is increased. With Q constant and C decreasing, the energy stored in the
capacitor, U = Q 2 2C, increases, making choice (a) false and choice (b) true. The potential
difference between the plates, ΔV = Q C = Q ⋅ d k ∈0 A, increases and the electric field between
the plates, E = ΔV d = Q k ∈0 A, is constant. This means that both choices (c) and (d) are false
and leaves choice (b) as the only correct response.
11.
Capacitances connected in parallel all have the same potential difference across them and the
equivalent capacitance, Ceq = C1 + C2 + C3 + … , is larger than the capacitance of any one of the
capacitors in the combination. Thus, choice (c) is a true statement. The charge on a capacitor is
Q = C(ΔV ), so with ΔV constant, but the capacitances different, the capacitors all store different
charges that are proportional to the capacitances, making choices (a), (b), (d), and (e) all false.
Therefore, (c) is the only correct answer.
12.
For a series combination of capacitors, the magnitude of the charge is the same on all plates of
capacitors in the combination. Also, the equivalent capacitance is always less than any individual
capacitance in the combination. Therefore, choice (a) is true while choices (b) and (c) are both false.
The potential difference across a capacitor is ΔV = Q C, so with Q constant, capacitors having
different capacitances will have different potential differences across them, with the largest potential
difference being across the capacitor with the smallest capacitance. This means that choices (d) and
(e) are false, and choice (f) is true. Thus, both choices (a) and (f) are true statements.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The potential energy between a pair of point charges separated by distance R is
PE = ke q1q2 R. Thus, the potential energy for each of the four systems is:
(a)
PEa = ke
Q ( 2Q )
Q2
= 2ke
r
r
(b)
PEb = ke
( −Q ) ( −Q ) = k
(c)
PEc = ke
Q ( −Q )
1 Q2
= − ke
2
2r
r
(d)
PEd = ke
( −Q ) ( −2Q ) = k
r
2r
Q2
r
e
e
Q2
r
Therefore, the correct ranking from largest to smallest is (a) > (b) = (d) > (c).
4.
To move like charges together from an infinite separation, at which the potential energy of the
system of two charges is zero, requires work to be done on the system by an outside agent. Hence
energy is stored, and potential energy is positive. As charges with opposite signs move together
from an infinite separation, energy is released, and the potential energy of the set of charges
becomes negative.
6.
A sharp point on a charged conductor would produce a large electric field in the region near the
point. An electric discharge could most easily take place at the point.
68719_16_ch16_p030-061.indd 33
1/7/11 2:32:40 PM
34
8.
Chapter 16
There are eight different combinations that use all three capacitors in the circuit. These combinations and their equivalent capacitances are:
⎛ 1
1
1⎞
All three capacitors in series: Ceq = ⎜
+
+ ⎟
C
C
C
⎝ 1
2
3⎠
−1
All three capacitors in parallel: Ceq = C1 + C2 + C3
One capacitor in series with a parallel combination of the other two:
−1
−1
⎛ 1
⎛ 1
⎛
1⎞
1 ⎞
1
1⎞
Ceq = ⎜
+ ⎟ , Ceq = ⎜
+ ⎟ , Ceq = ⎜
+ ⎟
⎝ C1 + C2 C3 ⎠
⎝ C3 + C1 C2 ⎠
⎝ C2 + C3 C1 ⎠
−1
One capacitor in parallel with a series combination of the other two:
⎛ CC ⎞
⎛ CC ⎞
⎛ CC ⎞
Ceq = ⎜ 1 2 ⎟ + C3, Ceq = ⎜ 3 1 ⎟ + C2, Ceq = ⎜ 2 3 ⎟ + C1
C
+
C
C
+
C
⎝ 1
⎝ 3
⎝ C 2 + C3 ⎠
2⎠
1⎠
10.
12.
(a)
If the wires are disconnected from the battery and not allowed to touch each other or
another object, the charge on the plates is unchanged.
(b)
If, after being disconnected from the battery, the wires are connected to each other, electrons will rapidly flow from the negatively charged plate to the positively charged plate to
leave the capacitor uncharged with both plates neutral.
The primary choice would be the dielectric. You would want to choose a dielectric that has a
large dielectric constant and dielectric strength, such as strontium titanate, where k ≈ 233 (Table
16.1). A convenient choice could be thick plastic or Mylar. Secondly, geometry would be a factor.
To maximize capacitance, one would want the individual plates as close as possible, since the
capacitance is proportional to the inverse of the plate separation—hence the need for a dielectric
with a high dielectric strength. Also, one would want to build, instead of a single parallel-plate
capacitor, several capacitors in parallel. This could be achieved through “stacking” the plates of
the capacitor. For example, you can alternately lay down sheets of a conducting material, such as
aluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none
of the conducting sheets are in contact with their nearest neighbors, connect every other plate
together as illustrated in the figure below.
Dielectric
Conductor
This technique is often used when “home-brewing” signal capacitors for radio applications, as
they can withstand huge potential differences without flashover (without either discharge between
plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich together flexible materials such as aluminum roof flashing and thick plastic, so the whole
product can be rolled up into a “capacitor burrito” and placed in an insulating tube, such as a
PVC pipe, and then filled with motor oil (again to prevent flashover).
68719_16_ch16_p030-061.indd 34
1/7/11 2:32:43 PM
Electrical Energy and Capacitance
14.
35
The material of the dielectric may be able to withstand a larger electric field than air can withstand before breaking down to pass a spark between the capacitor plates.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a)
6.16 × 10 −17 N
(b)
3.69 × 1010 m s 2 in the direction of the electric field
(c)
7.38 cm
4.
6.67 × 1011 electrons
6.
(a) 1.10 × 10 −2 N to the right
8.
(d)
− 49.5 V
(a)
− 2.31 kV
(b) 1.98 × 10 −3 J
(c)
−1.98 × 10 −3 J
(b)
Protons would require a greater potential difference.
(b)
+10.8 kV
(c) ΔVp ΔVe = − mp me
10.
40.2 kV
12.
(a)
14.
−9.08 J
16.
(a)
See Solution.
(b) 3ke q a
18.
(a)
See Solution.
(b)
2⎞
1
⎛
V = ( 22.5 V ⋅ m ) ⎜
− ⎟
⎝ 1.20 m − x x ⎠
(c)
−37.5 V
(d)
x = 0.800 m
(a)
Conservation of energy alone yields one equation with two unknowns.
(b)
Conservation of linear momentum
(c)
v p = 1.05 × 10 7 m s , vα = 2.64 × 10 6 m s
20.
+5.39 kV
22.
5.4 × 10 5 V
24.
(a) V = 4 2keQ a
26.
(a)
28.
(a) 1.00 mF
30.
31.0 Å
32.
1.23 kV
68719_16_ch16_p030-061.indd 35
3.00 mF
(c)
See Solution.
(b) W = 4 2ke qQ a
(b) 36.0 mC
(b) 100 V
1/7/11 3:11:16 PM
36
34.
Chapter 16
(a) 17.0 mF
(c)
(b)
9.00 V
45.0 mC on C1 , 108 mC on C2
36.
3.00 pF and 6.00 pF
38.
(a)
6.00 mF
(d)
Q4 = 144 mC, Q2 = 72.0 mC, Qrightmost = 216 mC
(b) 12.0 mF
(c)
432 mC
(e) Q24 = Q8 = 216 mC
branch
(f)
40.
9.00 V
(a) 2C
(g)
27.0 V
(b)
Q1 > Q3 > Q2
(c)
ΔV1 > ΔV2 = ΔV3
(d) Q1 and Q3 increase, Q2 decreases
42.
(a)
6.04 mF
44.
(a)
5.96 mF
(b) 83.6 mC
(b) 89.4 mC on the 20.0 mF capacitor, 63.0 mC on the 6.00 mF capacitor,
26.3 mC on the 15.0 mF capacitor, and 26.3 mC on the 3.00 mF capacitor
46.
(a) Ceq = 12.0 mF, Estored,total = 8.64 × 10 −4 J
(b)
Estored,1 = 5.76 × 10 −4 J, Estored,2 = 2.88 × 10 −4 J
It will always be true that Estored,1 + Estored,2 = Estored,total.
(c)
5.66 V; C2 , with the largest capacitance, stores the most energy.
48.
9.79 kg
50.
(a) 13.3 nC
52.
1.04 m
54.
0.443 mm
56.
(a)
13.5 mJ
(b)
Estored,2 = 3.60 mJ, Estored,3 = 5.40 mJ, Estored,4 = 1.80 mJ, Estored,6 = 2.70 mJ
(b)
272 nC
(c) The energy stored in the equivalent capacitance equals the sum of the energies stored in the
individual capacitors.
1
Cp ±
2
1 2
1
C p − C pCs , C2 = C p ∓
4
2
58.
C1 =
60.
(a) 1.8 × 10 4 V
(d)
62.
68719_16_ch16_p030-061.indd 36
1 2
C p − C pCs
4
(b)
−3.6 × 10 4 V
(b)
See Solution.
(c)
−1.8 × 10 4 V
−5.4 × 10 −2 J
(a) C =
ab
ke ( b − a )
1/7/11 2:32:46 PM
Electrical Energy and Capacitance
64.
k = 2.33
66.
(a)
(d)
68.
(a)
2ke q
d 5
(b)
4ke q 2
d 5
(b)
4.4 mm
37
4ke q 2
d 5
(c)
8ke q 2
md 5
0.1 mm
PROBLEM SOLUTIONS
16.1
(a)
Because the electron has a negative charge, it experiences a force in the direction opposite
to the field and, when released from rest, will move in the negative x-direction. The work
done on the electron by the field is
(
)
(
W = Fx ( Δx ) = ( qE x ) Δx = −1.60 × 10 −19 C ( 375 N C ) −3.20 × 10 −2 m
)
= 1.92 × 10 −18 J
(b)
The change in the electric potential energy is the negative of the work done on the particle
by the field. Thus,
ΔPE = −W = −1.92 × 10 −18 J
(c)
Since the Coulomb force is a conservative force, conservation of energy gives
ΔKE + ΔPE = 0, or KE f = 12 me v 2f = KEi − ΔPE = 0 − ΔPE, and
vf =
16.2
(
9.11× 10 −31 kg
)=
2.05 × 10 6 m s in the − x-direction
)
(a)
F = qE = 1.60 × 10 −19 C ( 385 N C ) = 6.16 × 10 −17 N
(b)
a=
(c)
Δx = v0 t +
F
6.16 × 10 −17 N
=
= 3.69 × 1010 m s 2 in the direction of the electric field
m p 1.67 × 10 −27 kg
(
)(
1 2
1
at = 0 + 3.69 × 1010 m s 2 2.00 × 10 −6 s
2
2
= 7.38 × 10
16.3
(
−2 −1.92 × 10 −18 J
−2 ( ΔPE )
=
me
−2
)
2
m = 7.38 cm
The work done by the agent moving the charge out of the cell is
Winput = −Wfield = − ( −ΔPEe ) = +q ( ΔV )
(
)(
)
= 1.60 × 10 −19 C + 90 × 10 −3 J C = 1.4 × 10 −20 J
16.4
Assuming the sphere is isolated, the excess charge on it is uniformly distributed over its surface.
Under this spherical symmetry, the electric field outside the sphere is the same as if all the excess
charge on the sphere were concentrated as a point charge located at the center of the sphere.
continued on next page
68719_16_ch16_p030-061.indd 37
1/7/11 2:32:50 PM
38
Chapter 16
Thus, at r = 8.00 cm > Rsphere = 5.00 cm, the electric field is E = ke Q r 2. The required charge
then has magnitude Q = Er 2 ke, and the number of electrons needed is
(
)(
)
2
1.50 × 10 5 N C 8.00 × 10 −2 m
Q Er 2
n=
= 6.67 × 1011 electrons
=
=
e
ke e
8.99 × 10 9 N ⋅ m 2 C2 1.60 × 10 −19 C
(
)(
)
ΔV
25 × 10 3 J C
=
= 1.7 × 10 6 N C
d
1.5 × 10 −2 m
16.5
E=
16.6
(a)
F = qE = +40.0 × 10 −6 C ( +275 N C ) = 1.10 × 10 −2 N directed toward the right
(b)
WAB = F ( Δx ) cosq = 1.10 × 10 −2 N ( 0.180 m ) cos 0° = 1.98 × 10 −3 J
(c)
ΔPE = −WAB = −1.98 × 10 −3 J
(d)
ΔV = VB − VA =
(a)
E=
(b)
F= qE=
(c)
W = F ⋅ s cosq
16.7
(
)
(
)
ΔPE −1.98 × 10 −3 J
=
= − 49.5 V
q
+40.0 × 10 −6 C
ΔV
600 J C
=
= 1.13 × 10 5 N C
d
5.33 × 10 −3 m
(
)
1.60 × 10 −19 C ( 600 J C )
q ΔV
=
= 1.80 × 10 −14 N
d
5.33 × 10 −3 m
(
)
= 1.80 × 10 −14 N ⎡⎣( 5.33 − 2.90 ) × 10 −3 m ⎤⎦ cos 0° = 4.37 × 10 −17 J
16.8
(a)
Using conservation of energy, ΔKE + ΔPE = 0, with KE f = 0 since the particle is “stopped,”
we have
1
1
⎛
⎞
ΔPE = −ΔKE = − ⎜ 0 − me vi2 ⎟ = + 9.11 × 10 −31 kg 2.85 × 10 7 m s
⎝
⎠
2
2
(
)(
)
2
= +3.70 × 10 −16 J
The required stopping potential is then
ΔV =
(b)
Being more massive than electrons, protons traveling at the same initial speed will have
more initial kinetic energy and require a greater magnitude stopping potential .
(c)
Since ΔVstopping = ΔPE q = ( −ΔKE ) q = − mv 2 2 q, the ratio of the stopping potential for a
proton to that for an electron having the same initial speed is
(
ΔVp
ΔVe
68719_16_ch16_p030-061.indd 38
ΔPE +3.70 × 10 −16 J
=
= −2.31 × 10 3 V = −2.31 kV
q
−1.60 × 10 −19 C
=
− mp vi2 2(+e)
− me vi2 2(−e)
)
= − mp me
1/7/11 2:32:51 PM
Electrical Energy and Capacitance
16.9
(a)
We use conservation of energy,
Δ ( KE ) + Δ ( PEs ) + Δ ( PEe ) = 0, recognizing
that Δ(KE) = 0 since the block is at rest at
both the beginning and end of the motion.
The change in the elastic potential energy is
2
given by Δ ( PEs ) = 12 kx max
− 0, where x max
is the maximum stretch of the spring. The
change in the electrical potential energy
is the negative of the work the electric field does, Δ ( PEe ) = −W = −Fe (Δx) = − (QE ) x max.
2
Thus, 0 + 12 kx max
− (QE ) x max = 0, which yields
x max =
(b)
39
(
)(
)
−6
4
2QE 2 35.0 × 10 C 4.86 × 10 V m
=
= 4.36 × 10 −2 m = 4.36 cm
k
78.0 N m
At equilibrium, ΣF = Fs + Fe = 0, or − kx eq + QE = 0. Therefore,
x eq =
QE 1
= x max = 2.18 cm
k
2
The amplitude is the distance from the equilibrium position to each of the turning points
( at x = 0 and x = 4.36 cm ), so A = 2.18 cm = xmax 2 .
(c)
2
From conservation of energy, Δ ( KE ) + Δ ( PEs ) + Δ ( PEe ) = 0, we have 0 + 12 kx max
+ QΔV = 0.
Since x max = 2A, this gives
2
kx max
k ( 2A )
=−
2Q
2Q
2
ΔV = −
16.10
ΔV = −
or
2kA2
Q
Using Δy = v0 y t + 12 a y t 2 for the full flight gives 0 = v0 y t f + 12 a y t 2f , or a y = −2v0 y t f , where t f is
the full time of the flight. Then, using v 2y = v02 y + 2a y (Δy) for the upward part of the flight gives
( Δy )max =
0 − v02 y
2a y
=
(
−v02 y
2 −2 v0 y t f
)
=
v0 y t f
4
=
( 20.1 m s ) ( 4.10 s ) = 20.6 m
4
From Newton’s second law,
ay =
ΣFy
m
=
−mg − qE
qE ⎞
⎛
= −⎜g +
⎟
⎝
m
m⎠
Equating this to the earlier result gives a y = − ( g + qE m ) = −2v0 y t f , so the electric field
strength is
⎤ ⎛ 2.00 kg ⎞ ⎡ 2 ( 20.1 m s )
⎤
⎛ m ⎞ ⎡ 2v0 y
E=⎜ ⎟⎢
− g⎥ = ⎜
− 9.80 m s 2 ⎥ = 1.95 × 10 3 N C
⎟
⎢
−6
⎝ q ⎠ ⎢⎣ t f
⎥⎦ ⎝ 5.00 × 10 C ⎠ ⎣ 4.10 s
⎦
(
)
Thus, ( ΔV )max = ( Δymax ) E = ( 20.6 m ) 1.95 × 10 3 N C = 4.02 × 10 4 V = 40.2 kV
16.11
(
)(
)
(
)(
)
(a)
VA =
8.99 × 10 9 N ⋅ m 2 C2 −1.60 × 10 −19 C
ke q
=
= −5.75 × 10 −7 V
0.250 × 10 −2 m
rA
(b)
VB =
8.99 × 10 9 N ⋅ m 2 C2 −1.60 × 10 −19 C
ke q
=
= −1.92 × 10 −7 V
0.750 × 10 −2 m
rB
continued on next page
68719_16_ch16_p030-061.indd 39
1/7/11 2:32:54 PM
40
Chapter 16
(
)
ΔV = VB − VA = −1.92 × 10 −7 V − −5.75 × 10 −7 V = +3.83 × 10 −7 V
(c)
16.12
(a)
No . The original electron will be repelled by the negatively charged particle which suddenly appears at point A. Unless the electron is fixed in place, it will move in the opposite
direction, away from points A and B, thereby lowering the potential difference between
these points.
VA = ∑
⎛ −15.0 × 10 −9 C 27.0 × 10 −9 C ⎞
ke qi
= 8.99 × 10 9 N ⋅ m 2 C2 ⎜
+
= +5.39 kV
ri
⎝ 2.00 × 10 −2 m 2.00 × 10 −2 m ⎟⎠
VB = ∑
⎛ −15.0 × 10 −9 C 27.0 × 10 −9 C ⎞
ke qi
= 8.99 × 10 9 N ⋅ m 2 C2 ⎜
+
= +10.8 kV
ri
⎝ 1.00 × 10 −2 m 1.00 × 10 −2 m ⎟⎠
i
(b)
i
16.13
(a)
(
)
(
)
Calling the 2.00 mC charge q3,
V =∑
i
⎛q q
ke qi
q
= ke ⎜ 1 + 2 + 2 3 2
⎜⎝ r1 r2
ri
r1 + r2
⎞
⎟
⎟⎠
⎛
⎛
N ⋅ m 2 ⎞ 8.00 × 10 −6 C 4.00 × 10 −6 C
⎜
= ⎜ 8.99 × 10 9
+
+
C2 ⎟⎠ ⎜ 0.060 0 m
0.030 0 m
⎝
⎝
2.00 × 10 −6 C
( 0.060 0 )2 + ( 0.030 0 )2
⎞
⎟
m ⎟⎠
V = 2.67 × 10 6 V
(b)
Replacing 2.00 × 10 −6 C by − 2.00 × 10 −6 C in part (a) yields
V = 2.13 × 10 6 V
16.14
(
)
W = q ( ΔV ) = q V f − Vi , and V f = 0 since the final location of the 8.00 mC is an infinite distance from other charges. The potential, due to the other charges, at the initial location of
the 8.00 mC is Vi = ke ( q1 r1 + q2 r2 ) . Thus, the required energy for the move is
⎡
⎛ q q ⎞⎤
W = q ⎢ 0 − ke ⎜ 1 + 2 ⎟ ⎥
⎝ r1 r2 ⎠ ⎥⎦
⎢⎣
⎛
⎛
N ⋅ m 2 ⎞ 2.00 × 10 −6 C
⎜
+
= − 8.00 × 10 −6 C ⎜ 8.99 × 10 9
C2 ⎟⎠ ⎜ 0.030 0 m
⎝
⎝
(
)
4.00 × 10 −6 C
( 0.030 0 )2 + ( 0.060 0 )2
⎞
⎟
m ⎟⎠
W = − 9.08 J
16.15
(a)
V =∑
i
(b)
ke qi ⎛
N ⋅ m 2 ⎞ ⎛ 5.00 × 10 −9 C 3.00 × 10 −9 C ⎞
= ⎜ 8.99 × 10 9
−
= 103 V
ri
C2 ⎟⎠ ⎜⎝ 0.175 m
0.175 m ⎟⎠
⎝
(
)(
)
2
5.00 × 10 −9 C − 3.00 × 10 −9 C
ke qi q2 ⎛
9 N⋅m ⎞
PE =
= ⎜ 8.99 × 10
= − 3.85 × 10 − 7 J
2
⎟
r12
C ⎠
0.350 m
⎝
The negative sign means that positive work must be done to separate the charges by an
infinite distance (that is, bring them up to a state of zero potential energy).
68719_16_ch16_p030-061.indd 40
1/7/11 2:32:57 PM
41
Electrical Energy and Capacitance
16.16
(a)
At the center of the triangle, each of the identical
charges produce a field contribution of magnitude
E1 = ke q a 2 . The three contributions are oriented
as shown at the right and the components of the
resultant field are:
y
E1
E1
30°
30°
x
E x = ΣE x = +E1 cos 30° − E1 cos 30° = 0
E y = ΣE y = +E1 sin 30° − E1 + E1 sin 30° = 0
E1
Thus, the resultant field has magnitude
E = E x2 + E y2 = 0
(b)
The total potential at the center of the triangle is
V = ΣVi = Σ
(c)
16.17
ke qi ke q ke q ke q 3ke q
=
+
+
=
ri
a
a
a
a
Imagine a test charge placed at the center of the triangle. Since the field is zero at the center,
the test charge will experience no electrical force at that point. The fact that the potential is
not zero at the center means that work would have to be done by an external agent to move
a test charge from infinity to the center.
The Pythagorean theorem gives the distance from the midpoint of the base to the charge at the
apex of the triangle as
r3 =
( 4.00 cm )2 − (1.00 cm )2 = 15 cm = 15 × 10 −2 m
Then, the potential at the midpoint of the base is V = ∑ ke qi ri, or
⎛
V = ⎜ 8.99 × 10 9
⎝
(
) (
i
) (
)
−9
−7.00 × 10 −9 C
+7.00 × 10 −9 C ⎞
N ⋅ m ⎞ ⎛ −7.00 × 10 C
+
+
⎜
⎟
C2 ⎟⎠ ⎝
0.010 0 m
0.010 0 m
15 × 10 −2 m ⎠
2
= −1.10 × 10 4 V = − 11.0 kV
16.18
(a)
See the sketch below:
1000
y
500
x
⫺1.2
⫺0.8
⫺0.4
0
0.4
0.8
1.2
1.6
2
⫺500
⫺1000
continued on next page
68719_16_ch16_p030-061.indd 41
1/7/11 2:32:59 PM
42
Chapter 16
(b)
At the point (x, 0), where 0 < x < 1.20 m, the potential is
V =∑
i
ke qi ke ( −2q )
ke q
1
2⎞
⎛
=
−
+
= ke q ⎜
⎝ 1.20 m − x x ⎟⎠
ri
x
1.20 m − x
or
⎛
N ⋅ m2 ⎞
1
2⎞
1
2⎞
⎛
⎛
V = ⎜ 8.99 × 10 9
= ( 22.5 V ⋅ m ) ⎜
−
−
2.50 × 10 −9 C ⎜
⎝ 1.20 m − x x ⎟⎠
⎝ 1.20 m − x x ⎟⎠
C2 ⎟⎠
⎝
(
(c)
)
At x = + 0.600 m, the potential is
2
22.5 V ⋅ m
1
⎛
⎞
V = ( 22.5 V ⋅ m ) ⎜
−
=−
= −37.5 V
⎝ 1.20 m − 0.600 m 0.600 m ⎟⎠
0.600 m
(d)
When 0 < x < 1.20 m and V = 0, we have 1 (1.20 m − x) − 2 x = 0, or x = 2.40 m − 2x.
This yields x = 2.40 m 3 = 0.800 m .
16.19
(a)
When the charge configuration consists of only the
two protons ( q1 and q2 in the sketch ), the potential
energy of the configuration is
(
)(
8.99 × 10 9 N ⋅ m 2 C2 1.60 × 10 −19 C
kqq
PEa = e 1 2 =
r12
6.00 × 10 −15 m
)
2
or PEa = 3.84 × 10 −14 J
(b)
When the alpha particle ( q3 in the sketch ) is added to the configuration, there are three distinct pairs of particles, each of which possesses potential energy. The total potential energy
of the configuration is now
( )
⎛ ke 2e 2 ⎞
ke q1q2 ke q1q3 ke q2 q3
+
+
= PEa + 2 ⎜
⎟
r12
r13
r23
⎝ r13 ⎠
where use has been made of the facts that q1q3 = q2 q3 = e ( 2e ) = 2e 2 and
PEb =
r13 = r23 = ( 3.00 fm ) + ( 3.00 fm ) = 18.0 fm = 18.0 × 10 −15 m. Also, note that the first
term in this computation is just the potential energy computed in part (a). Thus,
2
PEb = PEa +
4ke e 2
r13
= 3.84 × 10
(c)
2
−14
J+
(
)(
4 8.99 × 10 9 N ⋅ m 2 C2 1.60 × 10 −19 C
18.0 × 10 −15 m
)
2
= 2.55 × 10 −13 J
If we start with the three-particle system of part (b) and allow the alpha particle to escape to
infinity [thereby returning us to the two-particle system of part (a)], the change in electric
potential energy will be
ΔPE = PEa − PEb = 3.84 × 10 −14 J − 2.55 × 10 −13 J = −2.17 × 10 −13 J
(d)
Conservation of energy, ΔKE + ΔPE = 0, gives the speed of the alpha particle at infinity in
the situation of part (c) as 12 ma va2 − 0 = −ΔPE, or
continued on next page
68719_16_ch16_p030-061.indd 42
1/7/11 2:33:00 PM
43
Electrical Energy and Capacitance
va =
(e)
6.64 × 10
−27
kg
) = 8.08 × 10
6
ms
When, starting with the three-particle system, the two protons are both allowed to escape
to infinity, there will be no remaining pairs of particles and hence no remaining potential energy. Thus, ΔPE = 0 − PEb = −PEb, and conservation of energy gives the change
in kinetic energy as ΔKE = −ΔPE = +PEb. Since the protons are identical particles, this
increase in kinetic energy is split equally between them giving KEproton = 12 m p v 2p = 12 ( PEb ),
or
16.20
(
−2 −2.17 × 10 −13 J
−2 ( ΔPE )
=
ma
PEb
2.55 × 10 −13 J
=
= 1.24 × 10 7 m s
mp
1.67 × 10 −27 kg
vp =
(a)
If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and allowed to
recede to infinity, the final speeds of the two particles will differ because of the difference
in the masses of the particles. Thus, attempting to solve for the final speeds by use of conservation of energy alone leads to a situation of having one equation with two unknowns ,
and does not permit a solution.
(b)
In the situation described in part (a) above, one can obtain a second equation with the two
unknown final speeds by using conservation of linear momentum. Then, one would have
two equations which could be solved simultaneously for both unknowns.
(c)
From conservation of energy: ⎡⎣
ma v + m p v =
2
a
2
p
2ke qa q p
ri
(
=
1
2
)
ma va2 + 12 m p v 2p − 0 ⎤⎦ + ⎡⎣ 0 − ke qa q p ri ⎤⎦ = 0, or
(
)(
)(
2 8.99 × 10 9 N ⋅ m 2 C2 3.20 × 10 −19 C 1.60 × 10 −19 C
4.00 × 10
−15
)
m
yielding ma va2 + m p v 2p = 2.30 × 10 −13 J
[1]
From conservation of linear momentum,
ma va + m p v p = 0
or
⎛ mp ⎞
va = ⎜
vp
⎝ ma ⎟⎠
[2]
Substituting Equation [2] into Equation [1] gives
2
⎛ mp ⎞ 2
ma ⎜
v p + m p v 2p = 2.30 × 10 −13 J
⎝ ma ⎟⎠
or
⎛ mp
⎞
2
−13
⎜⎝ m + 1⎟⎠ m p v p = 2.30 × 10 J
a
and
vp =
2.30 × 10 −13 J
(m
p
)
ma + 1 m p
=
(1.67 × 10
−27
2.30 × 10 −13 J
= 1.05 × 10 7 m s
6.64 × 10 −27 +1 1.67 × 10 −27 kg
)(
)
Then, Equation [2] gives the final speed of the alpha particle as
⎛ mp ⎞
⎛ 1.67 × 10 −27 kg ⎞
va = ⎜
1.05 × 10 7 m s = 2.64 × 10 6 m s
vp = ⎜
⎟
⎝ 6.64 × 10 −27 kg ⎟⎠
⎝ ma ⎠
(
68719_16_ch16_p030-061.indd 43
)
1/7/11 2:33:03 PM
44
16.21
Chapter 16
(a)
Conservation of energy gives
⎛1 1⎞
KE f = KEi + PEi − PE f = 0 + ke q1q2 ⎜ − ⎟
⎝ ri rf ⎠
(
)
With q1 = +8.50 nC, q2 = −2.80 nC, ri = 1.60 mm, and rf = 0.500 mm, this becomes
⎛
N ⋅ m2 ⎞
1
1
⎛
⎞
KE f = ⎜ 8.99 × 10 9
−
8.50 × 10 −9 C −2.80 × 10 −9 C ⎜
⎟
2
−6
−6
⎟
⎝
C ⎠
1.60 × 10 m 0.500 × 10 m ⎠
⎝
(
When r = rf = 0.500 mm and KE = KE f = 0.294 J, the speed of the sphere having mass
m = 8.00 mg = 8.00 × 10 −6 kg is
vf =
16.22
)
KE f = 0.294 J
yielding
(b)
)(
(
2 KE f
m
)=
2 ( 0.294 J )
= 271 m s
8.00 × 10 −6 kg
The excess charge on the metal sphere will be uniformly distributed over its surface. In this
spherically symmetric situation, the electric field and the electric potential outside the sphere
is the same as if all the excess charge were concentrated as a point charge at the center of the
sphere. Thus, for points outside the sphere,
E = ke
Q
r2
V = ke
and
Q
= E ⋅r
r
If the sphere has a radius of r = 18 cm = 0.18 m and the air breaks down when E = 3.0 × 10 6 V m,
the electric potential at the surface of the sphere when breakdown occurs is
(
)
V = 3.0 × 10 6 V m ( 0.18 m ) = 5.4 × 10 5 V
16.23
From conservation of energy, ( KE + PEe ) f = ( KE + PEe )i , which gives
0 + keQq rf = 12 ma vi2 + 0, or
rf =
rf =
16.24
(a)
2 keQq 2 ke ( 79e )( 2e )
=
ma vi2
ma vi2
(
)
(
2 8.99 × 10 9 N ⋅ m 2 C2 (158 ) 1.60 × 10 −19 C
(6.64 × 10
)(
−27
kg 2.00 × 10 m s
7
)
2
)
2
= 2.74 × 10 −14 m
The distance from any one of the corners of the square to the point at the center is one half
the length of the diagonal of the square, or
r=
diagonal
=
2
a2 + a2 a 2
a
=
=
2
2
2
Since the charges have equal magnitudes and are all the same distance from the center of
the square, they make equal contributions to the total potential. Thus,
Vtotal = 4Vsingle = 4
charge
keQ
kQ
Q
= 4 e = 4 2ke
a
r
a 2
continued on next page
68719_16_ch16_p030-061.indd 44
1/7/11 2:33:05 PM
Electrical Energy and Capacitance
45
The work required to carry charge q from infinity to the point at the center of the square is
equal to the increase in the electric potential energy of the charge, or
(b)
Q⎞
qQ
⎛
W = PEcenter − PE∞ = qVtotal − 0 = q ⎜ 4 2ke ⎟ = 4 2ke
⎝
a⎠
a
16.25
(
)
(a)
6
2
C2 ⎞ 1.0 × 10 m
A ⎛
−12
C = ∈0 = ⎜ 8.85 × 10
= 1.1 × 10 −8 F
d ⎝
N ⋅ m 2 ⎟⎠
800
m
(
)
(b)
Qmax = C ( ΔV )max = C ( Emax d ) = ∈0
(
A
Emax d = ∈0 AEmax
d
(
)(
)
)(
)
= 8.85 × 10 −12 C2 N ⋅ m 2 1.0 × 10 6 m 2 3.0 × 10 6 N C = 27 C
16.26
16.27
C=
(b)
Q = C ( ΔV ) = ( 3.00 mF )(12.0 V ) = 36.0 mC
(a)
The capacitance of this air-filled ( dielectric constant, k = 1.00 ) parallel-plate capacitor is
C=
16.28
16.29
16.30
Q
27.0 mC
=
= 3.00 mF
ΔV
9.00 V
(a)
(
)(
)
−12
C2 N ⋅ m 2 2.30 × 10 −4 m 2
k ∈0 A (1.00 ) 8.85 × 10
=
= 1.36 × 10 −12 F = 1.36 pF
1.50 × 10 −3 m
d
(
)
(b)
Q = C ( ΔV ) = 1.36 × 10 −12 F (12.0 V ) = 1.63 × 10 −11 C = 16.3 × 10 −12 C = 16.3 pC
(c)
E=
ΔV
12.0 V
=
= 8.00 × 10 3 V m = 8.00 × 10 3 N C
d
1.50 × 10 −3 m
(a)
C=
Q 10.0 mC
=
= 1.00 mF
V
10.0 V
(b)
V=
Q 100 mC
=
= 100 V
C 1.00 mF
(a)
E=
ΔV
20.0 V
=
= 1.11 × 10 4 V m = 11.1 kV m toward the negative plate
d
1.80 × 10 −3 m
(b)
C=
8.85 × 10 −12 C2 N ⋅ m 2 7.60 × 10 −4 m 2
∈0 A
=
= 3.74 × 10 −12 F = 3.74 pF
1.80 × 10 −3 m
d
(c)
Q = C ( ΔV ) = 3.74 × 10 −12 F ( 20.0 V ) = 7.48 × 10 −11 C = 74.8 pC on one plate and
−74.8 pC on the other plate.
(
)(
(
)
)
C = ∈0 A d, so
d=
(
)(
)
8.85 × 10 −12 C2 N ⋅ m 2 21.0 × 10 −12 m 2
∈0 A
=
= 3.10 × 10 −9 m
60.0 × 10 −15 F
C
⎛ 1Å ⎞
d = 3.10 × 10 −9 m ⎜ −10 ⎟ = 31.0 Å
⎝ 10 m ⎠
(
68719_16_ch16_p030-061.indd 45
)
1/7/11 2:33:08 PM
46
16.31
Chapter 16
(a)
Assuming the capacitor is air-filled (k = 1), the capacitance is
C=
16.32
(
)(
)
8.85 × 10 −12 C2 N ⋅ m 2 0.200 m 2
∈0 A
=
= 5.90 × 10 −10 F
d
3.00 × 10 −3 m
(
)
(b)
Q = C ( ΔV ) = 5.90 × 10 −10 F ( 6.00 V ) = 3.54 × 10 −9 C
(c)
E=
ΔV
6.00 V
=
= 2.00 × 10 3 V m = 2.00 × 10 3 N C
d
3.00 × 10 −3 m
(d)
s =
Q 3.54 × 10 −9 C
=
= 1.77 × 10 −8 C m 2
A
0.200 m 2
(e)
Increasing the distance separating the plates decreases the capacitance, the charge
stored, and the electric field strength between the plates. This means that all of the
previous answers will be decreased .
ΣFy = 0 ⇒ T cos15.0° = mg or T =
mg
cos15.0°
ΣFx = 0 ⇒ qE = T sin15.0° = mg tan15.0°
or
E=
mg tan15.0°
q
ΔV = Ed =
mgd tan15.0°
q
(350 × 10
ΔV =
16.33
(a)
−6
)(
)
kg 9.80 m s 2 ( 0.040 0 m ) tan15.0°
30.0 × 10
−9
C
= 1.23 × 10 3 V = 1.23 kV
Capacitors in a series combination store the same charge, Q = Ceq (ΔV ), where Ceq is the
equivalent capacitance and ΔV is the potential difference maintained across the series combination. The equivalent capacitance for the given series combination is 1 Ceq = 1 C1 + 1 C2,
or Ceq = C1C2 (C1 + C2 ), giving
Ceq =
( 2.50 mF ) (6.25 mF ) = 1.79 mF
2.50 mF + 6.25 mF
and the charge stored on each capacitor in the series combination is
Q = Ceq ( ΔV ) = (1.79 mF ) ( 6.00 V ) = 10.7 mC
(b)
68719_16_ch16_p030-061.indd 46
When connected in parallel, each capacitor has the same potential difference, ΔV = 6.00 V,
maintained across it. The charge stored on each capacitor is then
For C1 = 2.50 mF:
Q1 = C1 ( ΔV ) = ( 2.50 mF )( 6.00 V ) = 15.0 mC
For C2 = 6.25 mF:
Q2 = C2 ( ΔV ) = ( 6.25 mF )( 6.00 V ) = 37.5 mC
1/7/11 2:33:10 PM
Electrical Energy and Capacitance
16.34
47
(a)
Ceq = C1 + C2 = 5.00 mF + 12.0 mF = 17.0 mF
(b)
In a parallel combination, the full potential difference maintained between the terminals of
the battery exists across each capacitor. Thus,
ΔV1 = ΔV2 = ΔVbattery = 9.00 V
(c)
Q1 = C1 ( ΔV1 ) = ( 5.00 mF )( 9.00 V ) = 45.0 mC
Q2 = C2 ( ΔV2 ) = (12.0 mF )( 9.00 V ) = 108 mC
16.35
(a)
First, we replace the parallel combination
between points b and c by its equivalent
capacitance, Cbc = 2.00 mF + 6.00 mF = 8.00 mF.
Then, we have three capacitors in series
between points a and d. The equivalent
capacitance for this circuit is therefore
1
1
1
3
1
=
+
+
=
Ceq Cab Cbc Ccd 8.00 mF
giving
(b)
Ceq =
8.00 mF
= 2.67 mF
3
The charge stored on each capacitor in the series combination is
Qab = Qbc = Qcd = Ceq ( ΔVad ) = ( 2.67 mF )( 9.00 V ) = 24.0 mC
Then, note that ΔVbc = Qbc Cbc = 24.0 mC 8.00 mF = 3.00 V. The charge on each capacitor
in the original circuit is:
(c)
On the 8.00 mF between a and b:
Q8 = Qab = 24.0 mC
On the 8.00 mF between c and d:
Q8 = Qcd = 24.0 mC
On the 2.00 mF between b and c:
Q2 = C2 ( ΔVbc ) = ( 2.00 mF )( 3.00 V ) = 6.00 mC
On the 6.00 mF between b and c:
Q6 = C6 ( ΔVbc ) = ( 6.00 mF )( 3.00 V ) = 18.0 mC
Note that ΔVab = Qab Cab = 24.0 mC 8.00 mF = 3.00 V, and that
ΔVcd = Qcd Ccd = 24.0 mC 8.00 mF = 3.00 V. We earlier found that ΔVbc = 3.00 V, so we
conclude that the potential difference across each capacitor in the circuit is
ΔV8 = ΔV2 = ΔV6 = ΔV8 = 3.00 V
16.36
Cparallel = C1 + C2 = 9.00 pF ⇒
1
Cseries
=
1
1
+
C1 C2
⇒
Cseries =
Thus, using Equation [1],
C1 = 9.00 pF − C2
[1]
C1C2
= 2.00 pF
C1 + C2
Cseries =
(9.00 pF − C2 ) C2
(9.00 pF − C2 ) + C2
= 2.00 pF which reduces to
continued on next page
68719_16_ch16_p030-061.indd 47
1/7/11 2:33:13 PM
48
Chapter 16
C22 − ( 9.00 pF ) C2 + 18.0 ( pF ) = 0 , or ( C2 − 6.00 pF ) (C2 − 3.00 pF ) = 0
2
Therefore, either C2 = 6.00 pF and, from Equation [1], C1 = 3.00 pF
or C2 = 3.00 pF
and
C1 = 6.00 pF.
We conclude that the two capacitances are 3.00 pF and 6.00 pF .
16.37
(a)
3.00 mF 6.00 mF
The equivalent capacitance of the series combination in the
upper branch is
1
=
Cupper
or
1
1
2 +1
+
=
3.00 mF 6.00 mF 6.00 mF
2.00 mF 4.00 mF
Cupper = 2.00 mF
Likewise, the equivalent capacitance of the series combination in the lower branch is
1
=
Clower
1
1
2 +1
+
=
2.00 mF 4.00 mF 4.00 mF
or
90.0 V
Clower = 1.33 mF
These two equivalent capacitances are connected in parallel with each other, so the equivalent capacitance for the entire circuit is
Ceq = Cupper + Clower = 2.00 mF + 1.33 mF = 3.33 mF
(b)
Note that the same potential difference, equal to the potential difference of the battery,
exists across both the upper and lower branches. The charge stored on each capacitor in the
series combination in the upper branch is
Q3 = Q6 = Qupper = Cupper ( ΔV ) = ( 2.00 mF )( 90.0 V ) = 180 mC
and the charge stored on each capacitor in the series combination in the lower branch is
Q2 = Q4 = Qlower = Clower ( ΔV ) = (1.33 mF )( 90.0 V ) = 120 mC
(c)
16.38
(a)
The potential difference across each of the capacitors in the circuit is:
ΔV2 =
Q2 120 mC
=
= 60.0 V
C2 2.00 mF
ΔV4 =
Q4 120 mC
=
= 30.0 V
C4 4.00 mF
ΔV3 =
Q3 180 mC
=
= 60.0 V
C3 3.00 mF
ΔV6 =
Q6 180 mC
=
= 30.0 V
C6 6.00 mF
The equivalent capacitance of the series
combination in the rightmost branch of
the circuit is
1
Cright
or
=
1
1
1+ 3
+
=
24.0 mF 8.00 mF 24.0 mF
Figure P16.38
Cright = 6.00 mF
continued on next page
68719_16_ch16_p030-061.indd 48
1/7/11 2:33:16 PM
Electrical Energy and Capacitance
(b)
49
The equivalent capacitance of the three
capacitors now connected in parallel with
each other and with the battery is
Ceq = 4.00 mF + 2.00 mF + 6.00 mF = 12.0 mF
(c)
The total charge stored in this circuit is
Diagram 1
Qtotal = Ceq ( ΔV ) = (12.0 mF )( 36.0 V )
or
(d)
Qtotal = 432 mC
The charges on the three capacitors shown in Diagram 1 are:
Diagram 2
Q4 = C4 ( ΔV ) = ( 4.00 mF )( 36.0 V ) = 144 mC
Q2 = C2 ( ΔV ) = ( 2.00 mF )( 36.0 V ) = 72 mC
Qright = Cright ( ΔV ) = ( 6.00 mF )( 36.0 V ) = 216 mC
Yes. Q4 + Q2 + Qright = Qtotal as it should.
(e)
The charge on each capacitor in the series combination in the rightmost branch of the original circuit (Figure P16.38) is
Q24 = Q8 = Qright = 216 mC
(f)
ΔV24 =
(g)
ΔV8 =
Q24 216 mC
=
= 9.00 V
C24 24.0 mF
Q8 216 mC
=
= 27.0 V
C8 8.00 mF
Note that ΔV8 + ΔV24 = ΔV = 36.0 V as it should.
16.39
The circuit may be reduced in steps as shown above.
Using Figure 3,
Qac = ( 4.00 mF )( 24.0 V ) = 96.0 mC
Then, in Figure 2, ( ΔV )ab =
Qac 96.0 mC
=
= 16.0 V
Cab 6.00 mF
continued on next page
68719_16_ch16_p030-061.indd 49
1/7/11 2:33:19 PM
50
Chapter 16
and
( ΔV )bc = ( ΔV )ac − ( ΔV )ab = 24.0 V − 16.0 V = 8.00 V
Finally, using Figure 1,
Q5 = ( 5.00 mF )( ΔV )ab = 80.0 mC
Q1 = C1 ( ΔV )ab = (1.00 mF )(16.0 V ) = 16.0 mC ,
Q8 = (8.00 mF )( ΔV )bc = 64.0 mC ,
16.40
(a)
Q4 = ( 4.00 mF )( ΔV )bc = 32.0 mC
and
Consider the simplification of the circuit as shown below:
C1
+
ΔV
C1
a
C3
C2
−
ΔV
+
a
ΔV
Cab
−
b
+
−
Ceq
b
Since C2 and C3 are connected in parallel, Cab = C2 + C3 = C + 5C = 6C.
Now observe that C1 and Cab are connected in series, giving
1
1
1
=
+
Ceq C1 Cab
(b)
or
Ceq =
C1Cab
( 3C )( 6C )
=
= 2C
C1 + Cab
3C + 6C
Since capacitors C1 and Cab are connected in series,
Q1 = Qab = Qeq = Ceq ( ΔV ) = 2C ( ΔV )
(c)
Qab 2C ( ΔV ) ΔV
=
=
, giving
Cab
6C
3
Then,
ΔVab =
Also,
Q3 = C3 ( ΔVab ) =
Q2 = C2 ( ΔVab ) =
5C ( ΔV )
. Therefore,
3
C ( ΔV )
3
Q1 > Q3 > Q2
Since capacitors C1 and Cab are in series with the battery,
ΔV1 = ΔV − ΔVab = ΔV −
ΔV 2
= ΔV
3
3
Also, with capacitors C2 and C3 in parallel between points a and b,
ΔV2 = ΔV3 = ΔVab =
Thus,
(d)
ΔV
3
ΔV1 > ΔV2 = ΔV3
Consider the following steps:
(i)
Increasing C3 while C1 and C2 remain constant will increase Cab = C2 + C3.
⎛ Cab ⎞
, will increase.
Therefore, the equivalent capacitance, Ceq = C1 ⎜
⎝ C1 + Cab ⎟⎠
continued on next page
68719_16_ch16_p030-061.indd 50
1/7/11 2:33:22 PM
Electrical Energy and Capacitance
(ii)
Since C1 and Cab are in series, Q1 = Qab = Ceq ( ΔV ). Thus, Q1 will increase as Ceq
increases. Also, Qab experiences the same increase.
(iii)
Because ΔV1 = Q1 C1, an increase in Q1 causes ΔV1 to increase and causes
ΔVab = ΔV − ΔV1 to decrease. Thus, since Q2 = C2 ( ΔVab ), Q2 will decrease .
51
(iv) With capacitors C2 and C3 in parallel between points a and b, we have Qab = Q2 + Q3 or
Q3 = Qab − Q2. Thus, with Qab increasing [see Step (ii)] while Q2 is decreasing
[see Step (iii)], we see that Q3 will increase .
16.41
(a)
From Q = C ( ΔV ), Q25 = ( 25.0 mF )( 50.0 V ) = 1.25 × 10 3 mC = 1.25 mC
and
(b)
Q40 = ( 40.0 mF )( 50.0 V ) = 2.00 × 10 3 mC = 2.00 mC
Since the negative plate of one capacitor was connected to the positive plate of the other,
the net charge stored in the new parallel combination is
Q = Q40 − Q25 = 2.00 × 10 3 mC − 1.25 × 10 3 mC = 750 mC
The two capacitors, now in parallel, have a common potential difference ΔV across them.
The new charges on each of the capacitors are Q25′ = C1 ( ΔV ) and Q40′ = C2 ( ΔV ). Thus,
Q25′ =
⎛ 25 mF ⎞
C1
5
Q40′ = Q40′
Q40′ = ⎜
⎟
8
C2
⎝ 40 mF ⎠
and the total change now stored in the combination may be written as
5
13
Q = Q40′ + Q25′ = Q40′ + Q40′ = Q40′ = 750 mC
8
8
giving Q40′ =
(c)
The potential difference across each capacitor in the new parallel combination is
ΔV =
16.42
(a)
8
( 750 mC) = 462 mC and Q25′ = Q − Q40′ = ( 750 − 462) mC = 288 mC
13
Q
Q
750 mC
=
=
= 11.5 V
Ceq C1 + C2 65.0 mF
The original circuit reduces to a single equivalent capacitor in the steps shown below.
eq
continued on next page
68719_16_ch16_p030-061.indd 51
1/7/11 2:33:25 PM
52
Chapter 16
⎛ 1
1 ⎞
Cs = ⎜ + ⎟
⎝ C1 C2 ⎠
−1
⎛
1
1 ⎞
=⎜
+
⎝ 5.00 mF 10.0 mF ⎟⎠
−1
= 3.33 mF
C p1 = Cs + C3 + Cs = 2 ( 3.33 mF ) + 2.00 mF = 8.66 mF
C p 2 = C2 + C2 = 2 (10.0 mF ) = 20.0 mF
⎛ 1
1 ⎞
Ceq = ⎜
+
⎟
⎝ C p1 C p 2 ⎠
(b)
−1
⎛
⎞
1
1
=⎜
+
⎝ 8.66 mF 20.0 mF ⎟⎠
−1
= 6.04 mF
The total charge stored between points a and b is
Qtotal = Ceq ( ΔV )ab = ( 6.04 mF )( 60.0 V ) = 362 mC
Then, looking at the third figure, observe that the charges of the series capacitors of that
figure are Q p1 = Q p 2 = Qtotal = 362 mC. Thus, the potential difference across the upper
parallel combination shown in the second figure is
( ΔV ) p1 =
Q p1
C p1
=
362 mC
= 41.8 V
8.66 mF
Finally, the charge on C3 is
Q3 = C3 ( ΔV ) p1 = ( 2.00 mF )( 41.8 V ) = 83.6 mC
16.43
From Q = C ( ΔV ), the initial charge of each capacitor is
Q1 = (1.00 mF )(10.0 V ) = 10.0 mC
and
Q2 = ( 2.00 mF ) ( 0 ) = 0
After the capacitors are connected in parallel, the potential difference across one is the same as
that across the other. This gives
ΔV =
Q1′
Q2′
=
1.00 mF 2.00 mF
or
Q2′ = 2 Q1′
[1]
From conservation of charge, Q1′ + Q2′ = Q1 + Q2 = 10.0 mC. Then, substituting from Equation [1],
this becomes
Q1′ + 2 Q1′ = 10.0 mC , giving
Q2′ = 20 mC 3 = 6.67 mC
Finally, from Equation [1],
16.44
(a)
Q1′ = 10 mC 3 = 3.33 mC
We simplify the circuit in stages as shown below:
15.0 mF 3.00 mF
Cs
a
b
a
b
20.0 mF
6.00 mF
20.0 mF
a
Cp
b
a
Ceq
b
20.0 mF
6.00 mF
continued on next page
68719_16_ch16_p030-061.indd 52
1/7/11 2:33:28 PM
Electrical Energy and Capacitance
1
1
1
=
+
Cs 15.0 mF 3.00 mF
or
Cs =
53
(15.0 mF ) (3.00 mF ) = 2.50 mF
15.0 mF + 3.00 mF
C p = Cs + 6.00 mF = 2.50 mF + 6.00 mF = 8.50 mF
1
1
1
=
+
Ceq C p 20.0 mF
(b)
or
Ceq =
(8.50 mF ) ( 20.0 mF ) =
8.50 mF + 20.0 mF
5.96 mF
Q20 = QC = Qeq = Ceq ( ΔVab ) = ( 5.96 mF )(15.0 V ) = 89.4 mC
p
ΔVC =
QC
p
Cp
p
(
=
89.4 mC
= 10.5 V
8.50 mF
) = (6.00 mF )(10.5 V ) = 63.0 mC
= Q = Q = C ( ΔV ) = ( 2.50 mF )(10.5 V ) = 26.3 mC
so
Q6 = C6 ΔVC
and
Q15
3
p
s
s
Cp
The charges are 89.4 mC on the 20 mF capacitor, 63.0 mC on the 6 mF capacitor, and
26.3 mC on both the 15 mF and 3 mF capacitors.
16.45
Energy stored =
16.46
(a)
Q2 1
1
2
2
= C ( ΔV ) = ( 4.50 × 10 −6 F )(12.0 V ) = 3.24 × 10 −4 J
2
2C 2
The equivalent capacitance of a series combination of C1 and C2 is
1
1
2 +1
1
=
+
=
Ceq 18.0 mF 36.0 mF 36.0 mF
or
Ceq = 12.0 mF
When this series combination is connected to a 12.0-V battery, the total stored energy is
Total energy stored =
(b)
1
1
2
2
Ceq ( ΔV ) = (12.0 × 10 −6 F )(12.0 V ) = 8.64 × 10 −4 J
2
2
The charge stored on each of the two capacitors in the series combination is
Q1 = Q2 = Qtotal = Ceq ( ΔV ) = (12.0 mF )(12.0 V ) = 144 mC = 1.44 × 10 −4 C
and the energy stored in each of the individual capacitors is
(1.44 × 10−4 C) = 5.76 × 10−4 J
Q2
Energy stored in C1 = 1 =
2C1 2 (18.0 × 10 −6 F )
2
(1.44 × 10−4 C) = 2.88 × 10−4 J
Q2
Energy stored in C2 = 2 =
2C2 2 ( 36.0 × 10 −6 F )
2
and
Energy stored in C1 + Energy stored in C2 = 5.76 × 10 −4 J + 2.88 × 10 −4 J = 8.64 × 10 −4 J,
which is the same as the total stored energy found in part (a). This must be true
if the computed equivalent capacitance is truly equivalent to the original combination.
continued on next page
68719_16_ch16_p030-061.indd 53
1/7/11 2:33:31 PM
54
Chapter 16
(c)
If C1 and C2 had been connected in parallel rather than in series, the equivalent capacitance
would have been Ceq = C1 + C2 = 18.0 mF + 36.0 mF = 54.0 mF. If the total energy stored
⎡⎣ 12 Ceq ( ΔV )2 ⎤⎦ in this parallel combination is to be the same as was stored in the original
series combination, it is necessary that
ΔV =
2 (8.64 × 10 −4 J )
2 ( Total energy stored )
=
Ceq
54.0 × 10 −6 F
= 5.66 V
Since the two capacitors in parallel have the same potential difference across them, the
2
energy stored in the individual capacitors ⎡⎣ 12 C ( ΔV ) ⎤⎦ is directly proportional to their
capacitances. The larger capacitor, C2 , stores the most energy in this case.
16.47
(a)
The energy initially stored in the capacitor is
( Energy stored )
1
(b)
=
Qi2
1
1
2
2
= Ci ( ΔV )i = ( 3.00 mF )( 6.00 V ) = 54.0 mJ
2
2Ci 2
When the capacitor is disconnected from the battery, the stored charge becomes isolated
with no way off the plates. Thus, the charge remains constant at the value Qi as long as
the capacitor remains disconnected. Since the capacitance of a parallel-plate capacitor
is C = k ∈0 A d, when the distance d separating the plates is doubled, the capacitance is
decreased by a factor of 2 C f = Ci 2 = 1.50 mF . The stored energy (with Q unchanged)
becomes
(
( Energy stored )
(c)
2
⎛ Q2 ⎞
Qi2
Qi2
=
= 2 ⎜ i ⎟ = 2 ( Energy stored )1 = 108 mJ
2C f 2 (Ci 2 )
⎝ 2Ci ⎠
=
When the capacitor is reconnected to the battery, the potential difference between the plates
is reestablished at the original value of ΔV = ( ΔV )i = 6.00 V, while the capacitance remains
at C f = Ci 2 = 1.50 mF. The energy stored under these conditions is
( Energy stored )
3
16.48
)
=
1
1
2
2
C f ( ΔV )i = (1.50 mF )( 6.00 V ) = 27.0 mJ
2
2
The energy transferred to the water is
W=
8
1 ⎡1
⎤ = ( 50.0 C )(1.00 × 10 V ) = 2.50 × 10 7 J
Q
ΔV
(
)
⎥⎦
100 ⎢⎣ 2
200
Thus, if m is the mass of water boiled away, W = m ⎡⎣c ( ΔT ) + Lv ⎤⎦ becomes
⎡⎛
⎤
J ⎞
2.50 × 10 7 J = m ⎢⎜ 4186
(100°C − 30.0°C ) + 2.26 × 106 J kg ⎥
⎟
kg
⋅°C
⎠
⎣⎝
⎦
giving
16.49
(a)
m=
2.50 × 10 7 J
= 9.79 kg
⎡⎣ 2.93 × 10 5 J kg + 2.26 × 10 6 J kg ⎤⎦
Note that the charge on the plates remains constant at the original value, Q0 , as the dielectric is inserted. Thus, the change in the potential difference, ΔV = Q C, is due to a change
in capacitance alone. The ratio of the final and initial capacitances is
Cf
Ci
=
k ∈0 A d
=k
∈0 A d
and
Cf
Ci
=
Q0 ( ΔV ) f
Q0 ( ΔV )i
=
( ΔV )i 85.0 V
=
= 3.40
( ΔV ) f 25.0 V
continued on next page
68719_16_ch16_p030-061.indd 54
1/7/11 2:33:34 PM
Electrical Energy and Capacitance
55
Thus, the dielectric constant of the inserted material is k = 3.40 , and the material is
probably nylon (see Table 16.1).
16.50
(b)
If the dielectric only partially filled the space between the plates, leaving the remaining
space air-filled, the equivalent dielectric constant would be somewhere between k = 1.00
(air) and k = 3.40. The resulting potential difference would then lie somewhere between
( ΔV )i = 85.0 V and ( ΔV ) f = 25.0 V.
(a)
If the maximum electric field that can exist between the plates before breakdown (i.e.,
the dielectric strength) is Em ax, the maximum potential difference across the plates is
ΔVm ax = Em ax ⋅ d, where d is the plate separation. The maximum change on either plate then
has magnitude
Qm ax = C ( ΔVm ax ) = C ( Em ax ⋅ d )
Since the capacitance of a parallel-plate capacitor is C = k ∈0 A d, the maximum charge is
⎛ k ∈0 A ⎞
Qm ax = ⎜
Em ax ⋅ d = k ∈0 AEm ax
⎝ d ⎟⎠
(
)
The area of each plate is A = 5.00 cm 2 = 5.00 × 10 −4 m 2, and when air is the dielectric,
k = 1.00 and Em ax = 3.00 × 10 6 V m (see Table 16.1). Thus,
Qm ax = (1.00 )(8.85 × 10 −12 C N ⋅ m 2 ) ( 5.00 × 10 −4 m 2 ) ( 3.00 × 10 6 V m )
= 1.33 × 10 −8 C = 13.3 nC
(b)
If the dielectric is now polystyrene (k = 2.56 and Em ax = 24.0 × 10 6 V m ), then
Qm ax = ( 2.56 )(8.85 × 10 −12 C N ⋅ m 2 ) ( 5.00 × 10 −4 m 2 ) ( 24.0 × 10 6 V m )
= 2.72 × 10 −7 C = 272 nC
16.51
(a)
The dielectric constant for Teflon® is k = 2.1, so the capacitance is
C=
−12
2
2
−4
2
k ∈0 A ( 2.1) (8.85 × 10 C N ⋅ m ) (175 × 10 m )
=
d
0.040 0 × 10 −3 m
C = 8.1× 10 −9 F = 8.1 nF
(b)
For Teflon®, the dielectric strength is Em ax = 60 × 10 6 V m, so the maximum voltage is
ΔVm ax = Em ax d = ( 60 × 10 6 V m ) ( 0.040 0 × 10 −3 m
)
ΔVm ax = 2.4 × 10 3 V = 2.4 kV
16.52
Before the capacitor is rolled, the capacitance of this parallel-plate capacitor is
C=
k ∈0 A k ∈0 ( w × L )
=
d
d
where A is the surface area of one side of a foil strip. Thus, the required length is
9.50 × 10 −8 F ) ( 0.025 0 × 10 −3 m )
(
C ⋅d
L=
=
= 1.04 m
k ∈0 w ( 3.70 )(8.85 × 10 −12 C2 N ⋅ m 2 ) ( 7.00 × 10 −2 m )
68719_16_ch16_p030-061.indd 55
1/7/11 2:33:36 PM
56
16.53
Chapter 16
(a)
V=
m 1.00 × 10 −12 kg
=
= 9.09 × 10 −16 m 3
r
1100 kg m 3
Since V = 4p r 3 3, the radius is r = [ 3V 4p ] , and the surface area is
13
3V ⎤
A = 4p r = 4p ⎡⎢
⎣ 4p ⎥⎦
2
2 3
⎡ 3 ( 9.09 × 10 −16 m 3 ) ⎤
= 4p ⎢
⎥
4p
⎢⎣
⎥⎦
2 3
= 4.54 × 10 −10 m 2
−12
2
2
−10
2
k ∈0 A ( 5.00 )(8.85 × 10 C N ⋅ m ) ( 4.54 × 10 m )
=
= 2.01× 10 −13 F
d
100 × 10 −9 m
(b)
C=
(c)
Q = C ( ΔV ) = ( 2.01× 10 −13 F ) (100 × 10 −3 V ) = 2.01 × 10 −14 C
and the number of electronic charges is
n=
16.54
(
)
For a parallel-plate capacitor, C = k ∈0 A d and Q = s A = C ( ΔV ). Thus, s A = k ∈0 A d ( ΔV ),
and d = (k ∈0 s )( ΔV ). With air as the dielectric material (k = 1.00 ), the separation of the plates
must be
d=
16.55
Q 2.01× 10 −14 C
=
= 1.26 × 10 5
e 1.60 × 10 −19 C
(1.00 )(8.85 × 10 −12 C2 N ⋅ m 2 )(150 V )
(3.00 × 10
−10
C cm 2 ) (10 4 cm 2 1 m 2 )
= 4.43 × 10 −4 m = 0.443 mm
Since the capacitors are in series, the equivalent capacitance is given by
d
d + d 2 + d3
1
1
1
d
1
d
=
+
+
= 1 + 2 + 3 = 1
Ceq C1 C2 C3 ∈0 A ∈0 A ∈0 A
∈0 A
16.56
∈0 A
where d = d1 + d 2 + d3
d
or
Ceq =
(a)
Please refer to the solution of Problem 16.37 where
the following results were obtained:
Ceq = 3.33 mF
Q3 = Q6 = 180 mC
3.00 mF 6.00 mF
Q2 = Q4 = 120 mC
2.00 mF 4.00 mF
The total energy stored in the full circuit is then
( Energy stored )
total
=
1
1
2
2
Ceq ( ΔV ) = ( 3.33 × 10 −6 F )( 90.0 V )
2
2
90.0 V
= 1.35 × 10 −2 J = 13.5 × 10 −3 J = 13.5 mJ
(b)
The energy stored in each individual capacitor is
(120 × 10 −6 C) = 3.60 × 10 −3 J = 3.60 mJ
Q2
= 2 =
2C2 2 ( 2.00 × 10 −6 F )
2
For 2.00 mF :
( Energy stored )
For 3.00 mF:
(180 × 10 −6 C) = 5.40 × 10 −3 J = 5.40 mJ
Q2
( Energy stored )3 = 3 =
2C3 2 ( 3.00 × 10 −6 F )
2
2
continued on next page
68719_16_ch16_p030-061.indd 56
1/7/11 2:33:39 PM
57
Electrical Energy and Capacitance
120 × 10 −6 C )
(
Q42
=
=
= 1.80 × 10 −3 J = 1.80 mJ
2C4 2 ( 4.00 × 10 −6 F )
2
For 4.00 mF:
( Energy stored )
For 6.00 mF:
( Energy stored )6 =
4
(180 × 10 −6 C) = 2.70 × 10−3 J = 2.70 mJ
Q62
=
2C6 2 ( 6.00 × 10 −6 F )
2
(c)
The total energy stored in the individual capacitors is
Energy stored = ( 3.60 + 5.40 + 1.80 + 2.70 ) mJ = 13.5 mJ = ( Energy stored )total
Thus, the sums of the energies stored in the individual capacitors equals the total energy
stored by the system.
16.57
In the absence of a dielectric, the capacitance of the
parallel-plate capacitor is
C0 = ∈0 A d.
1
d
3
1
d
3
k
k
c1
With the dielectric inserted,
2
d
it fills one-third of the gap
d
between the plates as shown 3
in sketch (a) at the right. We
2
d
c2
model this situation as con3
sisting of a pair of capacitors, C1 and C2, connected
(a)
in series as shown in sketch
(b) at the right. In reality,
the lower plate of C1 and the
(b)
upper plate of C2 are one
and the same, consisting of the lower surface of the dielectric shown in sketch (a). The capacitances in the model of sketch (b) are given by
C1 =
k ∈0 A 3k ∈0 A
=
d 3
d
and
C2 =
∈0 A 3 ∈0 A
=
2d 3
2d
The equivalent capacitance of the series combination is
d
1
2d
⎛1
⎞ ⎛ d ⎞ ⎛ 2k + 1 ⎞ d
⎛ 2k + 1 ⎞ d
⎛ 2k + 1 ⎞ 1
=
=
+
= ⎜ + 2⎟ ⎜
=
=
⎠ ⎝ 3 ∈0 A ⎟⎠ ⎜⎝ k ⎟⎠ 3 ∈0 A ⎜⎝ 3k ⎟⎠ ∈0 A ⎜⎝ 3k ⎟⎠ C0
Ceq 3k ∈0 A 3 ∈0 A ⎝ k
and Ceq = [ 3k (2k + 1) ]C0 .
16.58
For the parallel combination: C p = C1 + C2 which gives C2 = C p − C1
For the series combination:
Thus, we have C2 =
C p − C1 =
1
1
1
=
+
Cs C1 C2
or
[1]
1
1
1 C1 − Cs
=
−
=
C2 Cs C1
Cs C1
Cs C1
and equating this to Equation [1] above gives
C1 − Cs
Cs C1
C1 − Cs
We write this result as:
or
C p C1 − C p Cs − C12 + Cs C1 = Cs C1
C12 − C p C1 + C p Cs = 0
continued on next page
68719_16_ch16_p030-061.indd 57
1/7/11 2:33:42 PM
58
Chapter 16
C1 =
and use the quadratic formula to obtain
C2 =
Then, Equation [1] gives
16.59
1
1 2
Cp ±
C p − C p Cs
2
4
1
1 2
Cp ∓
C p − C p Cs
2
4
For a parallel-plate capacitor with plate separation d,
ΔVm ax = Em ax ⋅ d
d=
or
ΔVm ax
Em ax
The capacitance is then
C=
⎛ E
⎞
k ∈0 A
= k ∈0 A ⎜ m ax ⎟
d
⎝ ΔVm ax ⎠
and the needed area of the plates is A = C ⋅ ΔVm ax k ∈0 Em ax, or
A=
16.60
(a)
( 0.250 × 10
( 3.00 )(8.85 × 10
F ) ( 4.00 × 10 3 V )
C N⋅m
2
2
) ( 2.00 × 10
8
V m)
= 0.188 m 2
The 1.0-mC is located 0.50 m from point P, so its contribution to the potential at P is
V1 = ke
(b)
−12
−6
⎛ 1.0 × 10 −6 C ⎞
q1
= 1.8 × 10 4 V
= (8.99 × 10 9 N ⋅ m 2 C2 ) ⎜
r1
⎝ 0.50 m ⎟⎠
The potential at P due to the −2.0-mC charge located 0.50 m away is
V2 = ke
⎛ − 2.0 × 10 −6 C ⎞
q2
4
= (8.99 × 10 9 N ⋅ m 2 C2 ) ⎜
⎟⎠ = − 3.6 × 10 V
r2
0.50 m
⎝
(c)
The total potential at point P is VP = V1 + V2 = ( +1.8 − 3.6 ) × 10 4 V = − 1.8 × 10 4 V
(d)
The work required to move a charge q = 3.0 mC to point P from infinity is
W = qΔV = q (VP − V∞ ) = ( 3.0 × 10 −6 C ) ( −1.8 × 10 4 V − 0 ) = − 5.4 × 10 −2 J
16.61
The stages for the reduction of this circuit are shown below.
Thus, Ceq = 6.25 mF
68719_16_ch16_p030-061.indd 58
1/7/11 2:33:45 PM
Electrical Energy and Capacitance
16.62
(a)
59
Due to spherical symmetry, the charge on each of the concentric spherical shells will be
uniformly distributed over that shell. Inside a spherical surface having a uniform charge
distribution, the electric field due to the charge on that surface is zero. Thus, in this region,
the potential due to the charge on that surface is constant and equal to the potential at the
surface. Outside a spherical surface having a uniform charge distribution, the potential due
to the charge on that surface is given by V = ke q r, where r is the distance from the center
of that surface and q is the charge on that surface.
In the region between a pair of concentric spherical shells, with the inner shell having
charge + Q and the outer shell having radius b and charge − Q, the total electric potential at
distance r from the center is given by
V = Vdue to
+ Vdue to
inner shell
outer shell
=
ke Q ke ( −Q )
⎛ 1 1⎞
+
= ke Q ⎜ − ⎟
⎝ r b⎠
r
b
The potential difference between the two shells is therefore
⎛ 1 1⎞
⎛ 1 1⎞
⎛ b − a⎞
ΔV = V r = a − V r = b = ke Q ⎜ − ⎟ − ke Q ⎜ − ⎟ = ke Q ⎜
⎝ a b⎠
⎝ b b⎠
⎝ ab ⎟⎠
The capacitance of this device is given by
C=
(b)
Q
ab
=
ΔV
ke ( b − a )
When b >> a, then b − a ≈ b. Thus, in the limit as b → ∞, the capacitance found above
becomes
C→
16.63
The energy stored in a charged capacitor is Estored = 12 C ( ΔV ) . Hence,
2
ΔV =
16.64
ab
a
=
= 4p ∈0 a
ke ( b ) ke
2 Estored
=
C
2 ( 300 J )
= 4.47 × 10 3 V = 4.47 kV
30.0 × 10 −6 F
From Q = C ( ΔV ), the capacitance of the capacitor with air between the plates is
C0 =
Q0 150 mC
=
ΔV
ΔV
After the dielectric is inserted, the potential difference is held to the original value, but the charge
changes to Q = Q0 + 200 mC = 350 mC. Thus, the capacitance with the dielectric slab in place is
C=
Q
350 mC
=
ΔV
ΔV
The dielectric constant of the dielectric slab is therefore
k =
68719_16_ch16_p030-061.indd 59
C ⎛ 350 mC ⎞ ⎛ ΔV ⎞ 350
=
=
= 2.33
C0 ⎜⎝ ΔV ⎟⎠ ⎜⎝ 150 mC ⎟⎠ 150
1/7/11 2:33:47 PM
60
16.65
Chapter 16
The charges initially stored on the capacitors are
Q1 = C1 ( ΔV )i = ( 6.0 mF )( 250 V ) = 1.5 × 10 3 mC
and
Q2 = C2 ( ΔV )i = ( 2.0 mF )( 250 V ) = 5.0 × 10 2 mC
When the capacitors are connected in parallel, with the negative plate of one connected to the
positive pl ate of the other, the net stored charge is
Q = Q1 − Q2 = 1.5 × 10 3 mC − 5.0 × 10 2 mC = 1.0 × 10 3 mC
The equivalent capacitance of the parallel combination is Ceq = C1 + C2 = 8.0 mF. Thus, the final
potential difference across each of the capacitors is
Q 1.0 × 10 3 mC
=
= 125 V
Ceq
8.0 mF
( ΔV )′ =
and the final charge on each capacitor is
Q1′ = C1 ( ΔV )′ = ( 6.0 mF )(125 V ) = 750 mC = 0.75 mC
16.66
and
Q2′ = C2 ( ΔV )′ = ( 2.0 mF )(125 V ) = 250 mC = 0.25 mC
(a)
The distance from the charge 2q to either of the charges on the y-axis is r = d 2 + ( 2d ) = 5 d.
Thus,
2
V =∑
i
ke qi
kq
kq
2ke q
= e + e =
ri
5d
5d
5d
ke q1 q3 ke q2 q3 ke q ( 2q ) ke q ( 2q ) 4ke q 2
+
=
+
=
r1
r2
5d
5d
5d
(b)
PE2q =
(c)
From conservation of energy with PE = 0 at r = ∞,
KE f = KEi + PEi − PE f = 0 +
(d)
16.67
(
2 KE f
vf =
m
)=
4ke q 2
4ke q 2
−0=
5d
5d
1
2 ⎛ 4ke q 2 ⎞ ⎛ 8ke q 2 ⎞ 2
=
m ⎜⎝ 5 d ⎟⎠ ⎜⎝ 5 md ⎟⎠
When excess charge resides on a spherical surface that is far removed from any other charge, this
excess charge is uniformly distributed over the spherical surface, and the electric potential at the
surface is the same as if all the excess charge were concentrated at the center of the spherical surface.
In the given situation, we have two charged spheres, initially isolated from each other, with
charges and potentials of QA = +6.00 mC and VA = ke QA RA, where RA = 12.0 cm, QB = − 4.00 mC,
and VB = ke QB RB, with RB = 18.0 cm.
When these spheres are then connected by a long conducting thread, the charges are redistributed
( yielding charges of Q ′ and Q ′ , respectively ) until the two surfaces come to a common potential
(V ′ = kQ ′ R = V ′ = kQ ′ R ). When equilibrium is established, we have:
A
A
A
A
B
B
B
B
From conservation of charge:
QA′ + QB′ = QA + QB
⇒
QA′ + QB′ = +2.00 mC
[1]
continued on next page
68719_16_ch16_p030-061.indd 60
1/7/11 2:33:49 PM
Electrical Energy and Capacitance
From equal potentials: kQA′ = kQB′
RA
RB
16.68
⎛R ⎞
⇒ QB′ = ⎜ B ⎟ QA′
⎝ RB ⎠
QB′ = 1.50 QA′
or
61
[2]
+2.00 mC
= 0.800 mC
2.50
Substituting Equation [2] into [1] gives
QA′ =
Then, Equation [2] gives
QB′ = 1.50 ( 0.800 mC ) = 1.20 mC
The electric field between the plates is directed downward with magnitude
Ey =
ΔV
100 V
=
= 5.0 × 10 4 N m
D
2.0 × 10 −3 m
Since the gravitational force experienced by the electron is negligible in comparison to the electrical force acting on it, the vertical acceleration is
ay =
(a)
Fy
me
qE y
=
me
=
( −1.60 × 10
−19
C ) ( −5.0 × 10 4 N m )
9.11× 10 −31 kg
= + 8.8 × 1015 m s 2
At the closest approach to the bottom plate, v y = 0. Thus, the vertical displacement from
point O is found from v y2 = v02 y + 2 a y ( Δy ) as
Δy =
0 − ( v0 sinq 0 )
2 ay
2
=
− ⎡⎣ − ( 5.6 × 10 6 m s ) sin 45° ⎤⎦
2 (8.8 × 1015 m s 2 )
2
= − 8.9 × 10 −4 m = − 0.89 mm
The minimum distance above the bottom plate is then
d=
(b)
D
+ Δy = 1.0 mm − 0.89 mm = 0.1 mm
2
The time for the electron to go from point O to the upper plate (Δy = +1.0 mm) is found
from Δy = v0 y t + 12 a y t 2 as
m⎞
⎡ ⎛
⎤ 1⎛
15 m ⎞ 2
+1.0 × 10 −3 m = ⎢ − ⎜ 5.6 × 10 6
⎟⎠ sin 45° ⎥ t + ⎜⎝ 8.8 × 10
⎟t
⎝
s
s2 ⎠
2
⎣
⎦
Solving for t gives a positive solution of t = 1.1× 10 −9 s. The horizontal displacement from
point O at this time is
Δx = v0 x t = ⎡⎣( 5.6 × 10 6 m s ) cos 45° ⎤⎦ (1.1× 10 −9 s ) = 4.4 mm
68719_16_ch16_p030-061.indd 61
1/7/11 2:33:52 PM
17
Current and Resistance
QUICK QUIZZES
1.
Choice (d). Negative charges moving in one direction are equivalent to positive charges moving
in the opposite direction. Thus, I a , I b , I c , and I d are equivalent to the movement of 5, 3, 4, and
2 charges respectively, giving I d < I b < I c < I a.
2.
Choice (b). Under steady-state conditions, the current is the same in all parts of the wire. Thus,
the drift velocity, given by vd = I nqA, is inversely proportional to the cross-sectional area.
3.
Choices (c) and (d). Neither circuit (a) nor circuit (b) applies a difference in potential across the
bulb. Circuit (a) has both lead wires connected to the same battery terminal. Circuit (b) has a low
resistance path (a “short”) between the two battery terminals as well as between the bulb terminals.
4.
Choice (b). The slope of the line tangent to the curve at a point is the reciprocal of the resistance at that
point. Note that as ΔV increases, the slope (and hence 1 R) increases. Thus, the resistance decreases.
5.
Choice (b). From Ohm’s Law, R = ΔV I = 120 V 6.00 A = 20.0 Ω.
6.
L
L
= r 2 . Doubling all linear
A
pr
dimensions increases the numerator of this expression by a factor of 2 but increases the
denominator by a factor of 4. Thus, the net result is that the resistance will be reduced to
one-half of its original value.
Choice (b). Consider the expression for resistance: R = r
7.
Choice (a). The resistance of the shorter wire is half that of the longer wire. The power dissipated,
2
P = ( ΔV ) R, (and hence the rate of heating) will be greater for the shorter wire. Consideration of
the expression P = I 2 R might initially lead one to think that the reverse would be true. However,
one must realize that the currents will not be the same in the two wires.
8.
Choice (b). I a = I b > I c = I d > I e = I f . Charges constituting the current I a leave the positive
terminal of the battery and then split to flow through the two bulbs; thus, I a = I c + I e. Because the
potential difference ΔV is the same across the two bulbs and because the power delivered to a
device is P = I ( ΔV ), the 60-W bulb with the higher power rating must carry the greater current,
meaning that I c > I e . Because charge does not accumulate in the bulbs, all the charge flowing into
a bulb from the left has to flow out on the right; consequently I c = I d and I e = I f . The two currents
leaving the bulbs recombine to form the current back into the battery, I f + I d = I b.
9.
Choice (a). The power dissipated by a resistor may be expressed as P = I 2 R, where I is the current
carried by the resistor of resistance R. Since resistors connected in series carry the same current,
the resistor having the largest resistance will dissipate the most power.
10.
Choice (c). Increasing the diameter of a wire increases the cross-sectional area. Thus, the crosssectional area of A is greater than that of B, and from R = rL A, we see that RA < RB . Since the
power dissipated in a resistance may be expressed as P = (ΔV )2 R, the wire having the smallest
resistance dissipates the most power for a given potential difference.
62
68719_17_ch17_p062-082.indd 62
1/7/11 2:39:04 PM
Current and Resistance
63
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The average current in a conductor is the charge passing a given point per unit time, or
I = ΔQ Δt = e ( Δn ) Δt, so the number of electrons passing this point per second is
Δn I
1.6 C s
= 1.0 × 1019 electron s
= =
Δt e 1.6 × 10 −19 C electron
making (c) the correct choice.
2.
The drift velocity of charge carriers in a conductor is given by vd = I nqA. Wires A and B carry
the same current I. Also, they are made of the same material, so the density and charge of the
charge carriers (n and q) are the same for the two wires. Since the cross-sectional area is A = p r 2,
and rA = 2rB, wire A has a cross-section 4 times that of B. This makes v A = vB 4, and the correct
choice is (e).
3.
By cutting the wire of length L and cross-sectional area A into pieces and placing the pieces side
by side, a new wire with length L ′ = L 3, and cross-sectional area A′ = 3A is created. The new
resistance is
R′ =
r L ′ rL 3 1 ⎛ rL ⎞ 1
=
= ⎜
⎟= R
A′
3A
9⎝ A ⎠ 9
so the correct choice is (a).
4.
The resistance of a conductor having length L ′ and cross-sectional area A is R = r L ′ A.
Thus, the resistance of each wire is: R1 = rL / p r 2, R2 = rL / p (2r)2 = rL / 2p r 2 = 12 R1, and
R3 = r(2L) / p (3r)2 = 4rL / 9p r 2 = 49 R1. We see that wire 3 has the smallest resistance, and
choice (c) is the correct answer.
5.
The ampere is a unit of current (1 amp = 1coulomb second). Hence, the ampere-hour is a unit of
charge
1 ampere-hour = 1 ampere ⋅1 hour = (1 C s ) ( 3 600 s ) = 3 600 C
Thus, the ampere-hour rating is a measure of the charge the battery can supply, and choice (d) is
the correct answer.
6.
When the potential across the device is 2 V, the current is 2 A, so the resistance is
R = ΔV I = 2 V 2 A = 1 Ω, and (a) is the correct choice.
7.
The power consumption of the set is P = ( ΔV ) I = (120 V )( 2.5 A ) = 3.0 × 10 2 W = 0.30 kW.
Thus, the energy used in 8.0 h of operation is E = P ⋅ t = ( 0.30 kW )(8.0 h ) = 2.4 kWh, at a cost of
cost = ( 2.4 kWh )(8.0 cents kWh ) = 19 cents. The correct choice is (c).
8.
The temperature variation of resistance is given by R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦, where R0 is the
resistance of the conductor at the reference temperature T0, usually 20.0°C. Using the given
resistance at T = 90.0°C, the temperature coefficient of resistivity for this conducting material is
found to be
a=
68719_17_ch17_p062-082.indd 63
⎞
1 ⎛ R
1
⎛ 10.55 Ω ⎞
− 1⎟ =
− 1⎟ = 7.86 × 10 −4 °C−1
⎜
⎜
T − T0 ⎝ R0
⎠ 90.0°C − 20.0°C ⎝ 10.00 Ω ⎠
1/7/11 2:39:07 PM
64
Chapter 17
The resistance of this conductor at T = −20.0°C, where ΔT = −20.0°C − ( 20.0°C ) = −40.0°C, will be
R = (10.00 Ω ) ⎡⎣1+ ( 7.86 × 10 −4 °C−1 )( −40.0°C )⎤⎦ = 9.69 Ω
so (b) is the correct choice.
9.
When the potential difference across the device is 3.0 V, the current is 2.5 A, so the resistance is
R = ΔV I = 3.0 V 2.5 A = 1.2 Ω, and (b) is the correct choice.
10.
Resistors in a parallel combination all have the same potential difference across them. Thus, from
Ohm’s law, I = ΔV R, the resistor with the smallest resistance carries the largest current. Choice
(a) is the correct response.
11.
The current through the resistor is I = ΔV R = 1.0 V 10.0 Ω = 0.10 A, and the charge passing
through in a 20 s interval is ΔQ = I ⋅ Δt = ( 0.10 C s )( 20 s ) = 2.0 C. Thus, (c) is the correct choice.
12.
Resistors in a series combination all carry the same current. Thus, from Ohm’s law, ΔV = IR, the
resistor with the highest resistance has the greatest voltage drop across it. Choice (c) is the correct
response.
2
rLB (p d B2 4 ) ⎛ LB ⎞ ⎛ d A ⎞
PA ( ΔV ) RA RB rLB AB
⎛ 1⎞ 2
=
=
=
=
=⎜
= ⎜ ⎟ ( 2) = 2
2
2
⎟
⎜
⎟
⎝ 2⎠
PB ( ΔV ) RB RA rLA AA rLA (p d A 4 ) ⎝ LA ⎠ ⎝ d B ⎠
2
13.
Thus, PA PB = 2, and the correct choice is (c).
14.
The resistance of a conductor may be expressed as R = rL A, so the cross-sectional area is
A = rL R. Thus,
R
1
AA rL RA RB
=
=
= B =
AB rL RB RA 3RB 3
and choice (e) is the correct answer.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The amplitude of atomic vibrations increases with temperature, thereby scattering electrons more
efficiently and increasing the resistivity of the material.
4.
(a)
The number of cars would correspond to charge Q.
(b)
The rate of flow of cars past a point would correspond to current.
(a)
The 25 W bulb has the higher resistance. Because R = ( ΔV ) P, and both operate from
120 V, the bulb dissipating the least power has the higher resistance.
(b)
When the voltage is constant, the current and power are directly proportional to each other,
P = ( ΔV ) I = ( constant ) I . Thus, the higher power bulb (100 W) carries more current.
6.
2
8.
An electrical shock occurs when your body serves as a conductor between two points having a
difference in potential. The concept behind the admonition is to avoid simultaneously touching
points that are at different potentials.
10.
The knob is connected to a variable resistor. As you increase the magnitude of the resistance in
the circuit, the current is reduced, and the bulb dims.
68719_17_ch17_p062-082.indd 64
1/7/11 2:39:10 PM
Current and Resistance
65
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a) 5.57 × 10 −5 m s
4.
3.4 × 10 21 electrons
6.
159 mA
8.
0.130 mm s
10.
8.89 Ω
12.
(a) 1.8 m
14.
3.22 × 10 −8 Ω ⋅ m
16.
1.3
18.
(a)
2.8 × 108 A
20.
(a)
5.0 × 10 5 V
(b)
the drift speed is smaller
(b)
0.28 mm
(b) 1.8 × 10 7 A
(b) Rubber gloves and soles can increase resistance to current passing through the body.
22.
6.3 Ω
24.
(a) 1.6 × 10 2 °C
(b) The expansion of the cross-sectional area contributes slightly more than the expansion of
the length of the wire, so the answer would be slightly reduced.
26.
2.3 × 10 2 °C
28.
1.1× 10 −3 ( °C )
30.
(a) Yes, the design goal can be met with R0 ,C arbon = 4.4 Ω and R0 ,Nichrom e = 5.6 Ω.
−1
(b)
LC arbon = 0.89 m, LNichrom e = 26 m
32.
63.3°C
34.
(a) $0.29
(b)
$2.6
36.
(a) 50 MW
(b)
0.071 or 7.1%
38.
(a) 3.2 × 10 5 J
(b)
18 min
40.
(a)
184 W
(b) 461°C
42.
(a)
0.66 kWh
(b)
$0.079 = 7.9 cents
44.
(a) $1.61
(b)
0.582 cents
68719_17_ch17_p062-082.indd 65
(c)
$0.416 = 41.6 cents
1/7/11 2:39:11 PM
66
46.
48.
Chapter 17
(a) 3.5 × 10 2 W
(b)
41 Ω
0.38 mm
(d)
d = 4r0 L p R0
(e)
(a)
2.7 × 10 −3 Ω
(b) 1.45 × 10 3 K
(d) 1.1 V
(c)
4.0 × 101 Ω
(c) 1.7 × 10 −2 Ω
(e) They radiate only a small portion of the energy
consumed as visible light.
50.
15.0 h
52.
1.5 × 10 2 °C
54.
90 mV
56.
(a) 9.3 m
(b)
0.93 mm
58.
(a) 18 C
(b)
3.6 A
60.
(a)
9.1 Ω
(b) We assume the temperature coefficient of resistivity for Nichrome remains constant over
this temperature range.
62.
No, the fuse should limit the current to 3.87 A or less.
64.
(a)
66.
(a) 470 W
(b) 1.420 Ω versus 1.418 Ω
See Solution.
(b) 1.60 mm or more
(c)
2.93 mm or more
PROBLEM SOLUTIONS
17.1
(a)
The charge that moves past the cross section is ΔQ = I ( Δt ), and the number of electrons is
n=
=
17.2
ΔQ I ( Δt )
=
e
e
(80.0 × 10
−3
C s ) ⎡⎣(10.0 min )( 60.0 s min ) ⎤⎦
= 3.00 × 10 20 electrons
1.60 × 10 −19 C
(b)
The negatively charged electrons move in the direction
opposite to the conventional current flow.
(a)
From Example 17.2 in the textbook, the density of charge carriers (electrons) in a copper
wire is n = 8.46 × 10 28 electrons m 3. With A = p r 2 and q = e, the drift speed of electrons in
this wire is
vd =
I
3.70 C s
= 5.57 × 10 −5 m s
=
2
n q A (8.46 × 10 28 m −3 ) (1.60 × 10 −19 C ) p (1.25 × 10 −3 m )
continued on next page
68719_17_ch17_p062-082.indd 66
1/7/11 2:39:13 PM
Current and Resistance
(b)
17.3
The drift speed is smaller because more electrons are being conducted. With a greater
number of charge carriers in motion, they do not have to move as fast to have a specified
number of them passing a given point each second.
The period of the electron in its orbit is T = 2p r v, and the current represented by the orbiting
electron is I = ΔQ Δt = e T = v e 2p r, or
I=
17.4
( 2.19 × 10
m s ) (1.60 × 10 −19 C )
= 1.05 × 10 −3 C s = 1.05 mA
Since I = ΔQ Δt, we have ΔQ = I ( Δt ), and the number of electrons passing through in time Δt is
N = ΔQ e = I ( Δt ) e. Thus,
( 0.15
C s )(1 h )( 3 600 s 1 h )
= 3.4 × 10 21 electrons
1.60 × 10 −19 C electron
The resistance of a wire of length L and diameter d is R = rL A = rL (p d 2 4 ), giving
d 2 = 4rL p R. Using Table 17.1, the needed diameter is found to be
4 ( 2.82 × 10 −8 Ω ⋅ m )( 32.0 m )
4rL
=
pR
d=
17.6
6
2p ( 5.29 × 10 −11 m )
N=
17.5
67
p ( 2.50 Ω )
= 6.78 × 10 −4 m = 0.678 mm
The mass of a single gold atom is matom = M N A , where M is the molecular weight of gold and
N A is Avogadro’s number. Thus, the number of atoms deposited, and hence the number of ions
moving to the negative electrode, is
n=
−3
23
m
mN A ( 3.25 × 10 kg ) ( 6.02 × 10 atoms mol )
=
=
= 9.93 × 10 21
matom
M
(197 g mol ) (10 −3 kg g)
The current in the cell is then
I=
17.7
21
−19
ΔQ ne ( 9.93 × 10 ) (1.60 × 10 C )
=
=
= 0.159 A = 159 mA
Δt Δt
( 2.78 h )( 3 600 s 1 h )
The drift speed of electrons in the line is vd = I nqA = I n e (p d 2 4 ). The time to travel the
length of the 200-km line is then Δt = L vd = Ln e (p d 2 ) 4I, or
( 200 × 10 m ) ( 8.5 × 10
3
Δt =
17.8
28
m 3 ) (1.6 × 10 −19 C ) p ( 0.02 m )
4 (1 000 A ) ( 3.156 × 10 7 s yr )
2
= 27 yr
The mass of a single aluminum atom is matom = M N A, where the molecular weight (mass per
mole) of aluminum is M = 27.0 g mol, and Avogadro’s number (number of atoms per mole) is
N A = 6.02 × 10 23 mol. If each aluminum atom contributes one conduction electron, the number
of conduction electrons per unit volume is
n=
mass per unit volume density rN A
=
=
mass per atom
matom
M
and the drift speed is vd = I nqA = MI rN A qA. Thus, for the given case, we find
continued on next page
68719_17_ch17_p062-082.indd 67
1/7/11 2:39:15 PM
68
Chapter 17
vd =
17.9
( 27.0
g mol ) ( 5.00 C s )
⎡⎛
g ⎞ ⎛ 10 cm ⎞ ⎤ ⎛ 6.02 × 10 23 ⎞
−19
−6
2
⎢⎜⎝ 2.70
3 ⎟ ⎜
3
⎟⎠ ⎥ ⎜⎝
⎟⎠ (1.60 × 10 C ) ( 4.00 × 10 m )
⎠
1
m
mol
cm
⎝
⎣
⎦
6
3
or
vd = 1.30 × 10 −4 m s = 0.130 mm s .
(a)
Using the periodic table on the inside back cover of the textbook, we find
M Fe = 55.85 g mol = ( 55.85 g mol ) (1 kg 10 3 g ) = 55.85 × 10 −3 kg mol
(b)
From Table 9.1, the density of iron is rFe = 7.86 × 10 3 kg m 3 , so the molar density is
( molar density )
(c)
=
Fe
rFe
7.86 × 10 3 kg m 3
= 1.41× 10 5 mol m 3
=
M Fe 55.85 × 10 −3 kg mol
The density of iron atoms is
density of atoms = N A ( molar density )
atoms ⎞ ⎛
⎛
5 mol ⎞
28 atoms
= ⎜ 6.02 × 10 23
⎟ ⎜ 1.41× 10
⎟ = 8.49 × 10
⎝
mol ⎠ ⎝
m3 ⎠
m3
(d)
With two conduction electrons per iron atom, the density of charge carriers is
n = ( charge carriers atom ) ( density of atoms )
⎛ electrons ⎞ ⎛
28 atoms ⎞
29
3
= ⎜2
⎟ ⎜ 8.49 × 10
⎟ = 1.70 × 10 electrons m
⎝
atom ⎠ ⎝
m3 ⎠
(e)
With a current of I = 30.0 A and cross-sectional area A = 5.00 × 10 −6 m 2 , the drift speed of
the conduction electrons in this wire is
vd =
I
30.0 C s
=
= 2.21× 10 −4 m s
29
−3
nqA (1.70 × 10 m ) (1.60 × 10 −19 C ) ( 5.00 × 10 −6 m 3 )
R=
ΔV 1.20 × 10 2 V
= 8.89 Ω
=
13.5 A
I
17.10
From Ohm’s law,
17.11
( ΔV )m ax = I m ax R = (80 × 10 −6 A ) R
Thus, if R = 4.0 × 10 5 Ω, ( ΔV )m ax = 32 V
and if R = 2 000 Ω, ( ΔV )m ax = 0.16 V
17.12
The volume of the copper is
V=
m
1.00 × 10 −3 kg
=
= 1.12 × 10 −7 m 3
density 8.92 × 10 3 kg m 3
Since V = A ⋅ L, this gives A ⋅ L = 1.12 × 10 −7 m 3.
[1]
continued on next page
68719_17_ch17_p062-082.indd 68
1/7/11 2:39:18 PM
Current and Resistance
(a)
From R =
69
rL
, where r is the resistivity of copper, we find that
A
⎛ 1.7 × 10 −8 Ω ⋅ m ⎞
⎛ r⎞
−8
A=⎜ ⎟L=⎜
⎟⎠ L = ( 3.4 × 10 m ) L
⎝ R⎠
0.500 Ω
⎝
Inserting this expression for A into Equation [1] gives
(3.4 × 10
(b)
−8
m ) L2 = 1.12 × 10 −7 m 3, which yields
From Equation [1], A =
d=
L = 1.8 m
p d 2 1.12 × 10 −7 m 3
=
, or
4
L
4 (1.12 × 10 −7 m 3 )
pL
=
4 (1.12 × 10 −7 m 3 )
p (1.8 m )
= 2.8 × 10 −4 m = 0.28 mm
17.13
R=
ΔV 1.20 × 10 2 V
=
= 13.0 Ω
I
9.25 A
(a)
From Ohm’s law,
(b)
Using R = rL A and data from Table 17.1, the required length is found to be
2
−3
RA R (p r ) (13.0 Ω )p ( 0.791× 10 m )
L=
=
=
= 17.0 m
r
r
150 × 10 −8 Ω ⋅ m
2
17.14
From R = rL A = rL (p d 2 4 ), the resistivity is
−3
p Rd 2 p (1.60 Ω )( 0.800 × 10 m )
r=
=
= 3.22 × 10 −8 Ω ⋅ m
4L
4 ( 25.0 m )
2
17.15
ΔV
12 V
=
= 30 Ω
I
0.40 A
(a)
R=
(b)
From R = rL A,
2
−2
⎡
⎤
R ⋅ A ( 30 Ω ) ⎣p ( 0.40 × 10 m ) ⎦
= 4.7 × 10 − 4 Ω ⋅ m
r=
=
3.2 m
L
17.16
Using R = rL A and data from Table 17.1, we have rC u LC u p rC2u = rAl LAl p rAl2. This reduces to
rAl2 rC2u = rAl rC u, and yields
rAl
=
rC u
17.17
rAl
=
rC u
2.82 × 10 −8 Ω ⋅ m
= 1.3
1.7 × 10 −8 Ω ⋅ m
From Ohm’s law, R = ΔV I , and R = rL A = rL (p d 2 4 ), the resistivity is
−3
p Rd 2 p ( ΔV ) d 2 p ( 9.11 V )( 2.00 × 10 m )
r=
=
=
= 1.59 × 10 −8 Ω ⋅ m
4L
4IL
4 ( 36.0 A )( 50.0 m )
2
Then, from Table 17.1, we see that the wire is made of silver .
68719_17_ch17_p062-082.indd 69
1/7/11 2:39:20 PM
70
17.18
Chapter 17
With different orientations of the block, three different values of the ratio L A are possible.
These are:
10 cm
1
1
⎛ L⎞
=
=
,
⎜⎝ ⎟⎠ =
A 1 ( 20 cm )( 40 cm ) 80 cm 0.80 m
20 cm
1
1
⎛ L⎞
=
=
,
⎜⎝ ⎟⎠ =
A 2 (10 cm )( 40 cm ) 20 cm 0.20 m
17.19
and
40 cm
1
1
⎛ L⎞
=
=
⎜⎝ ⎟⎠ =
A 3 (10 cm )( 20 cm ) 5.0 cm 0.050 m
(a)
I m ax =
ΔV
ΔV
( 6.0 V )( 0.80 m )
=
=
= 2.8 × 108 A
Rm in r ( L A )m in
1.7 × 10 −8 Ω ⋅ m
(b)
I m in =
ΔV
ΔV
( 6.0 V )( 0.050 m )
=
=
= 1.8 × 10 7 A
Rm ax r ( L A )m ax
1.7 × 10 −8 Ω ⋅ m
The volume of material, V = AL0 = (p r02 ) L0 , in the wire is constant. Thus, as the wire is
stretched to decrease its radius, the length increases such that p rf2 L f = (p r02 ) L0 giving
2
2
2
L f = r0 rf L0 = ( r0 0.25r0 ) L0 = ( 4.0 ) L0 = 16L0 . The new resistance is then
(
)
Rf = r
17.20
(a)
(b)
17.21
( )
Lf
Af
=r
Lf
pr
2
f
=r
16L0
p ( 0.25r0 )
⎛ L ⎞
= 256 ⎜ r 02 ⎟ = 256R0 = 256 (1.00 Ω ) = 256 Ω
⎝ p r0 ⎠
From Ohm’s law, ΔV = IR = ( 500 × 10 −3 A ) (1.0 × 10 6 Ω ) = 5 × 10 5 V
Rubber-soled shoes and rubber gloves can increase the resistance to current and help reduce
the likelihood of a serious shock.
If a conductor of length L has a uniform electric field E maintained within it, the potential
difference between the ends of the conductor is ΔV = EL. But, from Ohm’s law, the
relation between the potential difference across a conductor and the current through it is
ΔV = IR, where R = r L A. Combining these relations, we obtain
ΔV = EL = IR = I ( r L A )
17.22
2
E = r ( I A ) = rJ
or
−1
Using R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦, with R0 = 6.00 Ω at T0 = 20.0°C and a silver = 3.8 × 10 −3 (°C ) (from
Table 17.1 in the textbook), the resistance at T = 34.0°C is
−1
R = ( 6.00 Ω ) ⎡⎣1+ 3.8 × 10 −3 (°C ) ( 34.0°C − 20.0°C )⎤⎦ = 6.3 Ω
17.23
From Ohm’s law, ΔV = I i Ri = I f R f , so the current in Antarctica is
⎛ R ⎡1+ a ( T − T ) ⎤ ⎞
⎛ 1+ ⎡3.9 × 10 −3 ( °C )−1 ⎤ ( 58.0°C − 20.0°C ) ⎞
⎛R ⎞
0 ⎣
i
0 ⎦
⎣
⎦
⎟ = (1.00 A ) ⎜
I f = Ii ⎜ i ⎟ = Ii ⎜
⎟
⎜⎝ 1+ ⎡3.9 × 10 −3 ( °C )−1 ⎤ ( −88.0°C − 20.0°C ) ⎟⎠
⎜ R0 ⎡1+ a T f − T0 ⎤ ⎟
⎝ Rf ⎠
⎣
⎦
⎝ ⎣
⎠
⎦
(
)
or I f = 2.0 A
68719_17_ch17_p062-082.indd 70
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Current and Resistance
17.24
(a)
Given: Aluminum wire with a = 3.9 × 10 −3 (°C ) (see Table 17.1 in textbook), and
R0 = 30.0 Ω at T0 = 20.0°C. If R = 46.2 Ω at temperature T, solving R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦
gives the final temperature as
−1
T = T0 +
(b)
17.25
( R R ) − 1 = 20.0°C + ( 46.2 Ω 30.0 Ω) − 1 = 1.6 × 10
0
3.9 × 10 −3 (°C )
a
−1
2
°C
The expansion of the cross-sectional area contributes slightly more than the expansion of
the length of the wire, so the answer would be slightly reduced.
The volume of the gold wire may be written as V = A ⋅ L = m rd , where rd is the density of gold.
Thus, the cross-sectional area is A = m rd L. The resistance of the wire is R = re L A, where re is
the electrical resistivity. Therefore,
−8
3
3
3
rL
r r L2 ( 2.44 × 10 Ω ⋅ m ) (19.3 × 10 kg m ) ( 2.40 × 10 m )
R= e
= e d =
m rd L
m
1.00 × 10 −3 kg
2
R = 2.71× 10 6 Ω = 2.71 MΩ
giving
17.26
71
For aluminum, the resistivity at room temperature is r0 ,Al = 2.82 × 10 −8 Ω ⋅ m, and the
temperature coefficient of resistivity is a Al = 3.9 × 10 −3 (°C)−1. Thus, if at some temperature,
the aluminum has a resistivity of rAl, solving rAl = r0 ,Al ⎡⎣1+ a Al ( T − T0 ) ⎤⎦ for that temperature
gives T = T0 + ⎡⎣ rAl r0 ,Al − 1⎤⎦ a Al , where T0 = 20°C.
{(
)
}
When rAl = 3r0 ,C u = 3 (1.7 × 10 −8 Ω ⋅ m ) = 5.1× 10 −8 Ω ⋅ m, the temperature must be
⎛ 5.1× 10 −8 Ω ⋅ m ⎞
⎜⎝ 2.82 × 10 −8 Ω ⋅ m ⎟⎠ − 1
T = 20°C +
= 2.3 × 10 2 °C
−1
3.9 × 10 −3 (°C )
17.27
At 80°C,
I=
or
17.28
I = 2.6 × 10 −2 A = 26 mA
If R = 41.0 Ω at T = 20.0°C and R = 41.4 Ω at T = 29.0°C, then R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ gives
the temperature coefficient of resistivity of the material making up this wire as
a=
17.29
ΔV
ΔV
5.0 V
=
=
2
R
R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ ( 2.0 × 10 Ω ) ⎡⎣1+ ( − 0.5 × 10 −3 °C −1 )(80°C − 20°C )⎤⎦
(a)
R − R0
41.4 Ω − 41.0 Ω
−1
=
= 1.1× 10 −3 (°C )
R0 ( T − T0 ) ( 41.0 Ω )( 29.0°C − 20.0°C )
The resistance at 20.0°C is
R0 = r
−8
L (1.7 × 10 Ω ⋅ m )( 34.5 m )
=
= 3.0 Ω
2
A
p ( 0.25 × 10 −3 m )
and the current will be I =
ΔV 9.0 V
=
= 3.0 A
R0
3.0 Ω
continued on next page
68719_17_ch17_p062-082.indd 71
1/7/11 2:39:26 PM
72
Chapter 17
(b)
At 30.0°C,
R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦
(
= ( 3.0 Ω ) ⎡1+ 3.9 × 10 −3 ( °C )
⎣
Thus, the current is I =
17.30
−1
)(30.0°C − 20.0°C)⎤⎦ = 3.1 Ω
ΔV 9.0 V
=
= 2.9 A .
R
3.1 Ω
We call the carbon resistor 1 and the Nichrome resistor 2. Then, connecting the resistors end to
end, the total resistance is
R = R1 + R2 = R01 [1+ a 1 ⋅ ΔT ] + R02 [1+ a 2 ⋅ ΔT ] = R01 + R02 + ( R01a 1 + R02a 2 ) ΔT
If this is not to vary with temperature, it is necessary that R01a 1 + R02a 2 = 0, or
⎛ 0.4 × 10 −3 ( °C )−1 ⎞
⎛a ⎞
R01 = − ⎜ 2 ⎟ R02 = − ⎜
R02 = 0.8R02
−1
−3
⎝ a1 ⎠
⎝ −0.5 × 10 ( °C ) ⎟⎠
If the constant total resistance is to be R = 10.0 Ω, it is necessary that
R01 + R02 = 0.8R02 + R02 = 10.0 Ω
or
(a)
(b)
R02 =
10.0 Ω
= 5.6 Ω
1.8
Yes it is possible for her to meet the design goal with this method.
From R = rL A = rL p r 2, we have L = (p r 2 ) ⋅ R r. This yields
L1 = (p r 2 )
2
R01
4.4 Ω
= p (1.50 × 10 −3 m )
= 0.89 m (carbon)
r01
3.5 × 10 −5 Ω ⋅ m
2
R02
5.6 Ω
= p (1.50 × 10 −3 m )
= 26 m (Nichrome)
r02
150 × 10 −8 Ω ⋅ m
and L2 = (p r 2 )
17.31
(a)
From R = rL A, the initial resistance of the mercury is
Ri =
(b)
R01 = 10.0 Ω − R02 = 4.4 Ω
and
−7
rLi ( 9.4 × 10 Ω ⋅ m ) (1.000 0 m )
= 1.2 Ω
=
2
Ai
p (1.00 × 10 −3 m ) 4
Since the volume of mercury is constant, V = A f ⋅ L f = Ai ⋅ Li gives the final cross-sectional
area as A f = Ai ⋅ Li L f . Thus, the final resistance is given by R f = r L f A f = r L2f Ai ⋅ Li.
The fractional change in the resistance is then
(
Δ=
R f − Ri
Ri
)
Rf
=
Ri
−1 =
r L2f
(A ⋅ L ) −1 = ⎛ L
i
r Li Ai
i
2
⎞
⎜⎝ L ⎟⎠ − 1
i
f
2
⎛ 100.04 ⎞
Δ=⎜
− 1 = 8.0 × 10 −4 or a 0.080% increase
⎝ 100.00 ⎟⎠
68719_17_ch17_p062-082.indd 72
1/7/11 2:39:29 PM
Current and Resistance
17.32
73
The resistance at the reference temperature of 20.0°C is
R0 =
R
200.0 Ω
=
= 217 Ω
1+ a ( T − T0 ) 1+ ⎡3.92 × 10 −3 ( °C )−1 ⎤ ( 0°C − 20.0°C )
⎣
⎦
Solving R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ for T gives the temperature of the melting potassium as
T = T0 +
17.33
(a)
R − R0
253.8 Ω − 217 Ω
= 20.0°C +
= 63.3°C
a R0
⎡⎣3.92 × 10 −3 ( °C )−1 ⎤⎦ ( 217 Ω )
The power consumed by the device is P = I ( ΔV ), so the current must be
I=
17.34
P
1.00 × 10 3 W
= 8.33 A
=
ΔV 1.20 × 10 2 V
R=
ΔV 1.20 × 10 2 V
=
= 14.4 Ω
I
8.33 A
(b)
From Ohm’s law, the resistance is
(a)
The energy used by a 100-W bulb in 24 h is
E = P ⋅ Δt = (100 W )( 24 h ) = ( 0.100 kW )( 24 h ) = 2.4 kWh
and the cost of this energy, at a rate of $0.12 per kilowatt-hour is
cost = E ⋅rate = ( 2.4 kWh )( $0.12 kWh ) = $0.29
(b)
The energy used by the oven in 5.0 h is
⎡
⎛ 1 kW ⎞ ⎤
E = P ⋅ Δt = [ I ( ΔV )] ⋅ Δt = ⎢( 20.0 C s ) ( 220 J C ) ⎜
⎥ ( 5.0 h ) = 22 kWh
⎝ 10 3 J s ⎟⎠ ⎦
⎣
and the cost of this energy, at a rate of $0.12 per kilowatt-hour is
cost = E ⋅rate = ( 22 kWh )( $0.12 kWh ) = $2.6
17.35
(a)
RC u =
and
(b)
RAl =
and
(c)
17.36
−8
rC u L 4rC u L 4 (1.7 × 10 Ω ⋅ m )(1.00 m )
=
=
= 5.2 × 10 −3 Ω
2
A
pd2
p ( 0.205 × 10 −2 m )
PC u = I 2 RC u = ( 20.0 A ) ( 5.2 × 10 −3 Ω ) = 2.1 W
2
−8
rAl L 4rAl L 4 ( 2.82 × 10 Ω ⋅ m )(1.00 m )
=
=
= 8.54 × 10 −3 Ω
2
−2
A
pd2
p ( 0.205 × 10 m )
PAl = I 2 RAl = ( 20.0 A ) (8.54 × 10 −3 Ω ) = 3.42 W
2
No , the aluminum wire would not be as safe. If surrounded by thermal insulation, it would
get much hotter than the copper wire.
(a)
The power loss in the line is Ploss = I 2 R = (1 000 A ) ⎡⎣( 0.31 Ω km )(160 km )⎤⎦, or
Ploss = 5.0 × 10 7 W = 50 MW .
(b)
The total power transmitted is Pinput = ( ΔV ) I = ( 700 × 10 3 V ) (1 000 A ) , or
Pinput = 7.0 × 108 W = 700 MW.
2
continued on next page
68719_17_ch17_p062-082.indd 73
1/7/11 2:39:32 PM
74
Chapter 17
Thus, the fraction of the total transmitted power represented by the line losses is
fraction loss =
17.37
Ploss
50 MW
=
= 0.071 or 7.1%
Pinput 700 MW
The energy required to bring the water to the boiling point is
E = m c ( ΔT ) = ( 0.500 kg ) ( 4 186 J kg ⋅°C )(100°C − 23.0°C ) = 1.61× 10 5 J
The power input by the heating element is
Pinput = ( ΔV ) I = (120 V )( 2.00 A ) = 240 W = 240 J s
Therefore, the time required is
t=
17.38
E
1.61× 10 5 J
⎛ 1 min ⎞
= 671 s ⎜
= 11.2 min
=
⎝ 60 s ⎟⎠
240 J s
Pinput
(a)
E = P ⋅ t = ( 90 W )(1 h ) = ( 90 J s ) ( 3 600 s ) = 3.2 × 10 5 J
(b)
The power consumption of the color set is
P = ( ΔV ) I = (120 V )( 2.50 A ) = 300 W
Therefore, the time required to consume the energy found in (a) is
t=
17.39
(a)
RA =
( ΔV )2
PA
=
(120 V )2
25.0 W
( ΔV )2
= 576 Ω
(120 V )2
RB =
(b)
Δt A =
Q
⎛R ⎞
⎛ 576 Ω ⎞
= Q ⎜ A ⎟ = (1.00 C ) ⎜
= 4.80 s
⎝ ΔV ⎠
⎝ 120 V ⎟⎠
IA
(c)
The charge is the same. However, as it leaves the bulb, it is at a lower potential than when it
entered the bulb.
(e)
(f )
PB
=
⎛ 1 min ⎞
s⎜
= 18 min
⎝ 60 s ⎟⎠
and
(d)
100 W
= 144 Ω
W
1.00 J
W,
so
Δt A =
=
= 0.040 0 s
P
25.0
Js
A
Δt
Energy enters the bulb by electrical transmission and leaves by heat and radiation.
P=
⎡
⎛ 24.0 h ⎞ ⎤
⎛ 1 kW ⎞ ⎤ ⎡
E = PA ( Δt ) = ⎢( 25.0 W ) ⎜ 3
= 18.0 kWh
30.0 d ) ⎜
(
⎟
⎢
⎥
⎝ 1 d ⎟⎠ ⎥⎦
⎝ 10 W ⎠ ⎦ ⎣
⎣
and
68719_17_ch17_p062-082.indd 74
E 3.2 × 10 5 J
=
= 1.1× 10 3
P
300 J s
cost = E × rate = (18.0 kWh )( $0.110 kWh ) = $1.98
1/7/11 2:39:34 PM
Current and Resistance
17.40
(a)
75
At the operating temperature,
P = ( ΔV ) I = (120 V )(1.53 A ) = 184 W
(b)
From R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ , the temperature T is given by T = T0 + (R − R0 ) a R0. The
resistances are given by Ohm’s law as
R=
( ΔV ) 120 V
I
=
1.53 A
and
R0 =
( ΔV )0
I0
=
120 V
1.80 A
Therefore, the operating temperature is
T = 20.0°C +
17.41
(120 1.53) − (120 1.80 )
=
( 0.400 × 10 (°C) )(120 1.80 )
−3
−1
461°C
The power loss per unit length of the cable is P L = (I 2 R) L = I 2 (R L). Thus, the resistance per
unit length of the cable is
R P L 2.00 W m
= 2 =
= 2.22 × 10 −5 Ω m
L
I
( 300 A )2
From R = rL A, the resistance per unit length is also given by R L = r A. Hence, the
cross-sectional area is p r 2 = A = r ( R L ), and the required radius is
r=
17.42
(a)
r
1.7 × 10 −8 Ω ⋅ m
=
= 0.016 m = 1.6 cm
p (R L)
p ( 2.22 × 10 −5 Ω m )
The rating of the 12-V battery is I ⋅ Δt = 55 A ⋅ h. Thus, the stored energy is
Energy stored = P ⋅ Δt = ( ΔV ) I ⋅ Δt = (12 V )( 55 A ⋅ h ) = 660 W ⋅ h = 0.66 kWh
(b)
value = ( 0.66 kWh )( $0.12 kWh ) = $0.079 = 7.9 cents
17.43
P = ( ΔV ) I = ( 75.0 × 10 −3 V ) ( 0.200 × 10 −3 A ) = 1.50 × 10 −5 W = 15.0 × 10 −6 W = 15.0 mW
17.44
(a)
E = P ⋅ t = ( 40.0 W )(14.0 d )( 24.0 h d ) = 1.34 × 10 4 Wh = 13.4 kWh
cost = E ⋅ ( rate ) = (13.4 kWh )( $0.120 kWh ) = $1.61
(b)
E = P ⋅ t = ( 0.970 kW )( 3.00 min )(1 h 60 min ) = 4.85 × 10 −2 kWh
cost = ( 4.85 × 10 −2 kWh ) ( $0.120 kWh ) = $0.005 82 = 0.582 cents
(c)
E = P ⋅ t = ( 5.20 kW )( 40.0 min )(1 h 60 min ) = 3.47 kWh
cost = E ⋅ ( rate ) = ( 3.47 kWh )( $0.120 kWh ) = $0.416 = 41.6 cents
17.45
Total length of transmission lines: L = 2 ( 50.0 m ) = 100 m. Thus, the resistance of these lines is
R = ( 0.108 Ω 300 m )(100 m ) = 3.60 × 10 −2 Ω.
(a)
The total potential drop along the transmission lines is ( ΔV )lines = IR, giving
( ΔV )house = ( ΔV )source − ( ΔV )lines = 120 V − (110 A )( 3.60 × 10 −2 Ω ) = 116 V
(b)
68719_17_ch17_p062-082.indd 75
Pdelivered = I ( ΔV )house = (110 A )(116 V ) = 1.28 × 10 4 W = 12.8 kW
1/7/11 2:39:36 PM
76
17.46
Chapter 17
(a)
The thermal energy needed to raise the water temperature by ΔT is E = mc(ΔT ). If
ΔT = 100°C − 20°C = 80°C and this energy is to be delivered in 4.00 min, the average
power required is
P=
(b)
The required resistance (at 100°C) of the heating element is then
R=
(c)
E mc ( ΔT ) ( 0.250 kg ) ( 4 186 J kg ⋅°C )(100°C − 20°C )
=
=
= 3.5 × 10 2 W
Δt
4.00
min
60
s
1
min
Δt
(
)(
)
( ΔV )2
P
=
(120 V )2
3.5 × 10 2 W
= 41 Ω
The resistance at 20.0°C would then be
R0 =
R
41 Ω
=
= 4.0 × 101 Ω
1+ a ( T − T0 ) 1+ ( 0.4 × 10 −3 °C−1 )(100°C − 20°C )
(d)
We find the needed dimensions of a Nichrome wire for this heating element from
R0 = r0 L A = r0 L (p d 2 4 ) = 4r0 L p d 2, where L is the length of the wire and d is its
diameter. This gives the diameter as d = 4r0 L p R0 .
(e)
If L = 3.00 m, the required diameter of the wire is
⎡ 4 (150 × 10 −8 Ω ⋅ m )( 3.00 m ) ⎤
⎡ 4r L ⎤
d=⎢ 0 ⎥ =⎢
⎥ = 3.8 × 10 −4 m = 0.38 mm
1
p
R
p
4.0
×
10
Ω
(
)
⎥⎦
0 ⎦
⎣
⎣⎢
1
2
1
2
17.47
The power dissipated in a conductor is P = ( ΔV ) R, so the resistance may be written as
2
R = ( ΔV ) P. Hence,
2
( ΔV )
RB
PA
P
=
⋅
= A =3
2
RA
PB
( ΔV ) PB
2
or
RB = 3RA
Since R = rL A = rL (p d 2 4 ), this result becomes
⎛ 4rL ⎞
= 3⎜
2
2 ⎟
p dB
⎝ p dA ⎠
4rL
or
d A2
=3
d B2
and yields d A d B = 3 .
17.48
(a)
For tungsten, Table 17.1 from the textbook gives the resistivity at T0 = 20.0°C = 293 K
as r0 = 5.6 × 10 −8 Ω ⋅ m and the temperature coefficient of resistivity as
−1
a = 4.5 × 10 −3 (°C ) = 4.5 × 10 −3 K −1. Thus, for a tungsten wire having a radius of 1.00 mm
and a length of 15.0 cm, the resistance at T0 = 293 K is
R0 = r0
(b)
(15.0 × 10−2 m ) = 2.7 × 10−3 Ω
L
L
−8
= r0
=
5.6
×
10
Ω
⋅
m
(
)
2
A
(p r 2 )
p (1.00 × 10 −3 m )
From Stefan’s law (see Section 11.5 of the textbook), the radiated power is P = s AeT 4 , where
A is the area of the radiating surface. Note that since we are computing the radiated power,
not the net energy gained or lost as a result of radiation, the ambient temperature is not needed
here. In the case of a wire, this is the cylindrical surface area A = 2p rL. The temperature of
1 4
the wire when it is radiating a power of P = 75.0 W is given by T = [ P s Ae ] as
continued on next page
68719_17_ch17_p062-082.indd 76
1/7/11 2:39:39 PM
Current and Resistance
77
1 4
⎡
⎤
75.0 W
T=⎢
⎥
−8
2
4
−3
⎢⎣ ( 5.669 × 10 W m ⋅K ) 2p (1.00 × 10 m )( 0.150 m )( 0.320 ) ⎥⎦
(c)
= 1.45 × 10 3 K
Assuming a linear temperature variation of resistance, the resistance of the wire at this
temperature is
R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ = ( 2.7 × 10 −3 Ω ) ⎡⎣1+ ( 4.5 × 10 −3 K −1 ) (1.45 × 10 3 K − 293 K ) ⎤⎦
giving
(d)
R = 1.7 × 10 −2 Ω
The voltage drop across the wire when it is radiating 75.0 W and has the resistance found in
2
part (c) above is given by P = ( ΔV ) R as
(1.7 × 10
ΔV = R ⋅ P =
(e)
17.49
−2
Ω )( 75.0 W ) = 1.1 V
Tungsten bulbs radiate little of the energy they consume in the form of visible light ,
making them inefficient sources of light.
The battery is rated to deliver the equivalent of 60.0 amperes of current (i.e., 60.0 C/s) for 1 hour.
This is
Q = I ⋅ Δt = ( 60.0 A )(1 h ) = ( 60.0 C s ) ( 3 600 s ) = 2.16 × 10 5 C
17.50
The energy available in the battery is
Energy stored = P ⋅ t = ( ΔV ) I ⋅ t = (12.0 V )( 90.0 A ⋅ h ) = 1.08 × 10 3 W ⋅ h
The two head lights together consume a total power of P = 2 ( 36.0 W ) = 72.0 W, so the time
required to completely discharge the battery is
Energy stored 1.08 × 10 3 W ⋅ h
= 15.0 h
=
72.0 W
P
Δt =
17.51
17.52
(a)
R=
(b)
I=
−8
rL
rL
4rL 4 ( 2.82 × 10 Ω ⋅ m )(15.0 m )
=
=
=
= 1.50 Ω
2
A pd2 4 pd2
p ( 0.600 × 10 −3 m )
ΔV 9.00 V
=
= 6.00 A
R
1.50 Ω
Using chemical symbols to denote the two different metals, the resistances are equal when
R0 ⎡⎣1+ a C u ( ΔT )⎤⎦ = R0 ⎣⎡1+ a W ( ΔT )⎤⎦
Cu
or
R0
Cu
R0
W
Thus,
W
⎡
⎤
⎛ R0 ⎞
− 1 = ⎢a W − ⎜
⎟ a C u ⎥ ( ΔT )
⎢⎣
⎥⎦
⎝ R0 ⎠
Cu
W
(R R ) −1
a − ( R R )a
(R R ) −1
− ( R R )a
0 Cu
ΔT = T − 20.0 °C =
W
or
T = 20.0 °C +
0 Cu
aW
0W
0 Cu
0W
Cu
0W
0 Cu
0W
Cu
continued on next page
68719_17_ch17_p062-082.indd 77
1/7/11 2:39:42 PM
78
Chapter 17
T = 20.0 °C +
(5.00 4.75) − 1
(
4.5 × 10 −3 ( °C ) − ( 5.00 4.75) 3.9 × 10 −3 ( °C )
−1
−1
)
= 20.0 °C + 1.3 × 10 2 °C = 1.5 × 10 2 °C
17.53
From P = ( ΔV ) R, the total resistance needed is R = (ΔV )2 P = (20 V)2 48 W = 8.3 Ω.
2
Thus, from R = rL A, the length of wire required is
L=
17.54
−6
2
R ⋅ A (8.3 Ω )( 4.0 × 10 m )
=
= 1.1× 10 3 m = 1.1 km
r
3.0 × 10 −8 Ω ⋅ m
The resistance of the 4.0 cm length of wire between the feet is
R=
−8
rL (1.7 × 10 Ω ⋅ m )( 0.040 m )
= 1.8 × 10 −6 Ω
=
2
A
p ( 0.022 m ) 4
so the potential difference is
ΔV = IR = ( 50 A )(1.8 × 10 −6 Ω ) = 9.0 × 10 −5 V = 90 mV
17.55
Ohm’s law gives the resistance as R = ( ΔV ) I . From R = rL A, the resistivity is given by
r = R ⋅ ( A L ). The results of these calculations for each of the three wires are summarized
in the table below.
L (m)
R (Ω)
r (Ω ⋅ m )
0.540
10.4
1.41× 10 −6
1.028
21.1
1.50 × 10 −6
1.543
31.8
1.50 × 10 −6
The average value found for the resistivity is
rav =
Σri
= 1.47 × 10 −6 Ω ⋅ m
3
which differs from the value of r = 150 × 10 −8 Ω ⋅ m = 1.50 × 10 −6 Ω ⋅ m given in Table 17.1
by 2.0% .
17.56
The volume of the material is
V=
mass
50.0 g ⎛ 1 m 3 ⎞
=
= 6.36 × 10 −6 m 3
density 7.86 g cm 3 ⎜⎝ 10 6 cm 3 ⎟⎠
Since V = A ⋅ L, the cross-sectional area of the wire is A = V L .
(a)
From R = rL A = rL (V L ) = rL2 V , the length of the wire is given by
L=
R ⋅V
=
r
(1.5 Ω )( 6.36 × 10 −6 m 3 )
11× 10 −8 Ω ⋅ m
= 9.3 m
continued on next page
68719_17_ch17_p062-082.indd 78
1/7/11 2:39:45 PM
Current and Resistance
(b)
The cross-sectional area of the wire is A = p d 2 4 = V L. Thus, the diameter is
(a)
4 ( 6.36 × 10 −6 m 3 )
4V
=
pL
d=
17.57
79
p ( 9.3 m )
= 9.3 × 10 −4 m = 0.93 mm
The total power you now use while cooking breakfast is
P = (1 200 + 500 ) W = 1.70 kW
The cost to use this power for 0.500 h each day for 30.0 days is
⎡
⎤
⎛
h ⎞
cost = [ P × ( Δt )] × rate = ⎢(1.70 kW ) ⎜ 0.500
(30.0 days )⎥ ($0.120 kWh ) = $3.06
⎟
day ⎠
⎝
⎣
⎦
(b)
If you upgraded, the new power requirement would be P = ( 2 400 + 500 ) W = 2 900 W, and
the required current would be I = P ΔV = 2 900 W 110 V = 26.4 A > 20 A.
No , your present circuit breaker cannot handle the upgrade.
17.58
(a)
The charge passing through the conductor in the interval
0 ≤ t ≤ 5.0 s is represented by the area under the I vs. t
graph given in Figure P17.58. This area consists of two
rectangles and two triangles. Thus,
ΔQ = Arectangle 1 + Arectangle 2 + Atriangle 1 + Atriangle 2
= ( 5.0 s − 0 )( 2.0 A − 0 ) + ( 4.0 s − 3.0 s )( 6.0 A − 2.0 A )
+
1
1
(3.0 s − 2.0 s ) (6.0 A − 2.0 A ) + (5.0 s − 4.0 s )(6.0 A − 2.0 A )
2
2
ΔQ = 18 C
(b)
The constant current that would pass the same charge in 5.0 s is
I=
17.59
(a)
The power input to the motor is Pinput = ( ΔV ) I = Poutput efficiency, so the required current is
I=
(b)
68719_17_ch17_p062-082.indd 79
Poutput
( ΔV )(efficiency)
E = Pinput ( Δt ) =
yielding
(c)
ΔQ 18 C
=
= 3.6 A
Δt 5.0 s
Poutput
efficiency
=
( 2.50 hp) ( 746 W 1 hp) =
( Δt ) =
(120 V )( 0.900 )
17.3 A
( 2.50 hp) ( 0.746 kW 1 hp) (3.00 h )
0.900
⎛ 3.60 MJ ⎞
E = 6.22 kWh = ( 6.22 kWh ) ⎜
= 22.4 MJ
⎝ 1 kWh ⎟⎠
cost = E × rate = ( 6.22 kWh )( $0.110 kWh ) = $0.684
1/7/11 2:39:47 PM
80
17.60
Chapter 17
R = R0 [1+ a ( ΔT )] = (8.0 Ω ) ⎡⎣1+ ( 0.4 × 10 −3 °C−1 )( 350°C − 20°C )⎤⎦ = 9.1 Ω
(a)
(b)
17.61
We assume that the temperature coefficient of resistivity for Nichrome remains constant
over this temperature range.
The current in the wire is I = ΔV R = 15.0 V 0.100 Ω = 150 A. Then, from the expression for the
drift velocity, vd = I nqA, the density of free electrons is
n=
or
17.62
I
150 A
=
2
vd e (p r 2 ) ( 3.17 × 10 −4 m s ) (1.60 × 10 −19 C ) p ( 5.00 × 10 −3 m )
n = 3.77 × 10 28 m 3
Each speaker has a resistance of R = 4.00 Ω and can handle a maximum power of 60.0 W. From
P = I 2 R, the maximum safe current is
Pm ax
=
R
I m ax =
60.0 W
= 3.87 A
4.00 Ω
Thus, the system is not adequately protected by a 4.00 A fuse.
17.63
2
2
− rinner
The cross-sectional area of the conducting material is A = p ( router
).
Thus,
R=
17.64
(a)
(3.5 × 105 Ω ⋅ m ) ( 4.0 × 10−2 m ) = 3.7 × 107 Ω = 37 MΩ
rL
=
A p ⎡(1.2 × 10 −2 m )2 − ( 0.50 × 10 −2 m )2 ⎤
⎣
⎦
At temperature T, the resistance is R =
rL
, where r = r0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ ,
A
L = L0 ⎡⎣1+ a ′ ( T − T0 ) ⎤⎦, and A = A0 ⎡⎣1+ 2a ′ ( T − T0 ) ⎤⎦
Thus,
R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ ⋅ ⎡⎣1+ a ′ ( T − T0 ) ⎤⎦
⎛ r L ⎞ ⎡1+ a ( T − T0 ) ⎤⎦ ⋅ ⎡⎣1+ a ′ ( T − T0 ) ⎤⎦
R=⎜ 0 0⎟ ⎣
=
⎡⎣1+ 2a ′ ( T − T0 ) ⎤⎦
⎡⎣1+ 2a ′ ( T − T0 ) ⎤⎦
⎝ A0 ⎠
(b)
R0 =
−8
r0 L0 (1.7 × 10 Ω ⋅ m ) ( 2.00 m )
=
= 1.082 Ω
2
A0
p ( 0.100 × 10 −3 )
Then R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ gives
R = (1.082 Ω ) ⎡⎣1+ ( 3.9 × 10 −3 °C−1 )(80.0°C )⎤⎦ = 1.420 Ω
The more complex formula gives
(1.420 Ω ) ⋅ ⎡⎣1+ (17 × 10 −6 °C−1 )(80.0°C )⎤⎦
R=
= 1.418 Ω
⎡1+ 2 (17 × 10 −6 °C−1 )(80.0°C )⎤
⎣
⎦
Note: Some rules for handing significant figures have been deliberately violated in this solution
in order to illustrate the very small difference in the results obtained with these two expressions.
68719_17_ch17_p062-082.indd 80
1/7/11 2:39:49 PM
Current and Resistance
17.65
81
The power the beam delivers to the target is
P = ( ΔV ) I = ( 4.0 × 10 6 V ) ( 25 × 10 −3 A ) = 1.0 × 10 5 W
The mass of cooling water that must flow through the tube each second if the rise in the water
temperature is not to exceed 50°C is found from P = ( Δm Δt ) c ( ΔT ) as
Δm
P
1.0 × 10 5 J s
=
=
= 0.48 kg s
Δt c ( ΔT ) ( 4 186 J kg ⋅°C )( 50°C )
17.66
Note: All potential differences in this solution have a value of ΔV = 120 V. First, we shall do a
symbolic solution for many parts of the problem and then enter the specified numeric values for
the cases of interest.
From the marked specifications on the cleaner, its internal resistance (assumed constant) is
Rvac =
( ΔV )2
P1
where P1 = 535 W
[1]
If each of the two conductors in the extension cord has resistance Rc, the total resistance in the
path of the current (outside of the power source) is
Rt = Rvac + 2Rc
[2]
so the current which will exist is I = ΔV Rt , and the power that is delivered to the cleaner is
2
Pdelivered = I Rvac
2
⎛ ΔV ⎞
⎛ ΔV ⎞
=⎜
Rvac = ⎜
⎟
⎝ Rt ⎠
⎝ Rt ⎟⎠
2
( ΔV )2
P1
=
( ΔV )4
[3]
Rt2 P1
The resistance of a copper conductor of length L and diameter d is
Rc = rC u
4r L
L
L
= rC u
= C u2
2
A
(p d 4 ) p d
Thus, if Rc , m ax is the maximum allowed value of Rc, the minimum acceptable diameter of the
conductor is
d m in =
(a)
4rC u L
p Rc , m ax
[4]
If Rc = 0.900 Ω, then from Equations [2] and [1],
Rt = Rvac + 2 ( 0.900 Ω ) =
( ΔV )2
P1
+ 1.80 Ω =
(120 V )2
535 W
+ 1.80 Ω
and, from Equation [3], the power delivered to the cleaner is
Pdelivered =
(120 V )4
⎡ (120 V )2
⎤
+ 1.80 Ω ⎥ ( 535 W )
⎢
535
W
⎣
⎦
2
= 470 W
continued on next page
68719_17_ch17_p062-082.indd 81
1/7/11 2:39:52 PM
82
Chapter 17
If the minimum acceptable power delivered to the cleaner is Pmin, then the maximum
allowable total resistance is given by Equations [2] and [3] as
Rt , m ax = Rvac + 2Rc , m ax =
( ΔV )4
Pm in P1
=
( ΔV )2
Pm in P1
so
Rc , m ax =
(b)
2
⎤ 1 ⎡ ( ΔV )2
1 ⎡ ( ΔV )
( ΔV )2
− Rvac ⎥ = ⎢
−
⎢
2 ⎢⎣ Pm in P1
P1
⎥⎦ 2 ⎢⎣ Pm in P1
⎤ ( ΔV )2
⎥=
2
⎥⎦
⎡ 1
1⎤
− ⎥
⎢
⎢⎣ Pm in P1 P1 ⎥⎦
When Pm in = 525 W, then
Rc , m ax =
(120 V )2 ⎡
2
⎢
⎢⎣
1 ⎤
1
−
⎥ = 0.128 Ω
( 525 W )( 535 W ) 535 W ⎥⎦
and, from Equation [4],
d m in =
(c)
p ( 0.128 Ω )
= 1.60 mm
When Pm in = 532 W, then
Rc , m ax =
and
68719_17_ch17_p062-082.indd 82
4 (1.7 × 10 −8 Ω ⋅ m )(15.0 m )
d m in =
(120 V )2 ⎡
2
⎢
⎢⎣
1
1 ⎤
−
⎥ = 0.037 9 Ω
( 532 W )( 535 W ) 535 W ⎥⎦
4 (1.7 × 10 −8 Ω ⋅ m )(15.0 m )
p ( 0.037 9 Ω )
= 2.93 mm
1/7/11 2:39:53 PM
18
Direct-Current Circuits
QUICK QUIZZES
1.
True. When a battery is delivering a current, there is a voltage drop within the battery due to
internal resistance, making the terminal voltage less than the emf.
2.
Because of internal resistance, power is dissipated into the battery material, raising its temperature.
3.
Choice (b). When the switch is opened, resistors R1 and R2 are in series, so that the total circuit
resistance is larger than when the switch was closed. As a result, the current decreases.
4.
Choice (a). When the switch is opened, resistors R1 and R2 are in series, so the total circuit resistance
is larger than and the current through R1 is less with the switch open than when it is closed. Since
the power delivered to R1 is P = I 2 R1, Po < Pc.
5.
Choice (a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit
resistance is smaller than when the switch was open. As a result, the current increases.
6.
Choice (b). Observe that the potential difference across R1 equals the terminal potential difference
of the battery. If the battery has negligible internal resistance, the terminal potential difference is
the same with the switch open or closed. Under these conditions, the power delivered to R1, equal
2
to P = ( ΔV ) R1, is unchanged when the switch is closed.
7.
The voltage drop across each bulb connected in parallel with each other and across the battery
equals the terminal potential difference of the battery. As more bulbs are added, the current
supplied by the battery increases. However, if the internal resistance is negligible, the terminal
potential difference is constant and the current through each bulb is the same regardless of the
number of bulbs connected. Under these conditions: (a) The brightness of a bulb, determined by
the current flowing in the bulb, is unchanged as bulbs are added. (b) The individual currents in the
bulbs, I = ΔV R, are constant as bulbs are added since ΔV does not change. (c) The total power
2
delivered by the battery increases by an amount ( ΔV ) R each time a bulb is added. (d) With the
total delivered power increasing, energy is drawn from the battery at an increasing rate and the
battery’s lifetime decreases.
8.
Adding bulbs in series with each other and the battery increases the total load resistance seen by
the battery. This means that the current supplied by the battery decreases with each new bulb that is
added. (a) The brightness of a bulb is determined by the power delivered to that bulb, Pbulb = I 2 R,
which decreases as bulbs are added and the current decreases. (b) For a series connection, the individual currents in the bulbs are the same and equal to the total current supplied by the battery. This
decreases as bulbs are added. (c) The total power delivered by the battery is given by Ptotal = ( ΔV ) I,
where ΔV is the terminal potential difference of the battery and I is the total current supplied by
the battery. With negligible internal resistance, ΔV is constant. Thus, with I decreasing as bulbs
are added, the total delivered power decreases. (d) With the delivered power decreasing, energy is
drawn from the battery at a decreasing rate, which increases the lifetime of the battery.
9.
Choice (c). After the capacitor is fully charged, current flows only around the outer loop of the
circuit. This path has a total resistance of 3 Ω, so the 6-V battery will supply a current of 2 amperes.
83
68719_18_ch18_p083-119.indd 83
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84
Chapter 18
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The same potential difference exists across all elements connected in parallel with each other,
while the current through each element is inversely proportional to the resistance of that element
(I = ΔV R). Thus, both (b) and (c) are true statements while the other choices are false.
2.
In a series connection, the same current exists in each element. The potential difference across a
resistor in this series connection is directly proportional to the resistance of that resistor, ΔV = IR,
and independent of its location within the series connection. The only true statement among the
listed choices is (c).
3.
For these parallel resistors,
1
1
2 +1
2.00 Ω
1
=
+
=
and Req =
= 0.667 Ω
3
Req 1.00 Ω 2.00 Ω 2.00 Ω
Choice (c) is the correct answer.
4.
The total power dissipated is Ptotal = P1 + P2 = 120 W + 60.0 W = 180 W, while the potential
difference across this series combination is ΔV = 120 V. The current drawn through the series
P
180 W
= 1.5 A, and (b) is the correct choice.
combination is then I = total =
120 V
ΔV
5.
The equation of choice (b) is the result of a correct application of Kirchhoff’s junction rule at either
of the two junctions in the circuit. The equation of choice (c) results from a correct application of
Kirchhoff’s loop rule to the lower loop in the circuit, while the equation of choice (d) is obtained by
correctly applying the loop rule to the loop forming the outer perimeter of the circuit. The equation
of choice (a) is the result of an incorrect application (involving 2 sign errors) of the loop rule to the
upper loop in the circuit. The correct answer is choice (a).
6.
The equivalent resistance of the parallel combination consisting of the 4.0-Ω, 6.0-Ω, and
10-Ω resistors is Rp = 1.9 Ω. This resistance is in series with a 2.0-Ω resistor, making the
total resistance of the circuit Rtotal = 3.9 Ω. The total current supplied by the battery is
I total = ΔV Rtotal = 12 V 3.9 Ω = 3.1 A. Thus, the potential difference across each resistor in the
parallel combination is ΔVp = Rp I total = (1.9 Ω )( 3.1 A ) = 5.9 V and the current through the 10 Ω
resistor is I10 = ΔVp 10 Ω = 5.9 V 10 Ω = 0.59 A. Choice (a) is the correct answer.
7.
The potential difference across each element of this parallel combination is ΔV = 120 V, and
the total power dissipated is Ptotal = 1 200 W + 600 W = 1 800 W. The total current through
the parallel combination, and hence, the current drawn from the external source is then
P
1 800 W
I = total =
= 15 A. The correct choice is (e).
120 V
ΔV
8.
When the two identical resistors are in series, the current supplied by the battery is I = ΔV 2R,
2
and the total power delivered is Ps = ( ΔV ) I = ( ΔV ) 2R. With the resistors connected in parallel,
the potential difference across each resistor is ΔV and the power delivered to each resistor is
2
P1 = ( ΔV ) R. Thus, the total power delivered in this case is
Pp = 2P1 = 2
( ΔV )2
R
⎡ ( ΔV )2
= 4⎢
⎣ 2R
⎤
⎥ = 4Ps = 4 (8.0 W ) = 32 W
⎦
and (b) is the correct choice.
68719_18_ch18_p083-119.indd 84
1/7/11 2:41:06 PM
Direct-Current Circuits
9.
85
The equivalent resistance for the series combination of five identical resistors is Req = 5R, and the
equivalent capacitance of five identical capacitors in parallel is Ceq = 5C. The time constant for
the circuit is therefore t = Req Ceq = ( 5R )( 5C ) = 25RC and (d) is the correct choice.
10.
When the switch is closed, the current has a large initial value but decreases exponentially in
time. The bulb will glow brightly at first, but fade rapidly as the capacitor charges. After a time
equal to many time constants of the circuit, the current is essentially zero and the bulb does not
glow. The correct answer is choice (c).
11.
The equivalent resistance of a group of resistors connected in parallel is the reciprocal of the
inverses of the individual resistances and is always less than the smallest resistance in the group.
Therefore, both (b) and (e) are true statements while all other choices are false.
12.
When resistors are connected in series, Kirchhoff’s junction rule requires that the current be the
same in all resistors at each instant in time. Thus, the charge entering each resistor in a given time
interval must be the same for all resistors, and both choices (b) and (d) are correct. The other
choices are false since the potential difference, ΔV = IR, and the power, P = I 2 R, must differ
when the resistances are different.
13.
With the capacitor initially uncharged, the potential difference across the capacitor, ΔV = q C,
starts at zero when the switch is first closed and rises exponentially toward the equilibrium
value of ( ΔV )m ax = ΔVbattery = 6.00 V. The time constant of the circuit is the time required for
the charge (and hence the potential difference) to increase from 0 to 63.2% of the maximum
equilibrium value. Thus, after one time constant, the potential difference across the capacitor
will be ΔV = 0.632 ( 6.00 V ) = 3.79 V. The correct answer is choice (d).
14.
The circuit breaker should be connected in series with the device so that an open circuit results
when the circuit breaker trips, causing the current through the device to cease. Thus, choice (a) is
the correct answer.
15.
The equivalent resistance of a series combination of resistors is the algebraic sum of the individual
resistances and is always greater than any individual resistance. Therefore, choices (a) and (d) are
true statements and all others are false.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
4.
68719_18_ch18_p083-119.indd 85
A short circuit can develop when the last bit of insulation frays away between the two conductors
in a lamp cord. Then the two conductors touch each other, creating a low resistance path in parallel
with the lamp. The lamp will immediately go out, carrying no current and presenting no danger. A
very large current will be produced in the power source, the house wiring, and the wire in the lamp
cord up to and through the short. The circuit breaker will interrupt the circuit quickly but not before
considerable heating and sparking is produced in the short-circuit path.
1/7/11 2:41:09 PM
86
Chapter 18
6.
A wire or cable in a transmission line is thick and made of material with very low resistivity.
Only when its length is very large does its resistance become significant. To transmit power over
a long distance it is most efficient to use low current at high voltage. The power loss per unit
length of the transmission line is Ploss L = I 2 ( R L ), where R L is the resistance per unit length
of the line. Thus, a low current is clearly desirable, but to transmit a significant amount of power
P = ( ΔV ) I with low current, a high voltage must be used.
8.
The bulbs of set A are wired in parallel. The bulbs of set B are wired in series, so removing one
bulb produces an open circuit with infinite resistance and zero current.
10.
(a) ii. The power delivered may be expressed as P = I 2 R, and while resistors connected in series
have the same current in each, they may have different values of resistance.
(b) ii. The power delivered may also be expressed as P = (ΔV )2 / R, and while resistors
connected in parallel have the same potential difference across them, they may have different
values of resistance.
12.
Compare two runs in series to two resistors connected in series. Compare three runs in parallel to
three resistors connected in parallel. Compare one chairlift followed by two runs in parallel to a
battery followed immediately by two resistors in parallel.
The junction rule for ski resorts says that the number of skiers coming into a junction must be
equal to the number of skiers leaving. The loop rule would be stated as the total change in altitude
must be zero for any skier completing a closed path.
14.
Because water is a good conductor, if you should become part of a short circuit when fumbling
with any electrical circuit while in a bathtub, the current would follow a pathway through you,
the water, and to ground. Electrocution would be the obvious result.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a)
27 Ω
(b)
0.44 A
4.
(a)
1.13 A
(b)
9.17 Ω
6.
(a)
35.0 Ω 3 = 11.7 Ω
(b)
1.00 A in the 12.0-Ω and 8.00-Ω resistors, 2.00 A in the 6.00-Ω and 4.00-Ω resistors, and
3.00 A in the 5.00-Ω resistor
(a)
3.33 Ω
(b)
7.33 Ω
(c)
2.13 Ω
(d)
4.13 Ω
(e)
1.94 A
(f )
3.88 V
(g)
4.12 V
(h)
1.37 A
(a)
ΔV1 = e 3, ΔV2 = 2e 9, ΔV3 = 4e 9, ΔV4 = 2e 3
(b)
I1 = I, I 2 = I 3 = I 3, I 4 = 2I 3
(a)
1.00 kΩ
(c)
3.00 kΩ
8.
10.
12.
68719_18_ch18_p083-119.indd 86
(b)
2.00 kΩ
(c)
3.0 Ω, 1.3 A
1/7/11 2:41:10 PM
Direct-Current Circuits
14.
(a) Yes, Req = 8.00 Ω
16.
(a)
(b)
2.25 A
(c)
87
40.5 W
I 6 = 3.0 A, I12 = 2.0 A, I 24 = 1.0 A
(b) Answers are the same as for part (a).
(a)
20.
I 2 = 2.0 A flowing from b toward a, I 3 = 1.0 A in direction shown
22.
(a)
60.0 Ω
(d)
0.50 W
24.
(a)
26.
28.
0.909 A
(b)
ΔVab = 1.82 V, with point b at the lower potential.
18.
(b)
0.20 A
4.59 Ω
(b)
0.081 6 or 8.16%
(a)
1.7 × 10 2 A
(b)
0.20 A
(a)
18.0I12 + 13.0I18 = 30.0
(b)
5.00I 36 − 18.0I12 = 24.0 (c)
(d)
I 36 = I18 − I12
(e)
5.00I18 − 23.0I12 = 24.0
(f )
I12 = −0.416 A, I18 = 2.88 A
(g)
I 36 = 3.30 A
(h)
I12 flows opposite to the assumed direction and has magnitude 0.416 A.
30.
See Solution.
32.
(a)
34.
1.3 × 10 2 mC
36.
587 kΩ
38.
(a)
8.0 A
(d)
5.8 × 10 2 W
(a)
I coffee = 10 A, I toaster = 9.2 A, I waffle = 12 A
40.
2.00 ms
m aker
(c)
2.4 W
I18 = I12 + I 36
(b)
180 mC
(c)
114 mC
(b)
120 V
(c)
0.80 A
m aker
(b)
I total = 31 A
(c)
No, I total > 15 A.
42.
(a)
4.1× 10 −11 J
(b)
0.56 mA
44.
Sixteen distinct resistances are possible – See Solution for how these are produced.
46.
(a)
2.0 kΩ
(b)
15 V
(c)
9.0 V
(d) Assumed negligible current through the voltmeter and negligible resistance in the ammeter.
48.
(a)
40.0 W
(b)
ΔV1 = 80.0 V, ΔV2 = ΔV3 = 40.0 V
50.
(a)
0.708 A
(b)
2.51 W
68719_18_ch18_p083-119.indd 87
1/7/11 2:41:12 PM
88
Chapter 18
(c)
Only the circuit of Figure P18.50c. In the other circuits, the batteries can be combined into
a single effective battery while the 5.00-Ω and 8.00-Ω resistors remain in parallel with each
other.
(d) The power is lowest in Figure P18.50c. The circuits in Figures P18.50b and P18.50d have in
effect 30-V batteries driving the current.
52.
14 Ω
54.
14 s
56.
R = 20 Ω or R = 98 Ω
58.
See Solution.
60.
q1 = ( 240 mC ) (1− e −1 000 t 6.0 s ) , q2 = ( 360 mC ) (1− e −1 000 t 6.0 s )
62.
(a)
Req = 0.099 9 Ω, I R = 50.0 A, I R = I R = I100 = 45.0 mA
(b)
Req = 1.09 Ω, I R = I R = 4.55 A, I R = I100 = 45.5 mA
(c)
Req = 9.99 Ω, I R = I R = I R = 0.450 A, I100 = 50.0 mA
(a)
In case 1, the two bulbs have the same current, power supplied, and brightness.
(b)
In case 2, the two bulbs have the same current, power supplied, and brightness.
64.
1
1
1
2
2
2
3
3
3
(c) The bulbs in case 2 are brighter, each having twice the current of a bulb in case 1.
66.
(d)
In case 1, both go out when one bulb fails. In case 2, the other bulb remains lit with
unchanged brightness when one bulb fails.
(a)
61.6 mA
(b)
0.235 mC
(c)
1.96 A
PROBLEM SOLUTIONS
18.1
From ΔV = I ( R + r ), the internal resistance is
r=
18.2
ΔV
9.00 V
−R=
− 72.0 Ω = 4.92 Ω
I
0.117 A
(a)
When the three resistors are in series, the equivalent resistance of the circuit is
Req = R1 + R2 + R3 = 3 (9.0 Ω) = 27 Ω .
(b)
The terminal potential difference of the battery is applied across the series combination of
the three 9.0-Ω resistors, so the current supplied by the battery and the current through each
resistor in the series combination is
I=
ΔV 12 V
=
= 0.44 A
Req 27 Ω
continued on next page
68719_18_ch18_p083-119.indd 88
1/7/11 2:41:16 PM
Direct-Current Circuits
(c)
89
If the three 9.0-Ω resistors are now connected in parallel with each other, the equivalent
resistance is
1
1
1
3
1
=
+
+
=
Req 9.0 Ω 9.0 Ω 9.0 Ω 9.0 Ω
or
Req =
9.0 Ω
= 3.0 Ω
3
When this parallel combination is connected to the battery, the potential difference across
each resistor in the combination is ΔV = 12 V, so the current through each of the resistors is
I=
18.3
ΔV 12 V
=
= 1.3 A
R
9.0 Ω
(a)
The bulb acts as a 192-Ω resistor (see below), so the circuit diagram is:
(b)
For the bulb in use as intended, Rbulb = ( ΔV ) P = (120 V ) 75.0 W = 192 Ω .
2
2
Now, assuming the bulb resistance is unchanged, the current in the circuit shown is
I=
ΔV
120 V
=
= 0.620 A
Req 0.800 Ω + 192 Ω + 0.800 Ω
and the actual power dissipated in the bulb is
P = I 2 Rbulb = ( 0.620 A ) (192 Ω ) = 73.8 W
2
18.4
(a)
When the 8.00-Ω resistor is connected across the 9.00-V terminal potential difference of the
battery, the current through both the resistor and the battery is
I=
(b)
ΔV 9.00 V
=
= 1.13 A
R
8.00 Ω
The relation between the emf and the terminal potential difference of a battery supplying
current I is ΔV = e − Ir, where r is the internal resistance of the battery. Thus, if the
battery has r = 0.15 Ω and maintains a terminal potential difference of ΔV = 9.00 V while
supplying the current found above, the emf of this battery must be
e = ΔV + Ir = 9.00 V + (1.13 A )( 0.15 Ω ) = ( 9.00 + 0.17 ) Ω = 9.17 Ω
18.5
(a)
The equivalent resistance of the two parallel
resistors is
1 ⎞
⎛ 1
Rp = ⎜
+
⎝ 7.00 Ω 10.0 Ω ⎟⎠
−1
= 4.12 Ω
Thus,
Rab = R4 + Rp + R9 = ( 4.00 + 4.12 + 9.00 ) Ω = 17.1 Ω
continued on next page
68719_18_ch18_p083-119.indd 89
1/7/11 2:41:18 PM
90
Chapter 18
(b)
18.6
(a)
( ΔV )ab
I ab =
=
Rab
34.0 V
= 1.99 A, so I 4 = I 9 = 1.99 A
17.1 Ω
Also,
( ΔV ) p = I ab Rp = (1.99 A )( 4.12 Ω ) = 8.20 V
Then,
I7 =
and
I10 =
( ΔV ) p
R7
( ΔV ) p
R10
=
8.20 V
= 1.17 A
7.00 Ω
=
8.20 V
= 0.820 A
10.0 Ω
The parallel combination of the 6.00-Ω and 12.0-Ω resistors has an equivalent resistance of
1
1
2 +1
1
=
+
=
Rp1 6.00 Ω 12.0 Ω 12.0 Ω
or
Rp1 =
12.0 Ω
= 4.00 Ω
3
Similarly, the equivalent resistance of the 4.00-Ω and 8.00-Ω parallel combination is
1
1
2 +1
1
=
+
=
Rp 2 4.00 Ω 8.00 Ω 8.00 Ω
or
Rp 2 =
8.00 Ω
3
The total resistance of the series combination between points a and b is then
Rab = Rp1 + 5.00 Ω + Rp 2 = 4.00 Ω + 5.00 Ω +
(b)
8.00
35.0
Ω=
Ω
3
3
If ΔVab = 35.0 V, the total current from a to b is
I ab = ΔVab Rab = 35.0 V ( 35.0 Ω 3) = 3.00 A
and the potential differences across the two parallel combinations are
ΔVp1 = I ab Rp1 = ( 3.00 A )( 4.00 Ω ) = 12.0 V, and
⎛ 8.00 ⎞
ΔVp 2 = I ab Rp 2 = ( 3.00 A ) ⎜
Ω⎟ = 8.00 V
⎝ 3
⎠
The individual currents through the various resistors are:
I12 = ΔVp1 12.0 Ω = 1.00 A ;
I 6 = ΔVp1 6.00 Ω = 2.00 A ;
I 5 = I ab = 3.00 A ;
I 8 = ΔVp 2 8.00 Ω = 1.00 A ;
and
18.7
I 4 = ΔVp 2 4.00 Ω = 2.00 A
When connected in series, we have R1 + R2 = 690 Ω
which we may rewrite as R2 = 690 Ω − R1
When in parallel,
1
1
1
+
=
R1 R2 150 Ω
[1]
[1a]
or
R1 R2
= 150 Ω
R1 + R2
[2]
continued on next page
68719_18_ch18_p083-119.indd 90
1/7/11 2:41:20 PM
Direct-Current Circuits
91
Substitute Equations [1] and [1a] into Equation [2] to obtain:
R1 ( 690 Ω − R1 )
= 150 Ω
690 Ω
R12 − ( 690 Ω ) R1 + ( 690 Ω )(150 Ω ) = 0
or
[3]
Using the quadratic formula to solve Equation [3] gives
R1 =
690 Ω ±
( 690 Ω )2 − 4 ( 690 Ω )(150 Ω )
2
with two solutions of
R1 = 470 Ω
and R1 = 220 Ω
Then Equation [1a] yields
R2 = 220 Ω
or
R2 = 470 Ω
Thus, the two resistors have resistances of 220 Ω and 470 Ω .
18.8
(a)
The equivalent resistance of this first parallel
combination is
1
1
1
=
+
Rp1 10.0 Ω 5.00 Ω
(b)
or
Rp1 = 3.33 Ω
For this series combination,
Rupper = Rp1 + 4.00 Ω = 7.33 Ω
(c)
For the second parallel combination,
1
1
1
1
1
=
+
=
+
Rp 2 Rupper 3.00 Ω 7.33 Ω 3.00 Ω
(d)
Rp 2 = 2.13 Ω
or
For the second series combination (and hence the entire resistor network)
Rtotal = 2.00 Ω + Rp 2 = 2.00 Ω + 2.13 Ω = 4.13 Ω
(e)
The total current supplied by the battery is
I total =
(f )
ΔV
8.00 V
=
= 1.94 A
Rtotal 4.13 Ω
The potential drop across the 2.00 Ω resistor is
ΔV2 = R2 I total = ( 2.00 Ω )(1.94 A ) = 3.88 V
(g)
The potential drop across the second parallel combination must be
ΔVp 2 = ΔV − ΔV2 = 8.00 V − 3.88 V = 4.12 V
(h)
68719_18_ch18_p083-119.indd 91
So the current through the 3.00 Ω resistor is
I total =
ΔVp 2
R3
=
4.12 V
= 1.37 A
3.00 Ω
1/7/11 2:41:23 PM
92
18.9
Chapter 18
(a)
Using the rules for combining resistors in series and parallel, the circuit reduces as shown
below:
25.0 V
10.0 Ω
10.0 Ω
− +
I
b
a
25.0 V
− +
10.0 Ω
I
b
a
25.0 V
− +
10.0 Ω
10.0 Ω
I
5.00 Ω
I20
c
20.0 Ω
5.00 Ω
I20
a
b
2.94 Ω
25.0 Ω
Step 2
5.00 Ω
Step 1
Step 3
From the figure of Step 3, observe that
I=
18.10
25.0 V
= 1.93 A and ΔVab = I ( 2.94 Ω ) = (1.93 A )( 2.94 Ω ) = 5.67 V
10.0 Ω + 2.94 Ω
ΔVab
5.67 V
=
= 0.227 A
25.0 Ω 25.0 Ω
(b)
From the figure of Step 1, observe that I 20 =
(a)
The figures below show the simplification of the circuit in stages:
a
I
I4
R1 = R
e
I2 = I3 = I23
R2 = 2R
R3 = 4R
e
I23
I4
R1
R4 = 3R
+
−
a
I
+
−
b
Figure 1
R4
a
I
I
R1
R23
e
b
Figure 2
R234
+
−
b
Figure 3
Note that R2 and R3 are in series with equivalent resistance R23 = R2 + R3 = 6R. Then,
R4 and R23 are in parallel with equivalent resistance
R234 =
R4 R23
( 3R )( 6R )
=
= 2R
R4 + R23
3R + 6R
The total current supplied by the battery is then
I=
e
e
e
=
=
R1 + R234 R + 2R 3R
⎛ e ⎞
The potential difference across R1 is ΔV1 = IR1 = ⎜
R= e 3
⎝ 3R ⎟⎠
continued on next page
68719_18_ch18_p083-119.indd 92
1/7/11 2:41:25 PM
93
Direct-Current Circuits
⎛ e ⎞
and that across R4 is ΔV4 = ΔVab = IR234 = ⎜
( 2R ) = 2e 3
⎝ 3R ⎟⎠
The current through R2 and R3 is I 23 = ΔVab R23 = ( 2e 3R ) 6R = e 9R
e ⎞
so the potential difference across R2 is ΔV2 = I 23 R2 = ⎛⎜
( 2R ) = 2e 9
⎝ 9R ⎟⎠
⎛ e ⎞
and that across R3 is ΔV3 = I 23 R3 = ⎜
( 4R ) = 4e 9
⎝ 9R ⎟⎠
From above, we have I1 = I and I 2 = I 3 = I 23 = e 9R = I 3
(b)
The current through R4 is I 4 = ΔV4 R4 = ( 2e 3) 3R = 2e 9R = 2I 3
18.11
The equivalent resistance is Req = R + Rp, where Rp is the total resistance of the three parallel
branches:
1
1
⎛ 1
⎞
Rp = ⎜
+
+
⎝ 120 Ω 40 Ω R + 5.0 Ω ⎟⎠
Thus,
75 Ω = R +
−1
1
⎛ 1
⎞
=⎜
+
⎝ 30 Ω R + 5.0 Ω ⎟⎠
−1
=
( 30 Ω )( R + 5.0 Ω )
R + 35 Ω
( 30 Ω )( R + 5.0 Ω ) R 2 + ( 65 Ω ) R + 150 Ω2
R + 35 Ω
=
R + 35 Ω
which reduces to R 2 − (10 Ω ) R − 2 475 Ω2 = 0 or ( R − 55 Ω )( R + 45 Ω ) = 0.
Only the positive solution is physically acceptable, so R = 55 Ω .
18.12
The sketch at the right shows the equivalent circuit
when the switch is in the open position. For this simple
series circuit,
R1 + R2 + R3 =
or
e
[1]
R2
e
R2
+ R3 = 5.00 kΩ
2
− R1
+
R2
R3
Ia
R
e
6.00 V
R1 + 2 + R3 = =
I a 1.20 × 10 −3 A
2
R1 +
R3
Figure 1
When the switch is closed in position a, the equivalent
circuit is shown in Figure 2. The equivalent resistance
of the two parallel resistors, R2, is Rp = R2 2 and the
total resistance of the circuit is Ra = R1 + ( R2 2 ) + R3.
Thus,
or
R2
Io
e
6.00 V
=
I o 1.00 × 10 −3 A
R1 + R2 + R3 = 6.00 kΩ
− R1
+
Figure 2
[2]
continued on next page
68719_18_ch18_p083-119.indd 93
1/7/11 2:41:27 PM
94
Chapter 18
When the switch is closed in position b, resistor R3 is
shorted out, leaving R1 and R2 in series with the
battery as shown in Figure 3. This gives
e
− R1
+
Ib
e
6.00 V
R1 + R2 = =
I b 2.00 × 10 −3 A
and
R2
Figure 3
R1 + R2 = 3.00 kΩ
[3]
Substitute Equation [3] into Equation [1] to obtain
3.00 kΩ + R3 = 6.00 kΩ
and
R3 = 3.00 kΩ
Now, Equation [1] minus Equation [2] gives R2 2 = 1.00 kΩ
or
R2 = 2.00 kΩ
Finally, Equation [3] tells that R1 + 2.00 kΩ = 3.00 kΩ,
or
R1 = 1.00 kΩ
In summary, we have
(a) R1 = 1.00 kΩ ,
18.13
(b) R2 = 2.00 kΩ ,
and
(c) R3 = 3.00 kΩ
The resistors in the circuit can be combined in the stages shown below to yield an equivalent
resistance of Rad = ( 63 11) Ω.
Figure 1
Figure 2
Figure 4
Figure 3
From Figure 5,
I=
( ΔV )ad
Rad
=
18 V
(63 11) Ω
Figure 5
= 3.1 A
continued on next page
68719_18_ch18_p083-119.indd 94
1/7/11 2:41:30 PM
Direct-Current Circuits
Then, from Figure 4,
95
( ΔV )bd = I Rbd = ( 3.1 A )( 30 11 Ω ) = 8.5 V
Now, look at Figure 2 and observe that
I2 =
so
( ΔV )bd
8.5 V
=
= 1.7 A
3.0 Ω + 2.0 Ω 5.0 Ω
( ΔV )be = I 2 Rbe = (1.7 A )( 3.0 Ω ) = 5.1 V
Finally, from Figure 1, I12 =
18.14
(a)
( ΔV )be
R12
=
5.1 V
= 0.43 A
12 Ω
The resistor network connected to the battery in
Figure P18.14 can be reduced to a single equivalent
resistance in the following steps. The equivalent
resistance of the parallel combination of the 3.00 Ω
and 6.00 Ω resistors is
1
1
3
1
=
+
=
Rp 3.00 Ω 6.00 Ω 6.00 Ω
or
Rp = 2.00 Ω
2.00 Ω
+
18.0 V
−
4.00 Ω
6.00 Ω
3.00 Ω
Figure P18.14
This resistance is in series with the 4.00 Ω and the other 2.00 Ω resistor, giving a total
equivalent resistance of Req = 2.00 Ω + Rp + 4.00 Ω = 8.00 Ω .
(b)
The current in the 2.00 Ω resistor is the total current supplied by the battery and is equal to
I total =
(c)
ΔV 18.0 V
=
= 2.25 A
Req 8.00 Ω
The power the battery delivers to the circuit is
P = ( ΔV ) I total = (18.0 V )( 2.25 A ) = 40.5 W
18.15
(a)
Connect two 50-Ω resistors in parallel to get 25 Ω. Then connect that parallel
combination in series with a 20-Ω resistor for a total resistance of 45 Ω.
(b)
Connect two 50-Ω resistors in parallel to get 25 Ω.
Also, connect two 20-Ω resistors in parallel to get 10 Ω.
Then, connect these two parallel combinations in series to obtain 35 Ω.
18.16
(a)
The equivalent resistance of the parallel
combination between points b and e is
1
1
1
=
+
Rbe 12 Ω 24 Ω
or
Rbe = 8.0 Ω
The total resistance between points a and e is then
Rae = Rab + Rbe = 6.0 Ω + 8.0 Ω = 14 Ω
continued on next page
68719_18_ch18_p083-119.indd 95
1/7/11 2:41:32 PM
96
Chapter 18
The total current supplied by the battery (and also the current in the 6.0-Ω resistor) is
I total = I 6 =
ΔVae 42 V
=
= 3.0 A
14 Ω
Rae
The potential difference between points b and e is
ΔVbe = Rbe I total = (8.0 Ω )( 3.0 A ) = 24 V
so
(b)
I12 =
ΔVbe 24 V
=
= 2.0 A
12 Ω
Rbce
I 24 =
and
Applying the junction rule at point b yields
ΔVbe 24 V
=
= 1.0 A
24 Ω
Rbde
I 6 − I12 − I 24 = 0
Using the loop rule on loop abdea gives +42 − 6I 6 − 24I 24 = 0
or
I 6 = 7.0 − 4I 24
[2]
and using the loop rule on loop bcedb gives −12I12 + 24I 24 = 0
or
I12 = 2I 24
[3]
Substituting Equations [2] and [3] into [1] yields 7I 24 = 7.0
Then, Equations [2] and [3] yield
18.17
[1]
I 6 = 3.0 A
and
or
I 24 = 1.0 A
I12 = 2.0 A
Going counterclockwise around the upper loop,
applying Kirchhoff’s loop rule, gives
+15.0 V − ( 7.00 ) I1 − ( 5.00 )( 2.00 A ) = 0
or
I1 =
15.0 V − 10.0 V
= 0.714 A
7.00 Ω
e
From Kirchhoff’s junction rule, I1 + I 2 − 2.00 A = 0
so
I 2 = 2.00 A − I1 = 2.00 A − 0.714 A = 1.29 A
Going around the lower loop in a clockwise direction gives
+ e − ( 2.00 ) I 2 − ( 5.00 )( 2.00 A ) = 0
or
18.18
(a)
e = ( 2.00 Ω )(1.29 A ) + ( 5.00 Ω ) ( 2.00 A ) = 12.6 V
Applying Kirchhoff’s junction rule at a gives
I2 = I4 − I6
Going counterclockwise around the lower loop,
and applying Kirchhoff’s loop rule, we obtain
+8.00 V − ( 6.00 Ω ) I 6 + ( 2.00 Ω ) I 2 = 0
or
I6 =
4 1
+ I2
3 3
12.0 V
+ −
[1]
4.00 Ω
I4
I6
b
I2
a
2.00 Ω
− +
8.00 V
6.00 Ω
[2]
continued on next page
68719_18_ch18_p083-119.indd 96
1/7/11 2:41:35 PM
97
Direct-Current Circuits
Applying the loop rule as we go counterclockwise around the upper loop:
− ( 2.00 Ω ) I 2 − ( 4.00 Ω ) I 4 + 12.0 V = 0
I 4 = 3.00 −
or
1
I2
2
[3]
Substituting Equations [2] and [3] into Equation [1] yields
1 1⎞
4
⎛
⎜⎝ 1+ + ⎟⎠ I 2 = 3.00 −
2 3
3
(b)
I 2 = 0.909 A
and
The potential difference between points a and b is
ΔVab = −I 2 ( 2.00 Ω ) = − ( 0.909 A )( 2.00 Ω ) = −1.82 V
or
18.19
ΔVab = 1.82 V with point b at the lower potential.
R
Consider the circuit diagram at the right, in which
Kirchhoff’s junction rule has already been
applied at points a and e.
+
−
e
Applying the loop rule around loop abca gives
I1 − I
e − R ( I1 − I ) − 4RI1 = 0
or
4R
I1
a
2R
c
b
d
+
− 2e
3R
I
I2 + I
I2
e
1⎛ e
⎞
I1 = ⎜ + I ⎟
5⎝ R ⎠
[1]
Next, applying the loop rule around loop cedc gives
−3RI 2 + 2e − 2R ( I 2 + I ) = 0
or
I2 =
2⎛ e
⎞
⎜ − I⎟
5⎝ R ⎠
[2]
Finally, applying the loop rule around loop caec gives
−4RI1 + 3RI 2 = 0
or
4I1 = 3I 2
Substituting Equations [1] and [2] into Equation [3] yields
[3]
I=
e
5R
Thus, if e = 250 V and R = 1.00 kΩ = 1.00 × 10 3 Ω, the current in the wire between a and e is
I=
18.20
250 V
= 50.0 × 10 −3 A = 50.0 mA flowing from a toward e.
5 (1.00 × 10 3 Ω )
Following the path of I1 from a to b and recording
changes in potential gives
Vb − Va = + 24 V − ( 6.0 Ω )( 3.0 A ) = + 6.0 V
Now, following the path of I 2 from a to b and
recording changes in potential gives
Vb − Va = − ( 3.0 Ω ) I 2 = + 6.0 V, or I 2 = − 2.0 A
continued on next page
68719_18_ch18_p083-119.indd 97
1/7/11 2:41:37 PM
98
Chapter 18
Thus, I 2 is directed from b toward a and has magnitude of 2.0 A.
Applying Kirchhoff’s junction rule at point a gives
I 3 = I1 + I 2 = 3.0 A + ( −2.0 A ) = 1.0 A
18.21
(a)
a
The circuit diagram at the
right shows the assumed
directions of the current in
each resistor. Note that the
total current flowing out of
the section of wire connecting
points g and f must equal the
current flowing into that
section. Thus,
b
c
d
40.0 V
+
− 360 V
−
80.0 V
+
200 Ω
80.0 Ω
20.0 Ω
I2
I3
I1
h
g
f
+
−
70.0 Ω
I4
e
I 3 = I1 + I 2 + I 4
[1]
Applying the loop rule around loop abgha gives
−200I1 − 40.0 + 80.0I 2 = 0
or
I2 =
1
(5I1 + 1.00 )
2
[2]
Next, applying the loop rule around loop bcfgb gives
+360 − 20.0I 3 − 80.0I 2 + 40.0 = 0
I 3 = 20.0 − 4I 2
or
[3]
Finally, applying the loop rule around the outer loop abcdefgha yields
−80.0 + 70I 4 − 200I1 = 0
or
I4 =
1
( 20I1 + 8.00 )
7
[4]
To solve this set of equations, we first substitute Equation [2] into Equation [3] to obtain
I 3 = 20.0 − 2 ( 5I1 + 1.00 )
or
I 3 = 18.0 − 10I1
[3ⴕ]
Now, substitute Equations [2], [4], and [3⬘] into Equation [1] to find
5
20 ⎞
1.00 8.00
⎛
−
⎜⎝ 10 + + 1+ ⎟⎠ I1 = 18.0 −
2
7
2
7
and
I1 = 1.00 A
Substituting this result back into Equations [2], [4], and [3⬘] gives
I 2 = 3.00 A
(b)
I 4 = 4.00 A
and
I 3 = 8.00 A
The potential difference across the 200-Ω resistor is
ΔV200 = I1 R200 = (1.00 A )( 200 Ω ) = 200 V
with point a at a lower potential than point h.
68719_18_ch18_p083-119.indd 98
1/7/11 2:41:40 PM
99
Direct-Current Circuits
18.22
(a)
The 30.0-Ω and 50.0-Ω resistors in the upper branch
are in series, and add to give a total
resistance of Rupper = 80.0 Ω for this path.
This 80.0-Ω resistance is in parallel with the 80.0-Ω
resistance of the middle branch, and
the rule for combining resistors in parallel
yields a total resistance of Rab = 40.0 Ω
between points a and b. This resistance is in
series with the 20.0-Ω resistor, so the total
equivalent resistance of the circuit is
Req = 20.0 Ω + Rab = 20.0 Ω + 40.0 Ω = 60.0 Ω
ΔV
12 V
=
= 0.20 A .
Req 60.0 Ω
(b)
The current supplied to this circuit by the battery is I total =
(c)
2
= ( 60.0 Ω )( 0.20 A ) = 2.4 W .
The power delivered by the battery is Ptotal = Req I total
(d)
The potential difference between points a and b is
2
ΔVab = Rab I total = ( 40.0 Ω )( 0.20 A ) = 8.0 V
ΔVab
8.0 V
and the current in the upper branch is I upper =
=
= 0.10 A, so the power
Rupper 80.0 Ω
delivered to the 50.0 Ω resistor is
2
P50 = R50 I upper
= ( 50.0 Ω )( 0.10 A ) = 0.50 W
2
18.23
(a)
We name the currents I1 , I 2 , and I 3 as shown.
e
Applying Kirchhoff’s loop rule to loop abcfa
gives + e 1 − e 2 − R2 I 2 − R1 I1 = 0
or
and
e
e
3I 2 + 2I1 = 10.0 mA
I1 = 5.00 mA − 1.50I 2
[1]
Applying the loop rule to loop edcfe yields
+ e 3 − R3 I 3 − e 2 − R2 I 2 = 0
and
or
3I 2 + 4I 3 = 20.0 mA
I 3 = 5.00 mA − 0.750I 2
[2]
Finally, applying Kirchhoff’s junction rule at junction c gives
I 2 = I1 + I 3
[3]
Substituting Equations [1] and [2] into [3] yields
I 2 = 5.00 mA − 1.50I 2 + 5.00 mA − 0.750I 2
or
3.25I 2 = 10.0 mA
This gives I 2 = (10.0 mA) 3.25 = 3.08 mA . Then, Equation [1] yields
⎛ 10.0 mA ⎞
I1 = 5.00 mA − 1.50 ⎜
= 0.385 mA
⎝ 3.25 ⎟⎠
continued on next page
68719_18_ch18_p083-119.indd 99
1/7/11 2:41:42 PM
100
Chapter 18
10.0 mA ⎞
and, from Equation [2], I 3 = 5.00 mA − 0.750 ⎛⎜
= 2.69 mA
⎝ 3.25 ⎟⎠
(b)
Start at point c and go to point f, recording changes in potential to obtain
V f − Vc = − e 2 − R2 I 2 = −60.0 V − ( 3.00 × 10 3 Ω ) ( 3.08 × 10 −3 A ) = −69.2 V
or
18.24
(a)
ΔV
cf
= 69.2 V and point c is at the higher potential .
Applying Kirchhoff’s loop rule to the circuit gives
+ 3.00 V − ( 0.255 Ω + 0.153 Ω + R )( 0.600 A ) = 0
or
(b)
R=
3.00 V
− ( 0.255 Ω + 0.153 Ω ) = 4.59 Ω
0.600 A
The total power input to the circuit is
Pinput = (e 1 + e 2 ) I = (1.50 V + 1.50 V )( 0.600 A ) = 1.80 W
The power loss by heating within the batteries is
Ploss = I 2 ( r1 + r2 ) = ( 0.600 A ) ( 0.255 Ω + 0.153 Ω ) = 0.147 W
2
Thus, the fraction of the power input that is dissipated internally is
Ploss
0.147 W
=
= 0.081 6 or 8.16%
1.80 W
Pinput
18.25
(a)
No. Some simplification could be made by recognizing that the 2.0-Ω and 4.0-Ω resistors
are in series, adding to give a total of 6.0 Ω; and the 5.0-Ω and 1.0-Ω resistors form a
series combination with a total resistance of 6.0 Ω. The circuit cannot be simplified any
further, and Kirchhoff’s rules must be used to analyze the circuit.
(b)
Applying Kirchhoff’s junction rule at junction a gives
I1 = I 2 + I 3
[1]
Using Kirchhoff’s loop rule on the upper loop yields
+ 24 V − ( 2.0 + 4.0 ) I1 − ( 3.0 ) I 3 = 0
or
I 3 = 8.0 A − 2 I1
For the lower loop:
[2]
+12 V + ( 3.0 ) I 3 − (1.0 + 5.0 ) I 2 = 0
Using Equation [2], this reduces to
I2 =
12 V + 3.0 (8.0 A − 2 I1 )
6.0
or
I 2 = 6.0 A − I1
[3]
Substituting Equations [2] and [3] into [1] gives I1 = 3.5 A .
Then, Equation [3] gives
68719_18_ch18_p083-119.indd 100
I 2 = 2.5 A , and Equation [2] yields I 3 = 1.0 A .
1/7/11 2:41:45 PM
Direct-Current Circuits
18.26
Using Kirchhoff’s loop rule on the outer
perimeter of the circuit gives
I1
0.01 Ω
+12 V − ( 0.01) I1 − ( 0.06 ) I 3 = 0
or
I1 = 1.2 × 10 A − 6.0 I 3
3
+
− 12 V
[1]
For the rightmost loop, the loop rule gives
Live
Battery
I2
101
I3
1.00 Ω
0.06 Ω
Starter
+
− 10 V
Dead
Battery
+10 V + (1.00 ) I 2 − ( 0.06 ) I 3 = 0
or
I 2 = 0.06 I 3 − 10 A
[2]
Applying Kirchhoff’s junction rule at either junction gives
I1 = I 2 + I 3
(a)
[3]
Substituting Equations [1] and [2] into [3] yields
7.1I 3 = 1.2 × 10 3 A and I 3 = 1.7 × 10 2 A ( in starter )
(b)
18.27
Then, Equation [2] gives I 2 = 0.20 A ( in dead battery ) .
(a)
No. This multi-loop circuit does not contain any resistors in series (i.e., connected so all
the current in one must pass through the other) nor in parallel (connected so the voltage
drop across one is always the same as that across the other). Thus, this circuit cannot be
simplified any further, and Kirchhoff’s rules must be used to analyze it.
(b)
Assume currents I1 , I 2 , and I 3 in the directions shown.
Then, using Kirchhoff’s junction rule at junction a gives
I 3 = I1 + I 2
[1]
Applying Kirchhoff’s loop rule on the lower loop,
+10.0 V − ( 5.00 ) I 2 − ( 20.0 ) I 3 = 0
or
I 2 = 2.00 A − 4 I 3
and for the loop around the perimeter of the circuit,
or
[2]
20.0 V − 30.0I1 − 20.0I 3 = 0
I1 = 0.667 A − 0.667I 3
[3]
Substituting Equations [2] and [3] into [1]: I 3 = 0.667 A − 0.667I 3 + 2.00 A − 4 I 3
which reduces to 5.67I 3 = 2.67 A and gives I 3 = 0.471 A .
Then, Equation [2] gives I 2 = 0.116 A , and from Equation [3], I1 = 0.353 A .
All currents are in the directions indicated in the circuit diagram given above.
68719_18_ch18_p083-119.indd 101
1/7/11 2:41:48 PM
102
18.28
Chapter 18
(a)
Going counterclockwise around the upper loop,
Kirchhoff’s loop rule gives
−11.0I12 + 12.0 − 7.00I12 − 5.00I18 + 18.0 − 8.00I18 = 0
or
(b)
18.0I12 + 13.0I18 = 30.0
[1]
Going counterclockwise around the lower loop:
−5.00I 36 + 36.0 + 7.00I12 − 12.0 + 11.0I12 = 0
or
(c)
5.00I 36 − 18.0I12 = 24.0
[2]
Applying the junction rule at the node in the left end of the circuit gives
I18 = I12 + I 36
[3]
(d)
Solving Equation [3] for I 36 yields I 36 = I18 − I12 .
(e)
Substituting Equation [4] into [2] gives 5.00 ( I18 − I12 ) − 18.0I12 = 24.0, or
[4]
5.00I18 − 23.0I12 = 24.0
(f )
18.29
[5]
Solving Equation [5] for I18 yields I18 = ( 24.0 + 23.0I12 ) 5.00. Substituting this into
Equation [1] and simplifying gives 389I12 = −162, and I12 = −0.416 A . Then, from
Equation [1], I18 = ( 30.0 − 18.0I12 ) 13.0 which yields I18 = 2.88 A .
(g)
Equation [4] gives I 36 = 2.88 A − ( −0.416 A ), or
(h)
The negative sign in the answer for I12 means that this current flows in the opposite
direction to that shown in the circuit diagram and assumed during this solution. That is,
the actual current in the middle branch of the circuit flows from right to left and has a
magnitude of 0.416 A.
I 36 = 3.30 A .
Applying Kirchhoff’s junction rule at
junction a gives
I 3 = I1 + I 2
[1]
Using Kirchhoff’s loop rule on the leftmost
loop yields
−3.00 V − ( 4.00 ) I 3 − ( 5.00 ) I1 + 12.0 V = 0
so
I1 = ( 9.00 A − 4.00I 3 ) 5.00
or
I1 = 1.80 A − 0.800 I 3
[2]
For the rightmost loop,
−3.00 V − ( 4.00 ) I 3 − ( 3.00 + 2.00 ) I 2 + 18.0 V = 0
and
I 2 = (15.0 A − 4.00I 3 ) 5.00
or
I 2 = 3.00 A − 0.800 I 3
[3]
continued on next page
68719_18_ch18_p083-119.indd 102
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Direct-Current Circuits
103
Substituting Equations [2] and [3] into [1] and simplifying gives 2.60I 3 = 4.80 and I 3 = 1.846 A.
Then Equations [2] and [3] yield I1 = 0.323 A and I 2 = 1.523 A.
Therefore, the potential differences across the resistors are
ΔV2 = I 2 ( 2.00 Ω ) = 3.05 V , ΔV3 = I 2 ( 3.00 Ω ) = 4.57 V
ΔV4 = I 3 ( 4.00 Ω ) = 7.38 V , and ΔV5 = I1 ( 5.00 Ω ) = 1.62 V
18.30
The time constant is t = RC. Considering units, we find
⎛ Volts ⎞ ⎛ Coulombs ⎞ ⎛ Coulombs ⎞
RC → ( Ohms )( Farads ) = ⎜
⎜
⎟=
⎝ Amperes ⎟⎠ ⎝ Volts ⎠ ⎜⎝ Amperes ⎟⎠
⎞
⎛
Coulombs
= Second
=⎜
⎝ Coulombs Second ⎟⎠
or
18.31
18.32
18.33
t = RC has units of time.
(a)
The time constant is: t = RC = ( 75.0 × 10 3 Ω ) ( 25.0 × 10 −6 F ) = 1.88 s .
(b)
At t = t , q = 0.632Qm ax = 0.632 (Ce ) = 0.632 ( 25.0 × 10 −6 F )(12.0 V ) = 1.90 × 10 −4 C .
(a)
t = RC = (100 Ω )( 20.0 × 10 −6 F ) = 2.00 × 10 −3 s = 2.00 ms
(b)
Qm ax = C e = ( 20.0 × 10 −6 F )( 9.00 V ) = 1.80 × 10 −4 C = 180 mC
(c)
⎛ 1⎞
Q = Qm ax (1− e − t t ) = Qm ax (1− e −t t ) = Qm ax ⎜ 1− ⎟ = 114 mC
⎝ e⎠
(a)
The time constant of an RC circuit is t = RC. Thus,
t = (1.00 × 10 6 Ω ) ( 5.00 × 10 −6 F ) = 5.00 s
(b)
(c)
Qm ax = Ce = ( 5.00 mF )( 30.0 V ) = 150 mC
To obtain the current through the resistor at time t after the switch
is closed, recall that the charge on the capacitor at that time is
q = Ce (1− e − t t ) and the potential difference across a capacitor is
VC = q C. Thus,
VC =
Ce (1− e
−t t
C
) =e
(1− e )
e
R
+
−
C
S
−t t
Then, considering switch S to have been closed at time t = 0, apply Kirchhoff’s loop rule
around the circuit shown above to obtain
+e − iR − VC = 0
or
i=
e − e (1− e − t t
)
R
The current in the circuit at time t after the switch is closed is then i = (e R ) e − t t , so the
current in the resistor at t = 10.0 s is
10.0 s
⎛ 30.0 V ⎞ − 5.00 s
i=⎜
= ( 30.0 mA ) e −2.00 = 4.06 mA
e
⎟
⎝ 1.00 × 10 6 Ω ⎠
68719_18_ch18_p083-119.indd 103
1/7/11 2:41:53 PM
104
18.34
Chapter 18
Assuming the capacitor is initially uncharged and the switch is closed at t = 0, the charge on the
capacitor at time t > 0 is q = Qm ax (1− e − t t ), where Qm ax = Ce and t is the time constant of the
circuit. When e = 30 V, C = 5.0 mF, and R = 1.0 MΩ, the time constant is
t = RC = (1.0 × 10 6 Ω ) ( 5.0 × 10 −6 F ) = 5.0 s
and we find the charge on the capacitor at t = 10 s to be
q = ( 5.0 mF )( 30 V )(1− e −10 s 5.0 s ) = 1.3 × 10 2 mC
18.35
The charge remaining on the capacitor after time t is q = Qe − t t .
(a)
Thus, if q = 0.750Q, then e − t t = 0.750 and −t t = ln ( 0.750 ),
t = −t ln ( 0.750 ) = − (1.50 s ) ln ( 0.750 ) = 0.432 s
or
t = RC , so
(b)
18.36
C=
t
1.50 s
=
= 6.00 × 10 −6 F = 6.00 mF
R 250 × 10 3 Ω
At time t after the switch is closed, the potential difference between the plates of the initially
uncharged capacitor is
ΔV = qC = CQm ax (1− e − t t ) = e (1 − e − t RC )
e − t RC = 1−
Thus,
giving
R=
ΔV
e
and
−
t
ΔV ⎞
⎛
= ln ⎜ 1−
⎟
⎝
RC
e ⎠
− (t C )
ln (1− ΔV e )
If e = 10.0 V, C = 10.0 mF, and ΔV = 4.00 V at t = 3.00 s after closing the switch, the resistance
must be
R=
18.37
ln (1− 4.00 V 10.0 V )
=
−3.00 × 10 5 Ω
= 5.87 × 10 5 Ω = 587 kΩ
ln ( 0.600 )
The current drawn by a single 75-W bulb connected to a 120-V source is I1 = P ΔV = 75 W 120 V.
Thus, the number of such bulbs that can be connected in parallel with this source before the total
current drawn will equal 30.0 A is
n=
18.38
− ( 3.00 s 10.0 × 10 −6 F )
(a)
30.0 A
⎛ 120 V ⎞
= ( 30.0 A ) ⎜
= 48
⎝ 75 W ⎟⎠
I1
The equivalent resistance of the parallel combination is
⎛ 1
1
1⎞
Req = ⎜ +
+ ⎟
R
R
R
⎝ 1
2
3 ⎠
−1
1
1 ⎞
⎛ 1
=⎜
+
+
⎝ 150 Ω 25 Ω 50 Ω ⎟⎠
−1
= 15 Ω
so the total current supplied to the circuit is
I total =
ΔV 120 V
=
= 8.0 A
R
15 Ω
continued on next page
68719_18_ch18_p083-119.indd 104
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Direct-Current Circuits
18.39
105
(b)
Since the appliances are connected in parallel, the voltage across each one is ΔV = 120 V .
(c)
I lam p =
(d)
Pheater =
ΔV
120 V
=
= 0.80 A
Rlam p 150 Ω
( ΔV )2
Rheater
=
(120 V )2
25 Ω
= 5.8 × 10 2 W
From P = ( ΔV ) R, the resistance of the element is
2
R=
( ΔV )2
P
=
( 240 V )2
3 000 W
=19.2 Ω
When the element is connected to a 120-V source, we find that
18.40
18.41
ΔV 120 V
=
= 6.25 A , and
R 19.2 Ω
(a)
I=
(b)
P = ( ΔV ) I = (120 V )( 6.25 A ) = 750 W
(a)
The current drawn by each appliance operating separately is
Coffee Maker:
I=
P
1 200 W
=
= 10 A
ΔV
120 V
Toaster:
I=
P
1100 W
=
= 9.2 A
ΔV
120 V
Waffle Maker:
I=
P
1 400 W
=
= 12 A
ΔV
120 V
(b)
If the three appliances are operated simultaneously, they will draw a total current of
I total = (10 + 9.2 + 12 ) A = 31 A .
(c)
No. The total current required exceeds the limit of the circuit breaker, so they cannot be
operated simultaneously. In fact, with a 15 A limit, no two of these appliances could be
operated at the same time without tripping the breaker.
(a)
The area of each surface of this axon membrane is
A = L ( 2p r ) = ( 0.10 m ) ⎡⎣ 2p (10 × 10 −6 m ) ⎤⎦ = 2p × 10 −6 m 2
and the capacitance is
C = k ∈0
⎛ 2p × 10 −6 m 2 ⎞
A
= 1.67 × 10 −8 F
= 3.0 (8.85 × 10 −12 C2 N ⋅ m 2 ) ⎜
d
⎝ 1.0 × 10 −8 m ⎟⎠
In the resting state, the charge on the outer surface of the membrane is
Qi = C ( ΔV )i = (1.67 × 10 −8 F ) ( 70 × 10 −3 V ) = 1.17 × 10 −9 C → 1.2 × 10 −9 C
The number of potassium ions required to produce this charge is
NK =
+
Qi 1.17 × 10 −9 C
=
= 7.3 × 10 9 K + ions
e
1.6 × 10 −19 C
continued on next page
68719_18_ch18_p083-119.indd 105
1/7/11 2:42:00 PM
106
Chapter 18
and the charge per unit area on this surface is
−20
2
Qi 1.17 × 10 −9 C ⎛
1e
1e
⎞ ⎛ 10 m ⎞
=
=
=
⎜
⎟
−6
2
−19
2
⎟
⎜
⎝
⎠
A 2p × 10 m 1.6 × 10 C ⎝ 1 Å ⎠ 8.6 × 10 4 Å 2
s =
1e
( 290 Å )
2
This corresponds to a low charge density of one electronic charge per square of side 290 Å,
compared to a normal atomic spacing of one atom every few Å.
(b)
In the resting state, the net charge on the inner surface of the membrane is
− Qi = −1.17 × 10 −9 C, and the net positive charge on this surface in the excited state is
Q f = C ( ΔV ) f = (1.67 × 10 −8 F ) ( +30 × 10 −3 V ) = + 5.0 × 10 −10 C
The total positive charge which must pass through the membrane to produce the excited
state is therefore
ΔQ = Q f − Qi
= + 5.0 × 10 −10 C − ( −1.17 × 10 −9 C ) = 1.67 × 10 −9 C → 1.7 × 10 −9 C
corresponding to
N Na =
+
(c)
If the sodium ions enter the axon in a time of Δt = 2.0 ms, the average current is
I=
(d)
ΔQ 1.67 × 10 −9 C
=
= 8.3 × 10 −7 A = 0.83 mA
Δt
2.0 × 10 −3 s
When the membrane becomes permeable to sodium ions, the initial influx of sodium ions
neutralizes the capacitor with no required energy input. The energy input required to charge
the now neutral capacitor to the potential difference of the excited state is
W=
18.42
ΔQ
1.67 × 10 −9 C
=
= 1.0 × 1010 Na + ions
e
1.6 × 10 −19 C Na + ion
2
1
1
2
C ( ΔV ) f = (1.67 × 10 −8 F ) ( 30 × 10 −3 V ) = 7.5 × 10 −12 J
2
2
The capacitance of the 10 cm length of axon was found to be C = 1.67 × 10 −8 F in the solution of
Problem 18.41.
(a)
When the membrane becomes permeable to potassium ions, these ions flow out of the axon
with no energy input required until the capacitor is neutralized. To maintain this outflow of
potassium ions and charge the now neutral capacitor to the resting action potential requires
an energy input of
W=
(b)
2
1
1
2
C ( ΔV ) = (1.67 × 10 −8 F ) ( 70 × 10 −3 V ) = 4.1× 10 −11 J
2
2
As found in the solution of Problem 18.41, the charge on the inner surface of the membrane
in the resting state is −1.17 × 10 −9 C, and the charge on this surface in the excited state is
+ 5.0 × 10 −10 C. Thus, the positive charge which must flow out of the axon as it goes from
the excited state to the resting state is
ΔQ = 5.0 × 10 −10 C + 1.17 × 10 −9 C = 1.67 × 10 −9 C
continued on next page
68719_18_ch18_p083-119.indd 106
1/7/11 2:42:01 PM
Direct-Current Circuits
107
and the average current during the 3.0 ms required to return to the resting state is
I=
18.43
From Figure 18.28, the duration of an action potential pulse is 4.5 ms. From the solution of
Problem 18.41, the energy input required to reach the excited state is W1 = 7.5 × 10 −12 J. The
energy input required during the return to the resting state is found in Problem 18.42 to be
W2 = 4.1× 10 −11 J. Therefore, the average power input required during an action potential pulse is
P=
18.44
ΔQ 1.67 × 10 −9 C
=
= 5.6 × 10 −7 A = 0.56 mA
Δt
3.0 × 10 −3 s
Wtotal W1 + W2 7.5 × 10 −12 J + 4.1× 10 −11 J
=
=
= 1.1× 10 −8 W = 11 nW
Δt
Δt
4.5 × 10 −3 s
Using a single resistor → 3 distinct values: R1 = 2.0 Ω, R2 = 4.0 Ω, R3 = 6.0 Ω
2 resistors in Series → 2 additional distinct values: R4 = 2.0 Ω + 6.0 Ω = 8.0 Ω, and
R5 = 4.0 Ω + 6.0 Ω = 10 Ω. Note: 2.0 Ω and 4.0 Ω in series duplicates R3 above.
2 resistors in Parallel → 3 additional distinct values:
R6 = 2.0 Ω and 4.0 Ω in parallel = 1.3 Ω
R7 = 2.0 Ω and 6.0 Ω in parallel = 1.5 Ω
R8 = 4.0 Ω and 6.0 Ω in parallel = 2.4 Ω
3 resistors in Series → 1 additional distinct value:
R9 = 2.0 Ω + 4.0 Ω + 6.0 Ω = 12 Ω
3 resistors in Parallel → 1 additional distinct value:
R10 = 2.0 Ω, 4.0 Ω, and 6.0 Ω in parallel = 1.1 Ω
1 resistor in Parallel with Series combination of the other 2: → 3 additional values:
(
= (R
= (R
)
= 4.0 Ω; 2.0 Ω and 6.0 Ω in series ) = 2.7 Ω
= 6.0 Ω; 2.0 Ω and 4.0 Ω in series ) = 3.0 Ω
R11 = Rp = 2.0 Ω; 4.0 Ω and 6.0 Ω in series = 1.7 Ω
R12
R12
p
p
1 resistor in Series with Parallel combination of the other 2: → 3 additional values:
R14 = ( Rs = 2.0 Ω; 4.0 Ω and 6.0 Ω in parallel ) = 4.4 Ω
R15 = ( Rs = 4.0 Ω; 2.0 Ω and 6.0 Ω in parallel ) = 5.5 Ω
R16 = ( Rs = 6.0 Ω; 2.0 Ω and 4.0 Ω in Parallel ) = 7.3 Ω
Thus, 16 distinct values of resistance are possible using these three resistors.
18.45
Since the circuit is open at points a and b, no current flows
through the 4.00-V battery or the 10.0-Ω resistor. A current I
will flow around the closed path through the 2.00-Ω resistor,
4.00-Ω resistor, and the 12.0-V battery as shown in the sketch
at the right. This current has magnitude
2.00 Ω
12.0 V
+
−
4.00 V
+ −
a
4.00 Ω
I
10.0 Ω
b
continued on next page
68719_18_ch18_p083-119.indd 107
1/7/11 2:42:03 PM
108
Chapter 18
I=
ΔV
12.0 V
=
= 2.00 A
Rpath 2.00 Ω + 4.00 Ω
Along the path from point a to point b, the change in potential that occurs is given by
ΔVab = +e 4 − IR4 = + 4.00 V − ( 2.00 A )( 4.00 Ω ) = − 4.00 V
(a)
The potential difference between points a and b has magnitude
ΔVab = 4.00 V
18.46
(b)
Since the change in potential in going from a to b was negative, we conclude that
point a is at the higher potential .
(a)
R=
(b)
The resistance in the circuit consists of a series
combination with an equivalent resistance of
Req = 2.0 kΩ + 3.0 kΩ = 5.0 kΩ. The emf of the
battery is then
ΔV
6.0 V
=
= 2.0 × 10 3 Ω = 2.0 kΩ
I
3.0 × 10 −3 A
e = IReq = ( 3.0 × 10 −3 A ) ( 5.0 × 10 3 Ω ) = 15 V
(c)
18.47
ΔV3 = IR3 = ( 3.0 × 10 −3 A ) ( 3.0 × 10 3 Ω ) = 9.0 V
(d)
In this solution, we have assumed that we have ideal devices in the circuit. In particular,
we have assumed that the battery has negligible internal resistance, the voltmeter has an
extremely large resistance and draws negligible current, and the ammeter has an extremely
low resistance and a negligible voltage drop across it.
(a)
The resistors combine to an equivalent resistance of Req = 15 Ω as shown.
Figure 1
Figure 3
Figure 2
Figure 4
Figure 5
continued on next page
68719_18_ch18_p083-119.indd 108
1/7/11 2:42:07 PM
Direct-Current Circuits
(b)
109
ΔVab 15 V
=
= 1.0 A
15 Ω
Req
From Figure 5, I1 =
Then, from Figure 4,
ΔVac = ΔVdb = I1 ( 6.0 Ω ) = 6.0 V and ΔVcd = I1 ( 3.0 Ω ) = 3.0 V
I2 = I3 =
From Figure 2,
ΔVed = I 3 ( 3.6 Ω ) = 1.8 V
Then, from Figure 1,
and
(c)
ΔVcd
3.0 V
=
= 0.50 A
6.0 Ω 6.0 Ω
From Figure 3,
I5 =
ΔV fd
9.0 Ω
=
I4 =
ΔVed
1.8 V
=
= 0.30 A
6.0 Ω 6.0 Ω
ΔVed
1.8 V
= 0.20 A
=
9.0 Ω 9.0 Ω
From Figure 2, ΔVce = I 3 ( 2.4 Ω ) = 1.2 V . All the other needed potential differences were
calculated above in part (b). The results were
ΔVac = ΔVdb = 6.0 V ; ΔVcd = 3.0 V ; and ΔV fd = ΔVed = 1.8 V
(d)
The power dissipated in each resistor is found from P = ( ΔV ) R with the following results:
2
Pac =
Ped =
Pcd =
18.48
(a)
( ΔV )2ac
Rac
( ΔV )ed2
Red
( ΔV )cd2
Rcd
=
=
=
( 6.0 V )2
6.0 Ω
(1.8 V )2
6.0 Ω
( 3.0 V )2
6.0 Ω
= 6.0 W
Pce =
= 0.54 W
Pfd =
= 1.5 W
Pdb =
( ΔV )ce2
Rce
( ΔV )2fd
R fd
( ΔV )2db
Rdb
=
=
=
(1.2 V )2
2.4 Ω
(1.8 V )2
9.0 Ω
( 6.0 V )2
6.0 Ω
= 0.60 W
= 0.36 W
= 6.0 W
From P = ( ΔV ) R, the resistance of each of the
three bulbs is given by
2
R=
( ΔV )2
P
=
(120 V )2
60.0 W
R1
120 V
= 240 Ω
R2
R3
As connected, the parallel combination of
R2 and R3 is in series with R1. Thus, the
equivalent resistance of the circuit is
⎛ 1
1⎞
Req = R1 + ⎜
+ ⎟
R
R
⎝ 2
3 ⎠
−1
1 ⎞
⎛ 1
= 240 Ω + ⎜
+
⎝ 240 Ω 240 Ω ⎟⎠
−1
= 360 Ω
The total power delivered to the circuit is
P=
( ΔV )2
Req
=
(120 V )2
360 Ω
= 40.0 W
continued on next page
68719_18_ch18_p083-119.indd 109
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110
Chapter 18
ΔV 120 V 1
The current supplied by the source is I =
=
= A. Thus, the potential
Req 360 Ω 3
difference across R1 is
(b)
( ΔV )1 = I R1 = ⎛⎜⎝ A ⎞⎟⎠ ( 240 Ω ) = 80.0 V
3
1
The potential difference across the parallel combination of R2 and R3 is then
( ΔV )2 = ( ΔV )3 = ( ΔV )source − ( ΔV ) 1 = 120 V − 80.0 V = 40.0 V
18.49
When the two resistors are connected in series, the equivalent resistance is Rs = R1 + R2 and the
power delivered when a current I = 5.00 A flows through the series combination is
Ps = I 2 Rs = ( 5.00 A ) ( R1 + R2 ) = 225 W
2
R1 + R2 =
Thus,
225 W
25.0 A 2
giving
R1 + R2 = 9.00 Ω
When the resistors are connected in parallel, the equivalent resistance is Rp = R1 R2
the power delivered by the same current ( I = 5.00 A ) is
[1]
(R
1
+ R2 ) and
RR ⎞
2 ⎛
Pp = I 2 Rp = ( 5.00 A ) ⎜ 1 2 ⎟ = 50.0 W
⎝ R1 + R2 ⎠
Rp =
giving
50.0 W
25.0 A 2
R1 R2
= 2.00 Ω
R1 + R2
or
[2]
Substituting Equation [1] into Equation [2] yields
R ( 9.00 Ω − R1 )
R1 R2
= 1
= 2.00 Ω
R1 + R2
9.00 Ω
or
R12 − ( 9.00 Ω ) R1 + 18.0 Ω2 = 0 . This quadratic equation factors as
(R
1
− 3.00 Ω ) ( R1 − 6.00 Ω ) = 0
Thus, either R1 = 3.00 Ω or R1 = 6.00 Ω, and from Equation [1], we find that either R2 = 6.00 Ω or
R2 = 3.00 Ω. Therefore, the pair contains one 3.00 Ω resistor and one 6.00 Ω resistor.
18.50
(a)
Recognize that the 5.00-Ω and the 8.00-Ω resistors are connected in parallel and that the
effective resistance of this parallel combination is
Rp =
( 5.00 Ω )(8.00 Ω )
5.00 Ω + 8.00 Ω
= 3.08 Ω
This resistance is in series with the 10.0-Ω resistor, giving a total resistance for the circuit
of Req = 10.0 Ω + 3.08 = 13.1 Ω. Thus, the current supplied by the battery is
I total =
e
15.0 V
=
= 1.15 A
Req 13.1 Ω
and the potential difference across the parallel combination is
ΔVp = I total Rp = (1.15 A )( 3.08 Ω ) = 3.54 V
continued on next page
68719_18_ch18_p083-119.indd 110
1/7/11 2:42:11 PM
111
Direct-Current Circuits
The current through the 5.00-Ω in this parallel combination is then
I5 =
(b)
ΔVp
R5
=
3.54 V
= 0.708 A
5.00 Ω
The power delivered to the 5.00-Ω resistor is
P5 = I 52 R5 = ( 0.708 A ) ( 5.00 Ω ) = 2.51 W
2
18.51
(c)
Only the circuit in Figure P18.50c requires the use of Kirchhoff’s rules for solution. In
the other circuits, the batteries can be combined into a single effective battery while the
5.00-Ω and 8.00-Ω resistors remain in parallel with each other.
(d)
The power delivered is lowest in Figure 18.50c. The circuits in Figures P18.50b and
P18.50d have in effect 30.0-V batteries driving current through the 10.0-Ω resistor, thus
delivering more power than the circuit in Figure 18.50a. In Figure 18.50c, the two 15.0-V
batteries tend to oppose each other’s efforts to drive current through the 10.0-Ω resistor,
making them less effective than the single 15.0-V battery of Figure 18.50a.
(a)
When switch S is open, all three bulbs are
in series and the equivalent resistance is
Reqopen = R + R + R = 3R .
When the switch is closed, bulb C is shorted
across and no current will flow through that
bulb. This leaves bulbs A and B in series with an
equivalent resistance of Reqclosed = R + R = 2R .
(b)
(c)
18.52
A
e
B
C
S
e2
e2
With the switch open, the power delivered by the battery is Popen = open =
, and with
Req
3R
the switch closed, Pclosed = e 2 Reqclosed = e 2 2R .
When the switch is open, the three bulbs have equal brightness. When S is closed,
bulb C goes out, while A and B remain equal at a greater brightness than they had when
the switch was open.
With the switch open, the circuit may be reduced as follows:
With the switch closed, the circuit reduces as shown below:
Since the equivalent resistance with the switch closed is one-half that when the switch is open,
we have
R + 18 Ω =
68719_18_ch18_p083-119.indd 111
1
( R + 50 Ω ), which yields R = 14 Ω
2
1/7/11 2:42:14 PM
112
18.53
Chapter 18
(a)
Note the assumed directions of the
three distinct currents in the circuit
diagram at the right. Applying the
junction rule at point c gives
I1 = I 3 − I 2
Applying Kirchhoff’s loop rule to
loop gbcfg gives
a
−
3.00 V
+
k
[1]
+
6.00 mF
−
h
+ 8.00 V + 3.00I 2 − 5.00I1 = 0
or
b
c
+ I1
I2
−
8.00 V
I1
d
5.00 Ω
I3
3.00 Ω
−
4.00 V
+
I1
I=0
g
5.00 Ω
I3
f
e
5.00I1 − 3.00I 2 = 8.00 V
[2]
Finally, applying Kirchhoff’s loop rule to loop fcdef yields
−3.00I 2 − 5.00I 3 + 4.00 V = 0
or
5.00I 3 + 3.00I 2 = 4.00 V
[3]
Substituting Equation [1] into Equation [2] gives
5.00I 3 − 8.00I 2 = 8.00 V
and subtracting this result from Equation [3] yields
I 2 = − 4.00 V 11.0
Equation [3] then gives the current in the 4.00-V battery as
I3 =
(b)
4.00 V − 3.00 ( −4.00 V 11.0 )
= +1.02 A
5.00
I 3 = 1.02 A down
From above, the current in the 3.00-Ω resistor is
I2 = −
(c)
or
4.00 V
= − 0.364 A
11.0
or
I 2 = 0.364 A down
Equation [1] now gives the current in the 8.00-V battery as
I1 = 1.02 A − ( −0.364 A ) = +1.38 A
or
I1 = 1.38 A up
(d)
Once the capacitor is charged, the current in the 3.00-V battery is I = 0 because of the
open circuit between the plates of the capacitor.
(e)
To obtain the potential difference between the plates of the capacitor, we start at the negative
plate and go to the positive plate (noting the changes in potential) along path hgbak. The result is
ΔVhk = +8.00 V + 3.00 V = 11.0 V
so the charge on the capacitor is
Qhk = Chk ( ΔVhk ) = ( 6.00 mF )(11.0 V ) = 66.0 mC
18.54
At time t, the charge on the capacitor will be q = Qm ax (1− e − t t ) , where
t = RC = ( 2.0 × 10 6 Ω ) ( 3.0 × 10 −6 F ) = 6.0 s
When q = 0.90Qm ax, this gives 0.90 = 1− e − t t or e − t t = 0.10. Thus, − t t = ln ( 0.10 ),
giving t = − ( 6.0 s ) ln ( 0.10 ) = 14 s .
68719_18_ch18_p083-119.indd 112
1/7/11 2:42:16 PM
Direct-Current Circuits
18.55
(a)
113
For the first measurement, the equivalent circuit is
as shown in Figure 1. From this,
Rab = R1 = Ry + Ry = 2 Ry
so
Ry =
1
R1
2
[1]
Figure 1
For the second measurement, the equivalent circuit
is shown in Figure 2. This gives
Rac = R2 =
1
Ry + Rx
2
[2]
Figure 2
Substitute [1] into [2] to obtain
R2 =
(b)
1⎛1 ⎞
1
⎜⎝ R1 ⎟⎠ + Rx, or Rx = R2 − R1
2 2
4
If R1 = 13 Ω and R2 = 6.0 Ω, then Rx = 2.8 Ω .
Since this exceeds the limit of 2.0 Ω, the antenna is inadequately grounded .
18.56
Assume a set of currents as shown in the
circuit diagram at the right. Applying
Kirchhoff’s loop rule to the leftmost loop
gives
+ 75 − ( 5.0 ) I − ( 30 )( I − I1 ) = 0
or
7 I − 6 I1 = 15
[1]
For the rightmost loop, the loop rule gives
− ( 40 + R ) I1 + ( 30 )( I − I1 ) = 0,
or
⎛7 R⎞
I = ⎜ + ⎟ I1
⎝ 3 30 ⎠
[2]
Substituting Equation [2] into [1] and simplifying gives
310 I1 + 7 ( I1 R ) = 450
[3]
Also, it is known that PR = I12 R = 20 W, so
I1 R =
20 W
I1
[4]
Substitution of Equation [4] into [3] yields
310 I1 +
140
= 450
I1
or
Using the quadratic formula: I1 =
310 I12 − 450 I1 + 140 = 0
− ( −450 ) ±
( −450 )2 − 4 ( 310 )(140 )
,
2 ( 310 )
20 W
, we find two possible values for
I12
R = 20 Ω or R = 98 Ω
yielding I1 = 1.0 A and I1 = 0.452 A. Then, from R =
the resistance R. These are:
68719_18_ch18_p083-119.indd 113
1/7/11 2:42:19 PM
114
18.57
Chapter 18
When connected in series, the equivalent resistance is Req = R1 + R2 + ⋅⋅⋅ + Rn = n R. Thus, the
current is I s = ( ΔV ) Req = ( ΔV ) n R, and the power consumed by the series configuration is
Ps = I s2 Req =
( ΔV )2
(n R)
2
(n R) =
( ΔV )2
nR
For the parallel connection, the power consumed by each individual resistor is P1 = ( ΔV ) R, and
the total power consumption is
2
n ( ΔV )
R
2
Pp = nP1 =
Ps ( ΔV )
1
1
R
=
= 2 or Ps = 2 Pp
⋅
2
n
n
Pp
n R n ( ΔV )
2
Therefore,
18.58
Consider a battery of emf e connected between
points a and b as shown. Applying Kirchhoff’s
loop rule to loop acbea gives
ε
− (1.0 ) I1 − (1.0 )( I1 − I 3 ) + e = 0
or
I 3 = 2 I1 − e
[1]
Applying the loop rule to loop adbea gives
− ( 3.0 ) I 2 − ( 5.0 )( I 2 + I 3 ) + e = 0
or
8 I2 + 5 I3 = e
[2]
For loop adca, the loop rule yields
I1 + I 3
3
[3]
I 2 = I1 − e 3
[4]
− ( 3.0 ) I 2 + (1.0 ) I 3 + (1.0 ) I1 = 0 or I 2 =
Substituting Equation [1] into [3] gives
26
13
Now, substitute Equations [1] and [4] into [2] to obtain 18I1 =
e , which reduces to I1 =
e.
3
27
13 9 ⎤
4
1
⎡
Then, Equation [4] gives I 2 = ⎢ − ⎥ e =
e , and [1] yields I 3 = − e .
27
27
⎣ 27 27 ⎦
Then, applying Kirchhoff’s junction rule at junction a gives
I = I1 + I 2 =
18.59
13
4
17
e
e
27
e+ e =
e . Therefore, Rab = =
=
Ω .
27
27
27
I (17e 27 )
17
(a) and (b): With R the value of the load resistor, the
current in a series circuit composed of a 12.0 V battery,
an internal resistance of 10.0 Ω, and a load resistor is
I=
12.0 V
R + 10.0 Ω
and the power delivered to the load resistor is
(144 V ) R
2
PL = I 2 R =
( R + 10.0 Ω )2
continued on next page
68719_18_ch18_p083-119.indd 114
1/7/11 2:42:22 PM
Direct-Current Circuits
Some typical data
values for the
graph are
3.50
3.00
R (Ω) PL (W)
1.19
5.00
3.20
10.0
3.60
15.0
3.46
20.0
3.20
25.0
2.94
30.0
2.70
2.50
PL(W)
1.00
2.00
1.50
1.00
0.500
The curve peaks at
PL = 3.60 W at a
load resistance of
R = 10.0 Ω.
18.60
115
0
5.00
10.0
15.0
R(Ω)
20.0
25.0
30.0
The total resistance in the circuit is
⎛ 1
1⎞
R=⎜ + ⎟
R
R
⎝ 1
2 ⎠
−1
⎛ 1
1 ⎞
=⎜
+
⎝ 2.0 kΩ 3.0 kΩ ⎟⎠
−1
= 1.2 kΩ
and the total capacitance is C = C1 + C2 = 2.0 mF + 3.0 mF = 5.0 mF.
Thus, Qm ax = Ce = ( 5.0 mF )(120 V ) = 600 mC
and
t = RC = (1.2 × 10 3 Ω ) ( 5.0 × 10 −6 F ) = 6.0 × 10 −3 s =
6.0 s
1 000
The total stored charge at any time t is then
q = q1 + q2 = Qm ax (1− e − t t
)
q1 + q2 = ( 600 mC ) (1− e −1 000 t 6.0 s )
or
[1]
Since the capacitors are in parallel with each other, the same potential difference exists across
both at any time.
Therefore,
( ΔV )C =
q1 q2
= ,
C1 C2
or
⎛C ⎞
q2 = ⎜ 2 ⎟ q1 = 1.5q1
⎝ C1 ⎠
[2]
Substituting Equation [2] into [1] gives
2.5q1 = ( 600 mC ) (1− e −1 000 t 6.0 s )
and
q1 =
( 240 mC) (1− e
Then, Equation [2] yields q2 = 1.5 ( 240 mC ) (1− e −1 000 t 6.0 s ) =
18.61
(a)
−1 000 t 6.0 s
(360 mC) (1− e
)
−1 000 t 6.0 s
)
With 4.0 × 10 3 cells, each with an emf of 150 mV, connected in series, the total terminal
potential difference is
ΔV = ( 4.0 × 10 3 ) (150 × 10 −3 V ) = 6.0 × 10 2 V
continued on next page
68719_18_ch18_p083-119.indd 115
1/7/11 2:42:25 PM
116
Chapter 18
When delivering a current of I = 1.0 A, the power output is
P = I ( ΔV ) = (1.0 A )( 6.0 × 10 2 V ) = 6.0 × 10 2 W
(b)
The energy released in one shock is
E1 = P ( Δt )1 = ( 6.0 × 10 2 W ) ( 2.0 × 10 −3 s ) = 1.2 J
(c)
The energy released in 300 such shocks is Etotal = 300E1 = 300 (1.2 J ) = 3.6 × 10 2 J. For a
1.0-kg object to be given a gravitational potential energy of this magnitude, the height the
object must be lifted above the reference level is
h=
18.62
(a)
PEg
=
mg
3.6 × 10 2 J
= 37 m
(1.0 kg) (9.80 m s2 )
When the power supply is connected to points A and B, the circuit reduces as shown below
to an equivalent resistance of Req = 0.099 9 Ω .
From the center figure above, observe that I R = I1 =
1
and
(b)
I R = I R = I100 =
2
3
5.00 V
= 50.0 A
0.100 Ω
5.00 V
= 0.045 0 A = 45.0 mA
111 Ω
When the power supply is connected to points A and C, the circuit reduces as shown below
to an equivalent resistance of Req = 1.09 Ω .
R3 + 100 Ω = 110 Ω
1.09 Ω
I100
R1 + R2 = 1.10 Ω
From the center figure above, observe that I R = I R = I1 =
1
and
(c)
I R = I100 =
3
2
5.00 V
= 4.55 A
1.10 Ω
5.00 V
= 0.045 5 A = 45.5 mA
110 Ω
When the power supply is connected to points A and D, the circuit reduces as shown below
to an equivalent resistance of Req = 9.99 Ω .
continued on next page
68719_18_ch18_p083-119.indd 116
1/7/11 2:42:27 PM
Direct-Current Circuits
From the center figure above, observe that I R = I R = I R = I1 =
1
and
18.63
I100 =
2
3
117
5.00 V
= 0.450 A
11.1 Ω
5.00 V
= 0.050 0 A = 50.0 mA
100 Ω
In the circuit diagram at the right, note that all points
labeled a are at the same potential and equivalent
to each other. Also, all points labeled c are equivalent.
To determine the voltmeter reading, go from point e
to point d along the path ecd, keeping track of all
changes in potential to find
ΔVed = Vd − Ve = −4.50 V + 6.00 V = + 1.50 V
Apply Kirchhoff’s loop rule around loop abcfa to find
− ( 6.00 Ω ) I + ( 6.00 Ω ) I 3 = 0
or
I3 = I
[1]
I 2 = 0.600 A − 0.600I
[2]
I1 = 0.900 A − 1.20I
[3]
Apply Kirchhoff’s loop rule around loop abcda to find
− ( 6.00 Ω ) I + 6.00 V − (10.0 Ω ) I 2 = 0
or
Apply Kirchhoff’s loop rule around loop abcea to find
− ( 6.00 Ω ) I + 4.50 V − ( 5.00 Ω ) I1 = 0
or
Finally, apply Kirchhoff’s junction rule at either point a or point c to obtain
I + I 3 = I1 + I 2
[4]
Substitute Equations [1], [2], and [3] into Equation [4] to obtain the current through the ammeter.
This gives
I + I = 0.900 A − 1.20I + 0.600 A − 0.600I
or
68719_18_ch18_p083-119.indd 117
3.80I = 1.50 A
and
I = 1.50 A 3.80 = 0.395 A
1/7/11 2:42:29 PM
118
18.64
18.65
Chapter 18
In the figure given below, note that all bulbs have the same resistance, R.
(a)
In the series situation, Case 1, the same current I1 flows through both bulbs. Thus, the same
power, P1 = I12 R, is supplied to each bulb. Since the brightness of a bulb is proportional to
the power supplied to it, they will have the same brightness. We conclude that
the bulbs have the same current, power supplied, and brightness .
(b)
In the parallel case, Case 2, the same potential difference ΔV is maintained across each
of the bulbs. Thus, the same current I 2 = ΔV R will flow in each branch of this parallel
circuit. This means that, again, the same power P2 = I 22 R is supplied to each bulb, and the
two bulbs will have equal brightness .
(c)
The total resistance of the single branch of the series circuit (Case 1) is 2R. Thus, the current
in this case is I1 = ΔV 2R. Note that this is one half of the current I 2 that flows through each
bulb in the parallel circuit (Case 2). Since the power supplied is proportional to the square
of the current, the power supplied to each bulb in Case 2 is four times that supplied to each
bulb in Case 1. Thus, the bulbs in Case 2 are much brighter than those in Case 1.
(d)
If either bulb goes out in Case 1, the only conducting path of the circuit is broken and all
current ceases. Thus, in the series case, the other bulb must also go out . If one bulb goes out
in Case 2, there is still a continuous conducting path through the other bulb. Neglecting any
internal resistance of the battery, the battery continues to maintain the same potential difference ΔV across this bulb as was present when both bulbs were lit. Thus, in the parallel case,
the second bulb remains lit with unchanged current and brightness when one bulb fails.
(a)
The equivalent capacitance of this parallel combination is
Ceq = C1 + C2 = 3.00 mF + 2.00 mF = 5.00 mF
When fully charged by a 12.0-V battery, the total stored charge
before the switch is closed is
Q0 = Ceq ( ΔV ) = ( 5.00 mF )(12.0 V ) = 60.0 mC
Once the switch is closed, the time constant of the resulting RC
circuit is
t = RCeq = ( 5.00 × 10 2 Ω ) ( 5.00 mF ) = 2.50 × 10 −3 s = 2.50 ms
Thus, at t = 1.00 ms after closing the switch, the remaining total stored charge is
q = Q0 e − t t = ( 60.0 mC ) e −1.00 m s 2.50 m s = ( 60.0 mC ) e −0.400 = 40.2 mC
continued on next page
68719_18_ch18_p083-119.indd 118
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Direct-Current Circuits
119
The potential difference across the parallel combination of capacitors is then
ΔV =
q
40.2 mC
=
= 8.04 V
Ceq 5.00 mF
and the charge remaining on the 3.00-mF capacitor will be
q3 = C3 ( ΔV ) = ( 3.00 mF )(8.04 V ) = 24.1 mC
(b)
The charge remaining on the 2.00-mF capacitor at this time is
q2 = q − q3 = 40.2 mC − 24.1 mC = 16.1 mC
or alternately,
(c)
Since the resistor is in parallel with the capacitors, it has the same potential difference
across it as do the capacitors at all times. Thus, Ohm’s law gives
I=
18.66
(a)
q2 = C2 ( ΔV ) = ( 2.00 mF )(8.04 V ) = 16.1 mC
ΔV
8.04 V
=
= 1.61× 10 −2 A = 16.1 mA
R
5.00 × 10 2 Ω
If the switch S in the circuit at the right is closed at t = 0, the charge
remaining on the capacitor at time t is q = Q0 e − t t , where Q0 = 5.10 mC
is the initial charge and
t = RC = (1.30 × 10 Ω ) ( 2.00 × 10
3
−9
F ) = 2.60 ms
+q −q
C
i
S
R
is the time constant of the circuit.
The potential difference across the capacitor, ΔvC , at time t is
ΔvC =
q Q0 e − t t
=
C
C
Applying Kirchhoff’s loop rule to the above circuit gives ΔvC − iR = 0 , or
i=
ΔvC Q0 e − t t ⎛ Q0 ⎞ − t t
=
=⎜ ⎟e
⎝ t ⎠
R
RC
The current through the resistor at time t = 9.00 ms is then
9.00 m s
⎛ 5.10 mC ⎞ − 2.60 m s
⎛Q ⎞
i = ⎜ 0 ⎟ e− t t = ⎜
e
= 6.16 × 10 −2 C s = 61.6 mA
⎝ t ⎠
⎝ 2.60 ms ⎟⎠
(b)
The charge remaining on the capacitor at t = 8.00 ms is
q = Q0 e − t t = ( 5.10 mC ) e
(c)
−
8.00 m s
2.60 m s
= 0.235 mC
The maximum current in the circuit occurs when the switch is first closed (at t = 0) and is
given by
5.10 m C
Q
⎛Q ⎞
i0 = ⎜ 0 ⎟ e −0 = 0 =
= 1.96 A
⎝ t ⎠
t
2.60 m s
68719_18_ch18_p083-119.indd 119
1/7/11 2:42:33 PM
19
Magnetism
QUICK QUIZZES
1.
Choice (b). The force that a magnetic field exerts on a charged particle moving through it is given
by F = qvB sinq = qvB⊥, where B⊥ is the component of the field perpendicular to the particle’s
velocity. Since the particle moves in a straight line, the magnetic force (and hence B⊥, since
qv ≠ 0) must be zero.
2.
Choice (c). The magnetic force exerted on a charge by a magnetic field is proportional to the
charge’s velocity relative to the field. If the charge is stationary, as in this situation, there is no
magnetic force.
3.
Choice (c). The torque that a planar current loop will experience when it is in a magnetic field is
given by t = BIAsinq . Note that this torque depends on the strength of the field, the current in the
coil, the area enclosed by the coil, and the orientation of the plane of the coil relative to the
direction of the field. However, it does not depend on the shape of the loop.
4.
Choice (a). The magnetic force acting on the particle is always perpendicular to the velocity of
the particle and hence to the displacement the particle is undergoing. Under these conditions, the
force does no work on the particle, and the particle’s kinetic energy remains constant.
5.
Choices (a) and (c). The magnitude of the force per unit length between two parallel current carrying wires is F = ( m 0 I1 I 2 ) ( 2p d ). The magnitude of this force can be doubled by doubling the
magnitude of the current in either wire. It can also be doubled by decreasing the distance between
them, d, by half. Thus, both choices (a) and (c) are correct.
6.
Choice (b). The two forces are an action-reaction pair. They act on different wires and have equal
magnitudes but opposite directions.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
If the magnitude of the magnetic force on the wire equals the weight of the wire, then BI sinq = w,
or B = w I sinq . The magnitude of the magnetic field is a minimum when q = 90° and sinq = 1.
Thus,
Bm in =
w
1.0 × 10 −2 N
=
= 0.20 T
I
( 0.10 A )( 0.50 m )
and (a) is the correct answer for this question.
2.
The electron moves in a horizontal plane in a direction of 35° N of E, which is the same as 55°
E of N. Since the magnetic field at this location is horizontal and directed due north, the angle
between the direction of the electron’s velocity and the direction of the magnetic field is 55°.
The magnitude of the magnetic force experienced by the electron is then
F = q vB sinq = (1.6 × 10 −19 C ) ( 2.5 × 10 6 m s ) ( 0.10 × 10 −4 T ) sin 55° = 3.3 × 10 −18 N
120
68719_19_ch19_p120-147.indd 120
1/7/11 2:43:34 PM
Magnetism
121
The right-hand rule number 1 predicts a force directed upward, away from the Earth’s surface
for a positively charged particle moving in the direction of the electron. However, the negatively
charged electron will experience a force in the opposite direction, downward toward the Earth’s
surface. Thus, the correct choice is (d).
3.
The z-axis is perpendicular to the plane of the loop, and the angle between the direction of this
normal line and the direction of the magnetic field is q = 30.0°. Thus, the magnitude of the torque
experienced by this coil containing N = 10 turns is
t = BIAN sinq = ( 0.010 T )( 2.0 A )[( 0.20 m )( 0.30 m )](10 ) sin 30.0° = 6.0 × 10 −3 N ⋅ m
meaning that (c) is the correct choice.
4.
The magnitude of the magnetic field at distance r from a long straight wire carrying current I is
B = m 0 I 2p r. Thus, for the described situation,
B=
( 4p × 10
−7
T⋅m A ) (1 A )
2p ( 2 m )
= 1× 10 −7 T
making (d) the correct response.
5.
Since the proton follows a semicircular path, not a helical path, it entered perpendicularly to the
field. A charged particle moving perpendicular to a magnetic field experiences a centripetal force
of magnitude Fc = m v 2 r = qvB and follows a circular path of radius r = mv qB. The speed of this
proton must be
−19
−3
qBr (1.6 × 10 C )( 0.050 T ) (1.0 × 10 m )
v=
=
= 4.8 × 10 3 m s
m
1.67 × 10 −27 kg
and choice (e) is the correct answer.
6.
The force per unit length between this pair of wires is
F
−7
m 0 I1 I 2 ( 4p × 10 T ⋅ m A )( 3 A )
=
= 9 × 10 −7 N ∼ 1× 10 −6 N
=
2p d
2p ( 2 m )
2
and (d) is the best choice for this question.
7.
The magnitude of the magnetic field inside the specified solenoid is
⎛ 120 ⎞
⎛ N⎞
B = m 0 nI = m 0 ⎜ ⎟ I = ( 4p × 10 −7 T ⋅ m A ) ⎜
( 2.0 A ) = 6.0 × 10 −4 T
⎝ 0.50 m ⎟⎠
⎝ ⎠
which is choice (e).
8.
68719_19_ch19_p120-147.indd 121
The force that a magnetic field exerts on a moving charge is always perpendicular to both the
direction of the field and the direction of the particle’s motion. Since the force is perpendicular
to the direction of motion, it does no work on the particle and hence does not alter its speed.
Because the speed is unchanged, both the kinetic energy and the magnitude of the linear
momentum will be constant. Correct answers among the list of choices are (d) and (e). All other
choices are false.
1/7/11 2:43:36 PM
122
9.
Chapter 19
The magnitude of the magnetic force experienced by a charged particle in a magnetic field is
given by F = qvB sinq , where v is the speed of the particle and q is the angle between the
direction of the particle’s velocity and the direction of the magnetic field. If either v = 0
[choice (e)] or sinq = 0 [choice (c)], this force has zero magnitude. All other choices are
false, so the correct answers are (c) and (e).
10.
By the right-hand rule number 1, when the proton first enters the field, it experiences a force
directed upward, toward the top of the page. This will deflect the proton upward, and as the
proton’s velocity changes direction, the force changes direction, always staying perpendicular to
the velocity. The force, being perpendicular to the motion, causes the particle to follow a circular
path, with no change in speed, as long as it is in the field. After completing a half circle, the
proton will exit the field traveling toward the left. The correct answer is choice (d).
11.
The contribution made to the magnetic field at point P by the lower wire is directed out of the page,
while the contribution due to the upper wire is directed into the page. Since point P is equidistant
from the two wires, and the wires carry the same magnitude currents, these two oppositely directed
contributions to the magnetic field have equal magnitudes and cancel each other. Therefore, the total
magnetic field at point P is zero, making (a) the correct answer for this question.
12.
The magnetic field due to the current in the vertical wire is directed into the page on the right side of
the wire and out of the page on the left side. The field due to the current in the horizontal wire is out
of the page above this wire and into the page below the wire. Thus, the two contributions to the total
magnetic field have the same directions at points B (both out of the page) and D (both contributions
into the page), while the two contributions have opposite directions at points A and C. The magnitude
of the total magnetic field will be greatest at points B and D where the two contributions are in the
same direction, and smallest at points A and C where the two contributions are in opposite directions
and tend to cancel. The correct choices for this question are (a) and (c).
13.
The torque exerted on a single turn coil carrying current I by a magnetic field B is t = BIAsinq .
The line perpendicular to the plane of each coil is also perpendicular to the direction of the
magnetic field (i.e., q = 90°). Since B and I are the same for all three coils, the torques exerted on
them are proportional to the area A enclosed by each of the coils. Coil A is rectangular with area
AA = (1 m )( 2 m ) = 2 m 2. Coil B is elliptical with semi-major axis a = 0.75 m and semi-minor axis
2
b = 0.5 m, giving an area AB = p ab or AB = p ( 0.75 m )( 0.5 m ) = 1.2 m . Coil C is triangular
1
1
with area AC = ( base )( height ), or AC = (1 m )( 3 m ) = 1.5 m 2. Thus, AA > AC > AB, meaning
2
2
that t A > t C > t B and choice (b) is the correct answer.
14.
Any point in region I is closer to the upper wire which carries the larger current. At all points in this
region, the outward-directed field due the upper wire will have a greater magnitude than will the
inward-directed field due to the lower wire. Thus, the resultant field in region I will be nonzero and
out of the page, meaning that choice (d) is a true statement and choice (a) is false. In region II, the
field due to each wire is directed into the page, so their magnitudes add and the resultant field
cannot be zero at any point in this region. This means that choice (b) is false. In region III, the field
due to the upper wire is directed into the page while that due to the lower wire is out of the page.
Since points in this region are closer to the wire carrying the smaller current, there are points in this
region where the magnitudes of the oppositely directed fields due to the two wires will possibly
have equal magnitudes, canceling each other and producing a zero resultant field. Thus, choice (c) is
true and choice (e) is false. The correct answers for this question are choices (c) and (d).
15.
According to right-hand rule number 2, the magnetic field at point P due to the current in the
wire is directed out of the page, meaning that choices (c) and (e) are false. The magnitude of this
field is given by B = m 0 I 2p r, so choices (b) and (d) are false. Choice (a) is correct about both
the magnitude and direction of the field and is the correct answer for the question.
68719_19_ch19_p120-147.indd 122
1/7/11 2:43:37 PM
Magnetism
16.
123
The magnetic field inside a solenoid, carrying current I, with N turns and length L, is
m ( 2N A ) I
m N I
m N I 1
⎛ N⎞
B = m 0 nI = m 0 ⎜ ⎟ I. Thus, BA = 0 A , BB = 0 A = BA, and BC = 0
= 4BA .
⎝ L⎠
LA
LA 2
2LA
2
Therefore, we see that BC > BA > BB, and choice (d) gives the correct rankings.
17.
Using right-hand rule number 1, we see that the magnetic force exerted on the positively
charged proton in the situation described is in the positive y-direction, making choice (c) the
correct answer.
18.
The correct answers to each of the parts of this question are as follows:
(a) Yes.
(b) No.
(c) Yes.
(d) Yes, unless the object moves parallel to the field.
(e) No.
(f) Yes, unless the current is parallel to the field.
(g) Yes.
(h) Yes, unless the electrons move parallel to the field.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
It should point straight down toward the surface of the Earth.
4.
No. The force that a constant magnetic field exerts on a charged particle is dependent on the
velocity of that particle. If the particle has zero velocity, it will experience no magnetic force and
cannot be set in motion by a constant magnetic field.
6.
Yes. Regardless of which pole is used, the magnetic field of the magnet induces magnetic poles in
the nail, with each end of the nail taking on a magnetic polarity opposite to that of the pole of the
magnet nearest to it. These opposite magnetic poles then attract each other.
8.
The magnet causes domain alignment in the iron, inducing magnetic poles such that the iron
becomes magnetic and attracted to the original magnet. Now that the iron is magnetic, it can
produce an identical effect in another piece of iron.
10.
No. The magnetic field created by a single current loop resembles that of a bar magnet—strongest
inside the loop, and decreasing in strength as you move away from the loop. Neither is the field
uniform in direction—the magnetic field lines form closed paths that curve around the conductor
forming the current loop and pass through the area enclosed by that current loop.
12.
Near the poles the magnetic field of Earth points almost straight downward (or straight upward),
in the direction (or opposite to the direction) the charges are moving. As a result, there is little or
no magnetic force exerted on the charged particles at the pole to deflect them away from Earth.
14.
The loop can be mounted on an axle so it is free to rotate. The current loop will tend to rotate in a
manner that brings the plane of the loop perpendicular to the direction of the field. As the current
in the loop is increased, the torque causing it to rotate will increase in magnitude.
16.
(a)
The blue magnet experiences an upward magnetic force equal to its weight. The yellow magnet
is repelled by the red magnets with a force whose magnitude equals the weight of the yellow
magnet plus the magnitude of the reaction force exerted on this magnet by the blue magnet.
(b)
The rods prevent motion to the side and prevent the magnets from rotating under their
mutual torques. Its constraint changes unstable equilibrium into stable.
68719_19_ch19_p120-147.indd 123
1/7/11 2:43:39 PM
124
Chapter 19
(c)
Most likely, the disks are magnetized perpendicular to their flat faces, making one face a
north pole and the other a south pole. The yellow magnet has a pole on its lower face which
is the same as the pole on the upper faces of the red magnets. The pole on the lower face of
the blue magnet is the same as that on the upper face of the yellow magnet.
(d)
If the upper magnet were inverted, the yellow and blue magnets would attract each other
and stick firmly together. The yellow magnet would continue to be repelled by and float
above the red magnets.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a)
a⬘) toward the left
⬘)d toward top of page
b⬘) into the page
e⬘) into the page
c⬘) out of the page
f⬘) out of the page
(b) The answer for each subpart is opposite to that given in part (a) above.
4.
(a)
toward top of page
(d)
into the page
(b)
out of page (q < 0)
6.
B = 2.09 ×10 −2 T in the negative y-direction
8.
(a)
7.90 × 10 −12 N
(b)
0
(c)
zero force
10.
807 N
12.
(a)
toward the left
(b)
into the page
(c)
out of the page
(d)
toward top of page
(e)
into the page
(f)
out of the page
(a)
0.12 N
(b)
Both the direction of the field and the direction of the current must be known before the
direction of the force can be determined.
14.
16.
Bm in = 0.245 T eastward
18.
(a)
9.0 × 10 −3 N at 15° above the horizontal in the northward direction
(b)
2.3 × 10 −3 N horizontal and due west
20.
(a)
0.109 A
22.
B = m k mg Id
24.
5.8 N into the page
26.
4.33 × 10 −3 N ⋅ m
28.
9.98 N·m, clockwise about the y-axis when viewed from above
30.
118 N·m
68719_19_ch19_p120-147.indd 124
(b)
toward the right
1/7/11 2:43:40 PM
Magnetism
125
(a)
+x-direction, zero torque about x-axis
(b)
−x-direction, zero torque about x-axis
(c)
No. The two forces are equal in magnitude and opposite in direction, canceling each other,
and can have no effect on the motion of the loop.
(d)
in the yz-plane at 130° counterclockwise from + y-direction. Torque is counterclockwise
about x-axis.
(e)
counterclockwise about x-axis
(f)
0.135 A ⋅ m 2
(g)
130°
34.
(a)
1.25 × 10 −13 N
(b)
7.49 × 1013 m s 2
36.
0.150 mm
38.
3.11 cm
40.
(a)
v = 2 ( KE ) qBR
(b)
m = q 2 B 2 R 2 2 ( KE )
42.
(a)
rd = 2 ⋅rp
(b)
ra = 2 ⋅rp
44.
(a)
toward the left
(b)
46.
675 A downward
48.
(a)
40.0 mT into the page
(b)
50.
(a)
I
Bnet = 4.00 mT toward the bottom of the page
(b)
I
Bnet = 6.67 mT upward at 77.0° to the left of vertical
(a)
in the –y-direction
32.
52.
(b)
(h)
0.155 N·m
out of the page
(c)
lower left to upper right
5.00 mT out of the page
(c)
1.67 mT out of the page
upward, in the positive z-direction
(c) The magnitude of the upward magnetic force must equal that of the downward gravitational
force.
54.
56.
(d)
d = qvm 0 I 2p mg
(e)
5.40 cm
(a)
in the + y-direction
(b)
BP = m 0 Ix p ( x 2 + d 2 )
(c)
BP x = 0 = 0. This is as expected since the two field contributions have equal magnitudes and
opposite directions at the point midway between the wires.
(a)
8.0 A
(c)
Reversing the direction of one current changes the force from repulsion to attraction.
Doubling the magnitude of one current doubles the magnitude of the force.
58.
2.70 × 10 −5 N toward the left
60.
4.8 × 10 4 turns
68719_19_ch19_p120-147.indd 125
(b)
opposite directions
1/7/11 3:11:50 PM
126
Chapter 19
62.
207 W
64.
(a)
B = mg NIw directed out of the page
(b) The forces on the two sides of the loop are equal in magnitude and opposite in direction,
canceling each other, and do not affect the state of balance of the system.
(c)
B = 0.26 T
66.
(a)
5.00 cm
68.
3.92 × 10 −2 T
70.
(a)
6.2 m s 2
(b)
0.40 s
72.
(a)
opposite directions
(b)
68 A
74.
0.59 T
76.
(a) 1.00 × 10 −5 T
(b) 8.00 × 10 −5 N directed toward wire 1
(c) 1.60 × 10 −5 T
(d) 8.00 × 10 −5 N directed toward wire 2
(b) 8.78 × 10 6 m s
PROBLEM SOLUTIONS
19.1
19.2
Remember that the direction of the magnetic force exerted on the negatively charged electron
is opposite to the direction predicted by right-hand rule number 1. The magnetic field near the
Earth’s equator is horizontal and directed toward the north. The magnetic force experienced by a
moving charged particle is always perpendicular to the plane formed by the vectors representing
the magnetic field and the particle’s velocity.
(a)
When the velocity of a positively charged particle is downward, right-hand rule number 1
predicts a magnetic force toward the east. Hence, the force experienced by the negatively
charged electron (and also the deflection of its velocity) is directed toward the west .
(b)
When the particle moves northward, its velocity is parallel to the magnetic field, and it will
experience zero force and zero deflection .
(c)
The direction of the force on the negatively charged electron (and the deflection of its
velocity) will be vertically upward .
(d)
The direction of the force on the negatively charged electron (and the deflection of its
velocity) will be vertically downward .
(a)
For a positively charged particle, the direction of the force is that predicted by the righthand rule number one. These are:
(a⬘)
in plane of page and to left
(b⬘)
into the page
(c⬘)
out of the page
(d⬘)
in plane of page and toward the top
(e⬘)
into the page
(f⬘)
out of the page
continued on next page
68719_19_ch19_p120-147.indd 126
1/7/11 2:43:44 PM
Magnetism
(b)
19.3
For a negatively charged particle, the direction of the force is exactly opposite what the
right-hand rule number 1 predicts for positive charges. Thus, the answers for part (b) are
reversed from those given in part (a) .
Since the particle is positively charged, use the right-hand rule number 1. In this case, start with
the fingers of the right hand in the direction of v and the thumb pointing in the direction of F. As
you start closing the hand, the fingers point in the direction of B after they have moved 90°. The
results are
(a)
19.4
into the page
(b)
toward the right
(c)
toward bottom of page
Hold the right hand with the fingers in the direction of v so that as you close your hand, the
fingers move toward the direction of B. The thumb will point in the direction of the force (and
hence the deflection) if the particle has a positive charge. The results are
(a)
toward top of page
(c) q = 180° ⇒ zero force
19.5
127
(a)
(b)
out of the page , since the charge is negative
(d)
into the page
The proton experiences maximum force when it moves perpendicular to the magnetic field,
and the magnitude of this maximum force is
Fm ax = qvB sin 90° = (1.60 × 10 −19 C ) ( 6.00 × 10 6 m s ) (1.50 T )(1) = 1.44 × 10 −12 N
19.6
Fm ax 1.44 × 10 −12 N
=
= 8.62 × 1014 m s 2
mp
1.67 × 10 −27 kg
(b)
am ax =
(c)
Since the magnitude of the charge of an electron is the same as that of a proton, the
force experienced by the electron would have the same magnitude but would be in the
opposite direction due to the negative charge of the electron.
(d)
The acceleration of the electron would have a much greater magnitude than that of the
proton because of the mass of the electron is much smaller.
+y
Since the acceleration (and hence the magnetic force) is in the positive
x-direction, the magnetic field must be in the negative y-direction (see
sketch at the right) according to right-hand rule number 1.
F
+x
v
The magnitude of the magnetic field is found from Fm = qvB sinq as
B=
or
B=
yielding
68719_19_ch19_p120-147.indd 127
+z
B
Fm
ma
=
qv sinq qv sinq
(1.67 × 10
(1.60 × 10
−19
−27
kg ) ( 2.00 × 1013 m s 2 )
C ) (1.00 × 10 7 m s ) sin 90.0°
= 2.09 × 10 −2 T
B = 2.09 ×10 −2 T in the negative y-direction .
1/7/11 2:43:46 PM
128
19.7
Chapter 19
The gravitational force is small enough to be ignored, so the magnetic force must supply the
needed centripetal acceleration. Thus,
m
v2
q Br
= qvB sin 90°, or v =
, where r = RE + 1000 km = 7.38 × 10 6 m
m
r
v=
(1.60 × 10
−19
C ) ( 4.00 × 10 −8 T ) ( 7.38 × 10 6 m )
1.67 × 10 −27 kg
= 2.83 × 10 7 m s
If v is toward the west and B is northward, F will be directed downward as required.
19.8
The speed attained by the electron is found from
v=
(a)
2e ( ΔV )
=
m
1 2
mv = q ( ΔV ), or
2
2 (1.60 × 10 −19 C ) ( 2 400 V )
9.11× 10 −31 kg
= 2.90 × 10 7 m s
Maximum force occurs when the electron moves perpendicular to the field.
Fm ax = q vB sin 90° = (1.60 × 10 −19 C ) ( 2.90 × 10 7 m s )(1.70 T ) = 7.90 × 10 −12 N
(b)
Minimum force occurs when the electron moves parallel to the field.
Fm in = q vB sin 0° = 0
19.9
The magnetic force experienced by a moving charged particle has magnitude Fm = qvB sinq ,
where q is the angle between the directions of the particle’s velocity and the magnetic field. Thus,
sinq =
and
19.10
Fm
8.20 × 10 −13 N
=
= 0.754
qvB (1.60 × 10 −19 C ) ( 4.00 × 10 6 m s )(1.70 T )
q = sin −1 ( 0.754 ) = 48.9°
or
q = 180° − 48.9° = 131°
The force on a single ion is
F1 = qvB sinq
= (1.60 × 10 −19 C ) ( 0.851 m s )( 0.254 T ) sin ( 51.0° ) = 2.69 × 10 −20 N
The total number of ions present is
ions ⎞
⎛
3
22
N = ⎜ 3.00 × 10 20
⎟ (100 cm ) = 3.00 × 10
⎝
cm 3 ⎠
Thus, assuming all ions move in the same direction through the field, the total force is
F = N ⋅ F1 = ( 3.00 × 10 22 ) ( 2.69 × 10 −20 N ) = 807 N
19.11
Gravitational force:
Fg = mg = ( 9.11× 10 −31 kg ) ( 9.80 m s 2 ) = 8.93 × 10 −30 N downward
continued on next page
68719_19_ch19_p120-147.indd 128
1/7/11 2:43:48 PM
Magnetism
129
Electric force:
Fe = qE = ( −1.60 × 10 −19 C ) ( −100 N C ) = 1.60 × 10 −17 N upward
Magnetic force:
Fm = qvB sinq = ( −1.60 × 10 −19 C ) ( 6.00 × 10 6 m s ) ( 50.0 × 10 −6 T ) sin ( 90.0° )
= 4.80 × 10 −17 N in direction opposite right hand rule prediction
Fm = 4.80 × 10 −17 N downward
19.12
19.13
Hold the right hand with the fingers in the direction of the current so, as you close the hand, the
fingers move toward the direction of the magnetic field. The thumb then points in the direction of
the force. The results are
(a)
to the left
(b)
into the page
(c)
out of the page
(d)
toward top of page
(e)
into the page
(f )
out of the page
From F = BIL sinq , the magnetic field is
B=
F L
0.12 N m
=
= 8.0 × 10 −3 T
I sinq (15 A ) sin 90°
The direction of B must be the + z-direction to have F in the –y-direction when I is in the
+x-direction.
19.14
19.15
(a)
F = BIL sinq = ( 0.28 T )( 3.0 A )( 0.14 m ) sin 90° = 0.12 N
(b)
Neither the direction of the magnetic field nor that of the current is given . Both must be
known before the direction of the force can be determined. In this problem, you can only
say that the force is perpendicular to both the wire and the field.
Use the right-hand rule number 1, holding your right hand with the fingers in the direction of the
current and the thumb pointing in the direction of the force. As you close your hand, the fingers
will move toward the direction of the magnetic field. The results are
(a)
19.16
into the page
(b)
toward the right
(c)
toward the bottom of the page
In order to just lift the wire, the magnetic force exerted on a unit length of the wire must be directed
upward and have a magnitude equal to the weight per unit length. That is, the magnitude is
F
⎛ m⎞
= BI sinq = ⎜ ⎟ g
⎝ ⎠
giving
⎛ m⎞ g
B=⎜ ⎟
⎝ ⎠ I sinq
To find the minimum possible field, the magnetic field should be perpendicular to the current
(q = 90.0°). Then,
⎡
g
g ⎛ 1 kg ⎞ ⎛ 10 2 cm ⎞ ⎤ 9.80 m s 2
⎛ m⎞
Bm in = ⎜ ⎟
= 0.245 T
= ⎢ 0.500
⎥
⎝ ⎠ I sin 90.0° ⎣
cm ⎜⎝ 10 3 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ ⎦ ( 2.00 A )(1)
To find the direction of the field, hold the right hand with the thumb pointing upward (direction of
the force) and the fingers pointing southward (direction of current). Then, as you close the hand,
the fingers point eastward. The magnetic field should be directed eastward .
68719_19_ch19_p120-147.indd 129
1/7/11 2:43:50 PM
130
Chapter 19
19.17
F = BIL sinq = ( 0.300 T )(10.0 A )( 5.00 m ) sin ( 30.0° ) = 7.50 N
19.18
(a)
The magnitude is
F = BIL sinq = ( 0.60 × 10 − 4 T )(15 A )(10.0 m ) sin ( 90° ) = 9.0 × 10 −3 N
F is perpendicular to B. Using the right-hand rule number 1, the orientation of F is found to
be 15° above the horizontal in the northward direction .
(b)
F = BIL sinq = ( 0.60 × 10 − 4 T )(15 A )(10.0 m ) sin (165° ) = 2.3 × 10 −3 N
and, from the right-hand rule number 1, the direction is horizontal and due west .
19.19
For minimum field, B should be perpendicular to the wire. If the force is to be northward, the
field must be directed downward .
To keep the wire moving, the magnitude of the magnetic force must equal that of the kinetic
friction force. Thus, BIL sin 90° = m k ( mg ), or
B=
19.20
(a)
2
m k ( m L ) g ( 0.200 )(1.00 g cm ) ( 9.80 m s ) ⎛ 1 kg ⎞ ⎛ 10 2 cm ⎞
=
⎜⎝ 10 3 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ = 0.131 T
I sin 90°
(1.50 A )(1.00 )
To have zero tension in the wires, the magnetic force per unit
length must be directed upward and equal to the weight per
unit length of the conductor. Thus,
Fm
L
I=
19.21
= BI =
mg
, or
L
( m L ) g = ( 0.040
B
kg m ) ( 9.80 m s 2 )
3.60 T
= 0.109 A
(b)
From the right-hand rule number 1, the current must be to the right if the force is to be
upward when the magnetic field is into the page.
(a)
The magnetic force must be directed upward and its magnitude must equal mg , the weight
of the wire. Then, the net force acting on the wire will be zero, and it can move upward at
constant speed.
(b)
The magnitude of the magnetic force must be BIL sinq = mg and for minimum field
q = 90°. Thus,
Bm in =
2
mg ( 0.015 kg ) ( 9.80 m s )
=
= 0.20 T
IL
( 5.0 A )( 0.15 m )
For the magnetic force to be directed upward when the current is toward the left, B must be
directed out of the page .
(c)
68719_19_ch19_p120-147.indd 130
If the field exceeds 0.20 T, the upward magnetic force exceeds the downward force of
gravity, so the wire accelerates upward .
1/7/11 2:43:52 PM
Magnetism
19.22
As shown in end view in the sketch at the right, the rod (and hence, the
current) is horizontal, while the magnetic field is vertical. For
the rod to move with constant velocity (zero acceleration), it is
necessary that
n = mg
Fm
fk
ΣFx = Fm − fk = BI sin 90.0° − m k n = 0
B=
or
reducing to
19.23
mg
m ( mg )
mk n
= k
I sin 90.0° Id (1.00 )
B = m k mg Id
For each segment, the magnitude of the force is given by F = BIL sinq , and the direction is given
by the right-hand rule number 1. The results of applying these to each of the four segments, with
B = 0.020 0 T and I = 5.00 A, are summarized below.
Segment
L (m)
q
F (N)
Direction
ab
0.400
180°
0
no direction
bc
0.400
90.0° 0.040 0 negative x
cd
0.400 2
45.0° 0.040 0 negative z
0.400 2
parallel to xzplane at 45° to
90.0° 0.056 6
both +x- and
+z-directions
da
19.24
131
y
B
a
d
z
I
c
b
x
B
The magnitude of the force is
F = BIL sinq = ( 5.0 × 10 −5 N ) ( 2.2 × 10 3 A )( 58 m ) sin 65° = 5.8 N
and the right-hand rule number 1 shows its direction to be into the page .
19.25
The torque exerted on a single turn current loop in a magnetic field is given by t = BIAsinq ,
where A is the area enclosed by the loop and q is the angle between the line normal to the plane
of the loop and the direction of the magnetic field B. For maximum torque, q = 90.0°, so we have
⎛ pd2 ⎞
1
2
sin 90.0° = ( 3.00 × 10 −3 T )( 5.00 A )p ( 0.100 m ) (1.00 )
t m ax = BI ⎜
4
⎝ 4 ⎟⎠
= 1.18 × 10 −4 N ⋅ m = 118 × 10 −6 N ⋅ m = 118 mN ⋅ m
19.26
The magnitude of the torque is t = NBIAsinq , where q is the angle between the field and the line
perpendicular to the plane of the loop. The circumference of the loop is 2p r = 2.00 m, so the
1.00 m
1 2
radius is r =
and the area is A = p r 2 =
m.
p
p
Thus,
68719_19_ch19_p120-147.indd 131
⎛1
⎞
t = (1)( 0.800 T ) (17.0 × 10 −3 A ) ⎜ m 2 ⎟ sin 90.0° = 4.33 × 10 −3 N ⋅ m
⎝p
⎠
1/7/11 2:43:55 PM
132
19.27
Chapter 19
The area of the loop is A = p ab, where a = 0.200 m and b = 0.150 m. Since the field is parallel to
the plane of the loop, q = 90.0° and the magnitude of the torque is
t = NBIAsinq
= 8 ( 2.00 × 10 −4 T )( 6.00 A )[p ( 0.200 m )( 0.150 m )] sin 90.0° = 9.05 × 10 −4 N ⋅ m
The torque is directed to make the left-hand side of the loop move toward you and the right-hand
side move away.
19.28
Note that the angle between the field and the line perpendicular to the plane of the loop is
q = 90.0° − 30.0° = 60.0°. Then, the magnitude of the torque is
t = NBIAsinq = 100 ( 0.800 T )(1.20 A )[( 0.400 m )( 0.300 m )] sin 60.0° = 9.98 N ⋅ m
With current in the –y-direction, the outside edge of the loop will experience a force directed out
of the page (+z-direction) according to the right-hand rule number 1. Thus, the loop will rotate
clockwise as viewed from above .
19.29
(a)
The torque exerted on a coil by a uniform magnetic field is t = NBIAsinq , with maximum
torque occurring when q = 90°. Thus, the current in the coil must be
I=
(b)
t m ax
0.15 N ⋅ m
= 0.56 A
=
NBA ( 200 )( 0.90 T ) ⎡( 3.0 × 10 −2 m ) ( 5.0 × 10 −2 m ) ⎤
⎣
⎦
If I has the value found above and q is now 25°, the torque on the coil is
t = NBIAsinq = ( 200 )( 0.90 T )( 0.56 A )[( 0.030 m )( 0.050 m )] sin 25° = 0.064 N ⋅ m
19.30
The resistance of the loop is
R=
−8
rL (1.70 × 10 Ω ⋅ m )(8.00 m )
=
= 1.36 × 10 −3 Ω
A
1.00 × 10 −4 m 2
and the current in the loop is I =
ΔV
0.100 V
=
= 73.5 A.
R 1.36 × 10 −3 Ω
The magnetic field exerts torque t = NBIAsinq on the loop, and this is a maximum when sinq = 1.
The wire forms a square loop with each side 2.00 m long. Thus,
t m ax = NBIA = (1)( 0.400 T )( 73.5 A )( 2.00 m ) = 118 N ⋅ m
2
19.31
(a)
Let q be the angle the plane of the loop makes
with the horizontal as shown in the sketch at the
right. Then, the angle it makes with the vertical is
f = 90.0° −q . The number of turns on the loop is
N=
L
4.00 m
=
= 10.0
circumference 4 ( 0.100 m )
The torque about the z-axis due to gravity is
⎛s
⎞
t g = mg ⎜ cosq ⎟ , where s = 0.100 m is the length
⎝2
⎠
of one side of the loop. This torque tends to rotate
the loop clockwise. The torque due to the magnetic
force tends to rotate the loop counterclockwise about
continued on next page
68719_19_ch19_p120-147.indd 132
1/7/11 2:43:57 PM
Magnetism
133
the z-axis and has magnitude t m = NBIAsinq . At equilibrium, t m = t g, or
NBI ( s 2 ) sinq = mg ( s cosq ) 2. This reduces to
tanq =
( 0.100 kg) (9.80 m s2 )
mg
=
= 14.4
2NBIs 2 (10.0 )( 0.010 0 T )( 3.40 A )( 0.100 m )
Since tanq = tan ( 90.0° − f ) = cotf , the angle the loop makes with the vertical at
equilibrium is f = cot −1 (14.4 ) = 3.97° .
(b)
At equilibrium,
t m = NBI ( s 2 ) sinq
= (10.0 )( 0.010 0 T )( 3.40 A )( 0.100 m ) sin ( 90.0° − 3.97° ) = 3.39 × 10 −3 N ⋅ m
2
19.32
(a)
The current in segment ab is in the + y-direction. Thus,
by right-hand rule 1, the magnetic force on it is in the
+x-direction . This force is parallel to the x-axis and
therefore has zero torque about that axis.
(b)
The current in segment cd is in the − y-direction, and
the right-hand rule 1 gives the direction of the magnetic
force as the −x-direction . This force is parallel to the
x-axis and can exert no torque about that axis.
(c)
No . These two forces are equal in magnitude and opposite in directions, so their sum is
zero. Further, each force has zero torque about the axis the loop is hinged on. Since the two
forces cancel each other and are both parallel to the rotation axis, they can have no effect on
the motion of the loop.
(d)
The magnetic force is perpendicular to both the direction of the current in bc
(the x-axis) and the magnetic field. As given by right-hand rule 1, this places it
in the yz-plane at 130° counterclockwise from the +y-axis . The force acting on
segment bc tends to rotate it counterclockwise about the x-axis, so the torque is in
the +x direction .
(e)
The loop tends to rotate counterclockwise about the x-axis .
(f )
m = IAN = ( 0.900 A )[( 0.500 m )( 0.300 m )](1) = 0.135 A ⋅ m 2
(g)
19.33
The magnetic moment vector is perpendicular to the plane of the loop (the xy-plane) and
is therefore parallel to the z-axis. Because the current flows clockwise around the loop, the
magnetic moment vector is directed downward, in the negative z-direction. This means that
the angle between it and the direction of the magnetic field is q = 90.0° + 40.0° = 130° .
(h)
t = mB sinq = ( 0.135 A ⋅ m 2 )(1.50 T ) sin (130° ) = 0.155 N ⋅ m
(a)
From KE = 12 me v 2, the speed of the electron is
v=
2 ( KE )
=
me
2 ( 3.30 × 10 −19 J )
9.11× 10 −31 kg
= 8.51× 10 5 m
continued on next page
68719_19_ch19_p120-147.indd 133
1/7/11 2:43:59 PM
134
Chapter 19
(b)
The magnetic force acting on the electron must provide the necessary centripetal acceleration.
Thus, me v 2 r = qvB sinq , which gives
r=
(9.11 × 10−31 kg) (8.51× 105 m s)
me v
=
qB sinq (1.60 × 10 −19 C )( 0.235 T ) sin 90.0°
= 2.06 × 10 −5 m = 20.6 × 10 −6 m = 20.6 mm
19.34
19.35
(a)
F = qvB sinq = (1.60 × 10 −19 C ) ( 5.02 × 10 6 m s )( 0.180 T ) sin ( 60.0° ) = 1.25 × 10 −13 N
(b)
a=
F 1.25 × 10 −13 N
=
= 7.49 × 1013 m s 2
m 1.67 × 10 −27 kg
For the particle to pass through with no deflection, the net force acting on it must be zero. Thus,
the magnetic force and the electric force must be in opposite directions and have equal magnitudes. This gives
Fm = Fe , or qvB = qE, which reduces to v = E B
19.36
The speed of the particles emerging from the velocity selector is v = E B (see Problem 35). In the
deflection chamber, the magnetic force supplies the centripetal acceleration, so
qvB =
mv 2
r
r=
or
mv m ( E B ) mE
=
= 2
qB
qB
qB
Using the given data, the radius of the path is found to be
( 2.18 × 10
(1.60 × 10
kg ) ( 950 V m )
−26
r=
19.37
−19
C )( 0.930 T )
2
= 1.50 × 10 −4 m = 0.150 mm
1
From conservation of energy, ( KE + PE ) f = ( KE + PE )i, we find that mv 2 + qV f = 0 + qVi , or the
2
speed of the particle is
(
2 q Vi − V f
v=
m
)=
2 q ( ΔV )
=
m
2 (1.60 × 10 −19 C )( 250 V )
2.50 × 10 −26 kg
= 5.66 × 10 4 m s
The magnetic force supplies the centripetal acceleration giving qvB =
or
19.38
r=
−26
4
mv ( 2.50 × 10 kg ) ( 5.66 × 10 m s )
=
= 1.77 × 10 −2 m = 1.77 cm
−19
qB
1.60
×
10
C
0.500
T
(
)
(
)
Since the centripetal acceleration is furnished by the magnetic force acting on the ions,
mv 2
mv
qvB =
, or the radius of the path is r =
. Thus, the distance between the impact points
qB
r
(that is, the difference in the diameters of the paths followed by the U 238 and the U 235 isotopes) is
Δd = 2 ( r238 − r235 ) =
=
or
68719_19_ch19_p120-147.indd 134
mv 2
r
2v
( m238 − m235 )
qB
2 ( 3.00 × 10 5 m s )
(1.60 × 10
−19
kg ⎞ ⎤
⎡
( 238 u − 235 u ) ⎛⎜⎝ 1.66 × 10 −27
⎟
⎢
u ⎠ ⎥⎦
C )( 0.600 T ) ⎣
Δd = 3.11× 10 −2 m = 3.11 cm
1/7/11 2:44:02 PM
Magnetism
19.39
135
In the perfectly elastic, head-on collision between the a -particle and the initially stationary proton, conservation of momentum requires that m p v p + ma va = ma v0 while conservation of kinetic
energy also requires that v0 − 0 = − va − v p or v p = va + v0 . Using the fact that ma = 4m p and
combining these equations gives
(
(
)
(
)
)
m p ( va + v0 ) + 4m p va = 4m p v0
and
Thus,
va = 3v0 5
or
v p = ( 3v0 5) + v0 = 8v0 5
va =
3
3⎛ 5 ⎞ 3
v0 = ⎜ v p ⎟ = v p
5
5⎝8 ⎠ 8
After the collision, each particle follows a circular path in the horizontal plane with the magnetic
force supplying the centripetal acceleration. If the radius of the proton’s trajectory is R, and that
of the alpha particle is r, we have
qp vp B = mp
19.40
R
va2
r
or
R=
or
r=
mp vp
qp B
=
mp vp
eB
(
)(
)
4m p 3v p 8
ma va
3
3 ⎛ mp vp ⎞
=
R
=
= ⎜
⎟
4
qa B
4 ⎝ eB ⎠
( 2e ) B
and
qa va B = ma
(a)
A charged particle follows a circular path when it moves perpendicular to the magnetic field.
The magnetic force acting on the particle provides the required centripetal acceleration.
Therefore, F = qvB sin 90° = m v 2 R. Since the kinetic energy is KE = mv 2 2, we rewrite the
force as F = qvB sin 90° = 2 ( KE ) R, and solving for the speed v gives v = 2 ( KE ) qBR .
(b)
From KE = mv 2 2, the mass of the particle is
m=
19.41
v 2p
(a)
2 ( KE )
2 ( KE )
q2 B2 R2
=
=
2
2
v
2 ( KE )
4 ( KE ) q 2 B 2 R 2
Within the velocity selector, the electric and magnetic fields exert forces in opposite
directions on charged particles passing through. For particles having a particular speed,
these forces have equal magnitudes, and the particles pass through without deflection. The
selected speed is found from Fe = qE = qvB = Fm, giving v = E B. In the deflection chamber,
the selected particles follow a circular path having a diameter of d = 2r = 2mv qB. Thus,
the mass to charge ratio for these particles is
Bd
B 2 d ( 0.093 1 T ) ( 0.396 m )
m Bd
=
=
=
= 2.08 × 10 −7 kg C
=
q 2v 2 ( E B ) 2E
2 (8 250 V m )
2
(b)
If the particle is doubly ionized (i.e., two electrons have been removed from the neutral
atom), then q = 2e, and the mass of the ion is
⎛ m⎞
m = ( 2e ) ⎜ ⎟ = 2 (1.60 × 10 −19 C ) ( 2.08 × 10 −7 kg C ) = 6.66 × 10 −26 kg
⎝ q⎠
(c)
Assuming this is an element, the mass of the ion should be approximately equal to the
atomic weight multiplied by the atomic mass unit (see Table C.5 in Appendix C of the
textbook). This would give the atomic weight as
At. wt. ≈
68719_19_ch19_p120-147.indd 135
m 6.66 × 10 −26 kg
=
= 40.1 , suggesting that the element is calcium .
1 u 1.66 × 10 −27 kg
1/7/11 2:44:05 PM
136
19.42
Chapter 19
Consider a particle of mass m and charge q accelerated from rest through a potential difference
ΔV = Vi − V f , Applying conservation of energy gives
KE f + PE f = 0 + PEi
KE f =
or
1 2
mv = PEi − PE f = qVi − qV f = q ( ΔV )
2
or the speed given the particle is v = 2q ( ΔV ) m.
If the particle now enters a magnetic field of strength B, moving perpendicular to the direction of
the field, it will follow a circular path of radius
r=
mv m
=
qB qB
2q ( ΔV )
m
which reduces to
r=
2m ( ΔV )
qB 2
[1]
For a proton (mass m p and charge q = e), Equation [1] gives
rp =
(a)
2m p ( ΔV )
eB 2
For the deuteron (mass md = 2m p and charge qd = e), Equation [1] gives
rd =
(b)
2ma ( ΔV )
=
qa B 2
(
)
2 4m p ( ΔV )
( 2e ) B
2
⎛ 2m p ( ΔV ) ⎞
= 2⎜
⎟=
eB 2
⎝
⎠
2 ⋅rp
−7
4
m 0 I ( 4p × 10 T ⋅ m A ) (1.00 × 10 A )
=
= 2.00 × 10 −5 T = 20.0 mT
2p r
2p (100 m )
toward the left
(b)
out of page
(c)
lower left to upper right
The magnetic field at distance r from a long conducting wire is B = m 0 I 2p r. Thus, if
B = 1.0 × 10 −15 T at r = 4.0 cm, the current must be
I=
19.46
2 ⋅rp
Imagine grasping the conductor with the right hand so the fingers curl around the conductor in the
direction of the magnetic field. The thumb then points along the conductor in the direction of the
current. The results are
(a)
19.45
eB 2
⎛ 2m p ( ΔV ) ⎞
= 2⎜
⎟=
eB 2
⎝
⎠
Treat the lightning bolt as a long, straight conductor. Then, the magnetic field is
B=
19.44
)
For the alpha particle (mass ma = 4m p and charge qa = 2e), we find
ra =
19.43
(
2 2m p ( ΔV )
2md ( ΔV )
=
qd B 2
−15
2p rB 2p ( 0.040 m )(1.0 × 10 T )
=
= 2.0 × 10 −10 A
m0
4p × 10 −7 T ⋅ m A
Model the tornado as a long, straight, vertical conductor and imagine grasping it with
the right hand so the fingers point northward on the western side of the tornado (that
is, at the observatory’s location.) The thumb is directed downward, meaning that the
conventional current is downward or negative charge flows upward .
continued on next page
68719_19_ch19_p120-147.indd 136
1/7/11 2:44:08 PM
Magnetism
137
The magnitude of the current is found from B = m 0 I 2p r as
I=
19.47
3
−8
2p rB 2p ( 9.00 × 10 m ) (1.50 × 10 T )
=
= 675 A
m0
4p × 10 −7 T ⋅ m A
From B = m 0 I 2p r, the required distance is
−7
m 0 I ( 4p × 10 T ⋅ m A )( 20 A )
r=
=
= 2.4 × 10 −3 m = 2.4 mm
−3
2p B
2p (1.7 × 10 T )
19.48
Assume that the wire on the right is wire 1 and that on the left is wire 2. Also, choose the positive
direction for the magnetic field to be out of the page and negative into the page.
(a)
At the point half way between the two wires,
⎡m I
m
m I ⎤
Bnet = −B1 − B2 = − ⎢ 0 1 + 0 2 ⎥ = − 0
2p
r
2p
r
2p
1
2 ⎦
⎣
=−
or
(b)
At point P1, Bnet = +B1 − B2 =
m0
2p
⎡ I1 I 2 ⎤ m 0 ⎡ I
I ⎤ m0 ⎡ I ⎤
⎢+ − ⎥ =
⎢⎣ + d − 2d ⎥⎦ = 2p ⎢⎣ + 2d ⎥⎦
r
r
2p
2 ⎦
⎣ 1
4p × 10 −7 T ⋅ m A ⎡ 5.00 A ⎤
−6
⎥ = 5.00 × 10 T = 5.00 mT out of page
⎢
2p
⎣ 2 ( 0.100 m ) ⎦
At point P2, Bnet = −B1 + B2 =
Bnet =
19.49
T ⋅ m A ) ⎡ 4 ( 5.00 A ) ⎤
−5
⎢ 0.100 m ⎥ = − 4.00 × 10 T
2p
⎦
⎣
−7
Bnet = 40.0 mT into the page
Bnet =
(c)
( 4p × 10
⎡ I
m 0 ⎡ 4I ⎤
I ⎤
⎢ d 2 + d 2 ⎥ = − 2p ⎢ d ⎥
⎣ ⎦
⎣
⎦
m0
2p
⎡ I1 I 2 ⎤ m 0 ⎡ I
I ⎤ m0 ⎡ I ⎤
−
+
⎢− + ⎥ =
⎢
⎥=
⎢ ⎥
⎣ r1 r2 ⎦ 2p ⎣ 3d 2d ⎦ 2p ⎣ 6d ⎦
4p ×10 −7 T ⋅ m A ⎡ 5.00 A ⎤
−6
⎢
⎥ = 1.67 ×10 T = 1.67 mT out of page
2p
⎣ 6 ( 0.100 m ) ⎦
The distance from each wire to point P is given by
r = 12 s 2 + s 2 = s 2, where s = 0.200 m.
At point P, the magnitude of the magnetic field
produced by each of the wires is
−7
m 0 I ( 4p × 10 T ⋅ m A )( 5.00 A ) 2
B=
=
= 7.07 mT
2p r
2p ( 0.200 m )
Carrying currents into the page, the field A produces at P
is directed to the left and down at –135°, while B creates
a field to the right and down at – 45°. Carrying currents
toward you, C produces a field downward and to the
right at – 45°, while D’s contribution is down and to the
left at –135°. The horizontal components of these equal magnitude contributions cancel in pairs,
while the vertical components all add. The total field is then
Bnet = 4 ( 7.07 mT ) sin 45.0° = 20.0 mT toward the bottom of the page
68719_19_ch19_p120-147.indd 137
1/7/11 2:44:11 PM
138
19.50
Chapter 19
Wire 1 has current I1 = 3.00 A and wire 2 has I 2 = 5.00 A, with both currents directed out of the
page. Choose the line running from wire 1 to wire 2 as the positive x-direction.
(a)
At the point midway between the wires, the field due to
each wire is parallel to the y-axis, and the net field is
Bnet = +B1 y − B2 y = m 0 ( I1 − I 2 ) ( 2p d 2 )
(b)
( 4p × 10
−7
T ⋅ m A)
( 3.00 A − 5.00 A ) = − 4.00 × 10 −6 T
Thus,
Bnet =
or
Bnet = 4.00 mT toward the bottom of the page
2p ( 0.100 m )
At point P, B1 is directed at q1 = +135°.
The magnitude of B1 is
4p × 10 −7 T ⋅ m A )( 3.00 A )
(
m 0 I1
m 0 I1
B1 =
=
=
= 2.12 mT
2p r1 2p d 2
2p 0.200 2 m
(
)
(
)
The contribution from wire 2 is in the –x-direction and
has magnitude
B2 =
−7
m 0 I 2 m 0 I 2 ( 4p × 10 T ⋅ m A )( 5.00 A )
=
= 5.00 mT
=
2p r2 2p d
2p ( 0.200 m )
Therefore, the components of the net field at point P are
Bx = B1 cos135° + B2 cos180°
= ( 2.12 mT ) cos135° + ( 5.00 mT ) cos180° = −6.50 mT
and
By = B1 sin135° + B2 sin180° = ( 2.12 mT ) sin135° + 0 = +1.50 mT
Thus, Bnet = Bx2 + By2 = 6.67 mT at
⎛ Bx ⎞
⎛ 6.50 mT ⎞
q = tan −1 ⎜
= tan −1 ⎜
= 77.0°
⎟
⎝ 1.50 mT ⎟⎠
⎝ By ⎠
or
19.51
Bnet = 6.67 mT upward at 77.0° to the left of vertical
Call the wire along the x-axis wire 1 and the other wire 2. Also, choose the positive direction for
the magnetic fields at point P to be out of the page.
At point P, Bnet = +B1 − B2 =
or
Bnet
( 4p × 10
=
−7
2p
m 0 I1 m 0 I 2 m 0 ⎛ I1 I 2 ⎞
−
=
−
2p r1 2p r2 2p ⎜⎝ r1 r2 ⎟⎠
T ⋅ m A ) ⎛ 7.00 A 6.00 A ⎞
−7
−
⎜⎝
⎟ = +1.67 × 10 T
3.00 m 4.00 m ⎠
Bnet = 0.167 mT out of the page
68719_19_ch19_p120-147.indd 138
1/7/11 2:44:13 PM
Magnetism
19.52
139
(a)
Imagine the horizontal xy-plane being perpendicular to the page, with the positive x-axis
coming out of the page toward you and the positive y-axis toward the right edge of the page.
Then, the vertically upward positive z-axis is directed toward the top of the page. With the
current in the wire flowing in the positive x-direction, the right-hand rule 2 gives the direction of the magnetic field above the wire as being toward the left, or in the −y-direction .
(b)
With the positively charged proton moving in the –x-direction (into the page), right-hand
rule 1 gives the direction of the magnetic force on the proton as being directed toward the
top of the page or upward, in the positive z-direction .
(c)
Since the proton moves with constant velocity, a zero net force acts on it. Thus, the
magnitude of the magnetic force must equal that of the gravitational force .
(d)
ΣFz = maz = 0 ⇒ Fm = Fg or qvB = mg, where B = m 0 I 2p d . This gives
qvm 0 I 2p d = mg, or the distance the proton is above the wire must be d = qvm 0 I 2p mg .
(e)
d=
−19
4
−7
−6
qvm 0 I (1.60 × 10 C ) ( 2.30 × 10 m s ) ( 4p × 10 T ⋅ m A ) (1.20 × 10 A )
=
2p mg
2p (1.67 × 10 −27 kg ) ( 9.80 m s 2 )
d = 5.40 × 10 −2 m = 5.40 cm
19.53
(a)
From B = m 0 I 2p r, observe that the field is inversely proportional to the distance from the
conductor. Thus, the field will have one-tenth its original value if the distance is increased
by a factor of 10. The required distance is then r ′ = 10r = 10 ( 0.400 m ) = 4.00 m .
(b)
A point in the plane of the conductors and 40.0 cm from the center of the cord is located
39.85 cm from the nearer wire and 40.15 cm from the far wire. Since the currents are in
opposite directions, so are their contributions to the net field. Therefore, Bnet = B1 − B2, or
Bnet =
−7
⎞
m 0 I ⎛ 1 1 ⎞ ( 4p × 10 T ⋅ m A ) ( 2.00 A ) ⎛
1
1
−
−
=
⎜
⎜
⎟
2p ⎝ r1 r2 ⎠
2p
⎝ 0.398 5 m 0.401 5 m ⎟⎠
= 7.50 × 10 −9 T = 7.50 nT
(c)
Call r the distance from cord center to field point P
and 2d = 3.00 mm the distance between centers of
the conductors.
Bnet =
m0 I ⎛ 1
1 ⎞ m 0 I ⎛ 2d ⎞
−
⎜⎝
⎟=
⎜
⎟
2p r − d r + d ⎠ 2p ⎝ r 2 − d 2 ⎠
7.50 × 10 −10 T =
so
( 4p × 10
−7
T ⋅ m A ) ( 2.00 A ) ⎛ 3.00 × 10 −3 m ⎞
⎜⎝ r 2 − 2.25 × 10 −6 m 2 ⎟⎠
2p
r = 1.26 m
The field of the two-conductor cord is weak to start with and falls off rapidly with distance.
(d)
68719_19_ch19_p120-147.indd 139
The cable creates zero field at exterior points, since a loop in Ampère’s law encloses zero
total current.
1/7/11 2:44:15 PM
140
19.54
Chapter 19
(a)
Point P is equidistant from the two wires which carry
identical currents. Thus, the contributions of the two
wires, B upper and Blower, to the magnetic field at P will
have equal magnitudes. The horizontal components
of these contributions will cancel, while the vertical
components add. The resultant field will be vertical
in the + y-direction .
(b)
The distance of each wire from point P is
r = x 2 + d 2 , and the cosine of the angle that
B upper and Blower make with the vertical is cosq = x r .
The magnitude of either B upper or Blower is
Bwire = m 0 I 2p r, and the vertical components of either of these contributions have values of
⎛ m0 I ⎞
m 0 Ix
( B ) = ( B ) cosq = ⎜⎝ 2p r ⎟⎠ rx = 2p r
wire
y
wire
2
The magnitude of the resultant field at point P is then
BP = 2 ( Bwire ) y =
19.55
m 0 Ix
m0 I x
=
2
pr
p (x2 + d 2 )
(c)
The point midway between the two wires is the origin (0, 0). From the above result for part (b),
the resultant field at this midpoint is BP x = 0 = 0 . This is as expected, because right-hand rule 2
shows that at the midpoint the field due to the upper wire is toward the right, while that due to
the lower wire is toward the left. Thus, the two fields cancel, yielding a zero resultant field.
(a)
The magnetic force per unit length on each of two parallel wires separated by the distance d
and carrying currents I1 and I 2 has the magnitude
F
=
m 0 I1 I 2
2p d
In this case, we have
F
19.56
=
( 4p × 10
−7
T ⋅ m A )(1.25 A )( 3.50 A )
2p ( 2.50 × 10 −2 m )
= 3.50 × 10 −5 N m
(b)
The magnetic forces two parallel wires exert on each other are attractive if their currents are
in the same direction and repulsive if the currents flow in opposite directions. In this case,
the currents in the two wires are in opposite directions, so the forces are repulsive .
(a)
The force per unit length that parallel conductors exert on each other is F = m 0 I1 I 2 2p d.
Thus, if F = 2.0 × 10 −4 N m, I1 = 5.0 A, and d = 4.0 cm, the current in the second wire
must be
I2 =
(b)
2p ( 4.0 × 10 −2 m )
2p d ⎛ F ⎞
=
( 2.0 × 10 −4 N m ) = 8.0 A
⎜ ⎟
m 0 I1 ⎝ ⎠ ( 4p × 10 −7 T ⋅ m A )( 5.0 A )
Since parallel conductors carrying currents in the same direction attract each other (see
Section 19.8 in the textbook), the currents in these conductors which repel each other must
be in opposite directions.
continued on next page
68719_19_ch19_p120-147.indd 140
1/7/11 2:44:18 PM
Magnetism
(c)
19.57
141
The result of reversing the direction of either of the currents would be that the
force of interaction would change from a force of repulsion to an attractive force . The
expression for the force per unit length, F = m 0 I1 I 2 2p d, shows that doubling either of
the currents would double the magnitude of the force of interaction .
In order for the system to be in equilibrium, the repulsive magnetic force per unit length on the
top wire must equal the weight per unit length of this wire.
Thus,
F
= m 0 I1 I 2 2p d = 0.080 N m, and the distance between the wires will be
4p × 10 −7 T ⋅ m A )( 60.0 A )( 30.0 A )
(
m 0 I1 I 2
d=
=
2p ( 0.080 N m )
2p ( 0.080 N m )
= 4.5 × 10 −3 m = 4.5 mm
19.58
The magnetic forces exerted on the top and bottom segments of the rectangular loop are equal
in magnitude and opposite in direction. Thus, these forces cancel, and we only need consider the
sum of the forces exerted on the right and left sides of the loop. Choosing to the left (toward the
long, straight wire) as the positive direction, the sum of these two forces is
Fnet = +
or
Fnet =
m 0 I1 I 2
m II
m I I ⎛1
1 ⎞
− 0 1 2 = 0 1 2 ⎜ −
⎟
2p c
2p ( c + a )
2p ⎝ c c + a ⎠
( 4p × 10
−7
T ⋅ m A )( 5.00 A )(10.0 A )( 0.450 m ) ⎛
1
1
⎞
−
⎜⎝
⎟
2p
0.100 m 0.250 m ⎠
= + 2.70 × 10 −5 N = 2.70 × 10 −5 N to the left
19.59
The magnetic field inside a solenoid which carries current I is given by B = m 0 nI, where n = N
is the number of turns of wire per unit length. Thus, the current in the windings of this solenoid
must be
1.00 × 10 −4 T )( 0.400 m )
(
B
B
I=
=
= 3.18 × 10 −2 A = 31.8 mA
=
m 0 n m 0 N ( 4p × 10 −7 T ⋅ m A ) (10 3 )
19.60
The magnetic field inside of a solenoid is B = m 0 nI = m 0 ( N L ) I. Thus, the number of turns on
this solenoid must be
N=
19.61
(a)
BL
( 9.0 T )( 0.50 m )
=
= 4.8 × 10 4 turns
m 0 I ( 4p × 10 −7 T ⋅ m A )( 75 A )
From R = rL A, the required length of wire to be used is
2
−3
⎡
⎤
R ⋅ A ( 5.00 Ω ) ⎣p ( 0.500 × 10 m ) 4 ⎦
L=
=
= 58 m
r
1.7 × 10 −8 Ω ⋅ m
The total number of turns on the solenoid (that is, the number of times this length of wire
will go around a 1.00-cm radius cylinder) is
N=
L
58 m
=
= 9.2 × 10 2 = 920
2p r 2p (1.00 × 10 −2 m )
continued on next page
68719_19_ch19_p120-147.indd 141
1/7/11 2:44:20 PM
142
Chapter 19
(b)
From B = m 0 nI, the number of turns per unit length on the solenoid is
n=
B
4.00 × 10 −2 T
=
= 7.96 × 10 3 turns m
m 0 I ( 4p × 10 −7 T ⋅ m A )( 4.00 A )
Thus, the required length of the solenoid is
=
19.62
N
9.2 ×10 2 turns
= 0.12 m = 12 cm
=
n 7.96 ×10 3 turns m
The wire used to wind the solenoid has diameter d wire = 0.100 cm = 1.00 × 10 −3 m. Thus, when
wound in a single layer with adjacent turns touching each other, the number of turns per unit
length on the solenoid is n = 1.00 × 10 3 per meter. The total number of turns on the solenoid,
N = n , will then be
N = (1.00 × 10 3 m )( 0.750 m ) = 7.50 × 10 2
If the magnetic field inside the solenoid is to be B = 8.00 mT, the required current in the
windings is
I=
B
8.00 × 10 −3 T
=
= 6.37 A
m 0 n ( 4p × 10 −7 T ⋅ m A ) (1.00 × 10 3 m )
The length of wire required for the solenoid is L = N ⋅ ( solenoid circumference ), or
L = N (p dsolenoid ) = ( 7.50 × 10 2 ) p ( 0.100 m ) = 75.0p meters
and its resistance will be
R=
4 (1.70 × 10 −8 Ω ⋅ m )( 75.0p m )
rC u L
rC u L
=
=
= 5.10 Ω
2
2
A
p ( d wire
4)
p (1.00 × 10 −3 m )
The power that must be delivered to the solenoid is then
P = I 2 R = ( 6.37 A ) ( 5.10 Ω ) = 207 W
2
19.63
(a)
The magnetic force supplies the centripetal acceleration, so qvB = mv 2 r . The magnetic
field inside the solenoid is then found to be
B=
(b)
−31
4
mv ( 9.11× 10 kg ) (1.0 × 10 m s )
=
= 2.8 × 10 −6 T = 2.8 mT
qr
(1.60 × 10 −19 C) ( 2.0 × 10−2 m )
From B = m 0 nI, the current is the solenoid is found to be
I=
B
2.8 × 10 −6 T
=
−7
m 0 n ( 4p × 10 T ⋅ m A ) ⎡⎣( 25 turns cm ) (100 cm 1 m ) ⎤⎦
= 8.9 × 10 −4 A = 0.89 mA
68719_19_ch19_p120-147.indd 142
1/7/11 2:44:22 PM
Magnetism
19.64
(a)
143
When switch S is closed, a total current NI (current I in a total
of N conductors) flows toward the right through the lower side
of the coil. This results in a downward force of magnitude
Fm = B ( NI ) w being exerted on the coil by the magnetic field,
with the requirement that the balance exert a upward force
F ′ = mg on the coil to bring the system back into balance.
In order for the magnetic force to be downward, the right-hand rule number 1 shows that
the magnetic field must be directed out of the page toward the reader. For the system to be
restored to balance, it is necessary that
Fm = F ′
19.65
or
B ( NI ) w = mg, giving B = mg NIw
(b)
The magnetic field exerts forces of equal magnitudes and opposite directions on the two
sides of the coil. These forces cancel each other and do not affect the balance of the coil.
Hence the dimension of the sides is not needed.
(c)
B=
(a)
The magnetic field at the center of a circular current loop of radius R and carrying current I
is B = m 0 I 2R. The direction of the field at this center is given by right-hand rule number 2.
Taking out of the page (toward the reader) as positive, the net magnetic field at the common
center of these coplanar loops has magnitude
−3
2
mg ( 20.0 × 10 kg ) ( 9.80 m s )
=
= 0.26 T
NIw ( 50 )( 0.30 A ) ( 5.0 × 10 −2 m )
Bnet = B2 − B1 =
−7
m 0 I 2 m 0 I1 ( 4p × 10 T ⋅ m A )
3.00 A
5.00 A
−
=
−
−2
2r2
2r1
2
9.00 × 10 m 12.0 × 10 −2 m
= −5.24 × 10 −6 T = 5.24 mT
(b)
Since we chose out of the page as the positive direction, and now find that Bnet < 0, we
conclude the net magnetic field at the center is into the page .
(c)
To have Bnet = 0, it is necessary that I 2 r2 = I1 r1, or
⎛I ⎞
⎛ 3.00 A ⎞
r2 = ⎜ 2 ⎟ r1 = ⎜
(12.0 cm ) = 7.20 cm
⎝ 5.00 A ⎟⎠
⎝ I1 ⎠
19.66
The angular momentum of a point mass moving in a circular path is
⎛ v⎞
L = Iw = ( mr 2 ) ⎜ ⎟ = mvr
⎝r⎠
where m is the mass of the particle, v is its speed, and r is the radius of its path.
(a)
The magnetic force experienced by the moving electron supplies the needed centripetal
acceleration, so
⎛ v2 ⎞
m ⎜ ⎟ = qvB sin 90.0°
⎝ r ⎠
or
mv = qBr
Thus, L = mvr = ( qBr ) r = qBr 2, and the radius of the path must be
continued on next page
68719_19_ch19_p120-147.indd 143
1/7/11 2:44:25 PM
144
Chapter 19
(b)
The speed of the electron may now be found from L = mvr as
v=
19.67
4.00 × 10 −25 kg ⋅ m 2 s
= 5.00 × 10 −2 m = 5.00 cm
−19
−3
C
1.00
×
10
T
1.60
×
10
(
)(
)
L
=
qB
r=
L
4.00 × 10 −25 kg ⋅ m 2 s
=
= 8.78 × 10 6 m s
mr ( 9.11× 10 −31 kg ) ( 5.00 × 10 −2 m )
Assume wire 1 is along the x-axis and wire 2 is along the y-axis.
(a)
Choosing out of the page as the positive field direction, the field at point P is
B = B1 − B2 =
−7
m 0 ⎛ I1 I 2 ⎞ ( 4p × 10 T ⋅ m A ) ⎛ 5.00 A
3.00 A ⎞
−
−
=
⎜⎝
⎟
⎜
⎟
2p ⎝ r1 r2 ⎠
2p
0.400 m 0.300 m ⎠
= 5.00 × 10 −7 T = 0.500 mT out of the page
(b)
At 30.0 cm above the intersection of the wires, the field
components are as shown at the right, where
By = −B1 = −
m 0 I1
2p r
( 4p × 10
=−
−7
T ⋅ m A )( 5.00 A )
2p ( 0.300 m )
and
Bx = B2 =
= −3.33 × 10 −6 T
−7
m 0 I 2 ( 4p × 10 T ⋅ m A )( 3.00 A )
=
= 2.00 × 10 −6 T
2p r
2p ( 0.300 m )
With Bz = 0, the resultant field is parallel to the xy-plane and
B = Bx2 + By2 = 3.88 × 10 −6 T
or
19.68
⎛ By ⎞
q = tan −1 ⎜ ⎟ = −59.0°
⎝ Bx ⎠
B = 3.88 mT parallel to the xy-plane and 59.0° clockwise from the +x-direction
For the rail to move at constant velocity, the net force acting on, it must be zero. Thus, the magnitude of the magnetic force must equal that of the friction force, giving BIL = m k ( mg ), or
B=
19.69
at
(a)
2
m k ( mg ) ( 0.100 )( 0.200 kg ) ( 9.80 m s )
=
= 3.92 × 10 −2 T
IL
(10.0 A )( 0.500 m )
Since the magnetic field is directed from N to S (that is, from left to right within the artery),
positive ions with velocity in the direction of the blood flow experience a magnetic deflection toward electrode A. Negative ions will experience a force deflecting them toward
electrode B. This separation of charges creates an electric field directed from A toward B.
At equilibrium, the electric force caused by this field must balance the magnetic force, so
qvB = qE = q ( ΔV d )
continued on next page
68719_19_ch19_p120-147.indd 144
1/7/11 2:44:27 PM
Magnetism
or
19.70
160 × 10 −6 V
ΔV
=
= 1.33 m s
Bd ( 0.040 0 T ) ( 3.00 × 10 −3 m )
v=
(b)
The magnetic field is directed from N to S. If the charge carriers are negative moving in the
direction of v, the magnetic force is directed toward point B. Negative charges build up at
point B, making the potential at A higher than that at B. If the charge carriers are positive
moving in the direction of v, the magnetic force is directed toward A, so positive charges
build up at A. This also makes the potential at A higher than that at B. Therefore the sign of
the potential difference does not depend on the charge of the ions .
(a)
The magnetic force acting on the wire is directed upward and of magnitude
Fm = BIL sin 90° = BI.
ay =
Thus,
ay =
(b)
ΣFy
m
=
Fm − mg BIL − mg
BI
=
=
− g, or
m
m
(m L )
( 4.0 × 10
−3
5.0 × 10
T ) ( 2.0 A )
−4
kg m
− 9.80 m s 2 = 6.2 m s 2
Using Δy = v0 y t + 12 a y t 2 , with v0 y = 0, gives
t=
19.71
145
2 ( Δy )
=
ay
2 ( 0.50 m )
= 0.40 s
6.2 m s 2
Label the wires 1, 2, and 3, as shown in Figure 1.
Also, let B1 , B2 , and B3 , respectively, represent the
magnitudes of the fields produced by the
currents in those wires, and observe that
q = 45°.
(
)
At point A, B1 = B2 = m 0 I 2p a 2 , or
B1 = B2 =
( 4p × 10
−7
T ⋅ m A ) ( 2.0 A )
2p ( 0.010 m ) 2
= 28 mT
FIGURE 1
( 4p × 10 T ⋅ m A ) ( 2.0 A ) = 13 mT
m0 I
=
2p ( 3a )
2p ( 0.030 m )
−7
and
B3 =
These field contributions are oriented as shown in Figure 2.
Observe that the horizontal components of B1 and B2 cancel
while their vertical components add to B3. The resultant
field at point A is then
BA = ( B1 + B2 ) cos 45° + B3 = 53 mT, or
FIGURE 2
B A = 53 mT directed toward the bottom of the page
At point B, B1 = B2 =
−7
m 0 I ( 4p × 10 T ⋅ m A ) ( 2.0 A )
=
= 40 mT
2p a
2p ( 0.010 m )
m0 I
= 20 mT. These contributions are oriented as
2p ( 2a )
shown in Figure 3. Thus, the resultant field at B is
and B3 =
FIGURE 3
continued on next page
68719_19_ch19_p120-147.indd 145
1/7/11 2:44:29 PM
146
Chapter 19
B B = B3 = 20 mT directed toward the bottom of the page
(
)
At point C, B1 = B2 = m 0 I 2p a 2 , while
B3 = m 0 I 2p a. These contributions are oriented
as shown in Figure 4. Observe that the horizontal
components of B1 and B2 cancel, while their vertical
components add to oppose B3 . The magnitude of the
resultant field at C is
FIGURE 4
m I ⎛ 2sin 45° ⎞
−1⎟ = 0
BC = ( B1 + B2 ) sin 45° − B3 = 0 ⎜
2p a ⎝
⎠
2
19.72
(a)
Since one wire repels the other, the currents must be in opposite directions .
(b)
Consider a free-body diagram of one of the wires as shown at
the right.
ΣFy = 0 ⇒ T cos8.0° = mg
or
T=
mg
cos8.0°
⎛ mg ⎞
ΣFx = 0 ⇒ Fm = T sin8.0° = ⎜
sin8.0°
⎝ cos8.0° ⎟⎠
or
Fm = ( mg ) tan8.0°. Thus,
I=
m0 I 2 L
= ( mg ) tan8.0°, which gives
2p d
d ⎡⎣( m L ) g ⎤⎦ tan8.0°
m 0 2p
Observe that the distance between the two wires is
d = 2 [( 6.0 cm ) sin8.0° ] = 1.7 cm, so
I=
19.73
(1.7 × 10
−2
m ) ( 0.040 kg m ) ( 9.80 m s 2 ) tan8.0°
2.0 × 10 −7 T ⋅ m A
= 68 A
Note: We solve part (b) before part (a) for this problem.
(b)
Since the magnetic force supplies the centripetal acceleration for this particle, qvB = mv 2 r,
or the radius of the path is r = mv qB = p qB , where
p = mv = 2m(KE) = 2 (1.67 × 10 −27 kg ) ( 5.00 × 10 6 eV ) (1.60 × 10 −19 J eV )
= 5.17 × 10 −20 kg ⋅ m s
Consider the circular path shown at the right and
observe that the desired angle is
⎡ (1.00 m ) qB ⎤
⎛ 1.00 m ⎞
a = sin −1 ⎜
= sin −1 ⎢
⎥
⎝ r ⎟⎠
p
⎣
⎦
or
⎡ (1.00 m )(1.60 × 10 −19 C ) ( 0.050 0 T ) ⎤
a = sin −1 ⎢
⎥ = 8.90°
5.17 × 10 −20 kg ⋅ m s
⎢⎣
⎥⎦
continued on next page
68719_19_ch19_p120-147.indd 146
1/7/11 2:44:31 PM
Magnetism
(a)
147
The linear momentum of the particle has constant magnitude p = mv, and
its vertical component as the particle leaves the field is py = − psina , or
py = − ( 5.17 × 10 −20 kg ⋅ m s ) sin (8.90° ) = − 8.00 × 10 −21 kg ⋅ m s
19.74
The force constant of the spring system is found from the elongation produced by the weight
acting alone.
F mg
=
x1
x1
k=
where
x1 = 0.50 cm
The total force stretching the springs when the field is turned on is
ΣFy = Fm + mg = kx total
where
x total = x1 + 0.30 cm = 0.80 cm
Thus, the downward magnetic force acting on the wire is
⎛ mg ⎞
⎛x
⎞
Fm = kx total − mg = ⎜
x total − mg = ⎜ total − 1⎟ mg
⎟
⎝ x1 ⎠
⎝ x1
⎠
⎛ 0.80 cm ⎞
=⎜
− 1 10.0 × 10 −3 kg ) ( 9.80 m s 2 ) = 5.9 × 10 −2 N
⎝ 0.50 cm ⎟⎠ (
Since the magnetic force is given by Fm = BIL sin 90°, the magnetic field is
B=
19.75
(12 Ω )( 5.9 × 10 −2 N )
Fm
Fm
=
= 0.59 T
=
IL ( ΔV R ) L ( 24 V ) ( 5.0 × 10 −2 m )
The magnetic force is very small in comparison to the weight of the ball, so we treat the motion as
that of a freely falling body. Then, as the ball approaches the ground, it has velocity components
with magnitudes of
vx = v0 x = 20.0 m s, and
v y = v02 y + 2a y ( Δy ) = 0 + 2 ( −9.80 m s 2 )( −20.0 m ) = 19.8 m s
The velocity of the ball is perpendicular to the magnetic field and, just before it reaches the
ground, has magnitude v = vx2 + v y2 = 28.1 m s. Thus, the magnitude of the magnetic force is
Fm = qvB sinq
= ( 5.00 × 10 −6 C ) ( 28.1 m s ) ( 0.010 0 T ) sin 90.0° = 1.41× 10 −6 N
19.76
(a)
(b)
(c)
(d)
68719_19_ch19_p120-147.indd 147
B1 =
F21
= B1 I 2 = (1.00 × 10 −5 T )(8.00 A ) = 8.00 × 10 −5 N directed toward wire 1
B2 =
F12
−7
m 0 I1 ( 4p × 10 T ⋅ m A )( 5.00 A )
=
= 1.00 × 10 −5 T
2p d
2p ( 0.100 m )
−7
m 0 I 2 ( 4p × 10 T ⋅ m A )(8.00 A )
=
= 1.60 × 10 −5 T
2p d
2p ( 0.100 m )
= B2 I1 = (1.60 × 10 −5 T )( 5.00 A ) = 8.00 × 10 −5 N directed toward wire 2
1/7/11 2:44:34 PM
20
Induced Voltages and Inductance
QUICK QUIZZES
1.
b, c, a. At each instant, the magnitude of the induced emf is proportional to the magnitude of the
rate of change of the magnetic field (hence, proportional to the absolute value of the slope of the
curve shown on the graph).
2.
Choice (c). Taking downward as the normal direction, the north pole approaching from above
produces an increasing positive flux through the area enclosed by the loop. To oppose this, the
induced current must generate a negative flux through the interior of the loop, or the induced
magnetic field must point upward through the enclosed area of the loop. Imagine gripping the
wire of the loop with your right hand so the fingers curl upward through the area enclosed by the
loop. You will find that your thumb, indicating the direction of the induced current, is directed
counterclockwise around the loop as viewed from above.
3.
Choice (b). If the positive z-direction is chosen as the normal direction, the increasing counterclockwise current in the left-hand loop produces an increasing positive flux through the area
enclosed by this loop. The magnetic field lines due to this current curl around and pass through
the area of the xy-plane outside the left-hand loop in the negative direction. Thus, the right-hand
loop has an increasing negative flux through it. To counteract this effect, the induced current must
produce positive flux, or generate a magnetic field in the positive z-direction, through the area
enclosed by the right-hand loop. Imagine gripping the wire of the right-hand loop with the right
hand so the fingers point in the positive z-direction as they penetrate the area enclosed by this
loop. You should find that the thumb is directed counterclockwise around the loop as viewed from
above the xy-plane.
4.
Choice (a). All charged particles within the metal bar move straight downward with the bar.
According to right-hand rule number 1, positive changes moving downward through a magnetic
field that is directed northward will experience magnetic forces toward the east. This means that
the free electrons (negative charges) within the metal will experience westward forces and will
drift toward the west end of the bar, leaving the east end with a net positive charge.
5.
Choice (b). According to Equation 20.3, when B and v are constant, the emf depends only on the
length of the wire cutting across the magnetic field lines. Thus, you want the long dimension of
the rectangular loop perpendicular to the velocity vector. This means that the short dimension
is parallel to the velocity vector, and (b) is the correct choice. From a more conceptual point of
view, you want the rate of change of area in the magnetic field to be the largest, which you do by
thrusting the long dimension into the field.
6.
Choice (b). When the iron rod is inserted into the solenoid, the inductance of the coil increases.
As a result, more potential difference appears across the coil than before. Consequently, less
potential difference appears across the bulb, and its brightness decreases.
148
68719_20_ch20_p148-170.indd 148
1/7/11 2:45:21 PM
Induced Voltages and Inductance
149
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The flux through a flat coil in a uniform magnetic field is a maximum when the plane of the coil
is perpendicular to the direction of the field. Thus, with the field parallel to the y-axis, the flux is a
maximum when the plane of the coil is parallel to the xz-plane, and (c) is the correct choice.
2.
Choose the positive z-direction to be the reference direction (q = 0°) for the normal to the plane
of the coil. Then, the change in flux through the coil is
(
)
ΔΦ B = B f cosq f − Bi cosq i A = [( 3.0 T ) cos 0° − (1.0 T ) cos180° ] ( 0.50 m ) = 1.0 Wb
2
and the magnitude of the induced emf is
e =N
ΔΦ B
Δt
⎛ 1.0 Wb ⎞
= 5.0 V
= (10 ) ⎜
⎝ 2.0 s ⎟⎠
The correct answer is choice (b).
3.
The magnitude of the voltage drop across the inductor is
eL = L
ΔI
= ( 5.00 H ) ( 2.00 A s ) = 10.0 V
Δt
and (d) is the correct choice.
4.
The angular velocity of the rotating coil is w = 10.0 rev s = 20p rad s and the maximum emf
induced in the coil is
(
)
e max = NBAw = (100 ) ( 0.050 0 T ) 0.100 m 2 ( 20p rad s ) = 31.4 V
showing the correct choice to be (a).
5.
The motional emf induced in a conductor of length C moving at speed v through a magnetic field
of magnitude B is ΔV = B⊥ Cv = ( B sinq ) Cv, where B⊥ = B sinq is the component of the field
perpendicular to the velocity of the conductor. In the described situation,
(
)
ΔV = B⊥ Cv = ( B sinq ) Cv = ⎡⎣ 60.0 × 10 −6 T sin 60.0° ⎤⎦ (12 m ) ( 60.0 m s )
= 3.7 × 10
−2
V = 37 mV
and the correct choice is (c).
6.
As the bar slides to the right along the rails, the magnetic flux through the conducting path formed
by the bar, rails, and the resistive element at the left end is directed out of the page and increasing
in magnitude. Thus, the induced current must generate a flux directed into the page through the area
enclosed by the current path. This means the induced current must be in the clockwise direction and
choice (b) is correct. Also, as the induced current flows, the rod will experience a magnetic force
that tends to impede the motion of the rod. Therefore, an external force must be exerted on the bar
to keep it moving at constant speed, and choice (d) is also correct.
7.
The amplitude of the induced emf in the coil of a generator is directly proportional to the angular
velocity of the coil (e max = NBAw ). Therefore, when the rate of rotation is doubled, the amplitude
of the induced emf is also doubled, and the correct choice is (b).
68719_20_ch20_p148-170.indd 149
1/7/11 2:45:21 PM
150
Chapter 20
8.
An emf is induced in the coil by any action which causes a change in the flux through the coil. The
actions described in choices (a), (b), (d), and (e) all change the flux through the coil and induce
an emf. However, moving the coil through the constant field without changing its orientation with
respect to the field will not cause a change of flux. Thus, choice (c) is the correct answer.
9.
With the current in the long wire flowing in the direction shown in Figure MCQ20.9, the magnetic
flux through the rectangular loop is directed into the page. If this current is decreasing in time, the
change in the flux is directed opposite to the flux itself (or out of the page). The induced current
will then flow clockwise around the loop, producing a flux directed into the page through the loop
and opposing the change in flux due to the decreasing current in the long wire. The correct choice
for this question is (b).
10.
A current flowing counterclockwise in the outer loop of Figure MCQ20.10 produces a magnetic
flux through the inner loop that is directed out of the page. If this current is increasing in time, the
change in the flux is in the same direction as the flux itself (or out of the page). The induced current in the inner loop will then flow clockwise around the loop, producing a flux through the loop
directed into the page, opposing the change in flux due to the increasing current in the outer loop.
The correct answer is choice (b).
11.
As the bar magnet approaches the loop from above, with its south end downward as shown
in Figure MCQ20.11, magnetic flux through the area enclosed by the loop is directed upward
and increasing in magnitude. To oppose this increasing upward flux, the induced current in the
loop will flow clockwise, as seen from above, producing a flux directed downward through the
area enclosed by the loop. After the bar magnet has passed through the plane of the loop and is
departing with its north end upward, a decreasing flux is directed upward through the loop. To
oppose this decreasing upward flux, the induced current in the loop flows counterclockwise as
seen from above, producing flux directed upward through the area enclosed by the loop. From
this analysis, we see that (a) is the only true statement among the listed choices.
12.
With the magnetic field perpendicular to the plane of the page in Figure MCQ20.12, the flux
through the closed loop to the left of the bar is given by Φ B = BA, where B is the magnitude of the
field and A is the area enclosed by the loop. Any action which produces a change in this product,
BA, will induce a current in the loop and cause the bulb to light. Such actions include increasing or
decreasing the magnitude of the field (B) and moving the bar to the right or left and changing the
enclosed area A. Thus, the bulb will light during all of the actions in choices (a), (b), (c), and (d).
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
Consider the copper tube to be a large set of rings stacked one on top of the other. As the magnet
falls toward or falls away from each ring, a current is induced in the ring. Thus, there is a current
in the copper tube around its circumference.
4.
(a)
The flux is calculated as Φ B = BA cosq = B⊥ A. The flux is therefore maximum when the
magnetic field vector is perpendicular to the plane of the loop.
(b)
The flux is zero when the magnetic field is parallel to the plane of the loop (and hence,
q = 90°, making B⊥ = B cosq = 0).
6.
68719_20_ch20_p148-170.indd 150
As water falls, it gains velocity and kinetic energy. It then pushes against the blades of a turbine,
transferring this energy to the rotor or coil of a large generator. The rotor moves in a strong
external magnetic field and a voltage is induced in the coil. This induced emf is the voltage
source for the current in our electric power lines.
1/7/11 2:45:23 PM
Induced Voltages and Inductance
8.
151
No. Once the bar is in motion and the charges are separated, no external force is necessary to
maintain the motion. During the initial acceleration of the bar, an external applied force will be
necessary to overcome both the inertia of the bar and a retarding magnetic force exerted on the bar.
10.
Let us assume the north pole of the magnet faces the ring. As the bar magnet falls toward the
conducting ring, a magnetic field is induced in the ring pointing upward. This upward directed
field will oppose the motion of the magnet, preventing it from moving as a freely falling body.
Try it for yourself to show that an upward force also acts on the falling magnet if the south end
faces the ring.
12.
The increasing counterclockwise current
in the solenoid coil produces an upward
magnetic field that increases rapidly. The
increasing upward flux of this field through
the ring induces an emf to produce a
clockwise current in the ring. At each point
on the ring, the field of the solenoid has a
radially outward component as well as an
upward component. This radial field component exerts an upward force on the current at each point in the ring. The resultant
magnetic force on the ring is upward and
exceeds the weight of the ring. Thus, the
ring accelerates upward off of the solenoid.
14.
B
I
Bsolenoid
F
F
Bring
I
Iring
Bsolenoid increasing
Isolenoid
As the magnet moves at high speed past the fixed coil, the magnetic flux through the coil changes
very rapidly, increasing as the magnet approaches the coil and decreasing as the magnet moves
away. The rapid change in flux through the coil induces a large emf, large enough to cause a spark
across the gap in the spark plug.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a) 1.00 × 10 −7 T ⋅ m 2
4.
zero
6.
(a) 6.98 mT
8.
0.10 mV
(b) 8.66 × 10 −8 T ⋅ m 2
0
(c)
right to left
(b) 1.97 × 10 −5 T ⋅ m 2
10.
34 mV
12.
2.26 mV
14.
into the page
16.
(a) left to right
(b)
18.
(a) 1.88 × 10 −7 T ⋅ m 2
(b) 6.27 × 10 −8 V
20.
8.8 A
68719_20_ch20_p148-170.indd 151
(c)
no induced current
1/7/11 2:45:24 PM
152
Chapter 20
e = NB0p r 2 t
22.
(a)
24.
1.20 mV, west end is positive
26.
(a) counterclockwise when viewed from the right end
28.
(b)
clockwise
(b)
clockwise when viewed from the right end
(c)
no induced current
(c)
I = B0p r 2 t R
(a) 6.0 m T
(b) The magnitude and direction of the Earth’s field varies from one location to another, so the
induced voltage in the wire will change.
30.
1.00 m s
32.
13 mV
34.
(a) 7.5 kV
(b)
when the plane of the coil is parallel to the magnetic field
36.
(a) 8.0 A
(b)
3.2 A
38.
(a) 5.9 mH
(b)
−24 V
40.
See Solution.
42.
1.92 × 10 −5 T ⋅ m 2
44.
(a)
1.00 kΩ
(b)
3.00 ms
46.
(a)
0
(b)
3.8 V
(c) 6.0 V
(d)
2.2 V
(a) 2.00 ms
(b)
0.176 A
(c)
(d)
3.22 ms
48.
1.50 A
(c)
60 V
50.
(a) 4.44 mH
(b)
0.555 mJ
52.
(a) 1.3 Ω
(b)
4.8 × 10 2 turns
(c)
0.48 m
(d)
0.76 mH
(e)
0.46 ms
(f)
3.6 A
(g)
3.2 ms
(h)
4.9 mJ
54.
(a)
increasing
(b) 62.2 mT s
56.
(a) 0.73 m s , counterclockwise
(c)
68719_20_ch20_p148-170.indd 152
(b)
0.65 mW
Work is being done on the bar by an external force to maintain constant speed.
1/7/11 2:45:24 PM
Induced Voltages and Inductance
58.
(a)
2.1 × 10 6 m s
(b)
153
(c) 1.7 × 1010 V
from side to side
(d) The very large induced emf would lead to powerful spontaneous electrical discharges.
The strong electric and magnetic fields would disrupt the flow of ions in their bodies.
60.
0.158 mV
62.
0.120 A, clockwise
64.
1.60 A
66.
(a)
e = NBCv
(b)
I = NBCv R
(c)
P = N 2 B 2 C2 v 2 R
(d)
F = N 2 B 2 C2 v R
(e)
clockwise
(f)
toward the left
PROBLEM SOLUTIONS
20.1
The angle between the direction of the constant field and the normal to the plane of the loop
is q = 0°, so
(
)(
)
Φ B = BA cosq = ( 0.50 T ) ⎡⎣ 8.0 × 10 −2 m 12 × 10 −2 m ⎤⎦ cos 0° = 4.8 × 10 −3 T ⋅ m 2
20.2
The magnetic flux through the loop is given by Φ B = BA cosq , where B is the magnitude of the
magnetic field, A is the area enclosed by the loop, and q is the angle the magnetic field makes
with the normal to the plane of the loop. Thus,
2
⎡
⎛ 10 −2 m ⎞ ⎤
Φ B = BA cosq = 5.00 × 10 −5 T ⎢ 20.0 cm 2 ⎜
⎥ cosq = 1.00 × 10 −7 T ⋅ m 2 cosq
⎝ 1 cm ⎟⎠ ⎥⎦
⎢⎣
I
(a) When B is perpendicular to the plane of the loop, q = 0° and Φ B = 1.00 × 10 −7 T ⋅ m 2 .
(
20.3
)
(
= (1.00 × 10
(
)
) cos 90.0° =
(b)
If q = 30.0°, then Φ B = 1.00 × 10 −7 T ⋅ m 2 cos 30.0° = 8.66 × 10 −8 T ⋅ m 2 .
(c)
If q = 90.0°, then Φ B
−7
T ⋅ m2
)
0.
Φ B = BA cosq = B(p r 2 ) cosq , where q is the angle between the direction of the field and the
normal to the plane of the loop.
(a)
If the field is perpendicular to the plane of the loop, q = 0°, and
B=
(b)
ΦB
8.00 × 10 −3 T ⋅ m 2
=
= 0.177 T
2
p r cosq p ( 0.12 m ) cos 0°
( )
2
If the field is directed parallel to the plane of the loop, q = 90°, and
Φ B = BA cosq = BA cos 90° = 0
20.4
68719_20_ch20_p148-170.indd 153
The magnetic field lines are everywhere parallel to the surface of the cylinder, so no magnetic
field lines penetrate the cylindrical surface. The total flux through the cylinder is zero .
1/7/11 2:45:26 PM
154
20.5
20.6
Chapter 20
(a)
Every field line that comes up through the area A on one side of the wire goes back down
through area A on the other side of the wire. Thus, the net flux through the coil is zero .
(b)
The magnetic field is parallel to the plane of the coil, so q = 90.0°. Therefore,
Φ B = BA cosq = BA cos 90.0° = 0 .
(a)
The magnitude of the field inside the solenoid is
⎛ N⎞
⎛ 400 ⎞
B = m 0 nI = m 0 ⎜ ⎟ I = 4p × 10 −7 T ⋅ m A ⎜
( 5.00 A )
⎝ C⎠
⎝ 0.360 m ⎟⎠
(
)
= 6.98 × 10 −3 T = 6.98 mT
(b)
The field inside a solenoid is directed perpendicular to the cross-sectional area, so q = 0°
and the flux through a loop of the solenoid is
( )
T ) p ( 3.00 × 10
Φ B = BA cosq = B p r 2 cosq
(
= 6.98 × 10 −3
20.7
(a)
−2
)
2
m cos 0° = 1.97 × 10 −5 T ⋅ m 2
The magnetic flux through an area A may be written as
Φ B = ( B cosq ) A
= ( component of B perpendicular to A ) ⋅ A
Thus, the flux through the shaded side of the cube is
(
Φ B = Bx ⋅ A = ( 5.0 T ) ⋅ 2.5 × 10 −2 m
(b)
20.8
20.9
20.10
68719_20_ch20_p148-170.indd 154
)
2
= 3.1 × 10 −3 T ⋅ m 2
Unlike electric field lines, magnetic field lines always form closed loops, without beginning
or end. Therefore, no magnetic field lines originate or terminate within the cube and any
line entering the cube at one point must emerge from the cube at some other point. The net
flux through the cube, and indeed through any closed surface, is zero .
(
)
2
−3
⎡
⎤
ΔΦ B ( ΔB ) A cosq (1.5 T − 0 ) ⎣p 1.6 × 10 m ⎦ cos 0°
e =
=
=
= 1.0 × 10 −4 V = 0.10 mV
Δt
Δt
120 × 10 −3 s
(a)
As loop A moves parallel to the long straight wire, the magnetic flux through loop A does
not change. Hence, there is no induced current in this loop.
(b)
As loop B moves to the left away from the straight wire, the magnetic flux through this
loop is directed out of the page, and is decreasing in magnitude. To oppose this change in
flux, the induced current flows counterclockwise around loop B producing a magnetic flux
directed out of the page through the area enclosed by loop B.
(c)
As loop C moves to the right away from the straight wire, the magnetic flux through this
loop is directed into the page and is decreasing in magnitude. In order to oppose this change
in flux, the induced current flows clockwise around loop C producing a magnetic flux
directed into the page through the area enclosed by loop C.
2
ΔΦ B
B ΔA cosq ( 0.15 T ) ⎡⎣p ( 0.12 m ) − 0 ⎤⎦ cos 0°
e =
=
=
= 3.4 × 10 −2 V = 34 mV
Δt
Δt
0.20 s
1/7/11 2:45:28 PM
Induced Voltages and Inductance
20.11
155
The magnitude of the induced emf is
e =
ΔΦ B
Δ ( B cosq ) A
=
Δt
Δt
If the normal to the plane of the loop is considered to point in the original direction of the
magnetic field, then q i = 0° and q f = 180°. Thus, we find
e =
20.12
1.5 s
= 9.4 × 10 −2 V = 94 mV
With the field directed perpendicular to the plane of the coil, the flux through the coil is
Φ B = BA cos 0° = BA. As the magnitude of the field decreases, the magnitude of the induced emf
in the coil is
e =
20.13
( 0.20 T ) cos180° − ( 0.30 T ) cos 0° p ( 0.30 m )2
ΔΦ B
ΔB
2
=
A = ( 0.050 0 T s ) ⎡⎣p ( 0.120 m ) ⎤⎦ = 2.26 × 10 −3 V = 2.26 mV
Δt
Δt
The magnetic field changes from zero to a magnitude of 2.5 T, directed at 45° to the plane of the
band, in a time of Δt = 0.18 s. If the band has a diameter of 6.5 cm, the magnitude of the average
induced emf in the metal band is
(
−2
ΔΦ B
ΔB
( 2.5 T − 0 ) 6.5 × 10 m
e =
=
A cosq =
p
Δt
Δt
0.18 s
4
)
2
cos 45° = 3.3 × 10 −2 V = 33 mV
20.14
When the switch is closed, the magnetic field due to the current from the battery will be directed
toward the left along the axis of the cylinder. To oppose this increasing leftward flux, the induced
current in the other loop must produce a field directed to the right through the area it encloses.
Thus, the induced current is directed out of the page through the resistor.
20.15
(a)
When the magnet moves to the left, the flux through the interior of the coil is directed
toward the right and is decreasing in magnitude. To oppose this change in flux, the
magnetic field generated by the induced current should be directed to the right along
the axis of the coil. The current must then be left to right through the resistor.
(b)
When the magnet moves to the right, the flux through the interior of the coil is directed
toward the right and is increasing in magnitude. To oppose this increasing flux, the
magnetic field generated by the induced current should be directed toward the left along
the axis of the coil. The current must then be right to left through the resistor.
20.16
68719_20_ch20_p148-170.indd 155
When the switch is closed, the current from the battery produces a magnetic field directed toward
the right along the axis of both coils.
(a)
As the battery current is growing in magnitude, the induced current in the rightmost coil
opposes the increasing rightward directed field by generating a field toward to the left along
the axis. Thus, the induced current must be left to right through the resistor.
(b)
Once the battery current, and the field it produces, have stabilized, the flux through the
rightmost coil is constant, and there is no induced current .
(c)
As the switch is opened, the battery current and the field it produces rapidly decrease in
magnitude. To oppose this decrease in the rightward directed field, the induced current
must produce a field toward the right along the axis, so the induced current is right to left
through the resistor.
1/7/11 2:45:30 PM
156
20.17
20.18
Chapter 20
(a)
The current is zero . The magnetic flux the current produces through the right side of the
loop is directed into the page, and is equal in magnitude to the outward-directed flux the
current produces through the left side of the loop. Thus, the net flux through the loop has a
constant value of zero and does not induce a current.
(b)
The flux through the loop due to the long wire is directed out of the page and is increasing
in magnitude. To oppose this increasing outward flux, the induced current must generate a
magnetic field that is directed into the page through the area enclosed by the loop. Thus, the
induced current in the loop must be clockwise .
The initial magnetic field inside the solenoid is
⎛ N⎞
⎛ 100 ⎞
B = m 0 nI = m 0 ⎜ ⎟ I = 4p × 10 −7 T ⋅ m A ⎜
( 3.00 A ) = 1.88 × 10 −3 T
⎝ C⎠
⎝ 0.200 m ⎟⎠
(
(
)(
)
2
(a)
Φ B = BAloop cosq = 1.88 × 10 −3 T 1.00 × 10 −2 m cos 0° = 1.88 × 10 −7 T ⋅ m 2
(b)
When the current is zero, the flux through the loop is Φ B = 0, and the average induced
emf has been
e =
20.19
)
(a)
ΔΦ B
Δt
=
0 − 1.88 × 10 −7 T ⋅ m 2
3.00 s
= 6.27 × 10 −8 V
The initial field inside the solenoid is
⎛ 300 ⎞
Bi = m 0 nI i = 4p × 10 −7 T ⋅ m A ⎜
( 2.00 A ) = 3.77 × 10 −3 T
⎝ 0.200 m ⎟⎠
(
(b)
)
The final field inside the solenoid is
⎛ 300 ⎞
B f = m 0 nI f = 4p × 10 −7 T ⋅ m A ⎜
( 5.00 A ) = 9.42 × 10 −3 T
⎝ 0.200 m ⎟⎠
(
)
(
)
2
(c)
The 4-turn coil encloses an area A = p r 2 = p 1.50 × 10 −2 m
(d)
The change in flux through each turn of the 4-turn coil during the 0.900-s period is
(
)(
= 7.07 × 10 −4 m 2
)
ΔΦ B = ( ΔB ) A = 9.42 × 10 −3 T − 3.77 × 10 −3 T 7.07 × 10 −4 m 2 = 3.99 × 10 −6 Wb
(e)
The average induced emf in the 4-turn coil is
⎛ 3.99 × 10 −6 Wb ⎞
⎛ ΔΦ B ⎞
−5
e = N2 ⎜
= 4⎜
⎟
⎟⎠ = 1.77 × 10 V
⎝ Δt ⎠
0.900 s
⎝
Since the current increases at a constant rate during this time interval, the induced emf at
any instant during the interval is the same as the average value given above.
(f)
68719_20_ch20_p148-170.indd 156
The induced emf is small, so the current in the 4-turn coil will also be very small.
This means that the magnetic field generated by this current will be negligibly small
in comparison to the field generated by the solenoid.
1/7/11 2:45:32 PM
Induced Voltages and Inductance
20.20
The magnitude of the average emf is
e =
=
N ΔΦ B
NBA Δ ( cosq )
=
Δt
Δt
(
)
200 (1.1 T ) 100 × 10 −4 m 2 cos180° − cos 0°
0.10 s
Therefore, the average induced current is I =
20.21
157
= 44 V
e
44 V
=
= 8.8 A .
R 5.0 Ω
If the magnetic field makes an angle of 28.0° with the plane of the coil, the angle it makes with
the normal to the plane of the coil is q = 62.0°. Thus,
e =
=
N ( ΔΦ B ) NB ( ΔA ) cosq
=
Δt
Δt
(
)(
)(
)
200 50.0 × 10 –6 T ⎡⎣ 39.0 cm 2 1 m 2 10 4 cm 2 ⎤⎦ cos 62.0°
1.80 s
= 1.02 × 10 −5 V = 10.2 mV
20.22
With the magnetic field perpendicular to the plane of the coil, the flux through each turn of
the coil is Φ B = BA = B p r 2 . Since the area remains constant, the change in flux due to the
changing magnitude of the magnetic field is ΔΦ B = ( ΔB ) p r 2 .
( )
The induced emf is e = N
(b)
When looking down on the coil from a location on the positive z-axis, the magnetic field
(in the positive z-direction) is directed up toward you and increasing in magnitude. This
means the change in the flux through the coil is directed upward. In order to oppose this
change in flux, the induced current must produce a magnetic field directed downward
through the area enclosed by the coil. Thus, the current must flow clockwise as seen from
your viewing location.
(c)
Since the turns of the coil are connected in series, the total resistance of the coil is Req = NR.
Thus, the magnitude of the induced current is
I=
20.23
⎡ B0 − 0 p r 2 ⎤ NB0p r 2
ΔΦ
=N⎢
.
⎥=
Δt
t
⎥⎦
⎣⎢ t − 0
(a)
B pr2
NB0p r 2 t
e
=
= 0
NR
Req
tR
The motional emf induced in a metallic object of length C moving through a magnetic field at
speed v is given by e = B⊥ vC, where B⊥ is the component of the magnetic field perpendicular to
the velocity of the object. Thus,
(
)
e = 35.0 × 10 −6 T ( 25.0 m s ) (15.0 m ) = 1.31 × 10 −2 V = 13.1 mV
20.24
The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of
the wire. Thus, the magnitude of the motional emf induced in the wire is
(
)
e = B⊥ Cv = 40.0 × 10 −6 T ( 2.00 m ) (15.0 m s ) = 1.20 × 10 −3 V = 1.20 mV
continued on next page
68719_20_ch20_p148-170.indd 157
1/7/11 2:45:33 PM
158
Chapter 20
Imagine holding your right hand horizontal with the fingers pointing north (the direction of
the wire’s velocity), such that when you close your hand the fingers curl downward (in the
direction of B⊥). Your thumb will then be pointing westward. By right-hand rule number 1, the
magnetic force on charges in the wire would tend to move positive charges westward. Thus,
the west end of the wire will be positive relative to the east end .
20.25
The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of
the metallic truck body. Thus, the motional emf induced across the width of the truck is
⎡
⎛ 1m ⎞⎤
−3
e = B⊥ Cv = 35 × 10 −6 T ⎢( 79.8 in ) ⎜
⎟⎠ ⎥ ( 37 m s ) = 2.6 × 10 V = 2.6 mV
⎝
39.37
in
⎣
⎦
(
20.26
20.27
)
(a)
As the loop passes position A, the flux through the area enclosed by the loop is
directed right to left and is increasing in magnitude. The induced current must flow
counterclockwise as seen from the right end in order to generate flux passing left to right
through the loop, opposing the increase in flux due to the magnet.
(b)
When the loop reaches position B, the flux through the enclosed area is again directed
right to left but is now decreasing in magnitude. The induced current must flow
clockwise as seen from the right end to generate additional flux passing right to left
through the loop, opposing the decrease in flux due to the magnet.
(c)
At position C, the flux through the loop is not changing so there is no induced emf, and
hence no induced current , in the loop.
(a)
Observe that only the horizontal component, Bh , of Earth’s magnetic field is effective
in exerting a vertical force on charged particles in the antenna. For the magnetic force,
Fm = qvBh sinq , on positive charges in the antenna to be directed upward and have
maximum magnitude (when q = 90°), the car should move eastward through the
northward horizontal component of the magnetic field.
(b)
e = Bh Cv, where Bh is the horizontal component of the magnetic field.
⎡⎛
km ⎞ ⎛ 0.278 m s ⎞ ⎤
e = ⎡⎣ 50.0 × 10 −6 T cos 65.0° ⎤⎦ (1.20 m ) ⎢⎜ 65.0
⎟
⎥
⎝
h ⎠ ⎜⎝ 1 km h ⎟⎠ ⎦
⎣
(
)
= 4.58 × 10 − 4 V
20.28
(a)
Since e = B⊥ Cv, the magnitude of the vertical component of the Earth’s magnetic field at
this location is
Bvertical = B⊥ =
(b)
20.29
e
0.45 V
=
= 6.0 × 10 −6 T = 6.0 m T
Cv ( 25 m ) 3.0 × 10 3 m s
(
)
Yes. The magnitude and direction of the Earth’s field varies from one location to the other,
so the induced voltage in the wire changes. Further, the voltage will change if the tether
cord changes its orientation relative to the Earth’s field.
The metal bar of length C and moving at speed v through the magnetic field experiences an
induced emf of magnitude e = B⊥ Cv, where B⊥ = B cosq is the component of the magnetic field
perpendicular to the velocity of the bar as shown in figure (a) below.
continued on next page
68719_20_ch20_p148-170.indd 158
1/7/11 2:45:35 PM
Induced Voltages and Inductance
159
y
end view
of bar
B
q
q
n
x
Fm
v
B
current
into page
mg
B||
q
q
(b)
(a)
As the bar slides down the rails, the magnetic flux through the conducting path formed by the
bar, the rails, and the resistor R is directed downward and is increasing in magnitude. Thus, the
induced current must flow counterclockwise around the conducting path to generate an upward
flux, opposing the increase in flux due to the field B. This current flows into the page as indicated
in figure (b) and has magnitude I = e R = BCv cosq R.
I
According to right-hand rule number 1, the bar will experience a magnetic force Fm directed
horizontally toward the left as shown in figure (b). The magnitude of this force is
B 2 C2 v cosq
⎛ BCv cosq ⎞
Fm = BIC = B ⎜
C=
⎟
⎝
⎠
R
R
I
Now, consider the free-body diagram of the bar in figure (b), where n is the normal force exerted
on the bar by the rails. If the bar is to move with constant velocity (i.e., be in equilibrium), it is
necessary that
and
mg
cosq
ΣFy = 0 ⇒ n cosq = mg
or
n=
ΣFx = 0 ⇒ Fm = n sinq
or
B 2 C2 v cosq ⎛ mg ⎞
=⎜
sinq = mg tanq
⎝ cosq ⎟⎠
R
Thus, the equilibrium speed of the bar is
v=
20.30
B C cosq
2 2
)
m s2 tan 25.0° (1.00 Ω )
( 0.500 T ) (1.20 m )2 cos 25.0°
2
(a)
e
IR ( 0.500 A ) ( 6.00 Ω )
=
=
= 1.00 m s
BC BC
( 2.50 T )(1.20 m )
In the initial orientation of the coil, the magnitude of the flux passing through the loop is
Φ B = NBA, where A is the area enclosed by the loop and N is the number of turns on the
loop. After the loop has rotated 90°, the magnetic field is now parallel to the plane of the loop
and the flux through the loop is zero. The average emf induced in the loop as it rotates is
(
−2
ΔΦ B
NBA − 0 28 (1.25 T ) 2.80 × 10 m
e=
=
=
Δt
Δt
0.335 s
(b)
68719_20_ch20_p148-170.indd 159
= 2.80 m s
From e = BCv, the required speed is
v=
20.31
( mg tanq ) R = ( 0.200 kg) (9.80
The average induced current is I =
)
2
= 8.19 × 10 −2 V = 81.9 mV
e 81.9 mV
=
= 105 mA .
R 0.780 Ω
1/7/11 2:45:38 PM
160
20.32
Chapter 20
Note that the vertical component of the magnetic field is always parallel to the plane of the coil,
and can never contribute to the flux through the coil. The maximum induced emf in the coil is then
rev ⎞ ⎛ 2p rad ⎞ ⎛ 1 min ⎞ ⎤
2 ⎡⎛
e max = NBhorizontal Aw = 100 2.0 × 10 −5 T ( 0.20 m ) ⎢⎜ 1500
⎟⎠ ⎜⎝
⎟⎜
⎟
⎝
min
1 rev ⎠ ⎝ 60 s ⎠ ⎥⎦
⎣
(
)
= 1.3 × 10 −2 V = 13 mV
20.33
Note the similarity between the situation in this problem and a generator. In a generator, one
normally has a loop rotating in a constant magnetic field so the flux through the loop varies
sinusoidally in time. In this problem, we have a stationary loop in an oscillating magnetic field,
and the flux through the loop varies sinusoidally in time. In both cases, a sinusoidal
emf e = e max sinwt, where e max = NBAw , is induced in the loop.
The loop in this case consists of a single band ( N = 1) around the perimeter of a red blood cell
with diameter d = 8.0 × 10 −6 m. The angular frequency of the oscillating flux through the area of
this loop is w = 2p f = 2p ( 60 Hz ) = 120p rad s. The maximum induced emf is then
e max
20.34
(a)
(
) (
1.0 × 10 −3 T p 8.0 × 10 −6 m
⎛ p d2 ⎞
= NBAw = B ⎜
w
=
4
⎝ 4 ⎟⎠
) (120p s ) =
2
–1
1.9 × 10 −11 V
Using e max = NBAw , we find
⎡⎛
rev ⎞ ⎛ 2p rad ⎞⎤
e max = 1 000 ( 0.20 T) 0.10 m 2 ⎢⎜60
⎟⎥
⎟⎜
s ⎠ ⎝ 1 rev ⎠⎦
⎣⎝
(
)
= 7.5×10 3 V = 7.5 kV
(b)
20.35
The maximum induced emf occurs when the flux through the coil is changing the most
rapidly. This is when the plane of the coil is parallel to the magnetic field .
rev ⎞ ⎛ 1 min ⎞ ⎛ 2p rad ⎞
⎛
The angular frequency is w = ⎜ 120
⎟⎜
⎟⎜
⎟ = 4p rad s.
⎝
min ⎠ ⎝ 60 s ⎠ ⎝ 1 rev ⎠
(a)
e max = NBAw = 500 ( 0.60 T ) [( 0.080 m ) ( 0.20 m )] ( 4p rad s ) = 60 V
(b)
Note that the calculator must be in radians mode for the next calculation.
⎡
⎛p
e = e max sin (w t ) = ( 60 V ) sin ⎢( 4p rad s ) ⎜
⎝ 32
⎣
(c)
The emf is first maximum when w t = p 2 radians, or when
t=
20.36
(a)
⎞⎤
s⎟ ⎥ = 57 V
⎠⎦
p 2 rad p 2 rad
=
= 0.13 s
w
4p rad s
Immediately after the switch is closed, the motor coils are still stationary and the back emf
is zero. Thus,
I=
e 240 V
=
= 8.0 A
R
30 Ω
continued on next page
68719_20_ch20_p148-170.indd 160
1/7/11 2:45:40 PM
Induced Voltages and Inductance
(b)
At maximum speed, e back = 145 V and
I=
20.37
161
e − e back 240 V − 145 V
=
= 3.2 A
R
30 Ω
(c)
e back = e − IR = 240 V − ( 6.0 A ) ( 30 Ω ) = 60 V
(a)
When a coil having N turns and enclosing area A rotates at angular frequency w in a
constant magnetic field, the emf induced in the coil is
e = e max sinw t, where e max = NB⊥ Aw
Here, B⊥ is the magnitude of the magnetic field perpendicular to the rotation axis of the coil.
In the given case, B⊥ = 55.0 mT; A = p ab, where a = (10.0 cm ) 2 and b = ( 4.00 cm ) 2; and
rev ⎞ ⎛ 1 min ⎞
⎛
w = 2p f = 2p ⎜ 100
⎟⎜
⎟ = 10.5 rad s
⎝
min ⎠ ⎝ 60.0 s ⎠
20.38
Thus,
p
e max = (10.0 ) 55.0 × 10 −6 T ⎡⎢ ( 0.100 m ) ( 0.040 0 m ) ⎤⎥ (10.5 rad s )
⎣4
⎦
or
e max = 1.81 × 10 −5 V = 18.1 mV
(
(b)
When the rotation axis is parallel to the field, then B⊥ = 0, giving e max = 0 . It is easily
understood that the induced emf is always zero in this case if you recognize that the
magnetic field lines are always parallel to the plane of the coil, and the flux through the
coil has a constant value of zero.
(a)
In terms of its cross-sectional area A, length C, and number of turns N, the self inductance
of a solenoid is given as L = m 0 N 2 A C. Thus, for the given solenoid,
L=
(
m0 N 2 p d 2 4
C
= 5.9 × 10
(b)
20.39
−3
) = ( 4p × 10
−7
)
(
T ⋅ m A ( 580 ) p 8.0 × 10 −2 m
2
4 ( 0.36 m )
)
2
H = 5.9 mH
⎛ ΔI ⎞
e = −L ⎜ ⎟ = − 5.9 × 10 −3 H ( +4.0 A s ) = −2.4 × 10 −2 V = −24 mV
⎝ Δt ⎠
(
)
From e = L ΔI Δt , we have
L=
20.40
)
(
)
−3
e
e ( Δt ) 12 × 10 V ( 0.50 s )
=
=
= 4.0 × 10 −3 H = 4.0 mH
ΔI Δt
ΔI
2.0 A − 3.5 A
The units of NΦ B I are T ⋅ m 2 A . From the force on a moving charged particle, F = qvB, the
magnetic field is B = F qv, and we find that
1T =1
N
N ⋅s
=1
C ⋅ ( m s)
C⋅m
⎛ N ⋅ s ⎞ 2 (N ⋅ m) ⋅ s ⎛ J ⎞
Thus, T ⋅ m 2 = ⎜
= ⎜ ⎟ ⋅ s = V ⋅ s, and T ⋅ m 2 A = V ⋅ s A.
⋅m =
⎝ C⎠
⎝ C ⋅ m ⎟⎠
C
The units of e ( ΔI Δt ) are V ( A s ) = V ⋅ s A , the same as the units of NΦ B I.
68719_20_ch20_p148-170.indd 161
1/7/11 2:45:42 PM
162
20.41
20.42
Chapter 20
(
)
(
)
(a)
2
2
4p × 10 −7 T ⋅ m A ( 400 ) ⎡p 2.5 × 10 −2 m ⎤
m0 N 2 A
⎣
⎦ = 2.0 × 10 −3 H = 2.0 mH
L=
=
0.20 m
C
(b)
From e = L ( ΔI Δt ),
ΔI e
75 × 10 −3 V
=
=
= 38 A s
Δt
L 2.0 × 10 −3 H
From e = L ( ΔI Δt ), the self-inductance is
L=
e
24.0 × 10 −3 V
=
= 2.40 × 10 −3 H
ΔI Δt
10.0 A s
Then, from L = NΦ B I, the magnetic flux through each turn is
ΦB =
20.43
(a)
(
In the series circuit of Figure P20.43, maximum current occurs after the switch has been
closed for a very long time, when current has stabilized and the back emf due to the
inductance has decreased to zero. This maximum current is given by
I max =
(b)
e 24.0 V
=
= 5.33 A
R 4.50 Ω
The time constant of the RL circuit is
t =
(c)
)
2.40 × 10 −3 H ( 4.00 A )
L⋅I
=
= 1.92 × 10 −5 T ⋅ m 2
N
500
L 12.0 H 12.0 Ω ⋅ s
=
=
= 2.67 s
R 4.50 Ω
4.50 Ω
If the switch in the RL circuit is closed at time t = 0, the current as a function of time is
given by I = I max 1 − e −t t .
(
)
Thus, e −t t = 1 − I I max, or t = −t ln (1 − I I max ) . With t = 2.67 s, the current in this circuit
will be I = 0.950I max at time
t = − ( 2.67 s ) ln (1 − 0.950 ) = 8.00 s
20.44
(a)
The time constant of the RL circuit is t = L R, and that of the RC circuit is t = RC. If the
two time constants have the same value, then RC = L R, or
R=
(b)
3.00 H
= 1.00 × 10 3 Ω = 1.00 kΩ
3.00 × 10 −6 F
The common value of the two time constants is
t =
20.45
L
=
C
L
3.00 H
=
= 3.00 × 10 −3 s = 3.00 ms
R 1.00 × 10 3 Ω
(a)
I max = e R , so e = I max R = (8.0 A ) ( 0.30 Ω ) = 2.4 V .
(b)
The time constant is t = L R, giving
L = t R = ( 0.25 s ) ( 0.30 Ω ) = 7.5 × 10 −2 H = 75 mH
continued on next page
68719_20_ch20_p148-170.indd 162
1/7/11 2:45:45 PM
Induced Voltages and Inductance
(c)
(
163
)
The current as a function of time is I = I max 1 − e −t t , so at t = t ,
(
)
I = I max 1 − e −1 = 0.632I max = 0.632 (8.0 A ) = 5.1 A
(d)
At t = t , I = 5.1 A, and the voltage drop across the resistor is
ΔVR = −IR = − ( 5.1 A ) ( 0.30 Ω ) = −1.5 V
(e)
Applying Kirchhoff’s loop rule to the circuit shown in Figure P20.43 gives e + ΔVR + ΔVL = 0.
Thus, at t = t , we have
ΔVL = − (e + ΔVR ) = − ( 2.4 V − 1.5 V ) = −0.90 V
20.46
(
resistor is ΔVR = RI = e 1 − e −t t
)
(
)
e
1 − e −t t . The potential difference across the
R
, and from Kirchhoff’s loop rule, the potential difference across
The current in the RL circuit at time t is I =
the inductor is
(
)
ΔVL = e − ΔVR = e ⎡⎣1 − 1 − e −t t ⎤⎦ = ee −t t
20.47
(
)
(
)
(a)
At t = 0, ΔVR = e 1 − e −0 = e (1 − 1) = 0 .
(b)
At t = t , ΔVR = e 1 − e −1 = ( 6.0 V ) (1 − 0.368 ) = 3.8 V .
(c)
At t = 0, ΔVL = e e −0 = e = 6.0 V .
(d)
At t = t , ΔVL = e e −1 = ( 6.0 V ) ( 0.368 ) = 2.2 V .
(
)
From I = I max 1 − e −t t , we obtain e −t t = 1 − I I max . If I I max = 0.900 at t = 3.00 s, then
e − (3.00 s) t = 0.100
or
t =
−3.00 s
= 1.30 s
ln ( 0.100 )
Since the time constant of an RL circuit is t = L R, the resistance is
20.48
R=
L 2.50 H
=
= 1.92 Ω
t
1.30 s
(a)
t =
L 8.00 mH
=
= 2.00 ms
R
4.00 Ω
(b)
I=
(c)
I max =
(d)
I = I max 1 − e −t t yields e −t t = 1 − I I max, and
V⎞
(1 − e ) = ⎛⎜⎝ 6.00
⎟ (1 − e
R
4.00 Ω ⎠
e
−t t
e
R
(
=
−250×10 −6 s 2.00×10 −3 s
)=
0.176 A
6.00 V
= 1.50 A
4.00 Ω
)
t = −t ln (1 − I I max ) = − ( 2.00 ms ) ln (1 − 0.800 ) = 3.22 ms
68719_20_ch20_p148-170.indd 163
1/7/11 2:45:49 PM
164
20.49
Chapter 20
(a)
The energy stored by an inductor is PE L = 12 LI 2 , so the self inductance is
(
(b)
If I = 3.0 A, the stored energy will be
PE L =
20.50
(a)
)
−3
2 ( PE L ) 2 0.300 × 10 J
=
= 2.08 × 10 −4 H = 0.208 mH
I2
(1.70 A )2
L=
(
)
1 2 1
2
LI = 2.08 ×10 −4 H (3.0 A ) = 9.36 ×10 −4 J = 0.936 mJ
2
2
The inductance of a solenoid is given by L = m 0 N 2 A C, where N is the number of turns on
the solenoid, A is its cross-sectional area, and C is its length. For the given solenoid,
L=
( ) = ( 4p × 10
m0 N 2 p r 2
−7
)
(
T ⋅ m A ( 300 ) p 5.00 × 10 −2 m
C
2
)
2
0.200 m
= 4.44 × 10 −3 H = 4.44 mH
(b)
When the solenoid described above carries a current of I = 0.500 A, the stored energy is
PE L =
20.51
1 2 1
2
LI = ( 4.44 mH ) ( 0.500 A ) = 0.555 mJ
2
2
(
(a)
As t → ∞, I → I max =
PE L →
(b)
(a)
e
24 V
=
= 3.0 A, and
R 8.0 Ω
1 2
1
2
LI max = ( 4.0 H ) (3.0 A ) = 18 J
2
2
(
)
At t = t , I = I max 1 − e −1 = ( 3.0 A ) (1 − 0.368 ) = 1.9 A, and
PE L =
20.52
)
The current in the circuit at time t is I = I max 1 − e −t t , where I max = e R , and the energy stored
in the inductor is PE L = 12 LI 2.
1
( 4.0 H )(1.9 A )2 = 7.2 J
2
Use Table 17.1 to obtain the resistivity of the copper wire and find
Rwire
(
)
1.7 × 10 −8 Ω ⋅ m ( 60.0 m )
rCu L rCu L
=
= 2 =
= 1.3 Ω
2
Awire p rwire
p 0.50 × 10 −3 m
(
)
L
L
60.0 m
=
=
= 4.8 ×10 2 turns
circumference of a loop 2p rsolenoid 2p 2.0 ×10 −2 m
(b)
N=
(c)
The length of the solenoid is
(
)
(
)
p ( 2.0 × 10
C = N ( diameter of wire ) = N ( 2rwire ) = ( 480 ) 2 0.50 × 10 −3 m = 0.48 m
(d)
(
)
2
4p × 10 −7 T ⋅ m A ( 480 )
m 0 N 2 Asolenoid m 0 N 2p rsolenoid
L=
=
=
C
C
0.48 m
giving
2
−2
m
)
2
L = 7.6 × 10 −4 H = 0.76 mH
continued on next page
68719_20_ch20_p148-170.indd 164
1/7/11 2:45:53 PM
Induced Voltages and Inductance
L
L
7.6 × 10 −4 H
=
=
= 4.6 × 10 −4 s = 0.46 ms
Rtotal Rwire + rinternal 1.3 Ω + 0.350 Ω
(e)
t =
(f)
I max =
(g)
I = I max 1 − e −t t , so when I = 0.999I max , we have e −t t = 1 − 0.999 = 0.001. Thus,
− t t = ln ( 0.001) , or t = −t ⋅ ln ( 0.001) = − ( 0.46 ms ) ⋅ ln ( 0.001) = 3.2 ms .
(h)
165
e
6.0 V
=
= 3.6 A
Rtotal 1.3 Ω + 0.350 Ω
(
)
( PEL )max =
(
)
1 2
1
2
LI max = 7.6 ×10 −4 H (3.6 A ) = 4.9 ×10 −3 J = 4.9 mJ
2
2
20.53
The flux due to the current in loop 1 passes from left to right through the area enclosed by
loop 2. As loop 1 moves closer to loop 2, the magnitude of this flux through loop 2 is increasing.
The induced current in loop 2 generates a magnetic field directed toward the left through the area
it encloses in order to oppose the increasing flux from loop 1. This means that the induced current
in loop 2 must flow counterclockwise as viewed from the left end of the rod.
20.54
(a)
The clockwise induced current in the loop produces a flux directed into the page through
the area enclosed by the loop. Since this flux opposes the change in flux due to the external
field, the outward-directed flux due to the external field must be increasing in magnitude.
This means that the magnitude of the external field itself must be increasing in time.
(b)
The induced emf in the loop must be
e = IR = ( 2.50 mA ) ( 0.500 Ω ) = 1.25 mV
Since e =
ΔΦ B Δ ( BA cos 0° ) ⎛ ΔB ⎞
=
=⎜
A, the rate of change of the field is
⎝ Δt ⎟⎠
Δt
Δt
ΔB e
e
1.25 × 10 −3 V
= = 2 =
Δt
A pr
p 8.00 × 10 −2 m
(
20.55
(a)
)
2
= 6.22 × 10 −2 T s = 62.2 mT s
After the right end of the coil has
entered the field, but the left end
has not, the flux through the area
enclosed by the coil is directed into
the page and is increasing in magnitude. This increasing flux induces
an emf of magnitude
e =
ΔΦ B NB ( ΔA )
=
= NBwv
Δt
Δt
in the loop. Note that in the above equation, ΔA = wv is the area enclosed by the coil that
enters the field in time Δt. This emf produces a counterclockwise current in the loop to oppose
the increasing inward flux. The magnitude of this current is I = e R = NBwv R. The right
end of the loop is now a conductor, of length Nw, carrying a current toward the top of the
page through a field directed into the page. The field exerts a magnetic force of magnitude
N 2 B2 w 2 v
⎛ NBwv ⎞
directed toward the left
F = BI ( Nw ) = B ⎜
Nw ) =
(
⎟
⎝ R ⎠
R
on this conductor, and hence, on the loop.
continued on next page
68719_20_ch20_p148-170.indd 165
1/7/11 2:45:56 PM
166
Chapter 20
(b)
When the loop is entirely within the magnetic field, the flux through the area enclosed by
the loop is constant. Hence, there is no induced emf or current in the loop, and the field
exerts zero force on the loop.
(c)
After the right end of the loop emerges from the field, and before the left end emerges,
the flux through the loop is directed into the page and is decreasing. This decreasing flux
induces an emf of magnitude e = NBwv in the loop, which produces an induced current
directed clockwise around the loop so as to oppose the decreasing flux. The current has
magnitude I = e R = NBwv R. This current flowing upward, through conductors of total
length Nw, in the left end of the loop, experiences a magnetic force given by
N 2 B2 w 2 v
⎛ NBwv ⎞
F = BI ( Nw ) = B ⎜
Nw ) =
(
⎟
⎝ R ⎠
R
20.56
(a)
directed toward the left
The motional emf induced in the bar must be e = IR, where I is the current in this series
circuit. Since e = B⊥ Cv, the speed of the moving bar must be
v=
(
)
8.5 × 10 −3 A ( 9.0 Ω )
e
IR
=
=
= 0.73 m s
B⊥ C B⊥ C
( 0.30 T )( 0.35 m )
The flux through the closed loop formed by the rails, the bar, and the resistor is directed
into the page and is increasing in magnitude. To oppose this increasing inward flux, the
induced current must generate a magnetic field directed out of the page through the area
enclosed by the loop. This means the current will flow countercloclockwise .
(b)
The rate at which energy is delivered to the resistor is
(
P = I 2 R = 8.5 × 10 −3 A
20.57
)
2
( 9.0 Ω ) = 6.5 × 10 −4 W = 0.65 mW
(c)
An external force directed to the right acts on the bar to balance the magnetic force to the
left. Hence, work is being done by the external force , which is transformed into internal
energy within the resistor.
(a)
The current in the solenoid reaches I sol = 0.632I max in a time of t = t = L R, where
L=
(
)
(
)
4p × 10 −7 T ⋅ m A (12 500 ) 1.00 × 10 −4 m 2
m0 N 2 A
=
= 0.280 H
C
7.00 × 10 −2 m
2
Thus, t = ( 0.280 H ) (14.0 Ω ) = 2.00 × 10 −2 s = 20.0 ms .
(b)
The change in the solenoid current during this time is
⎛ ΔV ⎞
⎛ 60.0 V ⎞
ΔI sol = 0.632I max − 0 = 0.632 ⎜
= 0.632 ⎜
= 2.71 A
⎝ R ⎟⎠
⎝ 14.0 Ω ⎟⎠
so the average back emf is
⎛ 2.71 A ⎞
⎛ ΔI ⎞
= 37.9 V
e back = L ⎜ sol ⎟ = ( 0.280 H ) ⎜
⎝ Δt ⎠
⎝ 2.00 × 10 −2 s ⎟⎠
(c)
The change in the magnitude of the magnetic field at the location of the coil is one-half the change
in the magnitude of the field at the center of the solenoid. Thus, ΔBcoil = 12 ⎡⎣ m 0 nsol ( ΔI sol ) ⎤⎦ , and
the average rate of change of flux through each turn of the coil is
continued on next page
68719_20_ch20_p148-170.indd 166
1/7/11 2:45:58 PM
Induced Voltages and Inductance
( ΔB )coil Acoil
⎛ ΔΦ B ⎞
=
⎜⎝
⎟⎠ =
Δt coil
Δt
=
(d)
I coil =
e coil
Rcoil
=
( 4p × 10
1
2
⎡⎣ m 0 nsol ( ΔI sol ) ⎤⎦ Acoil m 0 N sol ( ΔI sol ) Acoil
=
Δt
2 C sol ⋅ ( Δt )
)
(
T ⋅ m A (12 500 ) ( 2.71 A ) 1.00 × 10 −4 m 2
−7
(
)(
2 7.00 × 10 −2 m 2.00 × 10 −2 s
N coil ( ΔΦ B Δt )coil
Rcoil
=
167
)
)=
1.52 × 10 −3 V
(820 ) (1.52 × 10 −3 V )
24.0 Ω
= 0.051 9 A = 51.9 mA
20.58
(a)
The gravitational force exerted on the ship by the pulsar supplies the centripetal
GM pulsar mship mship v 2
acceleration needed to hold the ship in orbit. Thus, Fg =
, giving
=
2
rorbit
rorbit
GM pulsar
v=
rorbit
=
(6.67 × 10
−11
)(
N ⋅ m kg2 2.0 × 10 30 kg
3.0 × 10 m
7
)=
2.1 × 10 6 m s
(b)
The magnetic force acting on charged particles moving through a magnetic field is
perpendicular to both the magnetic field and the velocity of the particles (and therefore
perpendicular to the ship’s length). Thus, the charged particles in the materials making up
the spacecraft experience magnetic forces directed from one side of the ship to the other,
meaning that the induced emf is directed from side to side within the ship.
(c)
e = B⊥ Cv, where C = 2rship = 0.080 km = 80 m is the side-to-side dimension of the ship.
This yields
(
)
(
)
e = 1.0 × 10 2 T (80 m ) 2.1 × 10 6 m s = 1.7 × 1010 V
20.59
(d)
The very large induced emf would lead to powerful spontaneous electric discharges.
The strong electric and magnetic fields would disrupt the flow of ions in their bodies.
(a)
To move the bar at uniform speed, the magnitude of the applied force must equal that of the
magnetic force retarding the motion of the bar. Therefore, Fapp = BIC. The magnitude of the
induced current is
I=
( ΔΦ B Δt ) = B ( ΔA Δt ) = B Cv
e
=
R
R
R
R
so the field strength is B = IR Cv, giving Fapp = ( IR Cv ) IC = I 2 R v, and the current is
I=
20.60
Fapp ⋅ v
R
=
(1.00 N ) ( 2.00 m s )
8.00 Ω
= 0.500 A
(b)
Pdissipated = I 2 R = ( 0.500 A )2 (8.00 Ω ) = 2.00 W
(c)
Pinput = Fapp ⋅ v = (1.00 N ) ( 2.00 m s ) = 2.00 W
Since the magnetic field outside the solenoid is negligible in comparison to the field inside the
solenoid, we shall assume that the flux through the single-turn square loop is the same as that
through each turn of the solenoid. Then, the induced emf in the square loop is
continued on next page
68719_20_ch20_p148-170.indd 167
1/7/11 2:46:01 PM
168
Chapter 20
ΔΦ B
e=
=
Δt
or
⎛
⎞
Δ ⎜ Binside Asolenoid ⎟
⎝ solenoid
⎠ Δ ( m 0 nI solenoid ) Asolenoid
⎛ ΔI
⎞
= m 0 n ⎜ solenoid ⎟ Asolenoid
=
⎝ Δt ⎠
Δt
Δt
e = m 0 n ( ΔI solenoid Δt ) p r 2 , which gives
(
)(
)
(
e = 4p × 10 −7 T ⋅ m A 3.50 × 10 3 m ( 28.5 A s ) p 2.00 × 10 −2 m
)
2
= 1.58 × 10 −4 V = 0.158 × 10 −3 V = 0.158 mV
20.61
If d is the distance from the lightning bolt to the center of the coil, then
e av =
N ( ΔΦ B ) N ( ΔB ) A N ⎡⎣ m 0 ( ΔI ) 2p d ⎤⎦ A N m 0 ( ΔI ) A
=
=
=
Δt
Δt
Δt
2p d ( Δt )
(
)(
)
)
2
100 4p × 10 −7 T ⋅ m A 6.02 × 10 6 A − 0 ⎡⎣p ( 0.800 m ) ⎤⎦
=
2p ( 200 m ) 10.5 × 10 −6 s
(
= 1.15 × 10 5 V = 115 kV
20.62
When A and B are 3.00 m apart, the area enclosed
by the loop consists of four triangular sections,
each having hypotenuse of 3.00 m, altitude of
2
2
1.50 m, and base of ( 3.00 m ) − (1.50 m ) = 2.60 m.
The decrease in the enclosed area has been
1
2
ΔA = Ai − A f = ( 3.00 m ) − 4 ⎡⎢ (1.50 m ) ( 2.60 m )⎤⎥ = 1.20 m 2
⎣2
⎦
The average induced current has been
I av =
e av
R
=
( ΔΦ B
R
Δt )
=
(
)
2
B ( ΔA Δt ) ( 0.100 T ) 1.20 m 0.100 s
=
= 0.120 A
R
10.0 Ω
As the enclosed area decreases, the flux due to the external field (directed into the page) through
this area also decreases. Thus, the induced current will be directed clockwise around the loop to
create additional flux directed into the page through the enclosed area.
20.63
(a)
(
=
)
2
ΔΦ B
B ΔA B ⎡⎣ p d 4 − 0 ⎤⎦
=
=
Δt
Δt
Δt
e av =
( 25.0 mT ) p ( 2.00 × 10 –2 m )
(
4 50.0 × 10 −3 s
)
2
= 0.157 mV
As the inward-directed flux through the loop decreases, the induced current goes clockwise
around the loop to create additional inward flux through the enclosed area. With positive
charges accumulating at B, point B is positive relative to A .
continued on next page
68719_20_ch20_p148-170.indd 168
1/7/11 2:46:03 PM
Induced Voltages and Inductance
(b)
e av
(
–2
ΔΦ B ( ΔB ) A [(100 − 25.0 ) mT ] p 2.00 × 10 m
=
=
=
Δt
Δt
4 4.00 × 10 −3 s
(
)
)
169
2
= 5.89 mV
As the inward-directed flux through the enclosed area increases, the induced current goes
counterclockwise around the loop in to create flux directed outward through the enclosed
area. With positive charges now accumulating at A, point A is positive relative to B .
20.64
The induced emf in the ring is
e av =
=
ΔΦ B ( ΔB ) Asolenoid ( ΔBsolenoid 2 ) Asolenoid 1 ⎡
⎛ ΔI
⎞⎤
=
=
= ⎢ m 0 n ⎜ solenoid ⎟ ⎥ Asolenoid
⎝
Δt
Δt
Δt
2⎣
Δt ⎠ ⎦
(
)
(
)
2
1⎡
4p × 10 −7 T ⋅ m A (1 000 ) ( 270 A s ) p ⎡⎣3.00 × 10 −2 m ⎤⎦ ⎤ = 4.80 × 10 −4 V
⎢
⎥⎦
2⎣
Thus, the induced current in the ring is
I ring =
20.65
(a)
e av
R
4.80 × 10 −4 V
= 1.60 A
3.00 × 10 –4 Ω
As the rolling axle (of length C = 1.50 m) moves perpendicularly to the uniform magnetic
field, an induced emf of magnitude e = BCv will exist between its ends. The current
produced in the closed-loop circuit by this induced emf has magnitude
I=
(b)
=
e av
R
=
( ΔΦ B
R
Δt )
=
B ( Δ A Δt ) B Cv ( 0.800 T ) (1.50 m ) ( 3.00 m s )
=
=
= 9.00 A
R
R
0.400 Ω
The induced current through the axle will cause the magnetic field to exert a retarding force
I
of magnitude Fr = BIC on the axle. This force is directed opposite to the velocity v to oppose
the motion of the axle. If the axle is to continue moving at constant speed, an applied force
I
in the direction of v, and having magnitude Fapp = Fr , must be exerted on the axle.
Fapp = BIC = ( 0.800 T ) ( 9.00 A ) (1.50 m ) = 10.8 N
(c)
(d)
20.66
(a)
Using right-hand rule number 1, observe that positive charges within the moving axle
experience a magnetic force toward the rail containing point b, and negative charges
experience a force directed toward the rail containing point a. Thus, the rail containing
b is positive relative to the other rail, so b is at the higher potential .
I
I
No . Both the velocity v of the rolling axle and the magnetic field B are unchanged. Thus,
the polarity of the induced emf in the moving axle is unchanged, and the current continues
to be directed from b to a through the resistor R.
The time required for the coil to move distance C and exit the field is t = C v , where v is the
constant speed of the coil. Since the speed of the coil is constant, the flux through the area
enclosed by the coil decreases at a constant rate. Thus, the instantaneous induced emf is the
same as the average emf over the interval t seconds in duration, or
e = −N
(b)
ΔΦ
BC2 NBC2
( 0 − BA)
= −N
=N
=
= NBCv
Δt
t−0
t
Cv
The current induced in the coil is I = e R = NBCv R .
continued on next page
68719_20_ch20_p148-170.indd 169
1/7/11 2:46:04 PM
170
Chapter 20
(c)
Since the coil moves with constant velocity, the power delivered to the coil must equal the
power being dissipated within the coil. This is given by P = I 2 R, or
⎛ N 2 B 2 C2 v 2 ⎞
N 2 B 2 C2 v 2
P=⎜
R=
2
⎟
R
R
⎝
⎠
(d)
The rate that the applied force does work must equal the power delivered to the coil, so
Fapp ⋅ v = P, or
Fapp =
20.67
N 2 B 2 C2 v
P N 2 B 2 C2 v 2 R
=
=
v
R
v
(e)
As the coil is emerging from the field, the flux through the area it encloses is directed into
the page and is decreasing in magnitude. The induced current must flow clockwise around
the coil to generate a magnetic field directed into the page through the area enclosed by the
coil, opposing the decrease in the inward flux.
(f)
As the coil is emerging from the field, the left side of the coil carries an induced current
directed toward the top of the page through a magnetic field that is directed into the
page. Right-hand rule number 1 then shows that this side of the coil will experience
a magnetic force directed to the left , opposing the motion of the coil.
(a)
As the bottom conductor of the loop falls, it cuts across the
magnetic field lines coming out of the page. This induces
an emf of magnitude e = Bwv in this conductor, with the
left end at the higher potential. As a result, an induced current of magnitude
I=
e
Bwv
=
R
R
flows clockwise around the loop. The field then exerts an
upward force of magnitude
B2 w 2 v
⎛ Bwv ⎞
Fm = BIw = B ⎜
w=
⎟
⎝ R ⎠
R
on this current-carrying conductor forming the bottom of the loop. If the loop is falling at
terminal speed, the magnitude of this force must equal the downward gravitational force
acting on the loop. That is, when v = vt, we must have
Fm =
68719_20_ch20_p148-170.indd 170
B 2 w 2 vt
= Mg
R
or
vt =
MgR
B2 w 2
(b)
A larger resistance would make the current smaller, so the loop must reach higher speed
before the magnitude of the magnetic force will equal the gravitational force.
(c)
The magnetic force is proportional to the product of the field and the current, while the
current itself is proportional to the field. If B is cut in half, the speed must become four
times larger to compensate and yield a magnetic force with magnitude equal to the that
of the gravitational force.
1/7/11 2:46:07 PM
21
Alternating Current Circuits and
Electromagnetic Waves
QUICK QUIZZES
1.
Choice (c). The average power is proportional to the rms current which is non-zero even though
the average current is zero. (a) is only valid for an open circuit, for which R → ∞. (b) and (d) can
never be true because iav = 0 for AC currents.
2.
Choice (b). Choices (a) and (c) are incorrect because the unaligned sine curves in Figure 21.9
mean the voltages are out of phase, and so we cannot simply add the maximum (or rms) voltages
across the elements. (In other words, ΔV ≠ ΔVR + ΔVL + ΔVC even though Δv = ΔvR + ΔvL + ΔvC .)
3.
Choice (b). Closing switch A replaces a single resistor with a parallel combination of two
resistors. Since the equivalent resistance of a parallel combination is always less than the lowest
resistance in the combination, the total resistance of the circuit decreases, which causes the
2
+ ( X L − XC ) to decrease.
impedance Z = Rtotal
2
4.
Choice (a). Closing switch A replaces a single resistor with a parallel combination of two resistors. Since the equivalent resistance of a parallel combination is always less than the lowest
resistance in the combination, the total resistance of the circuit decreases, which causes the phase
angle, f = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ , to increase.
5.
Choice (a). Closing switch B replaces a single capacitor with a parallel combination of two
capacitors. Since the equivalent capacitance of a parallel combination is greater than that of either
of the individual capacitors, the total capacitance increases, which causes the capacitive reactance
XC = 1 2p fC to decrease. Thus, the net reactance, X L − XC , increases causing the phase angle,
f = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ , to increase.
6.
Choice (b). Inserting an iron core in the inductor increases both the self-inductance and the
inductive reactance, X L = 2p f L. This means the net reactance, X L − XC , and hence the impedance,
2
Z = Rtotal
+ ( X L − XC ) , increases, causing the current (and therefore, the bulb’s brightness) to
decrease.
2
7.
Choices (b) and (c). Since pressure is force per unit area, changing the size of the area without
altering the intensity of the radiation striking that area will not cause a change in radiation pressure. In (b), the smaller disk absorbs less radiation, resulting in a smaller force. For the same
reason, the momentum in (c) is reduced.
8.
Choices (b) and (d). The frequency and wavelength of light waves are related by the equation
l f = v or f = v l , where the speed of light v is a constant within a given medium. Thus, the
frequency and wavelength are inversely proportional to each other; when one increases the other
must decrease.
171
68719_21_ch21_p171-197.indd 171
1/7/11 2:46:50 PM
172
Chapter 21
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
2.
3.
I
I
In an electromagnetic wave, the electric field E, the magnetic field B, and the direction
I of
propagation of the wave are always mutually perpendicular to each other. Thus, with
the
B(in
I
−x-direction) and the direction of propagation (+ y-direction) both in the xy-plane, E must be
parallel to the z-axis, meaning that either (c) or (d) must be the correct answer. To choose between
these possible answers, recall the right-hand rule for electromagnetic waves (see Section 21.11 in
the textbook). Hold your right hand, with the fingers extended,
I so the thumb is in the direction of
B
propagation ( + y) and your palm is facing the direction of
I (−x-direction). Then the orientation of
the extended fingers is the direction of the electric field E. You should find this to be the negative
z-direction, so the correct choice is (d).
When the frequency doubles, the rms current I rm s = ΔVL , rm s X L = ΔVL , rm s 2p fL is cut in half.
Thus, the new current is I rm s = 3.0 A 2 = 1.5 A, and (e) is the correct answer.
At the resonance frequency, X L = XC and the impedance is Z = R 2 + ( X L − XC ) = R. Thus, the
rms current is I rm s = ΔVrm s Z = (120 V ) ( 20 Ω ) = 6.0 A , and (b) is the correct choice.
2
4.
ΔVL , rm s = I rm s X L = I rm s ( 2p fL) = ( 2.0 A ) 2p ( 60.0 Hz ) ⎡⎣ (1.0 2p ) H ⎤⎦ = 120 V, and the correct
answer is choice (c).
5.
ΔVC , rm s = I rm s XC =
6.
The battery produces a constant current in the primary coil, which generates a constant flux
through the secondary coil. With no change in flux through the secondary coil, there is no induced
voltage across the secondary coil, and choice (e) is the correct answer.
7.
The speed c, frequency f, and wavelength l of an electromagnetic wave are related by c = fl.
The wavelength of the waves in the oven is then
l=
I rm s
1.00 × 10 −3 A
=
= 16.7 V, so choice (a) is correct.
2p fC 2p ( 60.0 Hz ) ⎡⎣ (1.00 2p ) × 10 −6 F ⎤⎦
c 3.00 × 108 m s
=
= 0.122 m = 12.2 cm
f
2.45 × 10 9 Hz
and (b) is the correct choice.
8.
When a power source, AC or DC, is first connected to a RL combination, the presence of the
inductor impedes the buildup of a current in the circuit. The value of the current starts at zero and
increases as the back emf induced across the inductor decreases somewhat in magnitude. Thus,
the correct choice is (c).
9.
The voltage across the capacitor is proportional to the stored charge. This charge, and hence the
voltage ΔvC , is a maximum when the current has zero value and is in the process of reversing
direction after having been flowing in one direction for a half cycle. Thus, the voltage across the
capacitor lags behind the current by 90°, and (a) is the correct choice.
10.
68719_21_ch21_p171-197.indd 172
In an AC circuit, both an inductor and a capacitor store energy for one half of the cycle of the
current and return that energy to the circuit during the other half of the cycle. On the other
hand, a resistor converts electrical potential energy into thermal energy during all parts of the
cycle of the current. Thus, only the resistor has a non-zero average power loss. The power
delivered to a circuit by a generator is given by Pav = I rm s ΔVrm s cosf , where f is the phase
difference between the generator voltage and the current. Among the listed choices, the only
true statement is choice (c).
1/7/11 2:46:52 PM
Alternating Current Circuits and Electromagnetic Waves
173
11.
In an RLC circuit, the instantaneous voltages ΔvR , ΔvL , and ΔvC (across the resistance, inductance,
and capacitance respectively) are not in phase with each other. The instantaneous voltage ΔvR is in
phase with the current, ΔvL leads the current by 90°, while ΔvC lags behind the current by 90°. The
instantaneous values of these three voltages do add algebraically to give the instantaneous voltage
across the RLC combination, but the rms voltages across these components do not add algebraically.
The rms voltages across the three components must be added as vectors (or phasors) to obtain the
correct rms voltage across the RLC combination. Among the listed choices, choice (e) is the false
statement.
12.
If the voltage is a maximum when the current is zero, the voltage is either leading or lagging the current
by 90° (or a quarter cycle) in phase. Thus, the element could be either an inductor or a capacitor. It
could not be a resistor since the voltage across a resistor is always in phase with the current. If the
current and voltage were out of phase by 180°, one would be a maximum in one direction when the
other was a maximum value in the opposite direction. The correct choice for this question is (d).
13.
At resonance of the RLC series circuit, X L = XC , and the impedance becomes
Z = R 2 + ( X L − XC ) = R 2 + 0 = R
2
so choice (c) is correct.
14.
At a frequency of f = 5.0 × 10 2 Hz, the inductive reactance, capacitive reactance, and
impedance are X L = 2p f L, XC =
1
2
, and Z = R 2 + ( X L − XC ) , respectively. This yields
2p f C
2
⎡
⎤
1
Z = ( 20.0 Ω ) + ⎢ 2p ( 5.0 × 10 2 Hz ) ( 0.120 H ) −
⎥ = 51 Ω ,
2
−6
2p ( 5.0 × 10 Hz ) ( 0.75 × 10 F ) ⎥⎦
⎢⎣
ΔVrm s 120 V
and I rm s =
=
= 2.3 A. Choice (a) is the correct answer.
51 Ω
Z
2
15.
Choices (c) and (d) are the only true statements. Electromagnetic waves consist of
oscillating electric and magnetic fields that are perpendicular to each other and to the
direction of propagation. In a vacuum, all electromagnetic waves travel at the same speed,
c. The electromagnetic spectrum consists of waves having broad ranges of frequencies and
wavelengths, which are related by l f = c.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The phase angle in an RLC series circuit is given by
⎛ X − XC ⎞
−1 ⎛ 2p f L − 1 2p fC ⎞
f = tan −1 ⎜ L
⎟⎠
⎟⎠ = tan ⎜⎝
⎝
R
R
which is clearly frequency dependent. As f → 0, f → tan −1 ( − ∞ ) = − 90° and as f → ∞,
f → tan −1 ( + ∞ ) = + 90°. At resonance, X L = XC , and f = tan −1 ( 0 ) = 0°.
4.
68719_21_ch21_p171-197.indd 173
An antenna that is a conducting line responds to the electric field of the electromagnetic
wave—the oscillating electric field causes an electric force on electrons in the wire along its length.
The movement of electrons along the wire is detected as a current by the radio and is amplified.
Thus, a line antenna must have the same orientation as the broadcast antenna. A loop antenna
responds to the magnetic field in the radio wave. The varying magnetic field induces a varying
current in the loop (by Faraday’s law), and this signal is amplified. The loop should be in the
1/7/11 2:46:55 PM
174
Chapter 21
vertical plane containing the line of sight to the broadcast antenna, so the magnetic field lines go
through the area of the loop.
6.
(a)
In an electromagnetic wave, electric and magnetic fields oscillate at right angles to each
other and perpendicular to the direction of propagation of the wave.
(b)
Energy is the quantity transported by the oscillating electric and magnetic fields of the electromagnetic wave.
8.
Consider a typical metal rod antenna for a car radio. Charges in the rod respond to the electric
field portion of the carrier wave. Variations in the amplitude of the incoming radio wave cause
the electrons in the rod to vibrate with amplitudes emulating those of the carrier wave. Likewise,
for frequency modulation, the variations of the frequency of the carrier wave cause constantamplitude vibrations of the electrons in the rod but at frequencies that imitate those of the carrier.
10.
The brightest portion of your face shows where you radiate the most. Your nostrils and the openings of your ear canals are particularly bright. Brighter still are the pupils of your eyes.
12.
The changing magnetic field of the solenoid induces eddy currents in the conducting core. This
is accompanied by I 2 R conversion of electrically-transmitted energy into internal energy in the
conductor.
14.
The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams
throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across different
circuit elements are not simultaneously at their maximum values. Do not forget that an inductor can
induce an emf in itself and that the voltage across it is 90° in phase ahead of the current in the circuit.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a)
1.70 × 10 2 V
(b)
2.40 × 10 2 Ω
(c) The 100-W bulb will have the lower resistance.
4.
6.
(a)
I1, rm s = I 2 , rm s = 1.25 A, I 3 , rm s = 0.833 A
(b)
R1 = R2 = 96.0 Ω, R3 = 144 Ω
(a) 170 V
(b)
30.0 Hz
(c)
120 V
(d)
6.00 A
(e)
8.49 A
(f)
720 W
(g)
6.9 A
(h)
radians
8.
(a)
141 mA
(b)
235 mA
10.
(a)
221 Ω
(b)
0.163 A
(c)
0.231 A
(d) No. The charge is a maximum when the voltage is a maximum, but the voltage across a
capacitor is 90° out of phase with the current.
12.
68719_21_ch21_p171-197.indd 174
(a) 69.3 V
(b)
40 Hz
(d) 196 Ω
(e)
20.3 mF
(c)
0.354 A
1/7/11 2:46:58 PM
Alternating Current Circuits and Electromagnetic Waves
14.
(a)
12.6 Ω
(b)
6.19 A
(c)
8.75 A
16.
(a)
15.0 Hz
(b)
84.9 V
(c)
47.1 Ω
(e)
2.55 A
(f )
0
(d) 1.80 A
(g)
175
i = ( 2.55 A ) sin ( 30p t − p 2 )
(h) 20.9 ms
57.5 Ω
18.
(a)
20.
(a) 194 V
(b) The current leads the voltage by 49.9°.
22.
(a) 138 V
(b)
104 V
(d) 641 V
(e)
See Solution.
(a) 0.11 A
(b)
ΔVR, m ax = 1.3×10 2 V, ΔVL , m ax = 1.2 ×10 2 V
24.
26.
(b)
1.39 A
(c)
ΔvR = 1.3 × 10 2 V, ΔvL = 0, Δvsource = 1.3 × 10 2 V
(d)
ΔvR = 0, ΔvL = 1.2 × 10 2 V, Δvsource = 1.2 × 10 2 V
(a)
88.4 Ω
(b)
107 Ω
(c)
(c)
729 V
1.12 A
(d) The voltage lags the current by 55.8°.
(e) It changes the impedance, and therefore the current, in the circuit.
28.
2.79 kHz
30.
(a) 0.11 A
(b)
ΔVR, m ax = 1.3×10 2 V, ΔVC , m ax = 1.2 ×10 2 V
(c)
ΔvR = 0, ΔvC = 1.2 × 10 2 V, Δvsource = 1.2 × 10 2 V, qC = 300 mC
(d)
ΔvR = 1.3 × 10 2 V, ΔvC = 0, Δvsource = 1.3 × 10 2 V, qC = 0
32.
(a)
66.8 Ω
34.
88.0 W
36.
(a) No, the algebraic sum of the rms voltage drops is 20.9 V.
(b) to the resistor
38.
2.31 kHz
40.
(a)
Z = R = 15 Ω
(c) at resonance
68719_21_ch21_p171-197.indd 175
(b)
0.953 A
(c)
3.2 W
(b)
41 Hz
(d)
2.5 A
(c)
45.4 W
1/7/11 2:46:59 PM
176
42.
Chapter 21
(a) 480 W
(b)
0.192 W
(c) 30.7 mW
(d)
0.192 W
(e)
30.7 mW; Maximum power is delivered at the resonance frequency.
44.
(a) 9.23 V
(b)
30.0 W
46.
(a) fewer turns
(b)
25 mA
(c)
20 turns
48.
(a) 29.0 kW
(b)
0.577%
(c)
See Solution.
50.
(a)
(b)
8.31 min
(c)
2.56 s
52.
2.998 × 108 m s
54.
0.80 or 80%
56.
(a)
I = mgc 2A
(b)
1.46 × 10 9 W m 2
(c)
Propulsion by light pressure in a significant gravity field is impractical because of the
enormous power requirements. In addition, no material is perfectly reflecting, so the
absorbed energy would melt the reflecting surface.
58.
(a)
8.89 × 10 −8 W m 2
60.
11.0 m
62.
The radio listeners hear the news 8.4 ms before the studio audience because radio waves travel so
much faster than sound waves.
64.
(a)
66.
1.1× 10 7 m s
68.
∼ 10 6 J
70.
XC , i = 3R
72.
(a) 0.63 pF
74.
(a)
6.0 Ω
76.
(a)
6.2 mW cm 2, 24% higher than the maximum allowed leakage from a microwave
(b)
0.024 mW cm 2
6.80 × 10 2 y
(b)
6.003 6 × 1014 Hz
11.4 MW
(b)
3.6 × 1011 Hz
(b)
8.4 mm
(b)
12 mH
(c)
25 Ω
PROBLEM SOLUTIONS
21.1
For an AC circuit containing only resistance (the filament of the lightbulb), the power dissipated
2
2
is P = I rm2 s R = ( ΔVrm s R ) R = ΔVrm2 s R = ΔVm ax 2 R.
(
(a)
(
170 V 2
ΔVrm2 s
R=
=
P
75.0 W
)
)
2
= 193 Ω
continued on next page
68719_21_ch21_p171-197.indd 176
1/7/11 2:47:01 PM
Alternating Current Circuits and Electromagnetic Waves
21.2
(
)
2
(b)
170 V 2
ΔVrm2 s
R=
=
P
100.0 W
(a)
ΔVR , m ax = 2 ΔVR , rm s = 2 (1.20 × 10 2 V ) = 1.70 × 10 2 V
(b)
Pav = I
(c)
Because R =
(
177
= 145 Ω
)
2
ΔVrm2 s
ΔVrm2 s (1.20 × 10 V )
R=
⇒ R=
=
= 2.40 × 10 2 Ω
R
Pav
60.0 W
2
2
rm s
ΔVrm2 s
(see above), if the bulbs are designed to operate at the same voltage,
Pav
the 100 W will have the lower resistance .
21.3
For a simple resistance, i ( t ) = v ( t ) R = ( ΔVm ax sinwt ) R = I m ax sinwt. Thus, if i = 0.600I m ax at
t = 7.00 ms, we have
wt = 2p ft = sin −1 ( i I m ax )
giving f =
21.4
sin −1 ( i I m ax )
2p t
0.644 rad
( 0.600 )
=
= 14.6 s −1 = 14.6 Hz .
2p t
( 2p rad )( 7.00 × 10 −3 s )
−1
All lamps are connected in parallel with the voltage source, so ΔVrm s = 120 V for each lamp.
Also, for each bulb, the current is I rm s = Pav ΔVrm s and the resistance is R = ΔVrm s I rm s .
(a)
(b)
21.5
sin
f=
and
For bulbs 1 and 2:
I1, rm s = I 2 , rm s =
For bulb 3:
I 3 , rm s =
R1 = R2 =
150 W
= 1.25 A
120 V
100 W
= 0.833 A
120 V
120 V
= 96.0 Ω
1.25 A
R3 =
and
120 V
= 144 Ω
0.833 A
The total resistance (series connection) is Req = R1 + R2 = 8.20 Ω +10.4 Ω = 18.6 Ω, so the current
in the circuit is
I rm s =
ΔVrm s 15.0 V
=
= 0.806 A
18.6 Ω
Req
The power to the speaker is then Pav = I rm2 s Rspeaker = ( 0.806 A ) (10.4 Ω ) = 6.76 W .
2
21.6
The general form of the generator voltage is Δv = ( ΔVm ax ) sin (wt ), so by inspection
(a)
ΔVR , m ax = 170 V
(c)
ΔVR , rm s =
(d)
I rm s =
ΔVR , m ax
2
ΔVR , rm s
R
=
and
=
(b)
f=
w
60p rad s
=
= 30.0 Hz
2p
2p rad
170 V
= 120 V
2
120 V
= 6.00 A
20.0 Ω
continued on next page
68719_21_ch21_p171-197.indd 177
1/7/11 2:47:03 PM
178
Chapter 21
(e)
I m ax = 2I rm s = 2 ( 6.00 A ) = 8.49 A
(f )
Pav = I rm2 s R = ( 6.00 A ) ( 20.0 Ω ) = 720 W
(g)
At t = 0.005 0 s, the instantaneous current is
2
i=
21.7
Δv (170 V )
(170 V )
=
sin ⎡⎣( 60p rad s ) ( 0.005 0 s ) ⎤⎦ =
sin ( 0.94 rad ) = 6.9 A
R
20.0 Ω
20.0 Ω
(h)
The argument of the sine function has units of [wt ] = ( rad s )( s ) = radians .
(a)
The expression for capacitive reactance is XC = 1 2p fC. Thus, if XC < 175 Ω, it is
necessary that
f=
1
1
>
2p C XC 2p ( 22.0 ×10 −6 F ) (175 Ω)
or
f > 41.3 Hz
For C1 , the reactance is XC,1 = 1 2p fC1, while for C2 , XC, 2 = 1 2p fC 2. Thus, for the same
frequencies, the ratio of the reactance for the two capacitors is
(b)
XC, 2
XC,1
⎛ 1 ⎞ ⎛ 2p f C ⎞ C
1
⎟⎟ ⎜⎜
⎟⎟ = 1
= ⎜⎜
2p
f
C
1
⎝
⎠ C2
2
⎝
⎠
or
⎛C ⎞
XC, 2 = ⎜ 1 ⎟ XC,1
⎝ C2 ⎠
If C1 = 22.0 mF, C2 = 44.0 mF, and XC,1 < 175 Ω, we have
⎛ 22.0 mF ⎞
XC, 2 < ⎜
⎟ (175 Ω)
⎝ 44.0 mF ⎠
21.8
21.9
I m ax = 2 I rm s =
68719_21_ch21_p171-197.indd 178
XC
XC, 2 < 87.5 Ω
= 2 ( ΔVrm s ) 2p f C
(a)
I m ax = 2 (120 V ) 2p ( 60.0 Hz )( 2.20 × 10 − 6 C V ) = 0.141 A = 141 mA
(b)
I m ax = 2 ( 240 V ) 2p ( 50.0 Hz )( 2.20 × 10 − 6 C V ) = 0.235 A = 235 mA
I rm s =
ΔVrm s
=2p f C ( ΔVrm s ) , so
XC
f=
21.10
2 ( ΔVrm s )
or
I rm s
0.30 A
=
= 4.0 × 10 2 Hz
2p C ( ΔVrm s ) 2p ( 4.0 × 10 −6 F )( 30 V )
1
1
=
= 221 Ω
2p fC 2p ( 60.0 Hz )(12.0 × 10 −6 F )
(a)
XC =
(b)
I rm s =
(c)
I m ax = 2 I rm s = 2 ( 0.163 A ) = 0.231 A
(d)
No. The charge is a maximum when the voltage is a maximum, but the voltage across a
capacitor is 90° out of phase with the current.
ΔVC , rm s
XC
=
36.0 V
= 0.163 A
221 Ω
1/7/11 2:47:06 PM
Alternating Current Circuits and Electromagnetic Waves
21.11
21.12
21.13
I m ax =
ΔVm ax
= 2p fC ( ΔVm ax ) = 2p ( 90.0 Hz )( 3.70 × 10 −6 F )( 48.0 V )
XC
or
I m ax = 1.00 × 10 −1 A = 100 mA
(a)
By inspection, ΔVC, m ax = 98.0 V, so ΔVC , rm s =
(b)
Also by inspection, w = 80p rad s, so f =
(c)
I rm s =
(d)
XC =
(e)
XC =
ΔVC, m ax
2
=
179
98.0 V
= 69.3 V
2
w
80p rad s
=
= 40 Hz
2p
2p rad
I m ax 0.500 A
=
= 0.354 A
2
2
ΔVC , m ax
I m ax
=
98.0 V
= 196 Ω
0.500 A
1
1
=
,
2p fC wC
C=
so
X L = 2p ( 60.0 Hz ) L = 54.0 Ω
⇒
1
1
=
= 2.03 × 10 −5 F = 20.3 mF
w XC (80π rad s )(196 Ω )
2p L =
54.0 Ω
= 0.900 Ω ⋅s
60.0 s −1
Then, when ΔVrm s = 100 V and f = 50.0 Hz, the maximum current will be
I m ax =
21.14
21.15
ΔVm ax
=
XL
2 ( ΔVrm s )
( 2p L ) f
=
2 (100 V )
= 3.14 A
( 0.900 Ω ⋅s )( 50.0 Hz )
(a)
X L = 2p f L = 2p (80.0 Hz )( 25.0 × 10 −3 H ) = 12.6 Ω
(b)
I rm s =
(c)
I m ax = 2 I rm s = 2 ( 6.19 A ) = 8.75 A
(a)
I m ax =
L=
(b)
ΔVL , rm s
XL
=
78.0 V
= 6.19 A
12.6 Ω
ΔVm ax ΔVm ax
, so
=
XL
2p fL
ΔVm ax
100 V
=
= 4.24 × 10 −2 H = 42.4 mH
2p f I m ax 2p ( 50.0 Hz )( 7.50 A )
I m ax = ΔVm ax X L = ΔVm ax w L , or I m ax is inversely proportional to w . Thus,
I m ax,1 I m ax, 2 = w 2 w 1 , or
⎛I
⎞
⎛ 7.50 A ⎞
w 2 = ⎜ m ax,1 ⎟ w 1 = ⎜
2p ( 50.0 Hz )] = 942 rad s
⎝ 2.50 A ⎟⎠ [
⎝ I m ax,2 ⎠
21.16
Given: vL = (1.20 × 10 2 V ) sin ( 30p t ) and L = 0.500 H
(a)
By inspection, w = 30p rad s, so f =
w
30p rad s
=
= 15.0 Hz
2p
2p
continued on next page
68719_21_ch21_p171-197.indd 179
1/7/11 2:47:10 PM
180
Chapter 21
(b)
Also by inspection, ΔVL , m ax = 1.20 × 10 2 V, so
ΔVL, rm s =
ΔVL, m ax
2
=
1.20 ×10 2 V
= 84.9 V
2
(c)
X L = 2p fL = w L = ( 30p rad s )( 0.500 H ) = 47.1 Ω
(d)
I rm s =
(e)
I m ax = 2 I rm s = 2 (1.80 A ) = 2.55 A
(f )
The phase difference between the voltage across an inductor and the current through the
inductor is fL = 90°, so the average power delivered to the inductor is
ΔVL, rm s
XL
=
84.9 V
= 1.80 A
47.1 Ω
PL, av = I rm s ΔVL, rm s cosfL = I rm s ΔVL, rm s cos (90°) = 0
(g)
When a sinusoidal voltage with a peak value ΔVL, m ax is applied to an inductor, the current through the inductor also varies sinusoidally in time, with the same frequency as the
applied voltage, and has a maximum value of I m ax = ΔVL, m ax X L . However, the current lags
behind the voltage in phase by a quarter-cycle, or p 2 radians. Thus, if the voltage is given
by ΔvL = ΔVL, m ax sin (wt ) , the current as a function of time is i = I m ax sin (wt − p 2 ) . In the
case of the given inductor, the current through it will be i = ( 2.55 A ) sin ( 30p t − p 2 ) .
(h)
When i = +1.00 A, we have sin ( 30p t − p 2 ) = (1.00 A 2.55 A ) , or
30p t − p 2 = sin −1 (1.00 A 2.55 A ) = sin −1 ( 0.392 ) = 0.403 rad
t=
and
21.17
p 2 rad + 0.403 rad
= 2.09 × 10 −2 s = 20.9 ms
30p rad s
From L = N Φ B I (see Section 20.5 in the textbook), the total flux through the coil is
Φ B, total = N Φ B = L ⋅ I, where Φ B is the flux through a single turn on the coil. Thus,
(Φ
B, total
)
m ax
⎡ ΔV ⎤
= L ⋅ I m ax = L ⋅ ⎢ m ax ⎥
⎣ XL ⎦
=L
21.18
(a)
2 ( ΔVrm s )
2 (120 V )
=
= 0.450 T ⋅ m 2
2p fL
2p (60.0 Hz )
The applied voltage is Δv = ΔVm ax sin (wt ) = (80.0 V ) sin (150t ) , so ΔVm ax = 80.0 V and
w = 2p f = 150 rad s. The impedance for the circuit is
Z = R 2 + ( X L − XC ) = R 2 + ( 2p fL − 1 2p fC ) = R 2 + (w L − 1 w C )
2
or
2
2
⎡
⎤
1
Z = ( 40.0 Ω ) + ⎢(150 rad s ) (80.0 × 10 −3 H ) −
⎥
−6
(150 rad s ) (125 × 10 F ) ⎦⎥
⎢⎣
2
2
= 57.5 Ω
(b)
68719_21_ch21_p171-197.indd 180
I m ax =
ΔVm ax 80.0 V
=
= 1.39 A
Z
57.5 Ω
1/7/11 2:47:13 PM
Alternating Current Circuits and Electromagnetic Waves
21.19
XC =
181
1
1
=
= 66.3 Ω
2p f C 2p ( 60.0 Hz )( 40.0 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
( 50.0 Ω )2 + ( 0 − 66.3 Ω )2 = 83.0 Ω
ΔVrm s 30.0 V
=
= 0.361 A
Z
83.0 Ω
(a)
I rm s =
(b)
ΔVR , rm s = I rm s R = ( 0.361 A )( 50.0 Ω ) = 18.1 V
(c)
ΔVC , rm s = I rm s XC = ( 0.361 A )( 66.3 Ω ) = 23.9 V
(d)
⎛ X − XC ⎞
−1 ⎛ 0 − 66.3 Ω ⎞
f = tan −1 ⎜ L
⎟ = −53.0°
⎟ = tan ⎜⎝
⎝
R ⎠
50.0 Ω ⎠
so the voltage lags behind the current by 53° .
21.20
(a)
X L = 2p f L = 2p ( 50.0 Hz )( 400 × 10 −3 H ) = 126 Ω
XC =
1
1
=
= 719 Ω
2p f C 2p ( 50.0 Hz )( 4.43 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
m
( 500 Ω )2 + (126 Ω − 719 Ω )2 = 776 Ω
ΔVm ax = I m ax Z = ( 0.250 A ) ( 776 Ω ) = 194 V
(b)
⎛ X − XC ⎞
−1 ⎛ 126 Ω − 719 Ω ⎞
f = tan −1 ⎜ L
⎟⎠ = − 49.9°
⎟ = tan ⎜⎝
⎝
R ⎠
500 Ω
Thus, the current leads the voltage by 49.9° .
21.21
(a)
X L = 2p f L = 2p ( 240 Hz )( 2.50 H ) = 3.77 × 10 3 Ω
XC =
1
1
=
= 2.65 × 10 3 Ω
2p f C 2p ( 240 Hz )( 0.250 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
21.22
( 900 Ω )2 + ⎡⎣( 3.77 − 2.65) × 103 Ω ⎤⎦ = 1.44 kΩ
2
ΔVm ax
140 V
=
= 0.097 2 A
1.44 × 10 3 Ω
Z
(b)
I m ax =
(c)
⎡ ( 3.77 − 2.65) × 10 3 Ω ⎤
⎛ X − XC ⎞
f = tan −1 ⎜ L
= tan −1 ⎢
⎟
⎥ = 51.2°
⎝
900 Ω
R ⎠
⎣
⎦
(d)
f > 0 , so the voltage leads the current
X L = 2p fL = 2p ( 60.0 Hz )( 0.100 H ) = 37.7 Ω
XC =
1
1
=
= 265 Ω
2p fC 2p ( 60.0 Hz )(10.0 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
( 50.0 Ω )2 + ( 37.7 Ω − 265 Ω )2 = 233 Ω
(a)
ΔVR, rm s = I rm s R = ( 2.75 A ) (50.0 Ω) = 138 V
(b)
ΔVL, rm s = I rm s X L = ( 2.75 A ) (37.7 Ω) = 104 V
continued on next page
68719_21_ch21_p171-197.indd 181
1/7/11 2:47:16 PM
182
Chapter 21
(c)
ΔVC = I rm s XC = ( 2.75 A )( 265 Ω ) = 729 V
(d)
ΔVrm s = I rm s Z = ( 2.75 A )( 233 Ω ) = 641 V
(e)
21.23
(a)
X L = 2p f L = 2p ( 60.0 Hz )( 460 × 10 −3 H ) = 173 Ω
XC =
1
1
=
= 126 Ω
2p f C 2p ( 60.0 Hz )( 21.0 × 10 −6 F )
⎛ X − XC ⎞
−1 ⎛ 173 Ω − 126 Ω ⎞
Thus, f = tan −1 ⎜ L
⎟⎠ = + 17.4°
⎟ = tan ⎜⎝
⎝
R ⎠
150 Ω
(b)
21.24
X L > XC and f > 0, so the voltage leads the current. Hence, the voltage reaches its
maximum first.
X L = 2p fL = 2p ( 60 Hz ) ( 2.8 H ) = 1.1 × 10 3 Ω
Z = R 2 + ( X L − XC ) =
2
(1.2 × 10 Ω) + (1.1× 10
3
2
3
Ω − 0 ) = 1.6 × 10 3 Ω
2
ΔVm ax
170 V
=
= 0.11 A
1.6 × 10 3 Ω
Z
(a)
I m ax =
(b)
ΔVR, m ax = I m ax R = ( 0.11 A ) (1.2 ×10 3 Ω) = 1.3×10 2 V
ΔVL , m ax = I m ax X L = ( 0.11 A ) (1.1×10 3 Ω) = 1.2 ×10 2 V
(c)
When the instantaneous current is a maximum ( i = I m ax ) , the instantaneous voltage across
the resistor is ΔvR = iR = I m ax R = ΔVR, m ax = 1.3×10 2 V . The instantaneous voltage across
an inductor is always 90° or a quarter cycle out of phase with the instantaneous current.
Thus, when i = I m ax , ΔvL = 0 .
Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path.
Thus, for this series circuit Δvsource = ΔvR + ΔvL , and at this instant when i = I m ax , we have
Δvsource = I m ax R + 0 = 1.3 × 10 2 V .
(d)
68719_21_ch21_p171-197.indd 182
When the instantaneous current i is zero, the instantaneous voltage across the resistor is
ΔvR = iR = 0 . Again, the instantaneous voltage across an inductor is a quarter cycle out
of phase with the current. Thus, when i = 0, we must have ΔvL = ΔVL, m ax = 1.2 ×10 2 V .
Then, applying Kirchhoff’s loop rule to the instantaneous voltages around the series circuit
at the instant when i = 0 gives Δvsource = ΔvR + ΔvL = 0 + ΔVL, m ax = 1.2 ×10 2 V .
1/7/11 2:47:20 PM
Alternating Current Circuits and Electromagnetic Waves
21.25
XC =
1
1
=
= 1.33 × 108 Ω
2p f C 2p ( 60.0 Hz )( 20.0 × 10 −12 F )
(50.0 × 10 Ω) + (1.33 × 10 Ω)
2
Z RC = Rbody
+ XC2 =
and
183
I rm s =
3
( ΔV
secondary
)
rm s
Z RC
=
2
8
2
= 1.33 × 108 Ω
5 000 V
= 3.76 × 10 −5 A
1.33 × 108 Ω
Therefore, ΔVbody, rm s = I rm s Rbody = ( 3.76 × 10 −5 A ) ( 50.0 × 10 3 Ω ) = 1.88 V .
21.26
1
1
=
= 88.4 Ω
2p f C 2p ( 60.0 Hz )( 30.0 × 10 −6 F )
(a)
XC =
(b)
Z = R 2 + ( 0 − XC ) = R 2 + XC2 =
(c)
I m ax =
(d)
The phase angle in this RC circuit is
( 60.0 Ω )2 + (88.4 Ω )2 = 107 Ω
2
ΔVm ax 1.20 × 10 2 V
=
= 1.12 A
Z
107 Ω
⎛ X − XC ⎞
−1 ⎛ 0 − 88.4 Ω ⎞
f = tan −1 ⎜ L
⎟ = −55.8°
⎟ = tan ⎜⎝
⎝
R ⎠
60.0 Ω ⎠
Since f < 0,
21.27
(e)
Adding an inductor will change the impedance and the current in the circuit. If the added
inductive reactance is X L < 2XC , the impedance will be decreased and the current will increase.
However, if X L > 2XC , the impedance will be increased and the current will decrease.
(a)
X L = 2p fL = 2p ( 50.0 Hz )( 0.150 H ) = 47.1 Ω
(b)
XC =
(c)
Z=
(d)
Z = R 2 + ( X L − XC )
R=
(e)
21.28
the voltage lags behind the current by 55.8° .
1
1
=
= 637 Ω
2p f C 2p ( 50.0 Hz )( 5.00 × 10 −6 F )
ΔVm ax
240 V
=
= 2.40 × 10 3 Ω = 2.40 kΩ
1.00 × 10 −1 A
I m ax
( 2.40 × 10 Ω)
3
2
2
so
R = Z 2 − ( X L − XC )
2
− ( 47.1 Ω − 637 Ω ) = 2.33 × 10 3 Ω = 2.33 kΩ
2
⎛ X − XC ⎞
−1 ⎛ 47.1 Ω − 637 Ω ⎞
f = tan −1 ⎜ L
⎟ = − 14.2°
⎟⎠ = tan ⎜⎝
⎝
R
2.33 × 10 3 Ω ⎠
X L = w L and XC =1 wC , where w = 2p f . At resonance (i.e., when X L = XC ), we have
w 0 = 2p f0 =
1
LC
or
f0 =
1
2p LC
With L = 57.0 mH and C = 57.0 mF, this resonance frequency is
f0 =
68719_21_ch21_p171-197.indd 183
1
2p
(57.0 × 10
−6
H ) ( 57.0 × 10
−6
F)
= 2.79 × 10 3 Hz = 2.79 kHz
1/7/11 2:47:23 PM
184
21.29
Chapter 21
X L = 2p fL = 2p ( 50.0 Hz )( 0.185 H ) = 58.1 Ω
XC =
1
1
=
= 49.0 Ω
2p f C 2p ( 50.0 Hz )( 65.0 × 10 −6 F )
Z ad = R 2 + ( X L − XC ) =
2
21.30
(
( 40.0 Ω )2 + ( 58.1 Ω − 49.0 Ω )2 = 41.0 Ω
)
ΔVm ax 2
ΔVrm s
150 V
=
=
= 2.59 A
Z ad
Z ad
( 41.0 Ω ) 2
and
I rm s =
(a)
Z ab = R = 40.0 Ω, so ( ΔVrm s )ab = I rm s Z ab = ( 2.59 A )( 40.0 Ω ) = 104 V
(b)
Z bc = X L = 58.1 Ω, and ( ΔVrm s )bc = I rm s Z bc = ( 2.59 A )( 58.1 Ω ) = 150 V
(c)
Z cd = XC = 49.0 Ω, and ( ΔVrm s )cd = I rm s Z cd = ( 2.59 A )( 49.0 Ω ) = 127 V
(d)
Z bd = X L − XC = 9.10 Ω, so ( ΔVrm s )bd = I rm s Z bd = ( 2.59 A )( 9.10 Ω ) = 23.6 V
XC =
1
1
=
= 1.1× 10 3 Ω
2p fC 2p ( 60 Hz )( 2.5 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
(1.2 × 10 Ω) + ( 0 − 1.1× 10 Ω)
3
2
3
2
= 1.6 × 10 3 Ω
ΔVm ax
170 V
=
= 0.11 A
1.6 × 10 3 Ω
Z
(a)
I m ax =
(b)
ΔVR , m ax = I m ax R = ( 0.11 A )(1.2 × 10 3 Ω ) = 1.3 × 10 2 V
ΔVC , m ax = I m ax XC = ( 0.11 A )(1.1× 10 3 Ω ) = 1.2 × 10 2 V
(c)
When the instantaneous current i is zero, the instantaneous voltage across the
resistor is ΔvR = iR = 0 . The instantaneous voltage across a capacitor is always 90°,
or a quarter cycle, out of phase with the instantaneous current. Thus, when i = 0,
ΔvC = ΔVC, m ax = 1.2 ×10 2 V , and
qC = C ( ΔvC ) = ( 2.5 × 10 −6 F ) (1.2 × 10 2 V ) = 3.0 × 10 −4 C = 300 mC
Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path.
Thus, for this series circuit, Δvsource + ΔvR + ΔvC = 0, and at this instant when i = 0, we
have Δvsource = 0 + ΔvC = ΔVC , m ax = 1.2 ×10 2 V .
(d)
When the instantaneous current is a maximum (i = I m ax ), the instantaneous voltage across
the resistor is ΔvR = iR = I m ax R = ΔVR, m ax = 1.3×10 2 V . Again, the instantaneous voltage
across a capacitor is a quarter cycle out of phase with the current. Thus, when i = I m ax ,
we must have ΔvC = 0 and qC = C ΔvC = 0 . Then, applying Kirchhoff’s loop rule
to the instantaneous voltages around the series circuit at the instant when i = I m ax and
ΔvC = 0 gives
Δvsource + ΔvR + ΔvC = 0
68719_21_ch21_p171-197.indd 184
⇒
Δvsource = ΔvR = ΔVR, m ax = 1.3×10 2 V
1/7/11 2:47:26 PM
Alternating Current Circuits and Electromagnetic Waves
21.31
ΔVrm s
104 V
=
= 208 Ω
I rm s
0.500 A
(a)
Z=
(b)
Pav = I rm2 s R gives R =
(c)
Z = R 2 + X L2 , so X L = Z 2 − R 2 =
L=
and
21.32
185
Pav
10.0 W
=
= 40.0 Ω
2
I rm2 s ( 0.500 A )
( 208 Ω )2 − ( 40.0 Ω )2 = 204 Ω
XL
204 Ω
=
= 0.541 H
2p f 2p ( 60.0 Hz )
Given v = ΔVm ax sin (wt ) = ( 90.0 V ) sin ( 350t ) , observe that ΔVm ax = 90.0 V and w = 350 rad s.
Also, the net reactance is X L − XC = 2p fL − 1 2p fC = w L − 1 w C.
(a)
X L − XC = w L −
1
1
= ( 350 rad s )( 0.200 H ) −
= − 44.3 Ω
wC
(350 rad s ) ( 25.0 × 10 −6 F )
Z = R 2 ( X L − XC ) =
2
so the impedance is
ΔVrm s ΔVm ax
=
Z
Z
2
( 50.0 Ω )2 + ( − 44.3 Ω )2 = 66.8 Ω
90.0 V
= 0.953 A
2 ( 66.8 Ω )
(b)
I rm s =
=
(c)
The phase difference between the applied voltage and the current is
⎛ X − XC ⎞
−1 ⎛ − 44.3 Ω ⎞
f = tan −1 ⎜ L
⎟ = −41.5°
⎟⎠ = tan ⎜⎝
⎝
R
50.0 Ω ⎠
so the average power delivered to the circuit is
⎛ ΔV ⎞
⎛ 90.0 V ⎞
Pav = I rm s ΔVrm s cosf = I rm s ⎜ m ax ⎟ cosf = ( 0.953 A ) ⎜
⎟ cos ( −41.5° ) = 45.4 W
⎝ 2 ⎠
⎝
2 ⎠
21.33
If Δv = (100 V ) sin ⎡⎣(1 000 rad s ) t ⎤⎦ , then ΔVm ax = 100 V and w = 1 000 rad s.
Thus,
X L = w L = (1 000 rad s )( 0.500 H ) = 500 Ω
and
XC =
Therefore,
and
I rm s =
1
1
=
= 200 Ω
w C (1 000 rad s ) ( 5.00 × 10 −6 C )
Z = R 2 + ( X L − XC ) =
2
ΔVrm s ΔVm ax
=
Z
Z
2
=
( 400 Ω )2 + ( 300 Ω )2 = 500 Ω
100 V
0.200 A
=
2 ( 500 Ω )
2
The power delivered to the circuit equals the power dissipated in the resistor, or
2
⎛ 0.200 A ⎞
P = I rm2 s R = ⎜
⎟ ( 400 Ω ) = 8.00 W
⎝
2 ⎠
21.34
The rms current in the circuit is
I rm s =
ΔVrm s 160 V
=
= 2.00 A
80.0 Ω
Z
continued on next page
68719_21_ch21_p171-197.indd 185
1/7/11 2:47:30 PM
186
Chapter 21
and the average power delivered to the circuit is
Pav = I rm s ( ΔVrm s cosf ) = I rm s ΔVR , rm s = I rm s ( I rm s R ) = I rm2 s R = ( 2.00 A ) ( 22.0 Ω ) = 88.0 W
2
21.35
Pav = I rm2 s R = I rm s ( I rm s R) = I rm s ( ΔVR, rm s ), so I rm s =
(a)
R=
Thus,
ΔVR, rm s
50 V
= 1.8 ×10 2 Ω
0.28 A
=
I rm s
Pav
14 W
=
= 0.28 A
ΔVR , rm s 50 V
Z = R 2 + X L2 , which yields
(b)
2
2
⎛ ΔV ⎞
2
⎛ 90 V ⎞
X L = Z − R = ⎜ rm s ⎟ − R 2 = ⎜
− (1.8 × 10 2 Ω ) = 2.7 × 10 2 Ω
⎟
⎝
⎠
I
0.28
A
⎝ rm s ⎠
2
and
21.36
L=
2
XL
2.7 × 10 2 Ω
=
= 0.72 H
2p f 2p ( 60 Hz )
X L = 2p fL = 2p ( 600 Hz )( 6.0 × 10 −3 H ) = 23 Ω
XC =
1
1
=
= 11 Ω
2p fC 2p ( 600 Hz )( 25 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
(a)
( 25 Ω )2 + ( 23 Ω − 11 Ω )2 = 28 Ω
⎛ ΔV ⎞
⎛ 10 V ⎞
ΔVR, rm s = I rm s R = ⎜ rm s ⎟ R = ⎜
⎟ ( 25 Ω) = 8.9 V
⎝ 28 Ω ⎠
⎝ Z ⎠
⎛ ΔV ⎞
⎛ 10 V ⎞
ΔVL, rm s = I rm s X L = ⎜ rm s ⎟ X L = ⎜
⎟ ( 23 Ω) = 8.2 V
⎝ 28 Ω ⎠
⎝ Z ⎠
⎛ ΔV ⎞
⎛ 10 V ⎞
ΔVC, rm s = I rm s XC = ⎜ rm s ⎟ XC = ⎜
⎟ (11 Ω) = 3.9 V
⎝ 28 Ω ⎠
⎝ Z ⎠
No , ΔVR, rm s + ΔVL, rm s + ΔVC , rm s = 8.9 V + 8.2 V + 3.8 V = 20.9 V ≠ 10 V
(b)
The power delivered to the resistor is the greatest. No power losses occur in an ideal
capacitor or inductor.
⎛ ΔV ⎞
⎛ 10 V ⎞
Pav = I rm2 s R = ⎜ rm s ⎟ R = ⎜
( 25 Ω ) = 3.2 W
⎝ 28 Ω ⎟⎠
⎝ Z ⎠
2
(c)
21.37
(
)
The resonance frequency of a series RLC circuit is f0 = 1 2p LC . Thus, if L = 1.40 mH and
the desired resonance frequency is f0 = 99.7 MHz, the needed capacitance is
C=
21.38
2
1
1
=
= 1.82 × 10 −12 F = 1.82 pF
2
4p 2 f02 L 4p 2 ( 99.7 × 10 6 Hz ) (1.40 × 10 −6 H )
(
)
The resonance frequency of a series RLC circuit is f0 = 1 2p LC . Thus, the ratio of the resonance frequencies when the same inductance is used with two different capacitances in the circuit is
f0, 2
f0,1
⎛
1
= ⎜⎜
⎝ 2p LC2
⎞ ⎛ 2p LC ⎞
C1
1
⎟⎟ ⎜⎜
⎟⎟ =
1
C2
⎠⎝
⎠
continued on next page
68719_21_ch21_p171-197.indd 186
1/7/11 2:47:33 PM
Alternating Current Circuits and Electromagnetic Waves
187
If f0,1 = 2.84 kHz when C1 = 6.50 mF, the resonance frequency when the capacitance is
C2 = 9.80 mF will be
f0, 2 = f0,1
21.39
f0 =
C1
6.50 mF
= ( 2.84 kHz )
= 2.31 kHz
C2
9.80 mF
1
1
, so C =
4p 2 f02 L
2p LC
For f0 = ( f0 )m in = 500 kHz = 5.00 × 10 5 Hz
C = Cm ax =
1
4p
2
= 5.1× 10 −8 F = 51 nF
(5.00 × 10 Hz ) ( 2.0 × 10−6 H )
2
5
For f0 = ( f0 )m ax = 1600 kHz = 1.60 × 10 6 Hz
C = Cm in =
21.40
(a)
4p
2
(1.60 × 10
1
Hz ) ( 2.0 × 10 −6 H )
2
6
= 4.9 × 10 −9 F = 4.9 nF
At resonance, X L = XC , so the impedance will be
Z = R 2 + ( X L − XC ) = R 2 + 0 = R = 15 Ω
2
(b)
When X L = XC , we have 2p fL =
f=
1
2p LC
=
1
, which yields
2p fC
1
2p
( 0.20 H )( 75 × 10 −6 F )
= 41 Hz
(c)
The current is a maximum at resonance , where the impedance has its minimum value
of Z = R.
(d)
At f = 60 Hz, X L = 2p ( 60 Hz )( 0.20 H ) = 75 Ω, XC =
1
= 35 Ω,
2p ( 60 Hz )( 75 × 10 −6 F )
and
Z=
(15 Ω )2 + ( 75 Ω − 35 Ω )2 = 43 Ω
Thus, I rm s =
21.41
(
ΔVm ax
ΔVrm s
=
Z
Z
2
)=
150 V
= 2.5 A .
2 ( 43 Ω )
(
)
The resonance frequency of an LC circuit is f0 = 1 2p LC . If the capacitance remains constant
while the inductance increases by 1.000%, the new resonance frequency will be
f0′ =
f0
1
1
⎛ 1 ⎞
=⎜
⎟⎠ 2p LC = 1.010
⎝
1.010
2p (1.010L )C
The beat frequency detected in this case will be fbeat = f0 − f0′, or
1 ⎞
1 ⎞
⎛
⎛
fbeat = ⎜ 1−
f0 = ⎜ 1−
( 725 kHz ) = 3.60 kHz
⎟
⎝
⎝
1.010 ⎠
1.010 ⎟⎠
68719_21_ch21_p171-197.indd 187
1/7/11 2:47:36 PM
188
21.42
Chapter 21
The resonance frequency is w 0 = 2p f0 = 1
(a)
LC . Also, X L = w L and XC = 1 w C .
L
⎛ 1 ⎞
At resonance, XC = X L = w 0 L = ⎜
L=
=
⎝ LC ⎟⎠
C
3.00 H
= 1 000 Ω.
3.00 × 10 −6 F
Thus, Z = R 2 + 0 2 = R, I rm s = ΔVrm s Z = 120 V 30.0 Ω = 4.00 A, and
2
Pav = I rm2 s R = ( 4.00 A ) ( 30.0 Ω ) = 480 W .
(b)
(
1
1
At w = w 0 ; X L = X L
2
2
w0
) = 500 Ω, X = 2 ( X ) = 2 000 Ω,
C
Z = R 2 + ( X L − XC ) =
( 30.0 Ω )2 + ( 500 Ω − 2 000 Ω )2 = 1 500 Ω
2
and I rm s =
(c)
C w
0
120 V
2
= 0.080 0 A. Thus, Pav = I rm2 s R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W .
1 500 Ω
(
(
)
1
1
At w = w 0 ; X L = X L w = 250 Ω, XC = 4 XC
4
4
120 V
and I rm s =
= 0.032 0 A.
3 750 Ω
0
w0
) = 4 000 Ω,
Z = 3 750 Ω,
Therefore, Pav = I rm2 s R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW .
2
(d)
(
At w = 2w 0 ; X L = 2 X L
I rm s =
(e)
w0
) = 2 000 Ω, X
C
=
(
1
XC
2
w0
) = 500 Ω,
Z = 1 500 Ω, and
120 V
2
= 0.080 0 A. Then, Pav = I rm2 s R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W .
1 500 Ω
(
At w = 4w 0 ; X L = 4 X L
and I rm s =
w0
) = 4 000 Ω, X
C
=
(
1
XC
4
w0
) = 250 Ω,
Z = 3 750 Ω,
120 V
= 0.032 0 A. Hence,
3 750 Ω
Pav = I rm2 s R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW
2
The power delivered to the circuit is a maximum when the rms current is a maximum. This
occurs when the frequency of the source is equal to the resonance frequency of the circuit.
21.43
The maximum output voltage ( ΔVm ax )2 is related to the maximum input voltage ( ΔVm ax )1 by the
N
expression ( ΔVm ax )2 = 2 ( ΔVm ax )1, where N1 and N 2 are the number of turns on the primary coil
N1
and the secondary coil, respectively. Thus, for the given transformer,
( ΔV )
m ax
2
=
1 500
(170 V ) = 1.02 × 103 V
250
and the rms voltage across the secondary is ( ΔVrm s )2 =
21.44
(a)
( ΔV )
m ax
2
2
=
1.02 × 10 3 V
= 721 V .
2
The output voltage of the transformer is
⎛N ⎞
⎛ 1⎞
ΔV2 , rm s = ⎜ 2 ⎟ ΔV1, rm s = ⎜ ⎟ (120 V ) = 9.23 V
⎝ 13 ⎠
⎝ N1 ⎠
continued on next page
68719_21_ch21_p171-197.indd 188
1/7/11 2:47:40 PM
Alternating Current Circuits and Electromagnetic Waves
(b)
Assuming an ideal transformer, Poutput = Pinput , and the power delivered to the CD player is
(
(P ) = (P )
av
21.45
189
)
= I1, rm s ΔV1, rm s = ( 0.250 A )(120 V ) = 30.0 W
av 1
2
The power input to the transformer is
(P )
av
(
)
= ΔV1, rm s I1, rm s = ( 3 600 V )( 50 A ) = 1.8 × 10 5 W
input
For an ideal transformer, ( Pav )ouput = ( ΔV2, rm s ) I 2, rm s = ( Pav )input, so the current in the long-distance
power line is
I 2 , rm s =
(P )
av
(ΔV
=
input
2, rm s
)
1.8 ×10 5 W
= 1.8 A
100 000 V
The power dissipated as heat in the line is then
Plost = I 22, rm s Rline = (1.8 A ) (100 Ω ) = 3.2 × 10 2 W
2
The percentage of the power delivered by the generator that is lost in the line is
% Lost =
21.46
(a)
(b)
⎛ 3.2 × 10 2 W ⎞
Plost
× 100% = ⎜
× 100% = 0.18%
Pinput
⎝ 1.8 × 10 5 W ⎟⎠
Since the transformer is to step the voltage down from 120 volts to 6.0 volts, the secondary
must have fewer turns than the primary.
For an ideal transformer, ( Pav )input = ( Pav )ouput , or ( ΔV1, rm s ) I1, rm s = ( ΔV2, rm s ) I 2, rm s , so the
current in the primary will be
(ΔV
I1, rm s =
(c)
)I
2, rm s
2, rm s
ΔV1, rm s
=
(6.0 V ) (500 mA ) = 25 mA
120 V
The ratio of the secondary to primary voltages is the same as the ratio of the number of
turns on the secondary and primary coils, ΔV2 ΔV1 = N 2 N1. Thus, the number of turns
needed on the secondary coil of this step down transformer is
⎛ ΔV ⎞
⎛ 6.0 V ⎞
N 2 = N1 ⎜ 2 ⎟ = ( 400 ) ⎜
= 20 turns
⎝ 120 V ⎟⎠
⎝ ΔV1 ⎠
21.47
(a)
At 90% efficiency, ( Pav )output = 0.90 ( Pav )input. Thus, if ( Pav )output = 1 000 kW, the input power
to the primary is
(P )
av
68719_21_ch21_p171-197.indd 189
(b)
I1, rm s =
(c)
I 2, rm s =
input
(P )
av
=
(P )
input
ΔV1, rm s
(P )
av
output
ΔV2, rm s
av
output
0.90
=
=
1 000 kW
= 1.1× 10 3 kW
0.90
1.1×10 3 kW 1.1×10 6 W
= 3.1×10 2 A
=
ΔV1, rm s
3 600 V
=
1 000 kW 1.0 ×10 6 W
=
= 8.3×10 3 A
ΔV2, rm s
120 V
1/7/11 2:47:44 PM
190
21.48
Chapter 21
Rline = ( 4.50 × 10 −4 Ω m ) ( 6.44 × 10 5 m ) = 290 Ω
(a)
The power transmitted is ( Pav )transm itted = ( ΔVrm s ) I rm s, so
I rm s =
(P )
av
transm itted
ΔVrm s
=
5.00 × 10 6 W
= 10.0 A
500 × 10 3 V
Thus, ( Pav )loss = I rm2 s Rline = (10.0 A ) ( 290 Ω ) = 2.90 × 10 4 W = 29.0 kW .
2
(b)
The power input to the line is
(P )
av
input
= ( Pav )transm itted + ( Pav )loss = 5.00 × 10 6 W + 2.90 × 10 4 W = 5.03 × 10 6 W
and the fraction of input power lost during transmission is
fraction =
(P )
(P )
av
av
(c)
loss
=
input
2.90 × 10 4 W
= 0.005 77 or 0.577%
5.03 × 10 6 W
It is impossible to deliver the needed power at the generator voltage of 4.50 kV. The
maximum line current with an input voltage of 4.50 kV to the line occurs when the line is
shorted out at the customer’s end, and this current is
(I )
rm s
m ax
=
ΔVrm s 4 500 V
=
= 15.5 A
290 Ω
Rline
The maximum input power is then
(P )
input
m ax
= ( ΔVrm s ) ( I rm s )m ax
= ( 4.50 × 10 3 V )(15.5 A ) = 6.98 × 10 4 W = 69.8 kW
This is far short of meeting the customer’s request, and all of this power is lost in the
transmission line.
21.49
From v = l f , the wavelength is
l=
v 3.00 × 108 m s
=
= 4.00 × 10 6 m = 4 000 km
f
75 Hz
The required length of the antenna is then
L = l 4 = 1 000 km , or about 621 miles. Not very practical at all.
21.50
d
6.44 × 1018 m ⎛
1y
⎞
2
=
⎜
⎟ = 6.80 × 10 y
c 3.00 × 108 m s ⎝ 3.156 × 10 7 s ⎠
(a)
t=
(b)
From Table C.4 (in Appendix C of the textbook), the average Earth-Sun distance is
d = 1.496 × 1011 m, giving the transit time as
t=
d 1.496 × 1011 m ⎛ 1 min ⎞
=
⎜
⎟ = 8.31 min
c 3.00 × 108 m s ⎝ 60 s ⎠
continued on next page
68719_21_ch21_p171-197.indd 190
1/7/11 2:47:46 PM
Alternating Current Circuits and Electromagnetic Waves
(c)
Also from Table C.4, the average Earth-Moon distance is d = 3.84 × 108 m, giving the time
for the round trip as
8
2d 2 ( 3.84 × 10 m )
=
= 2.56 s
8
c
3.00 × 10 m s
t=
21.51
(a)
191
The solar energy incident each second on 1.00 m 2 of the surface of Earth’s atmosphere is
W
Js
N⋅m
N
= 1370 2 = 1370 2 = 1370
m2
m
m ⋅s
m ⋅s
U total = 1 370
Of this, 38.0% is reflected and 62.0% is absorbed. From Equation 21.29 and Equation 21.30
in the textbook, the radiation pressures P1 due to the reflected radiation and P2 due to the
absorbed radiation are given by
P1 =
2 U reflected 2 ( 0.380 U total ) 0.760 U total
=
=
c
c
c
P2 =
and
U absorbed 0.620 U total
=
c
c
The total radiation pressure is then
( 0.760 + 0.620 ) U total
Prad = P1 + P2 =
or
(b)
Prad =
c
1.38 (1 370 N m ⋅s )
= 6.30 × 10 −6 N m 2 = 6.30 × 10 −6 Pa
3.00 × 108 m s
In comparison, atmospheric pressure at the surface of the Earth is
Patm
101× 10 3 Pa
=
= 1.60 × 1010 times greater than the radiation pressure
Prad 6.30 × 10 −6 Pa
21.52
21.53
c=
1
=
m 0 ∈0
or
c = 2.998 × 108 m s
(a)
The frequency of an electromagnetic wave is f = c l, where c is the speed of light and l is
the wavelength of the wave. The frequencies of the two light sources are then
Red:
1
( 4p × 10
fred =
−7
N ⋅s C
2
2
) (8.854 × 10
−12
C2 N ⋅ m 2 )
c
3.00 × 108 m s
=
= 4.55 × 1014 Hz
lred
660 × 10 −9 m
and
Infrared:
(b)
fIR =
c
3.00 × 108 m s
=
= 3.19 × 1014 Hz
lIR
940 × 10 −9 m
The intensity of an electromagnetic wave is proportional to the square of its amplitude.
If 67% of the incident intensity of the red light is absorbed, then the intensity of the
continued on next page
68719_21_ch21_p171-197.indd 191
1/7/11 2:47:48 PM
192
Chapter 21
emerging wave is (100% − 67%) = 33% of the incident intensity, or I f = 0.33I i . Hence, we
must have
Em ax, f
Em ax, i
21.54
=
If
Ii
= 0.33 = 0.57
If I 0 is the incident intensity of a light beam, and I is the intensity of the beam after passing
through length L of a fluid having concentration C of absorbing molecules, the Beer-Lambert law
states that log10 ( I I 0 ) = − ∈CL, where ∈ is a constant.
For 660-nm light, the absorbing molecules are oxygenated hemoglobin. Thus, if 33% of this
wavelength light is transmitted through blood, the concentration of oxygenated hemoglobin in the
blood is
CHB O2 =
− log10 ( 0.33)
∈L
[1]
The absorbing molecules for 940-nm light are deoxygenated hemoglobin, so if 76% of this light
is transmitted through the blood, the concentration of these molecules in the blood is
CHB =
− log10 ( 0.76)
∈L
[2]
Dividing Equation [2] by Equation [1] gives the ratio of deoxygenated hemoglobin to oxygenated
hemoglobin in the blood as
log10 ( 0.76 )
CHB
=
= 0.25
CHB O2 log10 ( 0.33)
or
CHB = 0.25CHBO2
Since all the hemoglobin in the blood is either oxygenated or deoxygenated, it is necessary that
CHB + CHB O2 = 1.00, and we now have 0.25CHBO2 + CHB O2 = 1.0. The fraction of hemoglobin that is
oxygenated in this blood is then
CHB O2 =
1.0
= 0.80
1.0 + 0.25
or
80%
Someone with only 80% oxygenated hemoglobin in the blood is probably in serious trouble,
needing supplemental oxygen immediately.
21.55
From Intensity =
Em ax Bm ax
c Bm2 ax
E
and m ax = c, we find Intensity =
.
2 m0
Bm ax
2 m0
Thus,
Bm ax =
and
21.56
(a)
2 m0
( Intensity ) =
c
2 ( 4p × 10 −7 T ⋅ m A )
3.00 × 108 m s
(1 370
W m 2 ) = 3.39 × 10 −6 T
Em ax = Bm ax c = ( 3.39 × 10 −6 T ) ( 3.00 × 108 m s ) = 1.02 × 10 3 V m
To exert an upward force on the disk, the laser beam should be aimed vertically upward,
striking the lower surface of the disk. To just levitate the disk, the upward force exerted on
the disk by the beam should equal the weight of the disk.
continued on next page
68719_21_ch21_p171-197.indd 192
1/7/11 2:47:50 PM
Alternating Current Circuits and Electromagnetic Waves
193
The momentum that electromagnetic radiation of intensity I, incident normally on a perfectly
reflecting surface of area A, delivers to that surface in time Δt is given by Equation 21.30
in the textbook as Δp = 2U c = 2 ( IAΔt ) c. Thus, from the impulse-momentum theorem,
the average force exerted on the reflecting surface is F = Δp Δt = 2IA c. To just levitate
the surface, F = 2IA c = mg, and the required intensity of the incident radiation is
I = mgc 2A .
21.57
−3
2
8
mgc mgc ( 5.00 × 10 kg ) ( 9.80 m s ) ( 3.00 × 10 m s )
=
=
= 1.46 × 10 9 W m 2
2
−2
2A 2p r 2
2p ( 4.00 × 10 m )
(b)
I=
(c)
Propulsion by light pressure in a significant gravity field is impractical because of the
enormous power requirements. In addition, no material is perfectly reflecting, so the
absorbed energy would melt the reflecting surface.
The distance between adjacent antinodes in a standing wave is l 2.
Thus, l = 2 ( 6.00 cm ) = 12.0 cm = 0.120 m, and
c = lf = ( 0.120 m )( 2.45 × 10 9 Hz ) = 2.94 × 108 m s
21.58
(a)
The intensity of the radiation at distance r = 20.0 ly from the star is
I=
(b)
P
4.00 × 10 28 W
=
= 8.89 × 10 −8 W m 2
2
4p r 2 4p ⎡( 20.0 ly ) ( 9.461× 1015 m ly ) ⎤
⎣
⎦
The power of the starlight intercepted by Earth, with cross-sectional area of Acs = p RE2 , is
P = IAcs = I ⋅p RE2 = (8.89 × 10 −8 W m 2 ) p ( 6.38 × 10 6 m )
21.59
or
P = 1.14 × 10 7 W = 11.4 MW
(a)
l=
c 3.00 × 108 m s
=
= 6.00 × 10 −12 m = 6.00 pm
f 5.00 × 1019 Hz
(b)
l=
c 3.00 × 108 m s
=
= 7.50 × 10 −2 m = 7.50 cm
f
4.00 × 10 9 Hz
21.60
l=
21.61
(a)
c 3.00 × 108 m s
=
= 11.0 m
f 27.33 × 10 6 Hz
For the AM band,
lm in =
lm ax =
(b)
c
fm ax
c
fm in
=
3.00 × 108 m s
= 188 m
1 600 × 10 3 Hz
=
3.00 × 108 m s
= 556 m
540 × 10 3 Hz
For the FM band,
lm in =
lm ax =
68719_21_ch21_p171-197.indd 193
2
c
fm ax
c
fm in
=
3.00 × 108 m s
= 2.78 m
108 × 10 6 Hz
=
3.00 × 108 m s
= 3.4 m
88 × 10 6 Hz
1/7/11 2:47:53 PM
194
21.62
Chapter 21
The transit time for the radio wave is
tR =
dR
100 × 10 3 m
=
= 3.33 × 10 −4 s = 0.333 ms
c
3.00 × 108 m s
and that for the sound wave is
ds
3.0 m
=
= 8.7 × 10 −3 s = 8.7 ms
vsound 343 m s
ts =
Thus, the radio listeners hear the news 8.4 ms before the studio audience because radio waves
travel so much faster than sound waves.
21.63
If an object of mass m is attached to a spring of spring constant k, the natural frequency of vibration
of that system is f = k m 2p. Thus, the resonance frequency of the C=O double bond will be
f=
1
2p
k
1
=
moxygen 2p
2 800 N m
= 5.2 × 1013 Hz
2.66 × 10 −26 kg
atom
and the light with this frequency has wavelength
l=
c 3.00 × 108 m s
=
= 5.8 × 10 −6 m = 5.8 mm
f
5.2 × 1013 Hz
The infrared region of the electromagnetic spectrum ranges from lm ax ≈ 1 mm down to
lm in = 700 nm = 0.7 mm. Thus, the computed wavelength falls within the infrared region .
21.64
(a)
Since the space station and the ship are moving toward one another, the frequency after
being Doppler shifted is fO = fS (1+ u c ) , where u is the relative speed between the
observer and source. Thus,
⎡ 1.800 0 × 10 5 m s ⎤
14
fO = ( 6.000 0 × 1014 Hz ) ⎢1+
⎥ = 6.003 6 × 10 Hz
8
3.000
0
×
10
m
s
⎣
⎦
(b)
The change in frequency is
Δf = fO − fS = 6.003 6 × 1014 Hz − 6.000 0 × 1014 = 3.6 × 1011 Hz
21.65
Since you and the car ahead of you are moving away from each other (getting farther apart) at
a rate of u = 120 km h − 80 km h = 40 km h, the Doppler-shifted frequency you will detect is
fO = fS (1− u c ), and the change in frequency is
⎛ 40 km h ⎞ ⎛ 0.278 m s ⎞
⎛ u⎞
Δf = fO − fS = − fS ⎜ ⎟ = − ( 4.3 × 1014 Hz ) ⎜
= − 1.6 × 10 7 Hz
⎝ c⎠
⎝ 3.0 × 108 m s ⎟⎠ ⎜⎝ 1 km h ⎟⎠
The frequency you will detect will be
fO = fS + Δf = 4.3 × 1014 Hz − 1.6 × 10 7 Hz = 4.299 999 84 × 1014 Hz
21.66
The driver was driving toward the warning lights, so the correct form of the Doppler shift equation is fO = fS (1+ u c ) . The frequency emitted by the yellow warning light is
fS =
c 3.00 × 108 m s
=
= 5.17 × 1014 Hz
lS
580 × 10 −9 m
continued on next page
68719_21_ch21_p171-197.indd 194
1/7/11 2:47:55 PM
Alternating Current Circuits and Electromagnetic Waves
195
and the frequency the driver claims that she observed is
fO =
c
3.00 × 108 m s
= 5.36 × 1014 Hz
=
lO
560 × 10 −9 m
The speed with which she would have to approach the light for the Doppler effect to yield this
claimed shift is
⎛ f
⎞
⎛ 5.36 × 1014 Hz ⎞
− 1 = 1.1× 10 7 m s
u = c ⎜ O − 1⎟ = ( 3.00 × 108 m s ) ⎜
⎝ 5.17 × 1014 Hz ⎟⎠
⎝ fS
⎠
21.67
The energy incident on the mirror in time Δt is U = Plaser ⋅ Δt, where Plaser is the power transmitted
by the laser beam
Plaser = 25.0 × 10 −3 W = 25.0 × 10 −3 J s = 25.0 × 10 −3 N ⋅ m s
From Equation 21.30 in the textbook, the rate of change in the momentum of the mirror as the
beam reflects from it is
Δp 2U c 2Plaser
=
=
Δt
c
Δt
The impulse-momentum theorem then gives the force exerted on the mirror as F = Δp Δt = 2Plaser c ,
and the radiation pressure on the mirror is
Prad =
F 2Plaser c 2Plaser
=
=
A
A
cA
where A = p r 2 = p d 2 4 is the area of the mirror illuminated (i.e., the cross-sectional area of the
laser beam). Thus,
Prad =
Prad = 5.31× 10 −5 N m 2 = 5.31× 10 −5 Pa
or
21.68
8 ( 25.0 × 10 −3 N ⋅ m s )
2Plaser
8Plaser
=
=
2
c (p d 2 4 ) p cd 2 p ( 3.00 × 108 m s ) ( 2.00 × 10 −3 m )
Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the
Sun is 60°. Then the target area you fill in the Sun’s field of view is (1.7 m )( 0.3 m ) cos30° = 0.4 m 2 .
The intensity the radiation at Earth’s surface is I surface = 0.6 I incom ing , and only 50% of this is
absorbed. Since I = Pav A = ( ΔE Δt ) A, the absorbed energy is
(
)
ΔE = ( 0.5 I surface ) A ( Δt ) = ⎡⎣ 0.5 0.6 I incom ing ⎤⎦ A ( Δt )
= ( 0.5)( 0.6 )(1 370 W m 2 ) ( 0.4 m 2 ) ( 3 600 s ) = 6 × 10 5 J or
21.69
Z = R 2 + ( XC ) = R 2 + ( 2p f C )
2
=
−2
( 200 Ω )2 + ⎡⎣ 2p ( 60 Hz )( 5.0 × 10 −6 F )⎤⎦ = 5.7 × 10 2 Ω
−2
⎛ ΔV ⎞
⎛ 120 V ⎞
Pav = I rm2 s R = ⎜ rm s ⎟ R = ⎜
( 200 Ω ) = 8.9 W = 8.9 × 10 −3 kW
⎝ 5.7 × 10 2 Ω ⎟⎠
⎝ Z ⎠
2
Thus,
~10 6 J
2
continued on next page
68719_21_ch21_p171-197.indd 195
1/7/11 2:47:58 PM
196
Chapter 21
and
cost = ΔE ⋅ ( rate ) = Pav ⋅ Δt ⋅ ( rate )
= (8.9 × 10 −3 kW ) ( 24 h ) (8.0 cents kWh ) = 1.7 cents
21.70
For a parallel-plate capacitor, C = ∈ A d : cutting the plate separation d in half will double the
capacitance C f = 2Ci . Since the capacitive reactance is XC = 1 (2p f C), doubling the capacitance reduces the capacitive reactance by a factor of 2, giving XC, f = XC, i 2.
(
)
(
)
The current in the circuit is I rm s = ΔVrm s Z , so the impedance must be cut in half Z f = Z i 2 if
the current doubles while the applied voltage remains constant. With an inductive reactance equal
to the resistance ( X L = R ) , we then have
(
)
R 2 + R − XC, f
2
=
(
1
R 2 + R − XC, i
2
)
2
or
(
)
(
2
4 ⎡⎢ R 2 + R − XC, i 2 ⎤⎥ = R 2 + R − XC, i
⎣
⎦
)
2
Simplifying gives 6R 2 = 2RXC, i , and XC, i = 6R 2 2R = 3R .
21.71
R=
( ΔV )DC
I DC
=
12.0 V
= 19.0 Ω
0.630 A
ΔVrm s
24.0 V
=
= 42.1 Ω
I rm s
0.570 A
Z = R 2 + ( 2p f L ) =
2
Thus, L =
21.72
(a)
Z 2 − R2
=
2p f
( 42.1 Ω )2 − (19.0 Ω )2
= 9.97 × 10 −2 H = 99.7 mH
2p ( 60.0 Hz )
The frequency of a 3.0-cm radar wave, and hence the desired resonance frequency of the
tuning circuit, f0 = 1 2p LC , is
f0 = f =
c 3.00 × 108 m s
=
= 1.0 × 1010 Hz
l
3.00 × 10 −2 m
Therefore, the required capacitance is
C=
1
( 2p f )
2
0
L
=
( 2p × 10
10
1
Hz ) ( 400 × 10
2
−12
H)
= 6.3 × 10 −13 F = 0.63 pF
∈0 A ∈0 C
=
, so
d
d
2
(b)
C=
C=
21.73
C ⋅d
=
∈0
(6.3 × 10
−13
F ) (1.0 × 10 −3 m )
8.85 × 10 −12 C2 N ⋅ m
= 8.4 × 10 −3 m = 8.4 mm
(c)
XC = X L = ( 2p f0 ) L = 2p (1.0 × 1010 Hz ) ( 400 × 10 −12 H ) = 25 Ω
(a)
Em ax
E
0.20 × 10 −6 V m
= c, so Bm ax = m ax =
= 6.7 × 10 −16 T
3.00 × 108 m s
Bm ax
c
(b)
Intensity =
(c)
⎡pd2 ⎤
Pav = (Intensity) ⋅A = (Intensity) ⎢
⎥
⎣ 4 ⎦
−6
−16
Em ax Bm ax ( 0.20 × 10 V m ) ( 6.7 × 10 T )
=
= 5.3 × 10 −17 W m 2
2m 0
2 ( 4p × 10 −7 T ⋅ m A )
⎡ p ( 20.0 m )2 ⎤
−14
= ( 5.3 × 10 −17 W m 2 ) ⎢
⎥ = 1.7 × 10 W
4
⎣
⎦
68719_21_ch21_p171-197.indd 196
1/7/11 2:48:00 PM
Alternating Current Circuits and Electromagnetic Waves
21.74
(a)
Z=
ΔVrm s 12 V
=
= 6.0 Ω
I rm s
2.0 A
(b)
R=
ΔVDC 12 V
=
= 4.0 Ω
I DC
3.0 A
197
From Z = R 2 + X L2 = R 2 + ( 2p f L ) , we find
2
Z 2 − R2
=
2p f
L=
21.75
(a)
( 6.0 Ω )2 − ( 4.0 Ω )2
= 1.2 × 10 −2 H = 12 mH
2p ( 60 Hz )
From Equation 21.30 in the textbook, the momentum imparted in time Δt to a perfectly reflecting sail of area A by normally incident radiation of intensity I is Δp = 2U c = 2 ( IAΔt ) c.
From the impulse-momentum theorem, the average force exerted on the sail is then
Fav =
Fav 0.536 N
=
= 8.93 × 10 −5 m s 2
m 6 000 kg
(b)
aav =
(c)
From Δx = v0 t +
t=
21.76
2
4
2
Δp 2 ( IAΔt ) c 2IA 2 (1 340 W m ) ( 6.00 × 10 m )
=
=
=
= 0.536 N
8
Δt
Δt
c
3.00 × 10 m s
(a)
1 2
at , with v0 = 0, the time is
2
2 ( Δx )
=
aav
2 ( 3.84 × 108 m )
8.93 × 10
−5
1d
⎛
= ( 2.93 × 10 6 s ) ⎜
⎝
m s
8.64 × 10 4
2
⎞
⎟ = 33.9 d
s⎠
The intensity of radiation at distance r from a point source, which radiates total power P, is
I = P A = P 4p r 2. Thus, at distance r = 2.0 in from a cell phone radiating a total power of
P = 2.0 W = 2.0 × 10 3 mW, the intensity is
I=
2.0 × 10 3 mW
4p ⎡⎣( 2.0 in )( 2.54 cm 1 in ) ⎤⎦
2
= 6.2 mW cm 2
This intensity is 24% higher than the maximum allowed leakage from a microwave at this
distance of 2.0 inches.
(b)
If when using a Bluetooth headset (emitting 2.5 mW of power) in the ear at distance
rh = 2.0 in = 5.1 cm from the brain, the cell phone (emitting 2.0 W of power) is located in
the pocket at distance rp = 1.0 m = 1.0 × 10 2 cm from the brain, the total radiation intensity
at the brain is
I total = I phone + I headset =
or
68719_21_ch21_p171-197.indd 197
I total = 1.6 × 10 −2
2.0 × 10 3 mW
4p (1.0 × 10 cm )
2
2
+
2.5 mW
2
4p ( 5.1 cm )
mW
mW
mW
+ 7.6 × 10 −3
= 2.4 × 10 −2
= 0.024 mW cm 2
2
2
cm
cm
cm 2
1/7/11 2:48:03 PM
22
Reflection and Refraction of Light
QUICK QUIZZES
1.
Choice (a). In part (a), you can see clear reflections of the headlights and the lights on the top of
the truck. The reflection is specular. In part (b), although bright areas appear on the roadway in
front of the headlights, the reflection is not as clear, and no separate reflection of the lights from
the top of the truck is visible. The reflection in part (b) is mostly diffuse.
2.
Beams 2 and 4 are reflected; beams 3 and 5 are refracted.
3.
Choice (b). When light goes from one material into one having a higher index of refraction, it
refracts toward the normal line of the boundary between the two materials. If, as the light travels
through the new material, the index of refraction continues to increase, the light ray will refract
more and more toward the normal line.
4.
Choice (c). Both the wave speed and the wavelength decrease as the index of refraction increases.
The frequency is unchanged.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The index of refraction of a material in which light has speed v is n = c v = c l f. Thus,
nliquid
nair
=
c (lliquid f )
c (lair f )
=
lair
lliquid
⎛ l ⎞
⎛ 495 nm ⎞
nliquid = ⎜ air ⎟ nair = ⎜
(1.00 ) = 1.14
⎝ 434 nm ⎟⎠
⎝ lliquid ⎠
and
so the correct choice is (c).
2.
The critical angle as light passes from medium 1 into medium 2 is q c = sin −1 (n2 n1 ), so as light is
incident on crown glass (n2 = 1.52) from carbon disulfide (n1 = 1.63),
⎛ 1.52 ⎞
q c = sin −1 ⎜
= 68.8°
⎝ 1.63 ⎟⎠
and choice (b) is the correct response.
3.
The energy of a photon is E = hf = hc l. Thus, if n 800-nm photons have the same energy as four
200-nm photons, it is necessary that
n(hc 800 nm) = 4(hc 200 nm)
or
n = 4(800 nm 200 nm) = 16
Therefore, the correct answer is (e).
4.
Observe that the angle of refraction is greater than the angle of incidence as the light ray passes
from medium 1 into medium 2. Thus, the speed of light increases as the light crosses the boundary between these materials. Since n = c v , the index of refraction of medium 2 is less than that
of medium 1, or n1 > n2 . The correct choice is (d).
198
68719_22_ch22_p198-222.indd 198
1/7/11 2:48:49 PM
Reflection and Refraction of Light
5.
6.
7.
199
Total internal reflection will occur when light, in attempting to go from a medium with one index
of refraction n1 into a second medium where it travels faster than in the first medium (or where
n2 < n1), strikes the surface at an angle of incidence greater than or equal to the critical angle. The
correct choice is (b).
Water and air have different indices of refraction, with nwater ≈ 4nair 3. In passing from one of these
media into the other, light will be refracted (deviated in direction) unless the angle of incidence is
zero (in which case, the angle of refraction is also zero). Thus, rays B and D cannot be correct. In
refraction, the incident ray and the refracted ray are never on the same side of the line normal to
the surface at the point of contact, so ray A cannot be correct. Also in refraction, the ray makes a
smaller angle with the normal in the medium having the highest index of refraction. Therefore,
ray E cannot be correct, leaving only ray C as a likely path. Choice (c) is the correct answer.
When light is in water, the relationships between the values of its frequency, speed, and wavelength to the values of the same quantities in air are
fwater = fair ,
⎛ n ⎞
3
lwater = ⎜ air ⎟ lair ≈ lair ,
4
⎝ nwater ⎠
and
⎛ n ⎞
⎛ 3⎞
vwater = ⎜ air ⎟ vair = ⎜ ⎟ c
⎝ 4⎠
⎝ nwater ⎠
Therefore, only choice (b) is a completely true statement.
8.
In a dispersive medium, the index of refraction is largest for the shortest wavelength. Thus, the
violet light will be refracted (or bent) the most as it passes through a surface of the crown glass,
making (a) the correct choice.
9.
For any medium, other than vacuum, the index of refraction for red light is slightly lower than
that for blue light. This means that when light goes from vacuum (or air) into glass, the red light
deviates from its original direction less than does the blue light. Also, as the light reemerges from
the glass into vacuum (or air), the red light again deviates less than the blue light. If the two
surfaces of the glass are parallel to each other, the red and blue rays will emerge traveling parallel
to each other but displaced laterally from one another. The sketch that best illustrates this process
is C, so choice (c) is the best answer.
10.
Consider the sketch at the right and apply Snell’s law to
the refraction at each of the three surfaces. The resulting
equations are
and
(1.00)sinq = n1 sina
(1st surface)
n1 sina = n2 sin b
(2nd surface)
n2 sin b = (1.00 ) sinf
(3rd surface)
Combining these three equations yields
(1.00)sinf = (1.00)sinq , and f = q . Hence, choice (c) is the
correct answer.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
68719_22_ch22_p198-222.indd 199
(a)
From Snell’s law, sinq 2 = (n1 n2 )sinq1 . Thus, if n2 < n1, then sinq 2 > sinq1 and q 2 > q1, so
the ray refracts away from the normal.
1/7/11 2:48:51 PM
200
Chapter 22
(b)
The index of refraction is defined as n = c v = c l f . The wavelength may then be written
as l = c nf = (c f ) n = l0 n , where l0 is the wavelength of the light in vacuum. Thus, as
the ray moves into a medium of lower index of refraction, the wavelength will increase.
(c)
The frequency at which wavefronts move away from a boundary equals the frequency with
which they arrive at the boundary. That is, the frequency of the light stays the same as it
moves between the two materials.
4.
A mirage occurs when light changes direction as it moves between batches of air having different
indices of refraction. The different indices of refraction occur because the air has different densities at different temperatures. Two images are seen: one from a direct path from the object to you,
and the second arriving by rays originally heading toward Earth but refracted to your eye. On a
hot day, the Sun makes the surface of asphalt hot, so the air is hot directly above it, becoming
cooler as one moves higher into the sky. The “water” we see far in front of us is an image of the
blue sky. Adding to the effect is the fact that the image shimmers as the air changes in temperature, giving the appearance of moving water.
6.
The upright image of the hill is formed by light that has followed a direct path from the hill to the
eye of the observer. The second image is a result of refraction in the atmosphere. Some light is
reflected from the hill toward the water. As this light passes through warmer layers of air directly
above the water, it is refracted back up toward the eye of the observer, resulting in the observation
of an inverted image of the hill directly below the upright image.
8.
The index of refraction of water is 1.333, quite different from that of air, which has an index of
refraction of about 1. The boundary between the air and water is therefore easy to detect, because
of the differing refraction effects above and below the boundary. (Try looking at a glass half full
of water.) The index of refraction of liquid helium, however, happens to be much closer to that of
air. Consequently, the refractive differences above and below the helium-air boundary are harder
to see.
10.
Total internal reflection can only occur when light attempts to move from a medium of high index
of refraction to a medium of lower index of refraction. Thus, light moving from air (n = 1) to
water (n = 1.333) cannot undergo total internal reflection.
12.
With no water in the cup, light rays from the coin do not reach the eye because they are blocked
by the side of the cup. With water in the cup, light rays are bent away from the normal as they
leave the water so that some reach the eye.
In figure (a) below, ray a is blocked by the side of the cup so it cannot enter the eye, and ray b
misses the eye. In figure (b), ray a is still blocked by the side of the cup, but ray b refracts at the
water’s surface so that it reaches the eye. Ray b seems to come from position B, directly above
the coin at position A.
eye
b
b
a
a
B
A
(a)
68719_22_ch22_p198-222.indd 200
(b)
1/7/11 2:48:53 PM
Reflection and Refraction of Light
201
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
4.
(a) 1.99 × 10 −15 J
(b) 12.4 keV
(c)
decreased
(d)
increased
(a)
550 nm
(b)
366 nm
(c)
2.26 eV
(d) The energy does not change.
6.
(a) l = hc E
(b)
Higher energy photons have shorter wavelengths.
8.
(a)
(b)
50.0° above the horizontal
1.94 m
10.
See Solution.
12.
(a) The angle of refraction increases with wavelength, so the longest wavelength deviates the
least from the original path.
(b) l = 400 nm, q r = 16.0°; l = 500 nm, q r = 16.1°; l = 650 nm, q r = 16.3°
14.
80.0°
16.
(a) B
(b)
18.
(a) q r, top = 19.5°
(b) q i, bottom = 19.5°, q r , bottom = 30.0°
(c)
(d)
0.386 cm
A, B, and C
2.00 × 108 m s
(e) 1.06 × 10 −10 s
(f) The angle of incidence affects the distance the ray travels in the glass and hence affects the
travel time.
20.
1.22
22.
6.30 cm
24.
(a)
See Solution.
(b)
42.0°
(d)
26.9°
(e)
107 m
(b)
42.2°
26.
See Solution.
28.
0.39°
30.
q red = 34.9°, q violet = 34.5°
32.
4.5°
34.
(a)
36.
48.5°
38.
67.3°
68719_22_ch22_p198-222.indd 201
43.3°
(c)
63.1°
(c)
40.4°
1/7/11 2:48:54 PM
202
40.
Chapter 22
(a)
(b)
34.2°
(c) and (d)
42.
34.2°
Neither thickness nor index of refraction affects the result.
(a)
24.42°
(b)
(c)
33.44°
(d) Total internal reflection still occurs.
(e)
rotate clockwise
(f)
2.9°
(b)
q r → q i = 30.0°
(b)
q incidence ≤ 29.9°
44.
4.54 m
46.
(a)
48.
(a) q incidence ≤ 90.0°
(c)
23.7°
See Solution.
(c)
q r > q i = 30.0°
Total internal reflection is not possible since npolystyrene < ncarbon disulfide .
50.
77.5°
52.
(a)
Rm in = nd ( n − 1)
(b)
Rm in → 0 as d → 0; Rm in decreases as n increases; Rm in increases as n → 1. All are reasonable
behaviors.
(c)
Rm in = 350 mm
54.
f = 8.0°
56.
82 complete reflections
58.
1.33
60.
23.1°
PROBLEM SOLUTIONS
22.1
The total distance the light travels is
⎛
⎞
Δd = 2 ⎜ Dcenter to − REarth − RM oon ⎟
⎝ center
⎠
= 2 ( 3.84 × 108 − 6.38 × 10 6 − 1.76 × 10 6 ) m = 7.52 × 108 m
v=
Therefore,
22.2
(a)
Δd 7.52 × 108 m
=
= 3.00 × 108 m s
Δt
2.51 s
The energy of a photon is E = hf = hc l, where Planck’s constant is h = 6.63×10 −34 J ⋅ s,
and the speed of light in vacuum is c = 3.00 × 108 m s. If l = 1.00 × 10 −10 m,
(6.63 × 10
E=
−34
J ⋅ s ) ( 3.00 × 108 m s )
1.00 × 10 −10 m
= 1.99 × 10 −15 J
continued on next page
68719_22_ch22_p198-222.indd 202
1/7/11 2:48:55 PM
Reflection and Refraction of Light
(b)
203
1 eV
⎛
⎞
E = (1.99 × 10 −15 J ) ⎜
= 1.24 × 10 4 eV = 12.4 keV
⎝ 1.602 × 10 −19 J ⎟⎠
(c) and (d)
For the x-rays to be more penetrating, the photons should be more energetic. Since
the energy of a photon is directly proportional to the frequency and inversely
proportional to the wavelength, the wavelength should decrease , which is the same
as saying the frequency should increase .
22.3
(a)
1 eV
⎛
E = hf = ( 6.63 × 10 −34 J ⋅s ) ( 5.00 × 1017 Hz ) ⎜
⎝ 1.60 × 10 −19
(b)
E = hf =
⎞
3
⎟ = 2.07 × 10 eV = 2.07 keV
J⎠
−34
8
hc ( 6.63 × 10 J ⋅ s ) ( 3.00 × 10 m s ) ⎛ 1 nm ⎞
−19
=
⎜⎝ −9 ⎟⎠ = 6.63 × 10 J
l
3.00 × 10 2 nm
10 m
1 eV
⎛
⎞
E = 6.63 × 10 −19 J ⎜
= 4.14 eV
⎝ 1.60 × 10 −19 J ⎟⎠
22.4
c 3.00 × 108 m s
=
= 5.50 × 10 −7 m = 550 nm
f 5.45 × 1014 Hz
(a)
l0 =
(b)
From Table 22.1, the index of refraction for benzene is n = 1.501. Thus, the wavelength in
benzene is
ln =
(c)
(d)
22.5
22.6
l0 550 nm
=
= 366 nm
1.501
n
1 eV
⎛
⎞
E = hf = ( 6.63 × 10 −34 J ⋅s ) ( 5.45 × 1014 Hz ) ⎜
= 2.26 eV
⎝ 1.60 × 10 −19 J ⎟⎠
The energy of the photon is proportional to the frequency, which does not change as
the light goes from one medium to another. Thus, when the photon enters benzene,
the energy does not change .
The speed of light in a medium with index of refraction n is v = c n , where c is its speed in
vacuum.
3.00 × 108 m s
= 2.25 × 108 m s .
1.333
(a)
For water, n = 1.333, and v =
(b)
For crown glass, n = 1.52, and v =
(c)
For diamond, n = 2.419, and v =
(a)
The energy of a photon is
3.00 × 108 m s
= 1.97 × 108 m s .
1.52
3.00 × 108 m s
= 1.24 × 108 m s .
2.419
⎛ c ⎞ hc
E = hf = h ⎜ ⎟ =
⎝l⎠ l
Thus,
(b)
68719_22_ch22_p198-222.indd 203
l = hc E
Higher energy photons have shorter wavelengths.
1/7/11 2:48:57 PM
204
22.7
Chapter 22
From Snell’s law, n2 sinq 2 = n1 sinq1 . Thus, when q1 = 45.0° and the first medium is air (n1 = 1.00),
we have sinq 2 = (1.00 ) sin 45.0° n 2 .
(a)
⎛ (1.00 ) sin 45.0° ⎞
For quartz, n2 = 1.458, and q 2 = sin −1 ⎜
⎟⎠ = 29.0° .
⎝
1.458
(b)
⎛ (1.00 ) sin 45.0° ⎞
For carbon disulfide, n2 = 1.628, and q 2 = sin −1 ⎜
⎟⎠ = 25.7° .
⎝
1.628
(c)
⎛ (1.00 ) sin 45.0° ⎞
For water, n2 = 1.333, and q 2 = sin −1 ⎜
⎟⎠ = 32.0° .
⎝
1.333
(a)
From geometry, 1.25 m = d sin 40.0°, so d = 1.94 m .
(b)
50.0° above horizontal , or parallel to the incident ray
22.8
22.9
n1 sin q1 = n2 sin q 2
sin q1 = 1.333 sin 45.0°
q1 = sin −1 (1.333 sin 45.0° ) = 70.5°
Thus, the sun appears to be 19.5° above the horizontal .
22.10
In the sketch at the right, observe that the law of reflection is
obeyed as the ray reflects from each of the mirrors. Also, note
that the normal lines to the two mirrors intersect at a right
angle since the two mirrors are perpendicular to each other.
Considering the right triangle formed by the two normal lines
and the ray, and recalling that the sum of the interior angles of
any triangle is 180°, we find that
q1 +q 2 + 90.0° = 180°
or
q 2 = 90.0° −q1
Mirror 2
q2
q2
q1
q1
a
Mirror 1
Looking at the point where the incident ray strikes mirror 1, we see that a = 90° −q1 . Thus, both
the incident ray and the final outgoing reflected ray are at angle 90° −q1 above the horizontal and
hence, parallel to each other .
68719_22_ch22_p198-222.indd 204
1/7/11 2:49:00 PM
Reflection and Refraction of Light
22.11
22.12
(a)
From Snell’s law, n2 =
(b)
l2 =
(c)
f=
(d)
v2 =
(a)
205
n1 sinq1 (1.00 ) sin 30.0°
=
= 1.52 .
sinq 2
sin19.24°
l0 632.8 nm
=
= 416 nm
1.52
n2
c 3.00 × 108 m s
=
= 4.74 × 1014 Hz in the air and also in the solution
l0 632.8 × 10 −9 m
c 3.00 × 108 m s
=
= 1.97 × 108 m s
n2
1.52
When light refracts from air ( n1 = 1.00 ) into the
crown glass, Snell’s law gives the angle of
refraction as
q 2 = sin −1 (sin 25.0° ncrown glass )
For first quadrant angles, the sine of the angle increases as
the angle increases. Thus, from the above equation, note
that q 2 will increase when the index of refraction of the
crown glass decreases. From Figure 22.13, we see that the
angle of refraction will increase with wavelength, so the
longer wavelengths deviate the least from the original
path.
(b)
From Figure 22.13, observe that the index of refraction of
crown glass for the given wavelengths is
l = 400 nm, ncrown glass = 1.53;
l = 500 nm, ncrown glass = 1.52;
l
Figure 22.13
and l = 650 nm, ncrown glass = 1.51
Thus, Snell’s law gives
22.13
l = 400 nm:
q 2 = sin −1 ( sin 25.0° 1.53) = 16.0°
l = 500 nm:
q 2 = sin −1 ( sin 25.0° 1.52 ) = 16.1°
l = 650 nm:
q 2 = sin −1 ( sin 25.0° 1.51) = 16.3°
From Snell’s law,
⎡ n sinq1 ⎤
−1 ⎡ (1.00)sin 40.0° ⎤
q 2 = sin −1 ⎢ 1
⎥ = sin ⎢
⎥⎦ = 29.4°
1.309
⎣
⎣ n2 ⎦
and from the law of reflection, f = q1 = 40.0°.
Hence, the angle between the reflected and refracted rays is
a = 180° −q 2 − f = 180.0° − 29.4° − 40.0° = 110.6°
68719_22_ch22_p198-222.indd 205
1/7/11 2:49:02 PM
206
22.14
Chapter 22
Consider the sketch at the right and note that the incident horizontal ray is parallel to the surface of mirror 2. Thus, the angle
the incident ray makes with mirror 1 must be a = q1 = 50.0°.
Since the ray must obey the law of reflection at mirror 1,
the angle b must be b = a = 50.0°. Recalling that the
sum of the interior angles of a triangle is always 180.0°,
we find that
Mirror 1
a
b
q1
g
q2
g = 180.0° −q1 − b = 180.0° − 50.0° − 50.0° = 80.0°
Mirror 2
Hence, in order to obey the law of reflection at mirror 2, the angle the outgoing reflected ray
makes with the surface of mirror 2 must be q 2 = g = 80.0° .
22.15
22.16
The index of refraction of zircon is n = 1.923.
c 3.00 × 108 m s
=
= 1.56 × 108 m s
n
1.923
(a)
v=
(b)
The wavelength in the zircon is ln =
(c)
The frequency is f =
l0 632.8 nm
=
= 329.1 nm .
1.923
n
v
c 3.00 × 108 m s
=
=
= 4.74 × 1014 Hz .
ln l0 632.8 × 10 −9 m
The sketch at the right shows the path of the ray inside the glass slab.
Considering the reflections at points A and B, the law of reflection
tells us that a = q A and b = q B . Also, we observe that
q B = 90° − a = 90° −q A
and
q C = 90° − b = 90° −q B = 90° − ( 90° −q A ) = q A
C
b
qA
qC
Thus, if q A = 55°, then q B = 35° and q C = 55°.
a
b qB
In order for part of the ray to leave the glass slab and enter the
surrounding medium at a reflection point, the angle of incidence at
that point must be less than the critical angle, q c = sin −1 (n2 nglass ) = sin −1 (n2 1.52).
22.17
A
a
B
(a)
If the surrounding medium is air, then n2 = 1.00 and q c = sin −1 (1.00 1.52) = 41.1°. Thus, we
see that total internal reflection will occur at points A and C, but part of the ray can refract
into the surrounding air at point B .
(b)
When the surrounding medium is carbon disulfide, n2 = 1.628 > nglass . Thus, the critical
angle does not exist, and total internal reflection will not occur when the ray attempts to go
from the glass into the carbon disulfide. This means that part of the ray will enter the carbon
disulfide at points A, B, and C .
The incident light reaches the left-hand mirror at distance
d 2 = (1.00 m ) tan 5.00° = 0.087 5 m
above its bottom edge. The reflected light first reaches the right-hand mirror at height
d = 2 ( 0.087 5 m ) = 0.175 m
continued on next page
68719_22_ch22_p198-222.indd 206
1/7/11 2:49:04 PM
Reflection and Refraction of Light
207
It bounces between the mirrors with distance d
between points of contact with a given mirror.
Since the full 1.00 length of the right-hand mirror is
available for reflections, the number of reflections
from this mirror will be
N right =
1.00 m
= 5.71 →
0.175 m
5 full reflections
Since the first reflection from the left-hand mirror
occurs at a height of d 2 = 0.087 5 m, the total
number of that can occur from this mirror is
N left = 1+
22.18
(a)
1.00 m − 0.087 5 m
= 6.21 →
0.175 m
6 full reflections
From Snell’s law, the angle of refraction at the first surface is
⎡ n sinq1 ⎤
−1 ⎡ (1.00 ) sin 30.0° ⎤
q 2 = sin −1 ⎢ air
⎥ = sin ⎢
⎥ = 19.5°
n
1.50
⎦
⎣
⎢⎣ glass ⎥⎦
(b)
Since the upper and lower surfaces are parallel, the normal lines where the ray strikes these
surfaces are parallel. Hence, the angle of incidence at the lower surface will be q 2 = 19.5° .
The angle of refraction at this surface is then
⎡ nglass sinq glass
q 3 = sin −1 ⎢
nair
⎣
⎤
−1 ⎡ (1.50 ) sin19.5° ⎤
⎥ = sin ⎢
⎥ = 30.0°
1.00
⎦
⎣
⎦
Thus, the light emerges traveling parallel to the incident beam.
(c)
Consider the sketch at the right, and let h represent the
distance from point a to c (that is, the hypotenuse of
triangle abc). Then,
h=
2.00 cm 2.00 cm
=
= 2.12 cm
cosq 2
cos19.5°
Also, a = q1 −q 2 = 30.0° − 19.5° = 10.5°, so
d = h sina = ( 2.12 cm ) sin10.5° = 0.386 cm
(d)
The speed of the light in the glass is
v=
(e)
68719_22_ch22_p198-222.indd 207
nglass
=
3.00 × 108 m s
= 2.00 × 108 m s
1.50
The time required for the light to travel through the glass is
t=
(f )
c
h
2.12 cm
⎛ 1m ⎞
−10
=
⎜
⎟ = 1.06 × 10 s
v 2.00 × 108 m s ⎝ 10 2 cm ⎠
Changing the angle of incidence will change the angle of refraction and therefore the
distance h the light travels in the glass. Thus, the travel time will also change .
1/7/11 2:49:07 PM
208
22.19
Chapter 22
From Snell’s law, the angle of incidence at the air-oil interface is
⎡ n sinq oil ⎤
−1 ⎡ (1.48 ) sin 20.0° ⎤
q = sin −1 ⎢ oil
⎥ = sin ⎢
⎥ = 30.4°
n
1.00
⎦
⎣
air
⎣
⎦
and the angle of refraction as the light enters the water is
⎡ n sinq oil
q ′ = sin −1 ⎢ oil
⎣ nwater
22.20
⎤
−1 ⎡ (1.48 ) sin 20.0° ⎤
⎥ = sin ⎢
⎥ = 22.3°
1.333
⎦
⎣
⎦
Since the light ray strikes the first surface at normal incidence, it passes into the prism without deviation. Thus,
the angle of incidence at the second surface (hypotenuse
of the triangular prism) is q1 = 45.0° as shown in the
sketch at the right. The angle of refraction is
q 2 = 45.0° + 15.0° = 60.0°
and Snell’s law gives the index of refraction of the prism material as
n1 =
22.21
n2 sinq 2 (1.00 ) sin ( 60.0° )
=
= 1.22
sinq1
sin ( 45.0° )
Applying Snell’s law where the ray first enters the
glass gives
⎛ n sinq1 ⎞
−1 ⎡ (1.333 ) sin 42.0° ⎤
f = sin −1 ⎜ water
⎟ = sin ⎢
⎥ = 35.9°
n
1.52
⎦
⎣
⎝
⎠
glass
q1
water, n = 1.333
d
f
d
y
Thus, d = 90.0° − f = 90.0° − 35.9° = 54.1°.
(a)
The distance down to point P, where the ray
emerges from the glass, is now seen to be
d
Crown glass,
n = 1.52
P
q2
y = d tand = ( 3.50 cm ) tan 54.1° = 4.84 cm
(b)
The angle of refraction as the ray leaves the block
is given by Snell’s law as
⎛ nglass sind ⎞
⎡ (1.52 ) sin 54.1° ⎤
= sin −1 ⎢
q 2 = sin −1 ⎜
⎥ = 67.5°
1.333
⎝ nwater ⎟⎠
⎦
⎣
22.22
⎞
⎛n
From Snell’s law, sinq = ⎜ m edium ⎟ sin 50.0°.
⎝ nliver ⎠
But
so
nm edium c vm edium
v
=
= liver = 0.900
nliver
c vliver
vm edium
nmedium
nliver
q = sin −1 [( 0.900 ) sin 50.0° ] = 43.6°
continued on next page
68719_22_ch22_p198-222.indd 208
1/7/11 2:49:09 PM
Reflection and Refraction of Light
209
From the law of reflection,
d=
22.23
(a)
12.0 cm
= 6.00 cm
2
h=
and
d
6.00 cm
=
= 6.30 cm
tanq tan ( 43.6° )
Before the container is filled, the ray’s path is as
shown in figure (a) at the right. From this figure,
observe that
sinq1 =
d
=
s1
d
h2 + d 2
=
q1
h
1
(h d )
2
+1
d
After the container is filled, the ray’s path is shown
in figure (b). From this figure, we find that
d 2
=
s2
sinq 2 =
d 2
h + ( d 2)
2
2
s1
q1
(a)
q1
1
=
h q2
4 (h d ) + 1
2
s2
From Snell’s law, nair sinq1 = n sinq 2, or
1.00
(h d )
2
+1
=
d/2
n
4 (h d ) + 1
2
and
Simplifying, this gives ( 4 − n 2 ) ( h d ) = n 2 − 1
2
(b)
If d = 8.00 cm
h = (8.00 cm )
22.24
and
(b)
4 (h d ) + 1 = n2 (h d ) + n2
2
2
or
h
=
d
n2 − 1
4 − n2
n = nwater = 1.333, then
(1.333)2 − 1
= 4.73 cm
2
4 − (1.333)
(a)
A sketch illustrating the situation and the two triangles needed in the solution is given below:
(b)
The angle of incidence at the water surface is
⎛ 90.0 m ⎞
q1 = tan −1 ⎜
= 42.0°
⎝ 1.00 × 10 2 m ⎟⎠
continued on next page
68719_22_ch22_p198-222.indd 209
1/7/11 2:49:11 PM
210
Chapter 22
(c)
Snell’s law gives the angle of refraction as
⎛ n sinq1 ⎞
⎛ (1.333) sin 42.0° ⎞
= sin −1 ⎜
q 2 = sin −1 ⎜ water
⎟⎠ = 63.1°
⎟
⎝
n
1.00
⎝
⎠
air
(d)
The refracted beam makes angle f = 90.0° −q 2 = 26.9° with the horizontal.
(e)
Since tanf = h (2.10 × 10 2 m), the height of the target is
h = ( 2.10 × 10 2 m ) tan ( 26.9° ) = 107 m
22.25
As shown at the right, q1 + b +q 2 = 180°.
When b = 90°, this gives q 2 = 90° −q1.
Then, from Snell’s law
sinq1 =
ng sinq 2
nair
= ng sin ( 90° −q1 ) = ng cosq1
Thus, when b = 90° ,
22.26
22.27
( )
sinq1
= tanq1 = ng or q1 = tan −1 ng .
cosq1
The index of refraction of the atmosphere decreases
with increasing altitude because of the decrease in
density of the atmosphere with increasing altitude.
As indicated in the ray diagram at the right, the Sun
located at S below the horizon appears to be located
at S′.
Horizon
plane
Atmosphere
S´
S
Earth
When the Sun is 28.0° above the horizon, the angle of
incidence for sunlight at the air-water boundary is
q1 = 90.0° − 28.0° = 62.0°
Thus, the angle of refraction is
⎡ n sinq1 ⎤
q 2 = sin −1 ⎢ air
⎥
⎣ nwater ⎦
⎡ (1.00 ) sin 62.0° ⎤
= sin −1 ⎢
⎥ = 41.5°
1.333
⎦
⎣
The depth of the tank is then h =
68719_22_ch22_p198-222.indd 210
3.00 m
3.00 m
=
= 3.39 m .
tanq 2
tan ( 41.5° )
1/7/11 2:49:13 PM
Reflection and Refraction of Light
22.28
211
The angles of refraction for the two wavelengths are
⎛ n sinq1 ⎞
⎛ 1.000 sin 30.00° ⎞
= sin −1 ⎜
q red = sin −1 ⎜ air
⎟⎠ = 18.03°
⎟
⎝
n
1.615
⎝
⎠
red
and
⎛ n sinq1 ⎞
⎛ 1.000 sin 30.00° ⎞
= sin −1 ⎜
q blue = sin −1 ⎜ air
⎟⎠ = 17.64°
⎟
⎝
1.650
⎝ nblue ⎠
Thus, the angle between the two refracted rays is
Δq = q red −q blue = 18.03° − 17.64° = 0.39°
22.29
22.30
Using Snell’s law gives
(a)
⎛ n sinq i ⎞
⎛ (1.000)sin83.00° ⎞
= sin −1 ⎜
q blue = sin −1 ⎜ air
⎟⎠ = 47.79°
⎟
⎝
n
1.340
⎝
⎠
blue
(b)
⎛ n sinq i ⎞
⎛ (1.000)sin83.00° ⎞
= sin −1 ⎜
q red = sin −1 ⎜ air
⎟⎠ = 48.22°
⎟
⎝
1.331
⎝ nred ⎠
Using Snell’s law gives
⎛ n sinq i ⎞
⎛ (1.00)sin 60.0° ⎞
= sin −1 ⎜
q red = sin −1 ⎜ air
⎟⎠ = 34.9°
⎝
1.512
⎝ nred ⎟⎠
and
22.31
⎛ n sinq i ⎞
⎛ (1.00)sin 60.0° ⎞
= sin −1 ⎜
q violet = sin −1 ⎜ air
⎟⎠ = 34.5°
⎟
⎝
1.530
⎝ nviolet ⎠
Using Snell’s law gives
⎛ n sinq i ⎞
⎛ (1.000)sin 50.00° ⎞
= sin −1 ⎜
q red = sin −1 ⎜ air
⎟⎠ = 31.77°
⎝
1.455
⎝ nred ⎟⎠
and
⎛ n sinq i ⎞
⎛ (1.000)sin 50.00° ⎞
= sin −1 ⎜
q violet = sin −1 ⎜ air
⎟⎠ = 31.45°
⎟
⎝
1.468
⎝ nviolet ⎠
Thus, the dispersion is q red −q violet = 31.77° − 31.45° = 0.32° .
22.32
For the violet light, nglass = 1.66, and
⎛ n sinq1i ⎞
q1r = sin ⎜ air
⎟
⎝ nglass ⎠
nair = 1.00
60.0°
−1
a
q1i = 50.0°
⎛ 1.00 sin 50.0° ⎞
= sin ⎜
⎟⎠ = 27.5°
⎝
1.66
b
q1r q2i
−1
q2r
nglass
a = 90° −q1r = 62.5°, b = 180.0° − 60.0° − a = 57.5°,
and q 2i = 90.0° − b = 32.5°. The final angle of refraction of the violet light is
⎛ nglass sinq 2i ⎞
⎛ 1.66sin 32.5° ⎞
= sin −1 ⎜
q 2r = sin −1 ⎜
⎟⎠ = 63.1°
⎟
⎝
nair
1.00
⎝
⎠
continued on next page
68719_22_ch22_p198-222.indd 211
1/7/11 2:49:15 PM
212
Chapter 22
(
)
Following the same steps for the red light nglass = 1.62 gives
q1r = 28.2°, a = 61.8°, b = 58.2°, q 2i = 31.8°, and q 2r = 58.6°
Thus, the angular dispersion of the emerging light is
Dispersion = q 2 r
22.33
(a)
violet
−q 2 r
red
= 63.1° − 58.6° = 4.5°
The angle of incidence at the first surface is
q1i = 30° , and the angle of refraction is
⎛ n sinq1i ⎞
q1r = sin −1 ⎜ air
⎟
⎝ nglass ⎠
nair = 1.0
60°
⎛ 1.0 sin 30° ⎞
= sin −1 ⎜
⎟⎠ = 19°
⎝
1.5
b
a
q1i = 30°
q1r q2i
q2r
nglass = 1.5
Also, a = 90° −q1r = 71° and b = 180° − 60° − a = 49°.
Therefore, the angle of incidence at the second surface is q 2 i = 90° − b = 41° . The angle of
refraction at this surface is
⎛ n sinq 2i ⎞
−1 ⎛ 1.5sin 41° ⎞
q 2r = sin −1 ⎜ glass
⎟ = sin ⎜
⎟ = 80°
⎝ 1.0
⎠
nair
⎠
⎝
(b)
The angle of reflection at each surface equals the angle of incidence at that surface. Thus,
(q )
1 reflection
22.34
22.35
22.36
= q1i = 30° , and (q 2 )reflection = q 2i = 41°
As light goes from a medium having a refractive index n1 to a medium with refractive index
n2 < n1, the critical angle is given by the relation sinq c = n2 n1 . Table 22.1 gives the refractive
index for various substances at l0 = 589 nm.
(a)
For fused quartz surrounded by air, n1 = 1.458 and n2 = 1.00, giving
q c = sin −1 (1.00 1.458) = 43.3° .
(b)
In going from polystyrene (n1 = 1.49) to air, q c = sin −1 (1.00 1.49) = 42.2° .
(c)
From sodium chloride (n1 = 1.544) to air, q c = sin −1 (1.00 1.544 ) = 40.4° .
When light is coming from a medium of refractive index n1 into water (n2 = 1.333), the critical
angle is given by q c = sin −1 (1.333 n1 ).
(a)
For fused quartz, n1 = 1.458, giving q c = sin −1 (1.333 1.458 ) = 66.1° .
(b)
In going from polystyrene (n1 = 1.49) to water, q c = sin −1 (1.333 1.49 ) = 63.5° .
(c)
From sodium chloride (n1 = 1.544) to water, q c = sin −1 (1.333 1.544 ) = 59.7° .
Using Snell’s law, the index of refraction of the liquid is found to be
nliquid = (nair sinq i ) sinq r
continued on next page
68719_22_ch22_p198-222.indd 212
1/7/11 2:49:17 PM
Reflection and Refraction of Light
213
Thus, the critical angle for light going from this liquid into air is
⎡
⎤
⎛ n ⎞
nair
−1 ⎡ sin 22.0° ⎤
q c = sin −1 ⎜ air ⎟ = sin −1 ⎢
= 48.5°
⎥ = sin ⎢
n
(
n
sinq
)
sinq
⎣ sin 30.0° ⎥⎦
⎝ liquid ⎠
i
r ⎥
⎢⎣ air
⎦
22.37
22.38
When light attempts to cross a boundary from one medium of refractive index n1 into a new
medium of refractive index n2 < n1, total internal reflection will occur if the angle of incidence
exceeds the critical angle given by q c = sin −1 ( n2 n1 ).
(a)
⎛ 1.00 ⎞
If n1 = 1.53 and n2 = nair = 1.00, then q c = sin −1 ⎜
= 40.8° .
⎝ 1.53 ⎟⎠
(b)
⎛ 1.333 ⎞
If n1 = 1.53 and n2 = nwater = 1.333, then q c = sin −1 ⎜
= 60.6° .
⎝ 1.53 ⎟⎠
The critical angle for this material in air is
⎛ n ⎞
⎛ 1.00 ⎞
q c = sin −1 ⎜ air ⎟ = sin −1 ⎜
= 47.3°
⎝ 1.36 ⎟⎠
n
⎝ pipe ⎠
Thus, q r = 90.0° −q c = 42.7° and from Snell’s law,
⎛ npipe sinq r ⎞
⎛ (1.36 ) sin 42.7° ⎞
= sin −1 ⎜
q i = sin −1 ⎜
⎟⎠ = 67.3°
⎟
⎝
nair
1.00
⎝
⎠
22.39
The angle of incidence at each of the shorter faces of the prism is
45°, as shown in the figure at the right. For total internal reflection to
occur at these faces, it is necessary that the critical angle be less than
45°. With the prism surrounded by air, the critical angle is given by
sinq c = nair nprism = 1.00 nprism, so it is necessary that
sinq c =
or
22.40
(a)
nprism >
1.00
< sin 45°
nprism
1.00
1.00
=
=
sin 45°
2 2
2
The minimum angle of incidence for which
total internal reflection occurs is the critical angle. At the critical angle, the angle of
refraction is 90°, as shown in the figure at the
right. From Snell’s law, ng sinq i = na sin 90°,
the critical angle for the glass-air interface is
found to be
⎛ n sin 90° ⎞
−1 ⎛ 1.00 ⎞
q i = q c = sin −1 ⎜ a
⎟ = sin ⎜⎝ 1.78 ⎟⎠ = 34.2°
n
⎝
⎠
g
(b)
When the slab of glass has a layer of water
on top, we want the angle of incidence at
the water-air interface to equal the critical
angle for that combination of media. At this
angle, Snell’s law gives
air na = 1.00
glass ng = 1.78
qi
air na = 1.00
qc
qi
qc
water
nw = 1.333
glass
ng = 1.78
continued on next page
68719_22_ch22_p198-222.indd 213
1/7/11 2:49:21 PM
214
Chapter 22
nw sinq c = na sin 90° = 1.00
sinq c = 1.00 nw
and
Now, considering the refraction at the glass-water interface, Snell’s law gives
ng sinq i = ng sinq c. Combining this with the result for sinq c from above, we find the required
angle of incidence in the glass to be
⎛ n sinq c ⎞
⎛ n (1.00 nw ) ⎞
⎛ 1.00 ⎞
⎛ 1.00 ⎞
q i = sin −1 ⎜ w
= sin −1 ⎜ w
= sin −1 ⎜
= sin −1 ⎜
= 34.2°
⎟
⎟
⎟
⎝ 1.78 ⎟⎠
ng
⎠
⎝ ng ⎠
⎝
⎝ ng ⎠
(c) and (d)
22.41
(a)
Observe in the calculation of part (b) that all the physical properties of the intervening layer (water in this case) canceled, and the result of part (b) is identical to that of
part (a). This will always be true when the upper and lower surfaces of the intervening layer are parallel to each other. Neither the thickness nor the index of refraction
of the intervening layer affects the result.
Snell’s law can be written as sinq1 sinq 2 = v1 v2. At the critical angle of incidence (q1 = q c ),
the angle of refraction is 90°, and Snell’s law becomes sinq c = v1 v2. At the concrete-air
boundary,
⎛v ⎞
⎛ 343 m s ⎞
= 10.7°
q c = sin −1 ⎜ 1 ⎟ = sin −1 ⎜
⎝ 1 850 m s ⎟⎠
⎝ v2 ⎠
22.42
(b)
Sound can be totally reflected only if it is initially traveling in the slower medium. Hence,
at the concrete-air boundary, the sound must be traveling in air .
(c)
Sound in air falling on the wall from most directions is 100% reflected , so the wall is a
good mirror.
(a)
The index of refraction for diamond is ndiam ond = 2.419, and the critical angle at a diamondair boundary is
⎛ n
⎞
⎛ 1.000 ⎞
q c = sin −1 ⎜ air ⎟ = sin −1 ⎜
= 24.42°
⎝ 2.419 ⎟⎠
⎝ ndiam ond ⎠
(b)
With the face of the diamond tilted up 35.0° from the horizontal, the normal line to this
face at point P is tipped over 35.0° from the vertical. Thus, the angle of incidence at point
P is q i = 35.0° > q c = 24.42°. Since this angle of incidence exceeds the critical angle,
total internal reflection occurs at point P.
(c)
If the diamond is immersed in water (nwater = 1.333), the critical angle is
⎛ n
⎞
⎛ 1.333 ⎞
q c = sin −1 ⎜ water ⎟ = sin −1 ⎜
= 33.44°
⎝ 2.419 ⎟⎠
⎝ ndiam ond ⎠
(d)
The light continues to enter the top surface of the diamond at normal incidence, so
the angle of incidence at point P continues to be q i = 35.0° > q c = 33.44°. Since this
angle of incidence (barely) exceeds the critical angle for the diamond-water boundary,
total internal reflection still occurs at point P .
(e)
To have light exit the diamond at point P, we need to decrease the angle of incidence at
this point. Thus, we should rotate the diamond clockwise , thereby bringing the normal
line closer to the incident ray at point P.
continued on next page
68719_22_ch22_p198-222.indd 214
1/7/11 2:49:23 PM
Reflection and Refraction of Light
(f )
Rotating the diamond clockwise by an angle q
changes the angle of refraction at point A where the
ray enters the diamond. To find what this new angle
of refraction will be, we extend the line of the top of
the diamond and the line of the face containing point
P until they intersect at point B as shown at the right.
Summing the interior angles in triangle ABP gives
q
A
a
qr
and
35.0°
B
b
f
a + b + 35.0° = 180°
or
215
P
35.0°
(90.0° −q ) + (90.0° − f ) + 35.0° = 180°
r
q r = 35.0° − f
If the ray is to exit the diamond and enter the water at point P, the angle of incidence at
this point must be less than or equal to the critical angle found in part (c). Thus, we require
f ≤ 33.44° and see that we must have
q r ≥ 35.0° − 33.44° = 1.6°
Applying Snell’s law at point A gives the minimum required rotation as
⎛n
sinq r ⎞
⎡ 2.419sin (1.6° ) ⎤
= sin −1 ⎢
q m in = sin −1 ⎜ diam ond
⎥ = 2.9°
⎟
n
1.333
⎝
⎠
⎦
⎣
water
22.43
If q c = 42.0° at the boundary between the prism glass
and the surrounding medium, then sinq c = n2 n1 gives
q1
nm
= sin 42.0°
nglass
nglass
Surrounding
medium, nm
b 60.0°
qr
qc = 42.0° a
From the geometry shown in the figure at the right,
Surface 2
a = 90.0° − 42.0° = 48.0°, b = 180° − 60.0° − a = 72.0°
and q r = 90.0° − b = 18.0°. Thus, applying Snell’s law at the first
surface gives
⎛ sinq r ⎞
⎛ nglass sinq r ⎞
⎛ sin18.0° ⎞
= sin −1 ⎜
q1 = sin −1 ⎜
= sin −1 ⎜
= 27.5°
⎟
⎟
⎝ sin 42.0° ⎟⎠
nm
⎝
⎠
⎝ nm nglass ⎠
22.44
The circular raft must cover the area of the surface
through which light from the diamond could emerge.
Thus, it must form the base of a cone (with apex at the
diamond) whose half angle is q, where q is greater than
or equal to the critical angle.
q
q
h
q
rmin
q
The critical angle at the water-air boundary is
⎛ n ⎞
⎛ 1.00 ⎞
q c = sin −1 ⎜ air ⎟ = sin −1 ⎜
= 48.6°
⎝ 1.333 ⎟⎠
⎝ nwater ⎠
Thus, the minimum diameter of the raft is
2rm in = 2 h tanq m in = 2 h tanq c = 2 ( 2.00 m ) tan 48.6° = 4.54 m
68719_22_ch22_p198-222.indd 215
1/7/11 2:49:25 PM
216
22.45
Chapter 22
At the air-ice boundary, Snell’s law gives the angle of refraction in the ice as
sinq1r =
nair sinq1i
nice
Since the sides of the ice layer are parallel, the angle of incidence at the ice-water boundary is
q 2 i = q1r . Then, from Snell’s law, the angle of refraction in the water is
⎡ nice
⎛ n sinq 2 i ⎞
⎛ nice sinq1r ⎞
−1
−1
⎢
=
sin
=
sin
q 2 r = sin −1 ⎜ ice
⎜⎝ n
⎟⎠
⎝ nwater ⎟⎠
⎢⎣ nwater
water
or
⎛ n sinq ⎞ ⎤
air
1i
⎜
⎟⎥
⎝ nice ⎠ ⎥⎦
⎛ n sinq1i ⎞
⎡ (1.00 ) sin 30.0° ⎤
= sin −1 ⎢
q 2 r = sin −1 ⎜ air
⎥ = 22.0°
⎟
n
1.333
⎝
⎠
⎦
⎣
water
Note that all of the properties of the ice canceled out in the above calculation, and the result is the
same as if the ice had not been present. This will always be true when the intermediate medium
has parallel sides.
22.46
When light coming from the surrounding medium is
incident on the surface of the glass slab, Snell’s law
gives ng sinq r = ns sinq i , or
(
)
sinq r = ns ng sinq i
(a)
If q i = 30.0° and the surrounding medium is
water (ns = 1.333), the angle of refraction is
⎡ 1.333sin ( 30.0° ) ⎤
q r = sin −1 ⎢
⎥ = 23.7°
1.66
⎦
⎣
(b)
From Snell’s law given above, we see that as ns → ng we have sinq r → sinq i , or the
angle of refraction approaches the angle of incidence, q r → q i = 30.0° .
22.47
(c)
If ns > ng, then sinq r = (ns ng )sinq i > sinq i , or q r > q i .
(a)
Given that the angle q shown in the figure at the right is
30.0°, the maximum distance the observer can be from the
pool and continue to see the lower edge on the opposite
side of the pool is
d
q
d = (1.85 m ) tan 30.0° = 1.07 m
(b)
If the pool is now completely filled with water, the light
ray coming to the observer’s eye from the lower opposite
edge of the pool will refract at the surface of the water as
shown in the second figure. The angle of refraction is
⎛ n sin 30.0° ⎞
⎡ (1.333) sin 30.0° ⎤
= sin −1 ⎢
f = sin −1 ⎜ water
⎥ = 41.8°
⎟
nair
1.00
⎝
⎠
⎦
⎣
1.85 m
q
d´
air
nair
f
1.85 m
30.0°
nwater
The maximum distance the observer can now be from the
pool and still see the same boundary is
d ′ = (1.85 m ) tanf = (1.85 m ) tan 41.8° = 1.65 m
68719_22_ch22_p198-222.indd 216
1/7/11 2:49:27 PM
Reflection and Refraction of Light
22.48
(a)
217
For polystyrene surrounded by air, total internal
reflection at the left vertical face requires that
⎛n ⎞
⎛ 1.00 ⎞
q 3 ≥ q c = sin −1 ⎜ air ⎟ = sin −1 ⎜
= 42.2°
⎝ 1.49 ⎟⎠
n
⎝ p ⎠
From the geometry shown in the figure at the right,
q 2 = 90.0° −q 3 ≤ 90.0° − 42.2° = 47.8°
Thus, use of Snell’s law at the upper surface gives
sinq1 =
n p sinq 2
nair
≤
(1.49 ) sin 47.8°
1.00
= 1.10
so it is seen that any angle of incidence ≤ 90° at the upper surface will yield total internal
reflection at the left vertical face.
22.49
(b)
Repeating the steps of part (a) with the index of refraction of air replaced by that of water
yields q 3 ≥ 63.5°, q 2 ≤ 26.5°, sinq1 ≤ 0.499, and q1 ≤ 29.9° .
(c)
Total internal reflection is not possible since npolystyrene < ncarbon disulfide .
(a)
From the geometry of the figure at the
right, observe that q1 = 60.0°. Therefore,
a = 90.0° −q1 = 30.0°
(q
and
2
+ 90.0° ) + a + 30.0° = 180.0°
Thus, q 2 = 180.0° − 120.0° − a = 30.0°
nglass
Since the prism is immersed in water,
n2 = 1.333 and Snell’s law gives
⎛ nglass sinq 2 ⎞
−1 ⎛ (1.66 ) sin 30.0° ⎞
q 3 = sin −1 ⎜
⎟⎠ = 38.5°
⎟⎠ = sin ⎜⎝
n2
1.333
⎝
(b)
For refraction to occur at point P, it is necessary that q c > q1 .
Thus,
⎛ n ⎞
q c = sin −1 ⎜ 2 ⎟ > q1
⎝ nglass ⎠
which gives n2 > nglass sinq1 = (1.66 ) sin 60.0° = 1.44 .
22.50
Applying Snell’s law to this refraction, recognizing that nair = 1.00, gives
nglass sinq 2 = nair sinq1 = sinq1
If q1 = 2q 2 , this becomes
nglass sinq 2 = sin ( 2q 2 ) = 2sinq 2 cosq 2
continued on next page
68719_22_ch22_p198-222.indd 217
1/7/11 2:49:30 PM
218
Chapter 22
or
cosq 2 =
nglass
2
and
⎛ nglass ⎞
q 2 = cos −1 ⎜
⎝ 2 ⎟⎠
Then, the desired angle of incidence is
⎛ nglass ⎞
⎛ 1.56 ⎞
= 2 cos −1 ⎜
q1 = 2q 2 = 2 cos −1 ⎜
= 77.5°
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
22.51
In the figure at the right, observe that
b = 90° −q1 and a = 90° −q1 . Thus,
b =a.
Similarly, on the right side of the
prism, d = 90° −q 2 and g = 90° −q 2 ,
giving d = g .
Next, observe that the angle between
the reflected rays is B = (a + b ) + (g + d ) ,
so B = 2 (a + g ) . Finally, observe that
the left side of the prism is sloped at
angle a from the vertical, and the right
side is sloped at angle g . Thus, the
angle between the two sides is
A = a + g , and we obtain the result B = 2 (a + g ) = 2 A .
22.52
(a)
Observe in the sketch at the right that a ray originally
traveling along the inner edge will have the smallest angle
of incidence when it strikes the outer edge of the fiber
in the curve. Thus, if this ray is totally internally reflected,
all of the others are also totally reflected.
For this ray to be totally internally reflected it is necessary
that
q ≥ qc
But
sinq =
or
R−d
,
R
sinq ≥ sinq c =
nair
1
=
npipe n
so we must have
R−d 1
≥
n
R
which simplifies to R ≥ nd ( n − 1) , or Rm in = nd ( n − 1) .
(b)
As d → 0, Rm in → 0. This is reasonable behavior.
⎛
nd
d ⎞
As n increases, the minimum acceptable radius of curvature ⎜ Rm in =
=
n
−
1
1
−
1 n ⎟⎠
⎝
decreases. This is reasonable behavior.
As n → 1, Rm in increases. This is reasonable behavior.
(c)
68719_22_ch22_p198-222.indd 218
Rm in =
(1.40 )(100 mm )
nd
=
= 350 mm
n −1
1.40 − 1
1/7/11 2:49:32 PM
Reflection and Refraction of Light
22.53
219
Consider light which leaves the lower end of the wire and
travels parallel to the wire while in the benzene. If the wire
appears straight to an observer looking along the dry portion
of the wire, this ray from the lower end of the wire must enter
the observer’s eye as he sights along the wire. Thus, the ray
must refract and travel parallel to the wire in air. The angle of
refraction is then q 2 = 90.0° − 30.0° = 60.0°. From Snell’s law,
the angle of incidence was
⎛ n sinq 2 ⎞
⎛ (1.00 ) sin 60.0° ⎞
= sin −1 ⎜
q1 = sin −1 ⎜ air
⎟⎠ = 35.3°
⎝
1.50
⎝ nbenzene ⎟⎠
and the wire is bent by angle q = q 2 −q1 = 60.0° −q1 = 60.0° − 35.3° = 24.7° .
22.54
In the sketch at the right, the angle of incidence
at A is the same as the prism angle at point O.
This is true because tipping the line OA up
angle q from the horizontal necessarily tips its
normal line over at angle q from the vertical.
Given that q = 60.0°, application of Snell’s law
at point A gives
1.50 sin b = (1.00 ) sin 60.0° or
b = 35.3°
From triangle AOB, we calculate the angle of incidence and reflection, g , at point B:
q + ( 90.0° − b ) + ( 90.0° − g ) = 180°
or
g = q − b = 60.0° − 35.3° = 24.7°
Now, we find the angle of incidence at point C using triangle BCQ:
(90.0° − g ) + (90.0° − d ) + (90.0° −q ) = 180°
or
d = 90.0° − (q + g ) = 90.0° − 84.7° = 5.3°
Finally, application of Snell’s law at point C gives (1.00 ) sinf = (1.50 ) sin ( 5.3° ) ,
or f = sin −1 (1.50 sin 5.3° ) = 8.0° .
22.55
The path of a light ray during a reflection
and/or refraction process is always
reversible. Thus, if the emerging ray is
parallel to the incident ray, the path which the
light follows through this cylinder must be
symmetric about the center line as shown at
the right.
ncylinder
⎛ 1.00 m ⎞
⎛ d 2⎞
= 30.0°
Thus, q1 = sin −1 ⎜
= sin −1 ⎜
⎝ R ⎟⎠
⎝ 2.00 m ⎟⎠
Triangle ABC is isosceles, so g = a and b = 180° − a − g = 180° − 2a . Also, b = 180° −q1 , which
gives a = q1 2 = 15.0°. Then, from applying Snell’s law at point A,
ncylinder =
68719_22_ch22_p198-222.indd 219
nair sinq1 (1.00 ) sin 30.0°
=
= 1.93
sina
sin15.0°
1/7/11 2:49:35 PM
220
Chapter 22
22.56
The angle of refraction as the light enters the left end of the slab is
⎛ n sinq1 ⎞
⎛ (1.00 ) sin 50.0° ⎞
= sin −1 ⎜
q 2 = sin −1 ⎜ air
⎟⎠ = 31.2°
⎟
⎝
1.48
⎝ nslab ⎠
Observe from the figure that the first reflection occurs at x = d, the second reflection is at x = 3d,
the third is at x = 5d, and so forth. In general, the Nth reflection occurs at x = ( 2N − 1) d, where
d=
( 0.310 cm 2) =
tanq 2
0.310 cm
= 0.256 cm
2 tan 31.2°
Therefore, the number of reflections made before reaching the other end of the slab at
x = L = 42 cm is found from L = ( 2N − 1) d to be
N=
22.57
(a)
1 ⎛ L ⎞ 1 ⎛ 42 cm
⎞
+ 1⎟ = 82.5 or 82 complete reflections
⎜ + 1⎟ = ⎜
2 ⎝ d ⎠ 2 ⎝ 0.256 cm ⎠
If q1 = 45.0°, application of Snell’s law at the
point where the beam enters the plastic block
gives
(1.00 ) sin 45.0° = n sinf
[1]
Application of Snell’s law at the point where
the beam emerges from the plastic, with
q 2 = 76.0°, gives
n sin ( 90.0° − f ) = (1.00 ) sin 76.0°
or
(1.00 ) sin 76.0° = n cosf
[2]
Dividing Equation [1] by Equation [2], we obtain
tanf =
sin 45.0°
= 0.729
sin 76.0°
Thus, from Equation [1],
(b)
f = 36.1°
sin 45.0° sin 45.0°
=
= 1.20
sinf
sin 36.1°
Observe from the figure above that sinf = L d. Thus, the distance the light travels inside
the plastic is d = L sinf , and if L = 50.0 cm = 0.500 m, the time required is
Δt =
68719_22_ch22_p198-222.indd 220
n=
and
d L sinf
nL
1.20 ( 0.500 m )
=
=
=
= 3.39 × 10 −9 s = 3.39 ns
v
cn
c sinf ( 3.00 × 108 m s ) sin 36.1°
1/7/11 2:49:38 PM
Reflection and Refraction of Light
22.58
221
Snell’s law would predict that nair sinq i = nwater sinq r , or since nair = 1.00,
sinq i = nwater sinq r
Comparing this equation to the equation of a straight line, y = mx + b, shows that if Snell’s law is
valid, a graph of sinq i versus sinq r should yield a straight line that would pass through the origin
if extended and would have a slope equal to nwater .
q i ( deg ) q r ( deg ) sinq i
sinq r
10.0
7.50
0.174 0.131
20.0
15.1
0.342 0.261
30.0
22.3
0.500 0.379
40.0
28.7
0.643 0.480
50.0
35.2
0.766 0.576
60.0
40.3
0.866 0.647
70.0
45.3
0.940 0.711
80.0
47.7
0.985 0.740
The straightness of the graph line and the fact that its extension passes through the origin demonstrates the validity of Snell’s law. Using the end points of the graph line to calculate its slope
gives the value of the index of refraction of water as
nwater = slope =
22.59
0.985 − 0.174
= 1.33
0.740 − 0.131
Applying Snell’s law at points A, B, and C gives
and
1.40 sina = 1.60 sinq1
[1]
1.20 sin b = 1.40 sina
[2]
1.00 sinq 2 = 1.20 sin b
[3]
Combining Equations [1], [2], and [3] yields
sinq 2 = 1.60 sinq1
[4]
Note that Equation [4] is exactly what Snell’s law would yield if the second and third layers of
this “sandwich” were ignored. This will always be true if the surfaces of all the layers are parallel
to each other.
(a)
If q1 = 30.0°, then Equation [4] gives q 2 = sin −1 (1.60 sin 30.0° ) = 53.1° .
(b)
At the critical angle of incidence on the lowest surface, q 2 = 90.0°. Then, Equation [4] gives
⎛ sin 90.0° ⎞
⎛ sinq 2 ⎞
q1 = sin −1 ⎜
= 38.7°
= sin −1 ⎜
⎝ 1.60 ⎟⎠
⎝ 1.60 ⎟⎠
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222
Chapter 22
Given Conditions and Observed Results
22.60
Case 1
Case 2
For the first placement, Snell’s law gives
n2 =
Case 3
n1 sin 26.5°
sin 31.7°
In the second placement, application of Snell’s law yields
n sin 36.7°
⎛ n sin 26.5° ⎞
n3 sin 26.5° = n2 sin 36.7° = ⎜ 1
sin 36.7°, or n3 = 1
⎝ sin 31.7° ⎟⎠
sin 31.7°
Finally, using Snell’s law in the third placement gives
sinq R =
and
22.61
⎛ sin 31.7° ⎞
n1 sin 26.5°
= ( n1 sin 26.5° ) ⎜
= 0.392
n3
⎝ n1 sin 36.7° ⎟⎠
q R = 23.1°
From the right triangle PAC in the figure at the right,
observe that
⎛ 6.00 cm ⎞
⎛ h⎞
q1 = sin ⎜ ⎟ = sin −1 ⎜
= 30.0°
⎝ 12.0 cm ⎟⎠
⎝ R⎠
−1
nair = 1.00
nLucite = 1.50
q1
This is also the angle of incidence at point P where
the light ray enters the Lucite. Snell’s law then gives
⎛ n sinq1 ⎞
sin 30.0° ⎤
= sin −1 ⎡⎢
q 2 = sin ⎜ air
= 19.5°
⎟
⎣ 1.50 ⎥⎦
⎝ n
⎠
−1
P
q2
h
f
f
B
q
R
q1
A
C
Lucite
Now, by observing the vertical angles at point P, we find that q 2 + f = q1 , so the angle of incidence at point B where the ray exits from the Lucite is f = q1 −q 2 = 10.5°. The angle of refraction
at point B is then given by Snell’s law as
⎛n
sinf ⎞
−1
q = sin −1 ⎜ Lucite
⎟⎠ = sin [(1.50 ) sin10.5° ] = 15.9°
nair
⎝
68719_22_ch22_p198-222.indd 222
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23
Mirrors and Lenses
QUICK QUIZZES
1.
At C.
2.
Choice (c). Since nwater > nair, the virtual image of the fish formed by refraction at the flat water
surface is closer to the surface than is the fish. See Equation 23.9 in the textbook.
3.
(a)
False. A concave mirror forms an inverted image when the object distance is greater than
the focal length.
(b)
False. The magnitude of the magnification produced by a concave mirror is greater than 1 if
the object distance is less than the radius of curvature.
(c)
True.
4.
Choice (b). In this case, the index of refraction of the lens material is less than that of the surrounding medium. Under these conditions, a biconvex lens will be divergent.
5.
Although a ray diagram only uses 2 or 3 rays (those whose direction is easily determined using only
a straight edge), an infinite number of rays leaving the object will always pass through the lens.
6.
(a)
False. A virtual image is formed on the left side of the lens if p < f .
(b)
True. An upright, virtual image is formed when p < f , while an inverted, real image is
formed when p > f .
(c)
False. A magnified, real image is formed if 2 f > p > f , and a magnified, virtual image is
formed if p < f .
223
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224
Chapter 23
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The image formed by a flat mirror of a real object is always an upright, virtual image that is the
same size as the object and located as far behind the mirror as the object is in front of the mirror.
Thus, statements (b), (c), and (e) are all true, while statements (a) and (d) are false.
2.
Since the size of the image is 1.50 times the size of the object, the magnitude of the magnification is M = 1.50. Because the image is upright, M > 0. Thus, M = − q p = +1.50, which gives
q = −1.50 p. Then from the mirror equation, we have
f=
R
qp
( −1.50 p) p = +3.00 p = +3.00 ( +30.0 cm ) = +90.0 cm
=
=
2 p+q
p − 1.50 p
and the correct choice is (d).
3.
For a convergent lens, f > 0, and because the image is real, q > 0. The thin-lens equation,
1 p + 1 q = 1 f , then gives
p=
qf
(12.0 cm )(8.00 cm )
=
= +24.0 cm
q− f
12.0 cm − 8.00 cm
Since p > 0 , the object is in front (in this case, to the left) of the lens, and the correct
choice is (c).
4.
For a converging lens, the focal length is positive, or f > 0. Since the object is virtual, we know
that the object distance is negative, or p < 0 and p = − | p |. Thus, the thin-lens equation gives the
image distance as
q=
⎛ | p| ⎞
pf
−| p| f
f
=
= +⎜
p− f −| p|− f
⎝ | p | + f ⎟⎠
Since | p | and f are positive quantities, we see that q > 0 and the image is real. Also, since
| p | (| p | + f ) < 1, we see that q < f . Thus, we have shown that choices (a) and (d) are false statements, while choices (b), (c), and (e) are all true.
5.
From the mirror equation, 1 p + 1 q = 2 R = 1 f , with f < 0 since the mirror is convex, the image
distance is found to be
q=
pf
(16.0 cm )( −6.00 cm )
=
= − 4.36 cm
p − f 16.0 cm − ( −6.00 cm )
Since q < 0, the image is virtual and located 4.36 cm behind the mirror. Choice (d) is the correct
answer.
6.
For a divergent lens, f < 0, and because the object is real, p > 0. The thin-lens equation,
1 p + 1 q = 1 f , then gives
q=
pf
(10.0 cm )( −16.0 cm )
=
= − 6.15 cm
p − f 10.0 cm − ( −16.0 cm )
Since q < 0, the image is in front (in this case, to the left) of the lens, and the correct
choice is (b).
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Mirrors and Lenses
225
7.
A concave mirror forms inverted, real images of real objects located outside the focal point ( p > f ),
and upright, magnified, virtual images of real objects located inside the focal point ( p < f ) . Virtual
images, located behind the mirror, have negative image distances by the sign convention of
Table 23.1. Choices (d) and (e) are true statements, and all other choices are false.
8.
A convergent lens forms inverted, real images of real objects located outside the focal point
( p > f ). When p > 2 f , the real image is diminished in size, and the image is enlarged if
2 f > p > f . For real objects located inside the focal point ( p < f ) of the convergent lens, the
image is upright, virtual, and enlarged. In the given case, p > 2 f , so the image is real, inverted,
and diminished in size. Choice (c) is the correct answer.
9.
With a real object in front of a convex mirror, the image is always upright, virtual, diminished
in size, and located between the mirror and the focal point. Thus, of the available choices, only
choice (c) is a true statement.
10.
For a real object ( p > 0) and a diverging lens ( f < 0), the image distance given by the thin-lens
equation is
q=
p (− f )
p f
pf
=
=−
<0
p− f
p+ f
p − (− f )
and the magnification is
M=−
−q
q
=−
>0
p
p
Thus, the image is always virtual and upright, meaning that choice (b) is a true statement while
all other choices are false.
11.
Light rays reflecting from the fish, and passing through the flat refracting surface of the water
(nwater > nair) on the way to the fisherman’s eye, form a virtual image located closer to the surface
than the fish’s actual depth (see Example 23.6 in the textbook). Thus, the fisherman should aim
below the apparent location of the fish, and (b) is the correct choice.
12.
The image that a plane reflecting surface forms of a real object is an upright, virtual image located
as far behind the reflecting surface as the object is in front of it. Choices (a) and (d) must be eliminated because they portray inverted images. Also, the tip of the image arrow should be closer to the
mirror than the base of the image arrow, just as is true of the ends of the object arrow. Thus, choice
(c) must be eliminated, leaving (b) as best describing the image formed by the mirror.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The stream appears shallower than its true depth because refraction of light coming from pebbles
on the bottom of the stream forms virtual images located closer to the surface than the actual
objects.
4.
Chromatic aberration is produced when light passes through a material, as it does when passing
through the glass of a lens. A mirror, silvered on its front surface, never has light passing through
it, so this aberration cannot occur. This is only one of many reasons why large telescopes use
mirrors rather than lenses for their primary optical elements.
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226
Chapter 23
6.
Light rays diverge from the position of a virtual image just as they do from an actual object.
Thus, a virtual image can be as easily photographed as any object can. Of course, the camera
would have to be placed near the axis of the lens or mirror in order to intercept the light rays.
8.
Actually no physics is involved here. The design is chosen so your eyelashes will not brush
against the glass as you blink. A reason involving a little physics is that with this design, when
you direct your gaze near the outer circumference of the lens, you receive a ray that has passed
through glass with more nearly parallel surfaces of entry and exit. Then the lens minimally
distorts the direction to the object you are looking at.
10.
Both words are inverted. However, OXIDE looks the same right-side-up and upside-down. LEAD
does not.
12.
(a)
No. The screen is needed to reflect the light toward your eye.
(b)
Yes. The light is traveling toward your eye and diverging away from the position of the
image, the same as if the object were located at that position.
14.
The correct answer is choice (d). The entire image would appear because any portion of the lens
can form the image. The image would be dimmer because the card reduces the light intensity on
the screen by 50%.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a)
1.0 m behind nearest mirror
(b)
the palm
(c)
5.0 m behind nearest mirror
(d)
back of the hand
(e)
7.0 m behind nearest mirror
(f )
the palm
4.
4.58 m
6.
(a)
8.
R = − 0.790 cm
R = + 2.22 cm
(b)
(g)
all are virtual
M = +10.0
10.
The mirror is concave with R ≈ 60 cm and f ≈ 30 cm.
12.
(a)
p = +15.0 cm
14.
(a)
60.0 cm
(b)
q = 42.9 cm, M = −0.429, so the image is real, inverted, and 42.9 cm in front of mirror.
(c)
q = −15.0 cm, M = 1.50, so the image is virtual, upright, and 15.0 cm behind the mirror.
(a)
a real object located 16 cm in front of the mirror
(b)
upright and one-third the size of the object
(a)
p = 8.00 cm
16.
18.
68719_23_ch23_p223-255.indd 226
(b)
(b)
R = 60.0 cm
See Solution.
(c)
virtual
1/7/11 2:50:30 PM
Mirrors and Lenses
20.
227
(a)
As the ball moves from p = 3.00 m to p = 0.500 m, the image moves from q = + 0.600 m
to q = + ∞. As the ball moves from p = 0.500 m to p = 0, the image moves from q = − ∞ to
q = 0.
(b)
at t = 0.639 s when p = 1.00 m, and at t = 0.782 s when p = 0
22.
in the water, 9.00 cm inside the wall of the bowl
24.
(a)
26.
4.8 cm
28.
See Solution.
30.
(a)
at either 36.2 cm or 13.8 cm from the screen
(b)
When q = 36.2 cm, M = −2.62. When q = 13.8 cm, M = − 0.381.
(a)
See Solution.
(d)
The mirror equation yields q = − 40.0 cm and M = +2.00.
(a)
See Solution.
(c)
Graph (a) yields an upright, virtual image located 13.3 cm from the lens and one-third the
size of the object. Graph (b) yields an upright, virtual image located 6.7 cm from the lens
and two-thirds the size of the object. Both agree with the algebraic solutions.
32.
34.
1.50 m
(b)
(b)
(b)
1.75 m
40 cm behind the mirror
(c)
+2
See Solution.
36.
f = +5.68 cm
38.
(a)
12.3 cm to the left of the lens
(b)
M = +0.615
40.
(a)
q = p f (p − f )
(b)
q < 0 for all values of p > 0 and f < 0
(c)
q > 0 only if p > f when f > 0
(a)
f2 = −11.1 cm
(b)
M = +2.50
42.
(c)
See Solution.
(c)
virtual and upright
44.
(a)
13.3 cm
(b)
M = −5.91
46.
(a)
q1 = +30.0 cm
(b)
20.0 cm beyond the second lens
(c)
p2 = −20.0 cm
(d)
q2 = + 4.00 cm
(e)
M1 = −2.00
(f )
M 2 = + 0.200
(g)
M total = −0.400
(h)
real, inverted
48.
(a)
q = 5f 4
(b)
M = −1 4
(c)
real, inverted,
opposite side
50.
(a)
f = −12.0 cm
(b)
q = f = −12.0 cm
(c)
q = −9.00 cm
(d)
q = −6.00 cm
(e)
q = − 4.00 cm
68719_23_ch23_p223-255.indd 227
(c)
inverted
1/7/11 2:50:32 PM
228
52.
Chapter 23
(a)
25.3 cm to the right of the mirror
(b)
virtual
(c)
54.
Δx = 21.3 cm
56.
See the Solution.
58.
See the Solution.
60.
(a)
p=4f 3
(c)
M a = −3, M b = + 4
62.
(b)
upright
(d)
M = + 8.05
(b)
See Solution.
(c)
f = −20.0 cm
p = 3f 4
fm irror = 11.7 cm
64.
(a)
0.708 cm in front or the spherical ornament
66.
(a)
convex
(b)
at the 30.0-cm mark
PROBLEM SOLUTIONS
23.1
(a)
Due to the finite value of the speed of light, the light arriving at your eye must have
reflected from your face at a slightly earlier time. Thus, the image viewed in the mirror
shows you younger than your actual age .
(b)
If you stand 40 cm in front of the mirror, the time required for light scattered from your
face to travel to the mirror and back to your eye is
Δt =
23.2
(a)
2d
2 ( 0.40 m )
=
= 2.7 × 10 −9 s
c
3.0 × 108 m s
or
∼ 10 −9 s
With the palm located 1.0 m in front of the nearest mirror, that mirror forms an image, I P 1 ,
located 1.0 m behind the nearest mirror .
68719_23_ch23_p223-255.indd 228
(b)
The image I P 1 is an image of the palm .
(c)
The woman’s left hand is located 2.0 m in front of the farthest mirror, which forms an
image located 2.0 m behind this mirror (and hence, 5.0 m in front of the nearest mirror).
This image, I B1 , serves as an object for the nearest mirror, which then forms an image, I B 2 ,
located 5.0 m behind the nearest mirror .
(d)
The images I B1 and I B 2 are both images of the back of the hand .
(e)
The image I P 1 [see part (a)] serves as an object located 4.0 m in front of the farthest mirror,
which forms an image I P 2 , located 4.0 m behind that mirror and 7.0 m in front of the nearest mirror. This image then serves as an object for the nearest mirror, which forms an image
I P 3 , located 7.0 m behind the nearest mirror .
(f)
The images I P 2 and I P 3 are both images of the palm .
(g)
Since all images are located behind the mirrors, all are virtual images .
1/7/11 2:50:35 PM
Mirrors and Lenses
23.3
23.4
229
(1)
The first image in the left-hand mirror is 5.00 ft behind the mirror
or 10.0 ft from the person .
(2)
The first image in the right-hand mirror serves as an object for the left-hand mirror. It is
located 10.0 ft behind the right-hand mirror, which is 25.0 ft from the left-hand mirror.
Thus, the second image in the left-hand mirror is 25.0 ft behind the mirror
or 30.0 ft from the person .
(3)
The first image in the left-hand mirror serves as an object for the right-hand mirror. It is
located 20.0 ft in front of the right-hand mirror and forms an image 20.0 ft behind that
mirror. This image then serves as an object for the left-hand mirror. The distance from this
object to the left-hand mirror is 35.0 ft. Thus, the third image in the left-hand mirror is
35.0 ft behind the mirror or 40.0 ft from the person .
The virtual image is as far behind the mirror
as the choir is in front of the mirror. Thus,
the image is 5.30 m behind the mirror.
The image of the choir is
0.800 m + 5.30 m = 6.10 m
from the organist. Using similar triangles gives
6.10 m
h′
=
0.600 m 0.800 m
or
23.5
⎛ 6.10 m ⎞
h ′ = ( 0.600 m ) ⎜
= 4.58 m
⎝ 0.800 m ⎟⎠
When an object is in front of a plane mirror, that mirror forms an upright, virtual image that is
the same size as the object and as far behind the mirror as the object is in front of the mirror.
This statement is true even if the mirror is rotated, as shown in the ray diagrams given below. In
figure (a), a real object O1 is distance p1 in front of the upper mirror in the periscope. This mirror
forms the virtual image I1 at distance p1 behind the mirror. As shown in figure (b), this image
serves as the object for the lower mirror in the periscope, and is distance p2 = p1 + h in front
of the lower mirror. The lower mirror then forms the final image I2, an upright, virtual image,
located distance p2 = p1 + h behind this mirror.
O2 = I1
I1
Upper
mirror
Lower
mirror
| q1| = p1
O1
p2 = p1 + h
I2
p1
(a)
| q2| = p2 = p1 + h
(b)
continued on next page
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230
Chapter 23
(a)
As shown in the above ray diagrams, the final image is distance q2 = p1 + h behind the
lower mirror, where p1 is the distance from the original object to the upper mirror and h is
the vertical distance between the two mirrors in the periscope.
(b)
The final image is behind the mirror and is virtual .
(c)
As seen from the ray diagrams, the final image I2 is oriented the same way as the original
object, and is therefore upright .
(d)
The overall magnification is
⎛ q ⎞ ⎛ q ⎞ ⎛ −p ⎞ ⎛ −p ⎞
M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ( +1)( +1) = +1
⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p1 ⎠ ⎝ p2 ⎠
The final image is therefore upright and the same size as the original object.
23.6
(e)
No . The images formed by plane mirrors are upright in both directions. Just as plane
mirrors do not reverse up and down, neither do they reverse left and right.
(a)
Since the object is in front of the mirror, p > 0. With the image behind the mirror, q < 0.
The mirror equation gives the radius of curvature as
1
1
10 − 1
2 1 1
= + =
−
=
R p q 1.00 cm 10.0 cm 10.0 cm
or
23.7
⎛ 10.0 cm ⎞
R = 2⎜
⎟ = + 2.22 cm
⎝
9 ⎠
q
(−10.0 cm)
=−
= +10.0 .
p
1.00 cm
(b)
The magnification is M = −
(a)
The center of curvature of a convex mirror is behind the mirror. Therefore, the radius of
curvature, and hence the focal length f = R 2, is negative. With the image behind the
mirror, the image is virtual and q = −10.0 cm. The mirror equation then gives
p=
qf
( −10.0 cm )( −15.0 cm )
=
= +30.0 cm
q − f −10.0 cm − ( −15.0 cm )
The object should be placed 30.0 cm in front of the mirror .
(b)
The magnification of the mirror is
M=−
q
( −10.0 cm )
=−
= + 0.333
p
+30.0 cm
Therefore, the image is upright and one-third the size of the object.
23.8
The lateral magnification is given by M = − q p. Therefore, the image distance is
q = −Mp = − ( 0.013 0 )( 30.0 cm ) = −0.390 cm
The mirror equation,
2 pq
2 1 1
= + , or R =
, gives
p+q
R p q
continued on next page
68719_23_ch23_p223-255.indd 230
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Mirrors and Lenses
R=
231
2 ( 30.0 cm )( −0.390 cm )
= − 0.790 cm
30.0 cm − 0.390 cm
The negative sign tells us that the surface is convex, as expected.
23.9
(a)
The center of curvature of a concave mirror is in front of the mirror. Therefore, both the
radius of curvature and the focal length, f = R 2, are positive. Since the image is virtual,
the image distance is negative and q = −20.0 cm. With R = + 40.0 cm and f = +20.0 cm, the
mirror equation gives
p=
( −20.0 cm )( + 20.0 cm )
qf
=
= +10.0 cm
q − f −20.0 cm − ( + 20.0 cm )
Thus, the object should be placed 10.0 cm in front of the mirror .
(b)
The magnification of the mirror is
M=−
q
( −20.0 cm )
=−
= + 2.00
p
+10.0 cm
Therefore, the image is upright and twice the size of the object.
23.10
The image was initially upright but became inverted when Dina was more than 30 cm from the
mirror. From this information, we know that the mirror must be concave , because a convex
mirror will form only upright, virtual images of real objects.
When the object is located at the focal point of a concave mirror, the rays leaving the mirror are
parallel, and no image is formed. Since Dina observed that her image disappeared when she was
about 30 cm from the mirror, we know that the focal length must be f ≈ 30 cm . Also, for
spherical mirrors, R = 2 f . Thus, the radius of curvature of this concave mirror must be R ≈ 60 cm .
23.11
The enlarged, virtual images formed by a concave mirror are upright, so M > 0.
Thus, M = −
q h ′ 5.00 cm
= =
= + 2.50, giving
p h 2.00 cm
q = − 2.50 p = − 2.50 ( + 3.00 cm ) = −7.50 cm
The mirror equation then gives
f=
23.12
( 3.00 cm )( − 7.50 cm )
pq
=
= + 5.00 cm
p+q
3.00 cm − 7.50 cm
Realize that the magnitude of the radius of curvature, R , is the same for both sides of the
hubcap. For the convex side, R = − R ; and for the concave side, R = + R . The object distance
p is positive (real object) and has the same value in both cases. Also, we write the virtual image
distance as q = − q in each case. The mirror equation then gives:
For the convex side,
2
1
1
=
−
−q −R p
or
q=
Rp
R + 2p
[1]
For the concave side,
2 1
1
=
−
R p
−q
or
q=
Rp
R − 2p
[2]
continued on next page
68719_23_ch23_p223-255.indd 231
1/7/11 2:50:41 PM
232
Chapter 23
Comparing Equations [1] and [2], we observe that the smaller magnitude image distance,
q = 10.0 cm, occurs with the convex side of the mirror. Hence, we have
2
1
1
=
−
−10.0 cm − R p
[3]
and for the concave side, q = 30.0 cm gives
2 1
1
=
−
−30.0 cm R p
23.13
[4]
(a)
Adding Equations [3] and [4] yields
(b)
Subtracting [3] from [4] gives
3+1
2
=
, or p = +15.0 cm .
p 30.0 cm
3−1
4
=
, or R = 60.0 cm .
R 30.0 cm
The image is upright, so M > 0, and we have
M=−
q
= + 2.0, or q = − 2.0 p = − 2.0 ( 25 cm ) = − 50 cm
p
The radius of curvature is then found to be
1
1
2 −1
2 1 1
= + =
−
=
, or R = 2 ⎛ 0.50 m ⎞ = 1.0 m
⎜⎝ +1 ⎟⎠
R p q 25 cm 50 cm 50 cm
23.14
(a)
For spherical mirrors, both concave and convex, the radius of curvature is related to the
focal length by R = 2 f . Hence, the radius of curvature of this concave mirror, having
f = +30.0 cm, is R = + 60.0 cm .
(b)
When p = +100 cm (i.e., a real object located 100 cm in front of the mirror),
q=
and
pf
( +100 cm )( +30.0 cm )
=
= + 42.9 cm
p− f
100 cm − 30.0 cm
M=−
q
42.9 cm
=−
= −0.429
p
100 cm
Thus, the image is real (q > 0), inverted (M < 0), diminished in size ( | M |< 1), and
located 42.9 cm in front of the mirror .
(c)
When p = +10.0 cm (i.e., a real object located 10.0 cm in front of the mirror),
q=
and
pf
( +10.0 cm )( +30.0 cm )
=
= −15.0 cm
p− f
10.0 cm − 30.0 cm
M=−
q
( −15.0 cm )
=−
= +1.50
p
10.0 cm
Thus, the image is virtual (q < 0), upright (M > 0), enlarged ( | M |> 1), and
located 15.0 cm behind the mirror .
68719_23_ch23_p223-255.indd 232
1/7/11 2:50:43 PM
Mirrors and Lenses
23.15
233
The focal length of the mirror may be found from the given object and image distances as
1 f = 1 p + 1 q , or
f=
pq
(152 cm )(18.0 cm )
=
= +16.1 cm
p + q 152 cm + 18.0 cm
For an upright image twice the size of the object, the magnification is
M = − q p = +2.00 , giving q = − 2.00 p
Then, using the mirror equation again, 1 p + 1 q = 1 f becomes
1
2 −1
1
1 1 1
+ = −
=
=
p q p 2.00 p 2.00 p f
23.16
f
16.1 cm
=
= 8.05 cm
2.00
2.00
or
p=
(a)
The mirror is convex, so f < 0, and we have f = − f = −8.0 cm. The image is virtual, so
q < 0, or q = − q . Since we also know that q = p 3, the mirror equation gives
1 1 1 3 1
+ = − =
p q p p f
or
−
2
1
=
p − 8.0 cm
and
p = +16 cm
This means that we have a real object located 16 cm in front of the mirror .
23.17
(b)
The magnification is M = − q p = + q p = +1 3 . Thus, the image is upright and one-third
the size of the object.
(a)
Since the mirror is convex, R < 0. Thus, R = − 0.550 m and f = R 2 = − 0.275 m. The mirror
equation then yields
q=
pf
( +10.0 m )( −0.275 m )
=
= − 0.268 m = −26.8 cm
p − f +10.0 m − ( −0.275 m )
The image is located 26.8 cm behind the mirror .
23.18
(b)
The magnification of the image is M = − q p. Since p > 0 and q < 0 in this case, we see that
M > 0. Therefore, the image is upright .
(c)
M=−
(a)
Since the mirror is concave, R > 0, giving R = +24.0 cm and f = R 2 = +12.0 cm. Because
the image is upright (M > 0) and three times the size of the object ( M = 3.00 ) , we have
q
( −26.8 cm )
( −26.8 cm )
=−
=−
= +2.68 × 10 −2 = + 0.026 8
p
10.0 m
10.0 × 10 2 cm
M=−
q
= +3.00
p
and
q = −3p
The mirror equation then gives
1
2.00
1
1
−
=
=
p 3.00 p 3.00 p 12.0 cm
or
p = +8.00 cm
continued on next page
68719_23_ch23_p223-255.indd 233
1/7/11 2:50:46 PM
234
23.19
Chapter 23
(b)
The needed ray diagram, with the object 8.00 cm in front of the mirror, is shown below:
(c)
From a carefully drawn scale drawing, you should find that the image is
upright, virtual, 24.0 cm behind the mirror, and three times the size of the object .
(a)
An image formed on a screen is a real image. Thus, the mirror must be concave since, of
mirrors, only concave mirrors can form real images of real objects.
(b)
The magnified, real images formed by concave mirrors are inverted, so M < 0 and
M =−
q 5.0 m
q
= 1.0 m
= − 5 , or p = =
5
5
p
The object should be 1.0 m in front of the mirror .
(a — revisited) Now that we have both the object distance and image distance, we can use the
mirror equation to be more specific about the required mirror. The needed focal length of
the concave mirror is given by
f=
23.20
(a)
From
pq
(1.0 m )( 5.0 m )
=
= 0.83 m
p + q 1.0 m + 5.0 m
Rp
1 1 2
(1.00 m ) p
+ = , we find q =
=
.
2 p − R 2 p − 1.00 m
p q R
The table gives the image position at
a few critical points in the motion.
Between p = 3.00 m and p = 0.500 m,
the real image moves from 0.600 m to
positive infinity. From p = 0.500 cm
to p = 0, the virtual image moves from
negative infinity to 0.
Object Distance, p
Image Distance, q
3.00 m
0.500 m
0.600 m
±∞
0
0
Note the “jump” in the image position as the ball passes through the focal point of the mirror.
(b)
The ball and its image coincide when p = 0 and when 1 p + 1 p = 2 p = 2 R , or
p = R = 1.00 m.
From Δy = v0 y t + 12 a y t 2 , with v0 y = 0, the times for the ball to fall from y = +3.00 m to these
positions are found to be
68719_23_ch23_p223-255.indd 234
p = 1.00 m position:
t=
2 ( Δy )
=
ay
2 ( −2.00 m )
= 0.639 s and
−9.80 m s 2
p = 0 position:
t=
2 ( −3.00 m )
= 0.782 s
−9.80 m s 2
1/7/11 2:50:50 PM
Mirrors and Lenses
23.21
From
235
n1 n2 n2 − n1
+
=
, with R → ∞, the image position is found to be
p q
R
q =−
n2
p=
n1
⎛ 1.00 ⎞
−⎜
( 50.0 cm ) = − 38.2 cm
⎝ 1.309 ⎟⎠
or the virtual image is 38.2 cm below the upper surface of the ice .
23.22
The location of the image formed by refraction at this
spherical surface is described by Equation 23.7 from the
textbook, which states n1 p + n2 q = (n2 − n1 ) R and uses the
sign convention of Table 23.2. As light crosses this surface,
passing from the water into air, we have n1 = nwater = 1.333,
n2 = nair = 1.00, p = +10.0 cm (object is in front of the surface), and R = −15.0 cm (center of curvature is in front of the
nair = 1.00
surface). Thus, the image distance is found from
n2 n2 − n1 n1
=
−
q
R
p
or
p
R
nwater = 1.333
1.333
+0.666 − 3.999 −3.333
1.00 1.00 − 1.333
=
−
=
=
−15.0 cm
+10.0 cm
30.0 cm
30.0 cm
q
This yields q = 30.0 cm ( −3.333) = −9.00 cm. Thus, the goldfish appears to be in the water,
9.00 cm inside the wall of the bowl .
23.23
Since the center of curvature of the surface is on the side the light comes from, R < 0, giving
n n
n − n1
R = − 4.0 cm. Then, 1 + 2 = 2
becomes
p q
R
1.50
1.00 1.00 − 1.50
=
−
, or q = − 4.0 cm
− 4.0 cm
4.0 cm
q
Thus, the magnification M =
⎛n ⎞q
h′
= − ⎜ 1 ⎟ gives
h
⎝ n2 ⎠ p
⎛ nq⎞
1.50 ( −4.0 cm )
h′ = − ⎜ 1 ⎟ h = −
( 2.5 mm ) = 3.8 mm
1.00 ( 4.0 cm )
⎝ n2 p ⎠
23.24
For a plane refracting surface, R → ∞, and
(a)
n1 n2 n2 − n1
n
+
=
becomes q = − 2 p.
p q
R
n1
Considering the bottom of the pool as the object, p = 2.00 m when the pool is full, and
⎛ 1.00 ⎞
q =−⎜
( 2.00 m ) = −1.50 m
⎝ 1.333 ⎟⎠
or the pool appears to be 1.50 m deep.
(b)
68719_23_ch23_p223-255.indd 235
If the pool is half filled, then p = 1.00 m, and q = − 0.750 m. Thus, the bottom of the pool
appears to be 0.75 m below the water surface or 1.75 m below ground level.
1/7/11 2:50:53 PM
236
23.25
Chapter 23
As parallel rays from the Sun ( object distance, p → ∞ ) enter the transparent sphere from air
= 1.00 ) , the center of curvature of the surface is on the side the light is going toward (back
1
side). Thus, R > 0. It is observed that a real image is formed on the surface opposite the Sun,
giving the image distance as q = +2R. Then,
(n
n1 n2 n2 − n1
+
=
p q
R
becomes
0+
n
n − 1.00
=
2R
R
which reduces to n = 2n − 2.00, and gives n = 2.00 .
23.26
Light scattered from the bottom of the plate undergoes two refractions, once at the top of the plate
and once at the top of the water. All surfaces are planes ( R → ∞ ) , so the image distance for each
refraction is q = − ( n2 n1 ) p. At the top of the plate,
⎛n
⎞
⎛ 1.333 ⎞
q1 B = − ⎜ water ⎟ p1 B = − ⎜
(8.00 cm ) = − 6.42 cm
⎝ 1.66 ⎟⎠
⎝ nglass ⎠
or the first image is 6.42 cm below the top of the plate. This image serves as a real object for the
refraction at the top of the water, so the final image of the bottom of the plate is formed at
⎛ n ⎞
⎛ n ⎞
q2 B = − ⎜ air ⎟ p2B = − ⎜ air ⎟ 12.0 cm + q1B
n
⎝ water ⎠
⎝ nwater ⎠
(
)
⎛ 1.00 ⎞
= −⎜
(18.4 cm ) = −13.8 cm or 13.8 cm below the water surface.
⎝ 1.333 ⎟⎠
Now, consider light scattered from the top of the plate. It undergoes a single refraction, at the top
of the water. This refraction forms an image of the top of the plate at
⎛ n ⎞
⎛ 1.00 ⎞
qT = − ⎜ air ⎟ pT = − ⎜
(12.0 cm ) = − 9.00 cm
⎝ 1.333 ⎟⎠
⎝ nwater ⎠
or 9.00 cm below the water surface.
The apparent thickness of the plate is then Δy = q2B − qT = 13.8 cm − 9.00 cm = 4.8 cm .
23.27
In the drawing at the right, object O (the jellyfish)
is located distance p in front of a plane water-glass
interface. Refraction at that interface produces a
virtual image I ′ at distance q ′ in front it. This image
serves as the object for refraction at the glass-air
interface. This object is located distance p′ = q ′ + t
in front of the second interface, where t is the thickness of the layer of glass. Refraction at the glass-air
interface produces a final virtual image, I, located
distance q in front of this interface.
From n1 p + n2 q = (n2 − n1 ) R with R → ∞
for a plane, the relation between the object and
image distances for refraction at a flat surface is
q = − ( n2 n1 ) p. Thus, the image distance for the refraction at the water-glass interface is
q ′ = − ng nw p. This gives an object distance for the refraction at the glass-air interface of
p′ = (ng nw ) p + t and a final image position (measured from the glass-air interface) of
(
)
continued on next page
68719_23_ch23_p223-255.indd 236
1/7/11 2:50:56 PM
Mirrors and Lenses
q=−
(a)
na
n
p′ = − a
ng
ng
237
⎡⎛ n ⎞
⎛ na ⎞ ⎤
⎡ ⎛ ng ⎞
⎤
a
⎢⎜ ⎟ p + t ⎥ = − ⎢⎜ ⎟ p + ⎜ ⎟ t ⎥
⎝ ng ⎠ ⎥⎦
⎢⎣ ⎝ nw ⎠
⎥⎦
⎢⎣⎝ nw ⎠
If the jellyfish is located 1.00 m (or 100 cm) in front of a 6.00-cm thick pane of glass, then
p = +100 cm, t = 6.00 cm, and the position of the final image relative to the glass-air interface is
⎡⎛ 1.00 ⎞
⎛ 1.00 ⎞
q = − ⎢⎜
100 cm ) + ⎜
(
( 6.00 cm )⎤⎥ = −79.0 cm
⎟
⎝ 1.50 ⎟⎠
⎣⎝ 1.333 ⎠
⎦
It appears to be in the water, 79.0 cm back of the outer surface of the glass pane .
(b)
If the thickness of the glass is negligible (t → 0), the distance of the final image from the
glass-air interface is
q=−
na
ng
⎡ ⎛ ng ⎞
⎤
⎛ na ⎞
⎛ 1.00 ⎞
⎢ ⎜ ⎟ p + 0 ⎥ = − ⎜ ⎟ p = − ⎜⎝
⎟ (100 cm ) = −75.0 cm
1.333 ⎠
⎝ nw ⎠
⎢⎣ ⎝ nw ⎠
⎥⎦
Now, it appears to be 75.0 cm back of the outer surface of the glass pane .
(c)
23.28
Comparing the results of parts (a) and (b), we see that the 6.00-cm thickness of the glass
in part (a) made a 4.00-cm difference in the apparent position of the jellyfish. We conclude
that the thicker the glass, the greater the distance between the final image and the outer
surface of the glass.
We assume the ray shown in the diagram
at the right is a paraxial ray so q1 and q 2
are both sufficiently small to allow us to
write Snell’s law as
n1 tanq1 = n2 tanq 2
n1
| h | = +h
q1
P
n2
q2
O
p
I
| h′ | = –h′
q
Also, note that we are assuming transverse
distances measured upward from the axis OPI are positive, while those measured downward
from this axis are negative. Looking at the right triangle having distance OP as its base, we see
that tanq1 = h p, and looking at the triangle having PI as a base, we have tanq 2 = h ′ q = − h ′ q.
Thus, Snell’s law becomes n1 (h p) = −n2 (h ′ q), and the magnification becomes
M=
23.29
nq
h′
=− 1
n2 p
h
With R1 = + 2.00 cm and R2 = + 2.50 cm, the lens maker’s equation gives the focal length as
⎛ 1
⎞
⎛
1⎞
1
1
1
= + 0.050 0 cm −1
= ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜
−
f
⎝ 2.00 cm 2.50 cm ⎟⎠
⎝ R1 R2 ⎠
or
68719_23_ch23_p223-255.indd 237
f=
1
= + 20.0 cm
0.050 0 cm −1
1/7/11 2:50:59 PM
23.30
Chapter 23
Consider the figure at the right showing an object O
located distance L in front of a screen. A convergent lens
is positioned to focus an image I on the screen. Observe
that the sum of the object and image distances is p + q = L,
giving p = L − q. Also, from the thin-lens equation, we
have
p=
fq
q− f
L−q =
giving
fq
q− f
L
screen
238
O
I
p
q
Simplifying the last result gives (L − q)(q − f ) = f q, which reduces to q 2 − Lq + Lf = 0. Solving by
use of the quadratic formula, with L = 50.0 cm and f = 10.0 cm, gives
q=
(a)
(b)
23.31
L ± L2 − 4Lf 50.0 cm ± 2500 cm 2 − 2000 cm 2
=
= 25.0 cm ± 11.2 cm
2
2
Thus, there are 2 locations, generally known as “conjugate positions,” where the lens could
be positioned to form a clear image on the screen. These positions are:
(i)
36.2 cm from the screen
→
( q = 36.2 cm and p = 13.8 cm )
(ii)
13.8 cm from the screen
→
( q = 13.8 cm and p = 36.2 cm )
If q = 36.2 cm, the magnification is M = −
q
36.2 cm
=−
= − 2.62 .
p
13.8 cm
If q = 13.8 cm, the magnification is M = −
q
13.8 cm
=−
= − 0.381 .
p
36.2 cm
The focal length of a converging lens is positive, so f = +10.0 cm. The thin-lens equation then
p (10.0 cm )
pf
.
yields an image distance of q =
=
p − f p −10.0 cm
(a)
When p = +20.0 cm, q =
( 20.0 cm )(10.0 cm )
20.0 cm − 10.0 cm
= +20.0 cm, and M = −
q
= −1.00, so the
p
image is located 20.0 cm beyond the lens , is real (q > 0) , is inverted (M < 0) , and is
the same size as the object ( M = 1.00 ) .
(b)
When p = f = +10.0 cm, the object is at the focal point and no image is formed . Instead,
parallel rays emerge from the lens .
(c)
When p = 5.00 cm, q =
( 5.00 cm )(10.0 cm )
q
= +2.00, so the
p
5.00 cm − 10.0 cm
image is located 10.0 cm in front of the lens , is virtual (q < 0) , is upright (M > 0) , and is
= −10.0 cm, and M = −
twice the size of the object ( M = 2.00 ) .
68719_23_ch23_p223-255.indd 238
1/7/11 2:51:01 PM
Mirrors and Lenses
23.32
(a)
To approximate paraxial rays, the rays should be drawn so they reflect at the vertical plane
that passes through the vertex of the mirror, rather than at the mirror’s surface as is generally done in the textbook. For this reason, the concave surface of the mirror appears flat in
the ray diagram given below.
C
F
I
h
O
p
h′
q
(b)
In the diagram above, each square of the grid represents 5 cm. Thus, we see that the image I
is located 40 cm behind the mirror, or q = − 40 cm .
(c)
From the above ray diagram, the height h ′ of the upright image I is seen to be twice the
height h of the object O. Therefore, the magnification is M = h ′ h = + 2 .
(d)
From the mirror equation, the image distance is q = pf ( p − f ), or
q=
( +20.0 cm )( + 40.0 cm )
20.0 cm − 40.0 cm
M=−
and the magnification is
23.33
239
= − 40.0 cm
q
( − 40.0 cm ) = +2.00
=−
p
+20.0 cm
For a divergent lens, the focal length is negative. Hence, f = −20.0 cm in this case. The
thin-lens equation gives the image distance as q = pf ( p − f ) , and the magnification is given by
M = − q p.
(i)
(ii)
( 40.0 cm )( −20.0 cm )
= −13.3 cm
40.0 cm − ( −20.0 cm )
(a)
q=
(b)
q<0
(c)
M = −q p > 0
(d)
M=−
(a)
q=
(b)
q<0
(c)
M = −q p > 0
(d)
M=−
⇒
13.3 cm in front of the lens.
⇒
10.0 cm in front of the lens.
virtual image
⇒
( −13.3 cm )
+ 40.0 cm
upright image
= + 0.333
( 20.0 cm )( −20.0 cm )
= −10.0 cm
20.0 cm − ( −20.0 cm )
⇒
⇒
virtual image
⇒
( −10.0 cm )
+ 20.0 cm
upright image
= + 0.500
continued on next page
68719_23_ch23_p223-255.indd 239
1/7/11 2:51:04 PM
240
Chapter 23
(iii) (a)
23.34
q=
(10.0 cm ) (−20.0 cm ) = −6.67 cm
10.0 cm − (−20.0 cm )
(b)
q<0
⇒
(c)
M = −q p > 0
(d)
M=−
⇒
6.67 cm in front of the lens
virtual image
⇒
( − 6.67 cm ) =
+10.0 cm
upright image
+ 0.667
(a) and (b) Your scale drawings should look similar to those given below:
Figure (a)
Figure (b)
A carefully drawn-to-scale version of Figure (a) should yield an upright, virtual image
located 13.3 cm in front of the lens and one-third the size of the object. Similarly, a carefully drawn-to-scale version of Figure (b) should yield an upright, virtual image located
6.7 cm in front of the lens and two-thirds the size of the object.
23.35
(c)
The results of the graphical solution are consistent with the algebraic answers found in
Problem 23.33, allowing for small deviances due to uncertainties in measurement. Graphical
answers may vary, depending on the size of the graph paper and accuracy of the drawing.
(a)
The real image case is shown in the ray
diagram. Notice that p + q = 12.9 cm, or
q = 12.9 cm − p. The thin-lens equation, with
f = 2.44 cm, then gives
1
1
1
+
=
p 12.9 cm − p 2.44 cm
or
p2 − (12.9 cm ) p + 31.5 cm 2 = 0
continued on next page
68719_23_ch23_p223-255.indd 240
1/7/11 2:51:06 PM
Mirrors and Lenses
241
Using the quadratic formula to solve gives
p = 9.63 cm or p = 3.27 cm
Both are valid solutions for the real image case.
(b)
The virtual image case is shown in the second
diagram. Note that in this case, q = − (12.9 cm + p ) ,
so the thin-lens equation gives
1
1
1
−
=
p 12.9 cm + p 2.44 cm
or
p2 + (12.9 cm ) p − 31.5 cm 2 = 0
The quadratic formula then gives p = 2.10 cm or p = −15.0 cm .
Since the object is real, the negative solution must be rejected, leaving p = 2.10 cm .
23.36
We must first realize that we are looking at an upright, enlarged, virtual image. Thus, we
have a real object located between a converging lens and its front-side focal point, so
q < 0, p > 0, and f > 0.
q
The magnification is M = − = + 2, giving q = − 2 p. Then, from the thin-lens equation,
p
1
1
1
1
−
=+
= , or f = 2 p = 2 ( 2.84 cm ) = 5.68 cm .
p 2p
2p f
23.37
It is desired to form a magnified, real image on the screen using a single thin lens. To do this, a
converging lens must be used, and the image will be inverted. The magnification then gives
M=
−1.80 m
q
h′
=
= − , or q = 75.0 p
p
h 24.0 × 10 −3 m
Also, we know that p + q = 3.00 m. Therefore, p + 75.0 p = 3.00 m , giving
3.00 m
= 3.95 × 10 − 2 m = 39.5 mm
76.0
(b)
p=
(a)
The thin-lens equation then gives
1
1
76.0
1
+
=
= , or
p 75.0 p 75.0 p f
⎛ 75.0 ⎞
⎛ 75.0 ⎞
f =⎜
p=⎜
( 39.5 mm ) = 39.0 mm
⎟
⎝ 76.0 ⎠
⎝ 76.0 ⎟⎠
23.38
(a)
This is a real object, so the object distance is p = +20.0 cm. The thin-lens equation
gives the image distance as
q=
pf
( 20.0 cm )( −32.0 cm )
=
= −12.3 cm
p − f 20.0 cm − ( −32.0 cm )
F
O
I
so the image is 12.3 cm to the left of the lens .
68719_23_ch23_p223-255.indd 241
q
( −12.3 cm )
=−
= + 0.615 .
p
+20.0 cm
(b)
The magnification is M = −
(c)
The ray diagram for this arrangement is shown above.
1/7/11 2:51:08 PM
242
23.39
Chapter 23
Since the light rays incident to the first lens are parallel, p1 = ∞, and the thin-lens equation gives
q1 = f1 = −10.0 cm.
The virtual image formed by the first lens serves as the object for the second lens, so
p2 = 30.0 cm + q1 = 40.0 cm. If the light rays leaving the second lens are parallel, then q2 = ∞,
and the thin-lens equation gives f2 = p2 = 40.0 cm .
23.40
(a)
Solving the thin-lens equation for the image distance q gives
1 1 1 p− f
= − =
pf
q f p
(b)
or
q=
pf
p− f
For a real object, p > 0 and p = p . Also, for a diverging lens, f < 0 and f = − f . The result
of part (a) then becomes
q=
p (− f
)
p − (− f
)
=−
p f
p+ f
Thus, we see that q < 0 for all numeric values of p and f . Since negative image
distances mean virtual images, we conclude that a diverging lens will always form virtual
images of real objects.
(c)
For a real object, p > 0 and p = p . Also, for a converging lens, f > 0 and f = f . The result
of part (a) then becomes
q=
p f
>0
p− f
if
p − f >0
Since q must be positive for a real image, we see that a converging lens will form real
images of real objects only when p > f (or p > f since both p and f are positive in this
situation).
23.41
The thin-lens equation gives the image position for the first lens as
q1 =
p1 f1
( 30.0 cm )(15.0 cm )
=
= + 30.0 cm
p1 − f1
30.0 cm − 15.0 cm
and the magnification by this lens is M1 = −
q1
30.0 cm
=−
= −1.00.
30.0 cm
p1
The real image formed by the first lens serves as the object for the second lens, so
p2 = 40.0 cm − q1 = +10.0 cm. Then, the thin-lens equation gives
q2 =
p2 f2
(10.0 cm )(15.0 cm )
=
= − 30.0 cm
p2 − f2
10.0 cm − 15.0 cm
and the magnification by the second lens is
M2 = −
q2
( − 30.0 cm ) = + 3.00
=−
p2
10.0 cm
Thus, the final, virtual image is located 30.0 cm in front of the second lens ,
and the overall magnification is M = M1 M 2 = ( −1.00 ) ( + 3.00 ) = − 3.00 .
68719_23_ch23_p223-255.indd 242
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Mirrors and Lenses
23.42
(a)
243
With p1 = +15.0 cm, the thin-lens equation gives the position of the image formed by the
first lens as
q1 =
p1 f1
(15.0 cm )(10.0 cm )
=
= + 30.0 cm
p1 − f1
15.0 cm − 10.0 cm
This image serves as the object for the second lens, with an object distance of
p2 = 10.0 cm − q1 = 10.0 cm − 30.0 cm = − 20.0 cm (a virtual object). If the image
formed by this lens is at the position of O1 , the image distance is
q2 = − (10.0 cm + p 1 ) = − (10.0 cm + 15.0 cm ) = − 25.0 cm
The thin-lens equation then gives the focal length of the second lens as
f2 =
(b)
p2 q2
( −20.0 cm )( −25.0 cm )
=
= −11.1 cm
p2 + q2
−20.0 cm − 25.0 cm
The overall magnification is
⎛ q ⎞ ⎛ q ⎞ ⎛ 30.0 cm ⎞ ⎡ ( −25.0 cm ) ⎤
M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ −
⎟ ⎢−
⎥ = + 2.50
⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ 15.0 cm ⎠ ⎣ ( −20.0 cm ) ⎦
(c)
23.43
Since q2 < 0, the final image is virtual ; and since M > 0, it is upright .
From the thin-lens equation, q1 =
p1 f1
( 4.00 cm )(8.00 cm )
=
= − 8.00 cm .
p1 − f1
4.00 cm − 8.00 cm
The magnification by the first lens is M1 = −
q1
( − 8.00 cm ) = + 2.00.
=−
p1
4.00 cm
The virtual image formed by the first lens is the object for the second lens, so
p2 = 6.00 cm + q1 = +14.0 cm, and the thin-lens equation gives
q2 =
(14.0 cm )( −16.0 cm )
p2 f2
=
= − 7.47 cm
p2 − f2 14.0 cm − ( −16.0 cm )
q2
( − 7.47 cm ) = + 0.533, so the overall
=−
p2
14.0 cm
magnification is M = M1 M 2 = ( + 2.00 ) ( + 0.533) = +1.07.
The magnification by the second lens is M 2 = −
The position of the final image is 7.47 cm in front of the second lens , and its height is
h ′ = M h = ( +1.07 )(1.00 cm ) = 1.07 cm .
Since M > 0, the final image is upright ; and since q2 < 0, this image is virtual .
23.44
(a)
We start with the final image and work backward. From Figure P23.44, observe that
q2 = − ( 50.0 cm − 31.0 cm ) = −19.0 cm. The thin-lens equation then gives
p2 =
q2 f 2
( −19.0 cm )( 20.0 cm ) = + 9.74 cm
=
q2 − f2
−19.0 cm − 20.0 cm
continued on next page
68719_23_ch23_p223-255.indd 243
1/7/11 2:51:13 PM
244
Chapter 23
The image formed by the first lens serves as the object for the second lens and is located
9.74 cm in front of the second lens.
Thus, the image distance for the first lens is q1 = 50.0 cm − 9.74 cm = 40.3 cm, and the
thin-lens equation gives
p1 =
q1 f1
( 40.3 cm )(10.0 cm )
=
= +13.3 cm
q1 − f1
40.3 cm − 10.0 cm
The original object should be located 13.3 cm in front of the first lens.
(b)
The overall magnification is
⎛ q ⎞ ⎛ q ⎞ ⎛ 40.3 cm ⎞ ⎛ ( −19.0 cm ) ⎞
M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ −
= − 5.91
⎟ −
9.74 cm ⎟⎠
⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ 13.3 cm ⎠ ⎜⎝
(c)
23.45
Since M < 0, the final image is inverted .
Note: Final answers to this problem are highly sensitive to round-off error. To avoid this, we
retain extra digits in intermediate answers and round only the final answers to the correct number
of significant figures.
Since the final image is to be real and in the film plane, q2 = + d .
Then, the thin-lens equation gives p2 =
q2 f 2
d (13.0 cm )
=
.
q2 − f2 d − 13.0 cm
The object of the second lens ( L2 ) is the image formed by the first lens ( L1 ) , so
13.0 cm ⎞
d2
⎛
q1 = (12.0 cm − d ) − p2 = 12.0 cm − d ⎜ 1+
=
12.0
cm
−
⎟
⎝
d − 13.0 cm ⎠
d − 13.0 cm
If d = 5.00 cm, then q1 = +15.125 cm; and when d = 10.0 cm, q1 = + 45.3 cm .
From the thin-lens equation, p1 =
q1 f1
q (15.0 cm )
= 1
.
q1 − f1 q1 − 15.0 cm
When q1 = +15.125 cm ( d = 5.00 cm ) , then p1 = 1.82 × 10 3 cm = 18.2 m .
When q1 = + 45.3 cm ( d = 10.0 cm ) , then p1 = 22.4 cm = 0.224 m .
Thus, the range of focal distances for this camera is 0.224 m to 18.2 m .
23.46
(a)
From the thin-lens equation, the image distance for the first lens is
q1 =
(b)
p1 f1
(15.0 cm )(10.0 cm )
=
= +30.0 cm
p1 − f1
15.0 cm − 10.0 cm
With q1 = +30.0 cm, the image of the first lens is located 30.0 cm in back of that lens. Since
the second lens is only 10.0 cm beyond the first lens, this means that the first lens is trying
to form its image at a location 20.0 cm beyond the second lens .
continued on next page
68719_23_ch23_p223-255.indd 244
1/7/11 2:51:16 PM
Mirrors and Lenses
(c)
The image the first lens forms (or would form if allowed to do so) serves as the object for
the second lens. Considering the answer to part (b) above, we see that this will be a virtual
object, with object distance p2 = −20.0 cm .
(d)
From the thin-lens equation, the image distance for the second lens is
q2 =
23.47
245
p2 f2
( −20.0 cm )( 5.00 cm )
=
= + 4.00 cm
p2 − f2
−20.0 cm − 5.00 cm
(e)
M1 = −
q1
30.0 cm
=−
= − 2.00
15.0 cm
p1
(f )
M2 = −
q2
4.00 cm
=−
= + 0.200
p2
( −20.0 cm )
(g)
M total = M1 M 2 = ( −2.00 )( +0.200 ) = − 0.400
(h)
Since q2 > 0, the final image is real , and since M total < 0, that image is inverted relative to
the original object.
Since q = + 8.00 cm when p = +10.0 cm, we find that
1
1
18.0
1 1 1
= + =
+
=
f p q 10.0 cm 8.00 cm 80.0 cm
Then, when p = 20.0 cm,
18.0
1
18.0 − 4.00
14.0
1 1 1
= − =
−
=
=
80.0 cm
80.0 cm
q f p 80.0 cm 20.0 cm
or
q=
80.0 cm
= + 5.71 cm
14.0
Thus, a real image is formed 5.71 cm in front of the mirror .
23.48
(a)
We are given that p = 5 f , with both p and f being positive. The thin-lens equation
then gives
q=
68719_23_ch23_p223-255.indd 245
pf
(5 f ) f = 5 f
=
p− f 5f − f
4
q
(5 f 4 ) = − 1
=−
p
5f
4
(b)
M=−
(c)
Since q > 0, the image is real . Because M < 0, the image is inverted . Since the object is
real, it is located in front of the lens, and with q > 0, the image is located in back of the lens.
Thus, the image is on the opposite side of the lens from the object.
1/7/11 2:51:19 PM
246
23.49
Chapter 23
incoming
In the sketch at the right, the center of curvature of
the left side of the biconcave lens is on the side of the light
incoming light. Thus, by the convention of Table 23.2,
the radius of curvature of this side is negative while
the radius of curvature of the right side is positive.
If R1 = −32.5 cm and R2 = 42.5 cm, the lens maker’s
equation gives the focal length of the lens as
| R1|
R2
⎞
⎛
1
1
1
= (1− n )( 5.43 × 10 −2 cm −1 )
= ( n − 1) ⎜
−
f
⎝ − 32.5 cm 42.5 cm ⎟⎠
(a)
For a very distant object ( p → ∞), the thin-lens equation gives the image distance as q = f .
Thus, if the index of refraction of the lens material is n = 1.53 for violet light,
q= f =
1
= −34.7 cm
(1 − 1.53)( 5.43 × 10 −2 cm −1 )
and the image of violet light is formed 34.7 cm to the left of the lens .
(b)
If n = 1.51 for red light, the image distance for very distance source emitting red light is
q= f =
1
= −36.1 cm
(1− 1.51)( 5.43 × 10 −2 cm −1 )
The image of the very distant red light source is formed 36.1 cm to the left of the lens .
23.50
(a)
Using the sign convention from Table 23.2, the radii of curvature of the surfaces are
R1 = −15.0 cm and R2 = +10.0 cm. The lens maker’s equation then gives
⎛ 1
⎛
1⎞
1
1
1 ⎞
or f = −12.0 cm
= ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜
−
R
f
R
−15.0
cm
10.0
cm ⎟⎠
⎝
⎝ 1
2 ⎠
(b)
If p → ∞, then q = f = −12.0 cm .
The thin-lens equation gives q =
23.51
p ( −12.0 cm )
pf
=
and the following results:
p− f
p + 12.0 cm
(c)
If p = 3 f = + 36.0 cm, q = − 9.00 cm .
(d)
If p = f = +12.0 cm, q = − 6.00 cm .
(e)
If p = f 2 = + 6.00 cm, q = − 4.00 cm .
As light passes left to right through the lens, the image position is given by
q1 =
p1 f1
(100 cm )(80.0 cm )
=
= + 400 cm
p1 − f1
100 cm − 80.0 cm
This image serves as an object for the mirror with an object distance of p2 = 100 cm − q1 = −300 cm
(virtual object). From the mirror equation, the position of the image formed by the mirror is
q2 =
p2 f2
( − 300 cm ) ( − 50.0 cm ) = − 60.0 cm
=
p2 − f2 −300 cm − ( − 50.0 cm )
continued on next page
68719_23_ch23_p223-255.indd 246
1/7/11 2:51:21 PM
Mirrors and Lenses
247
This image is the object for the lens as light now passes through it going right to left. The object
distance for the lens is p3 = 100 cm − q2 = 100 cm − ( − 60.0 cm ) , or p3 = 160 cm. From the
thin- lens equation,
q3 =
p3 f3
(160 cm )(80.0 cm )
=
= +160 cm
p3 − f3
160 cm − 80.0 cm
Thus, the final image is located 160 cm to the left of the lens .
⎛ q ⎞⎛ q ⎞⎛ q ⎞
The overall magnification is M = M1 M 2 M3 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎜ − 3 ⎟ , or
⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p3 ⎠
⎛ 400 cm ⎞ ⎡ ( − 60.0 cm ) ⎤ ⎛ 160 cm ⎞
M = ⎜−
−
= − 0.800
−
⎝ 100 cm ⎟⎠ ⎢⎣ ( −300 cm ) ⎥⎦ ⎜⎝ 160 cm ⎟⎠
Since M < 0, the final image is inverted .
23.52
(a)
Since the object is midway between the lens and mirror, the object distance for the mirror is
p1 = +12.5 cm. The mirror equation gives the image position as
1 2 1
2
1
5−4
1
= − =
−
=
=
q1 R p1 +20.0 cm 12.5 cm 50.0 cm 50.0 cm
or
q1 = + 50.0 cm
This image serves as the object for the lens, so p2 = 25.0 cm − q1 = −25.0 cm. Note that
since p2 < 0, this is a virtual object. The thin-lens equation gives the image position for the
lens as
q2 =
p2 f2
( − 25.0 cm ) ( −16.7 cm ) = − 50.3 cm
=
p2 − f2 −25.0 cm − ( −16.7 cm )
Since q2 < 0, this image is located 50.3 cm in front of (to the right of) the lens or
25.3 cm behind (to the right of) the mirror .
(b)
With q2 < 0, the final image is a virtual image.
(c) and (d) The overall magnification is
⎛ q ⎞ ⎛ q ⎞ ⎛ 50.0 cm ⎞ ⎡ ( − 50.3 cm ) ⎤
M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ −
⎢−
⎥ = +8.05
⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ 12.5 cm ⎟⎠ ⎣ ( −25.0 cm ) ⎦
Since M > 0, the final image is upright .
23.53
A hemisphere is too thick to be described
as a thin lens. The light is undeviated on
entry into the flat face. We next consider
the light’s exit from the curved surface,
for which R = − 6.00 cm.
The incident rays are parallel, so p = ∞.
Then,
68719_23_ch23_p223-255.indd 247
n1 n2 n2 − n1
1.00 1.00 − 1.56
+
=
, from which q = 10.7 cm .
=
becomes 0 +
q
− 6.00 cm
p q
R
1/7/11 2:51:24 PM
248
23.54
Chapter 23
The diagram at the right shows a light ray traveling
parallel to the principle axis of a plano-convex lens
at distance h off the axis. This ray strikes the plane
surface at normal incidence and passes into the glass
undeviated. The angle of incidence at the spherical
surface (of radius R) is q1 , where sinq1 = h R. If the
index of refraction of the lens material is n, Snell’s law
gives the angle of refraction at the spherical surface as
q2
q1
h
R
q1
f
h
f
C
h/tanf
¥ R2 − h2
⎛ n sinq1 ⎞
⎛ nh ⎞
= sin −1 ⎜ ⎟
q 2 = sin −1 ⎜
⎟
⎝ R⎠
⎝ nair ⎠
x
The distance x from the center of curvature of the spherical surface to the point where the
refracted ray crosses the principle axis of the lens is
x = R2 − h2 +
h
tanf
f = q 2 −q1
where
If R = 20.0 cm, n = 1.60, and the first ray is distance h1 = 0.500 cm off the axis, we find
⎛ 0.500 cm ⎞
q1 = sin −1 ⎜
= 1.43°
⎝ 20.0 cm ⎟⎠
so
x1 =
( 20.0 cm )2 − ( 0.500 cm )2 +
and
⎡ 1.60 ( 0.500 cm ) ⎤
q 2 = sin −1 ⎢
⎥ = 2.29°
20.0 cm
⎦
⎣
0.500 cm
= 53.3 cm
tan ( 2.29° − 1.43° )
If the second ray is distance h2 = 12.0 cm from the axis, then
⎛ 12.0 cm ⎞
q1 = sin −1 ⎜
= 36.9°
⎝ 20.0 cm ⎟⎠
giving
x2 =
and
( 20.0 cm )2 − (12.0 cm )2 +
⎡ 1.60 (12.0 cm ) ⎤
q 2 = sin −1 ⎢
⎥ = 73.7°
⎣ 20.0 cm ⎦
12.0 cm
= 32.0 cm
tan ( 73.7° − 36.9° )
The distance between the points where these two rays cross the principle axis is then
Δx = x1 − x 2 = 53.3 cm − 32.0 cm = 21.3 cm
23.55
(a)
With light going through the piece of glass from left to right, the radius of the first surface
is positive and that of the second surface is negative according to the sign convention of
Table 23.2. Thus, R1 = + 2.00 cm, and R2 = − 4.00 cm.
1.50 1.50 − 1.00
n1 n2 n2 − n1
1.00
+
+
=
,
=
to the first surface gives
q1
+ 2.00 cm
p q
R
1.00 cm
which yields q1 = − 2.00 cm. The first surface forms a virtual image 2.00 cm to the left of
that surface, and 16.0 cm to the left of the second surface.
Applying
The image formed by the first surface is the object for the second surface, so p2 = +16.0 cm,
1.00 1.00 − 1.50
1.50
n n
n − n1
+
=
, or q2 = + 32.0 cm.
and 1 + 2 = 2
gives
q2
− 4.00 cm
16.0 cm
p q
R
The final image is located 32.0 cm to the right of the second surface .
(b)
68719_23_ch23_p223-255.indd 248
Since q2 > 0, the final image formed by the piece of glass is a real image .
1/7/11 2:51:27 PM
Mirrors and Lenses
23.56
249
Consider an object O1 at distance p1 in
front of the first lens. The thin-lens equation
gives the image position for this lens as
1 1
1
= − .
q1 f1 p1
The image, I1 , formed by the first lens serves as the object, O2 , for the second lens. With the
lenses in contact, this will be a virtual object if I1 is real and will be a real object if I1 is virtual. In
either case, if the thicknesses of the lenses may be ignored,
p2 = −q1
and
1
1 1
1
=− =− +
q1
f1 p1
p2
Applying the thin-lens equation to the second lens,
−
1 1
1
1
+ +
=
f1 p1 q2
f2
or
1
1
1
+
=
becomes
p2 q2
f2
1
1 1
1
+
= +
f1 f2
p1 q2
Observe that this result is a thin-lens-type equation relating the position of the original object, O1 ,
and the position of the final image, I 2 , formed by this two-lens combination. Thus, we see that we
may treat two thin lenses in contact as a single lens having a focal length, f, given by
1 1 1
= +
.
f1 f2
f
23.57
From the thin-lens equation, the image distance for the first lens is
q1 =
p1 f1
( 40.0 cm )( 30.0 cm )
=
= +120 cm
p1 − f1
40.0 cm − 30.0 cm
and the magnification by this lens is M1 = −
q1
120 cm
=−
= − 3.00.
40.0 cm
p1
The real image formed by the first lens serves as the object for the second lens, with object distance of p2 = 110 cm − q1 = −10.0 cm (a virtual object). The thin-lens equation gives the image
distance for the second lens as
q2 =
(a)
p2 f2
( −10.0 cm ) f2
=
p2 − f2 −10.0 cm − f2
If f2 = − 20.0 cm, then q2 = + 20.0 cm, and the magnification by the second lens is
M 2 = − q2 p2 = − ( 20.0 cm ) ( −10.0 cm ) = + 2.00.
The final image is located 20.0 cm to the right of the second lens , and the overall
magnification is M = M1 M 2 = ( −3.00 )( + 2.00 ) = − 6.00 .
(b)
Since M < 0, the final image is inverted .
continued on next page
68719_23_ch23_p223-255.indd 249
1/7/11 2:51:30 PM
250
Chapter 23
(c)
If f2 = + 20.0 cm, then q2 = + 6.67 cm, and M 2 = −
q2
6.67 cm
=−
= + 0.667.
p2
( −10.0 cm )
The final image is 6.67 cm to the right of the second lens , and the overall magnification is
M = M1 M 2 = ( −3.00 )( + 0.667 ) = − 2.00 .
Since M < 0, the final image is inverted .
23.58
The diagram at the right gives a ray diagram showing
how this mirror system forms a final image I2 located
just above the opening in the upper mirror. The focal
point of each mirror is at the center of the opposite
mirror, so their focal length f is the vertical distance
between the centers of the mirrors. The original object
O1 sits just above the surface at the center of the lower
mirror. This places it on the principle axis and just
inside the focal point of the upper mirror.
I2
O1
Therefore, the object distance for the upper mirror is p1 = f − e , where e is a very small number
approaching zero. The mirror equation then gives
q1 =
p1 f
( f − e ) f = − f 2 − e f = −d
=
f −e − f
p1 − f
e
and
d → ∞ when e → 0
Thus, the upper mirror forms a virtual image, on the principle axis, far above this mirror. This
image serves as the object for the lower mirror and is located far in front of that mirror. With
p2 ≈ q1 = d
f , the image distance for the lower mirror is
q2 =
p2 f
df
=
p2 − f d − f
≈+
df
=+f
d
Since q2 > 0, the final image I2 is a real image located just above the focal point of the lower
mirror (at the opening in the upper mirror). The overall magnification for this mirror system is
⎛ q ⎞⎛ q ⎞
M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟
⎝ p1 ⎠ ⎝ p2 ⎠
⎛ −d ⎞ ⎛ f ⎞
⎜ − ⎟ = +1
⎝ f ⎟⎠ ⎝ d ⎠
≈ ⎜−
This means that the final image is upright and the same size as the original object .
When a flashlight beam is aimed at the final image,
it passes through the opening in the upper mirror and
reflects as shown in the ray diagram at the right. It
emerges from the mirror as if it had reflected from
that image, even obeying the law of reflection as it
did so!
68719_23_ch23_p223-255.indd 250
1/7/11 2:51:33 PM
Mirrors and Lenses
23.59
(a)
251
⎛ 1
1⎞
1
= ( n − 1) ⎜ − ⎟ , gives
f
⎝ R1 R2 ⎠
The lens maker’s equation,
⎞
⎛
1
1
1
= ( n − 1) ⎜
−
⎟⎠
5.00 cm
9.00
cm
−11.0
cm
⎝
which simplifies to n = 1+
(b)
1 ⎛
99.0 ⎞
⎜⎝
⎟ = 1.99 .
5.00 11.0 + 9.00 ⎠
As light passes from left to right through the lens, the thin-lens equation gives the image
distance as
q1 =
p1 f
(8.00 cm )( 5.00 cm )
=
= +13.3 cm
p1 − f
8.00 cm − 5.00 cm
This image formed by the lens serves as an object for the mirror with object distance
p2 = 20.0 cm − q1 = + 6.67 cm. The mirror equation then gives
q2 =
p2 R
( 6.67 cm )(8.00 cm )
=
= +10.0 cm
2 p2 − R 2 ( 6.67 cm ) − 8.00 cm
This real image, formed 10.0 cm to the left of the mirror, serves as an object
for the lens as light passes through it from right to left. The object distance is
p3 = 20.0 cm − q2 = +10.0 cm, and the thin-lens equation gives
q3 =
p1 f
(10.0 cm )( 5.00 cm )
=
= +10.0 cm
p3 − f
10.0 cm − 5.00 cm
The final image is located 10.0 cm to the left of the lens and its overall magnification is
⎛ q ⎞ ⎛ q ⎞ ⎛ q ⎞ ⎛ 13.3 ⎞ ⎛ 10.0 ⎞ ⎛ 10.0 ⎞
M = M1 M 2 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎜ − 3 ⎟ = ⎜ −
⎟ ⎜−
⎟ ⎜−
⎟ = − 2.50
⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p3 ⎠ ⎝ 8.00 ⎠ ⎝ 6.67 ⎠ ⎝ 10.0 ⎠
(c)
23.60
23.61
Since M < 0, the final image is inverted .
From the thin-lens equation, the object distance is p =
(4 f ) f
(a)
If q = + 4 f , then p =
(b)
When q = − 3 f , we find p =
(c)
In case (a), M = −
4f − f
=
qf
.
q− f
4f
or 1.33 f .
3
( − 3 f ) f = 3 f 4 , or 0.750 f .
−3f − f
q
4f
q
−3f
=−
= − 3 , and in case (b), M = − = −
= +4 .
p
4f 3
p
3f 4
If R1 = −3.00 m and R2 = −6.00 m, the focal length is given by
⎞⎛
1 ⎛ n1
1
1 ⎞ ⎛ n1 − n2 ⎞ ⎛ −1 ⎞
=
− 1⎟ ⎜
=
+
⎜
⎟
f ⎜⎝ n2
−
3.00
m
6.00
m ⎟⎠ ⎜⎝ n2 ⎟⎠ ⎝ 6.00 m ⎠
⎝
⎠
continued on next page
68719_23_ch23_p223-255.indd 251
1/7/11 2:51:35 PM
252
Chapter 23
( 6.00 m ) n2
or
f=
(a)
If n1 = 1.50 and n2 = 1.00, then f =
[1]
n2 − n1
The thin-lens equation gives q =
( 6.00 m )(1.00 )
= −12.0 m .
1.00 − 1.50
(10.0 m )( −12.0 m )
pf
=
= − 5.45 m .
p− f
10.0 m + 12.0 m
A virtual image is formed 5.45 m to the left of the lens .
(b)
If n1 = 1.50 and n2 = 1.33, the focal length is f =
q=
( 6.00 m )(1.33)
1.33 − 1.50
= − 46.9 m, and
(10.0 m )( − 46.9 m )
pf
=
= − 8.24 m
p− f
10.0 m + 46.9 m
The image is located 8.24 m to the left of the lens .
(c)
When n1 = 1.50 and n2 = 2.00, f =
q=
( 6.00 m )( 2.00 )
2.00 − 1.50
= + 24.0 m, and
pf
(10.0 m )( 24.0 m )
=
= −17.1 m
p− f
10.0 m − 24.0 m
The image is 17.1 m to the left of the lens .
(d)
23.62
Observe from Equation [1] that f < 0 if n1 > n2 and f > 0 when n1 < n2. Thus, a diverging
lens can be changed to converging by surrounding it with a medium whose index of refraction exceeds that of the lens material.
The inverted image is formed by light that leaves the object and goes directly through the lens,
never having reflected from the mirror. For the formation of this inverted image, we have
M=−
q1
= −1.50
p1
giving
q1 = +1.50 p1
The thin-lens equation then gives
1
1
1
+
=
p1 1.50 p1 10.0 cm
or
1 ⎞
⎛
p1 = (10.0 cm ) ⎜ 1+
= 16.7 cm
⎝ 1.50 ⎟⎠
The upright image is formed by light that passes through the lens after reflecting from the mirror.
The object for the lens in this upright image formation is the image formed by the mirror. In order
for the lens to form the upright image at the same location as the inverted image, the image formed
by the mirror must be located at the position of the original object (so the object distances, and
hence image distances, are the same for both the inverted and upright images formed by the lens).
Therefore, the object distance and the image distance for the mirror are equal, and their common
value is
qm irror = pm irror = 40.0 cm − p1 = 40.0 cm − 16.7 cm = +23.3 cm
continued on next page
68719_23_ch23_p223-255.indd 252
1/7/11 2:51:37 PM
Mirrors and Lenses
The mirror equation,
1
fm irror
23.63
(a)
=
253
1
1
2
1
+
= =
, then gives
pm irror qm irror R fm irror
1
1
+2
+
=
23.3 cm 23.3 cm 23.3 cm
fm irror = +
or
23.3 cm
= + 11.7 cm
2
The lens-maker’s equation for a lens made of material with refractive index n1 = 1.55 and
immersed in a medium having refractive index n2 is
⎞⎛ 1
1 ⎞ ⎛ 1.55 − n2 ⎞ ⎛ 1
1⎞
1 ⎛ n1
= ⎜ − 1⎟ ⎜ − ⎟ = ⎜
− ⎟
⎜
⎟
f ⎝ n2
n2
⎠ ⎝ R1 R2 ⎠
⎠ ⎝ R1 R2 ⎠ ⎝
Thus, when the lens is in air, we have
1⎞
1 ⎛ 1.55 − 1.00 ⎞ ⎛ 1
=⎜
⎟⎠ ⎜ − ⎟
⎝
1.00
fair
⎝ R1 R2 ⎠
[1]
1⎞
⎛ 1.55 − 1.33 ⎞ ⎛ 1
=⎜
− ⎟
⎟
⎜
⎝ 1.33 ⎠ ⎝ R1 R2 ⎠
[2]
1
and when it is immersed in water,
fwater
Dividing Equation [1] by Equation [2] gives
fwater ⎛ 1.33 ⎞ ⎛ 1.55 − 1.00 ⎞
=⎜
= 3.33
⎝ 1.00 ⎟⎠ ⎜⎝ 1.55 − 1.33 ⎟⎠
fair
If fair = 79.0 cm, the focal length when immersed in water is
fwater = 3.33 ( 79.0 cm ) = 263 cm
(b)
The focal length for a mirror is determined by the law of reflection, which is independent
of the material of which the mirror is made and of the surrounding medium. Thus, the
focal length depends only on the radius of curvature and not on the material making up the
mirror or the surrounding medium. This means that, for the mirror,
fwater = fair = 79.0 cm
23.64
(a)
The spherical ornament serves as a convex mirror, forming an image that is three-fourths
the size of the object. Since convex mirrors only form upright, virtual images for real
objects, the magnification is positive. Thus,
M=
q
3
h′
=− =+
p
4
h
and
q=
− 3p
4
With R < 0 (convex mirror) and a diameter of 8.50 cm, we have
R = − (8.50 cm) 2 = − 4.25 cm
The mirror equation, 1 p + 1 q = 2 R , then gives
2
1 4
−
=
p 3p R
or
p=−
R
( − 4.25 cm ) = + 0.708 cm
=−
6
6
Hence, the object is 0.708 cm in front of the spherical ornament .
continued on next page
68719_23_ch23_p223-255.indd 253
1/7/11 2:51:40 PM
254
Chapter 23
(b)
The ray diagram at the right shows the
formation of the upright, virtual image
by this convex mirror.
O
I
F
23.65
The diagram at the right shows a bubble O, located 5.00 cm
above the center of a glass sphere having radius R = 15.0 cm.
This bubble is an actual distance of p = 10.0 cm below the
refracting surface separating the glass and air. Refraction at
this surface forms a virtual image I at distance q below the
surface as shown. The object distance p and image distance q
are related by
n1 n2 n2 − n1
+
=
p q
R
or
C
n2 = 1.00
p
| q|
I
O
5.00 cm
R
n1 = 1.50
n2 n2 − n1 n1
=
−
q
R
p
Thus, with R < 0 for the concave surface the light crosses going from glass into the air, we have
R = −15.0 cm, and
1.50
+1.00 − 4.50
−3.50
1.00 1.00 − 1.50
=
−
=
=
−15.0 cm 10.0 cm
30.0 cm
30.0 cm
q
or
q=
30.0 cm
= − 8.57 cm
− 3.50
Therefore, the bubble has an apparent depth of 8.57 cm below the glass surface.
23.66
(a)
The only upright images that spherical mirrors can form of real objects are virtual images.
Therefore, we know that the image described is virtual and diminished in size (4.00 cm
high in comparison to the 10.0-cm height of the object). Since the virtual images formed by
concave mirrors are always enlarged, we must conclude that this image is formed by a
convex mirror .
(b)
The sketch at the right shows an upright, diminished,
virtual image formed by a convex mirror. The distance
between the object and this virtual image in this case is
known to be p + q = 42.0 cm. The image distance for a
virtual image is negative, so q < 0 and q = −q. Thus, we
have
Convex
surface
I
O
p − q = 42.0 cm
p
| q|
42.0 cm
[1]
We also know that the magnification is positive (upright image) and
M=
q 4.00 cm
h′
=− =
= 0.400
p 10.0 cm
h
or
q = − 0.400 p
[2]
continued on next page
68719_23_ch23_p223-255.indd 254
1/7/11 2:51:42 PM
Mirrors and Lenses
255
Substituting Equation [2] into Equation [1] gives
p − ( −0.400 ) p = 42.0 cm
or
p=
42.0 cm
= 30.0 cm
1.40
With the object at the zero end of the meter stick, the mirror is at the 30.0-cm mark .
(c)
The image distance is now seen to be q = −0.400 p = −0.400 ( 30.0 cm ) = −12.0 cm. The
mirror equation (1 p + 1 q = 2 R = 1 f ) then gives the focal length of this mirror as
f=
68719_23_ch23_p223-255.indd 255
pq
( 30.0 cm )( −12.0 cm )
=
= −20.0 cm
p+q
30.0 cm − 12.0 cm
1/7/11 2:51:44 PM
24
Wave Optics
QUICK QUIZZES
1.
Choice (c). The fringes on the screen are equally spaced only at small angles, where tanq ≈ sinq
is a valid approximation.
2.
Choice (c). The screen locations of the dark fringes of order m are given by
(ydark )m = (lL d)(m + 12 ), with m = 0 corresponding to the first dark fringe on either side of the
central maximum. The width of the central maximum is then 2(ydark )0 = 2(lL d)( 12 ) = lL d .
Thus, doubling the distance d between the slits will cut the width of the central maximum in half.
3.
Choice (c). The screen locations of the bright fringes of order m are given by (ybright )m = (lL d)m ,
and the distance between successive bright fringes for a given wavelength is
(
Δybright = ybright
)
m +1
(
− ybright
)
m
= ( lL d )m +1 − ( lL d )m = lL d
Observe that this spacing is directly proportional to the wavelength. Thus, arranged from smallest
to largest spacing between bright fringes, the order of the colors will be blue, green, red.
4.
Choice (b). The space between successive bright fringes is proportional to the wavelength of the
light. Since the wavelength in water is less than that in air, the bright fringes are closer together in
the second experiment.
5.
Choice (b). The outer edges of the central maximum occur where sinq = ± l a. Thus, as the
width of the slit, a, becomes smaller, the width of the central maximum will increase.
6.
The compact disc. The tracks of information on a compact disc are much closer together than on
a phonograph record. As a result, the diffraction maxima from the compact disc will be farther
apart than those from the record.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
With phase changes occurring in the reflections at both the air-oil boundary and the oil-water
boundary, the condition for constructive interference in the reflected light is 2noil t = ml, where
m is any integer. Thus, the minimum nonzero thickness of the oil which will strongly reflect the
530-nm light is tm in = l 2noil = (530 nm) 2(1.25) = 212 nm, and (d) is the proper choice.
2.
The bright fringe of order m occurs where d = d sinq = ml. For small angles, the sine of the angle
is approximately equal to the angle expressed in radians. Thus, the angular position of the second
order bright fringe in the case described is
⎛ 5.0 × 10 −7 m ⎞
⎛l⎞
= 0.050 radians
q = 2⎜ ⎟ = 2⎜
⎝d⎠
⎝ 2.0 × 10 −5 m ⎟⎠
making (a) the correct choice.
256
68719_24_ch24_p256-281.indd 256
1/7/11 2:52:56 PM
Wave Optics
3.
257
In a single-slit diffraction pattern, formed on a screen at distance L from the slit, dark fringes
occur where sinq dark = m(l a) ≈ tanq dark = ydark L and m is any nonzero integer. Thus, the width
of the slit, a, in the described situation must be
a=
(1) lL
(y )
dark
1
=
(5.00 × 10
−7
m ) (1.00 m )
5.00 × 10 −3 m
= 1.00 × 10 −4 m = 0.100 mm
and the correct answer is seen to be choice (b).
4.
From Malus’s law, the intensity of the light transmitted through a polarizer having its
transmission axis oriented at angle q to the plane of polarization of the incident polarized
light is I = I 0 cos 2 q . Therefore, the intensity transmitted through the first polarizer having
q = 45° − 0 = 45° is I1 = I 0 cos 2 (45°) = 0.50I 0 , and the intensity passing through the second
polarizer having q = 90° − 45° = 45° is I 2 = (0.50I 0 ) cos 2 (45°) = 0.25I 0 . The fraction of the
original intensity making it through both polarizers is then I 2 I 0 = 0.25, which is choice (b).
5.
The spacing between successive bright fringes in a double-slit interference pattern is given by
Δybright = ( lL d ) ( m + 1) − ( lL d ) m = lL d , where d is the slit separation. As d decreases, the
spacing between the bright fringes will increase and choice (b) is the correct answer.
6.
As discussed in question 5 above, the fringe spacing in a double-slit interference pattern is
Δybright = lL d . Therefore, as the distance L between the screen and the plane of the slits is
increased, the spacing between the bright fringes will increase, and (a) is the correct choice.
7.
The fringe spacing in a double-slit interference pattern is Δybright = lL d . Note that this spacing is
directly proportional to the wavelength of the light illuminating the slits. Thus, with l2 > l1 , the
spacing is greater in the second experiment, and choice (c) gives the correct answer.
8.
The reflected light tends to be partially polarized, with the degree of polarization depending
on the angle of incidence on the reflecting surface. Only if the angle of incidence equals the
polarizing angle (or Brewster’s angle) will the reflected light be completely polarized. The better
answer for this question is choice (d).
9.
The bright colored patterns are the result of interference between light reflected from the upper
surface of the oil and light reflected from the lower surface of the oil film. Thus, the best answer
is choice (e).
10.
In a single-slit diffraction pattern, dark fringes occur where ydark L ≈ sinq dark = m(l a). The width
of the central maximum is the distance between the locations of the first dark fringes on either
side of the center (i.e., between the m = ±1 dark fringes), giving
width of central maximum = ( +1)
l
l 2l
− ( −1) =
a
a
a
Thus, if the width of the slit, a, is made half as large, the width of the central maximum will
double and (d) is the correct choice.
11.
68719_24_ch24_p256-281.indd 257
The colors observed on the oil film are the result of constructive interference of waves reflected
from the upper and lower surfaces of the film. For this to produce bright colored fringes, the
thickness of the film must be on the same order of magnitude as the wavelength of the light. If
the thickness of the oil film were smaller than half of the wavelengths of visible light, no colors
would appear. If the thickness of the oil film were much larger, the constructive fringes for
different colors would overlap and mix to white or gray. The best choice for this question is (b).
1/7/11 2:52:58 PM
258
12.
Chapter 24
In principle, the ribs (or any set of parallel openings that waves pass through) can act as a
diffraction grating. However, for any significant diffraction to occur as the waves pass through
the openings, and also for the interference fringes to be separated sufficiently to be observed,
the spacing between the slits must be on the same order of magnitude as the wavelength of the
radiation. The spacing between the ribs is very large in comparison to the wavelength of x-rays,
so the correct choice is (b).
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The wavelength of light is extremely small in comparison to the dimensions of your hand, so
the diffraction of light around obstacles the size of your hand is totally negligible. However,
sound waves have wavelengths that are comparable to the dimensions of the hand or even larger.
Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.
4.
The spacing between double-slit interference fringes on a screen is proportional to the wavelength of the light (Δybright = lL d ). Light having a wavelength l0 in vacuum will have wavelength l = l0 n in a medium with index of refraction n. Thus, the wavelength of the light is
smaller in water than it is in air, meaning the interference fringes will be more closely spaced
when the experiment is performed in water.
6.
Every color produces its own interference pattern, and we see them superimposed. The central
maximum is white. The first maximum is a full spectrum with violet on the inside and red on the
outside. The second maximum is also a full spectrum, with red in it overlapping with violet in
the third maximum. At larger angles, the light soon starts mixing to white again.
8.
The skin on the tip of a finger has a series of closely spaced ridges and swirls on it. When the
finger touches a smooth surface, the oils from the skin will be deposited on the surface in the
pattern of the closely spaced ridges. The clear spaces between the lines of deposited oil can serve
as the slits in a crude diffraction grating and produce a colored spectrum of the light passing
through or reflecting from the glass surface.
10.
Suppose the index of refraction of the coating is intermediate between vacuum and the glass.
When the coating is very thin, light reflected from its top and bottom surfaces will interfere
constructively, so you see the surface white and brighter. Once the thickness reaches one-quarter
of the wavelength of violet light in the coating, destructive interference for violet light will make
the surface look red. Then other colors in spectral order (blue, green, yellow, orange, and red) will
interfere destructively, making the surface look red, magenta, and then blue. As the coating gets
thicker, constructive interference is observed for violet light and then for other colors in spectral
order. Even thicker coatings give constructive and destructive interference for several visible
wavelengths, so the reflected light starts looking white again.
12.
The reflected light is partially polarized, with the component parallel to the reflecting surface
being the most intense. Therefore, the polarizing material should have its transmission axis
oriented in the vertical direction in order to minimize the intensity of the reflected light from
horizontal surfaces.
14.
Due to gravity, the soap film tends to sag in its holder, being quite thin at the top and becoming
thicker as one moves toward the bottom of the holding ring. Because light reflecting from the front
surface of the film experiences a phase change, and light reflecting from the back surface of the
film does not (see Figure 24.7 in the textbook), the film must be a minimum of a half wavelength
thick before it can produce constructive interference in the reflected light. Thus, the light must
be striking the film at some distance from the top of the ring before the thickness is sufficient to
produce constructive interference for any wavelength in the visible portion of the spectrum.
68719_24_ch24_p256-281.indd 258
1/7/11 2:53:00 PM
Wave Optics
259
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a) 1.77 mm
4.
2.40 mm
6.
(a)
8.
2.9 mm
1.52 cm
(b) 1.47 mm
(b)
2.13 cm
d = 3.00l
10.
(a)
1.93 mm
(b)
12.
(a)
2.3°, 4.6°, and 6.9°
(b) 1.1°, 3.4°, and 5.7°
(c)
At small angles, q ≈ sinq . The approximation breaks down at larger angles.
(a)
See Solution.
(c)
0.144°, 2.51× 10 −3 , sinq ≈ tanq only when q is small
(d)
603 nm
(a)
640 nm
(b)
make use of a higher order constructive interference (i.e., a larger value of m)
(c)
360 nm (m = 1), 600 nm (m = 2)
18.
(a)
512 nm
20.
233 nm
22.
(a)
541 nm
(b)
406 nm; The most strongly transmitted wavelengths are those which suffer destructive
interference in reflection.
14.
16.
(b)
(e)
(b)
(c)
a maximum
tanq = 2.51× 10 −3
0.720°
(f )
2.26 cm
2.5m1 = 2m2 + 1
24.
8 dark fringes, including the one at the line of contact
26.
6.5 × 10 2 nm
28.
99.6 nm
30.
(a)
97.8 nm
(b) Yes, some of which are 293 nm, 489 nm, and 685 nm.
32.
(a)
2.3 mm
(b)
34.
91.1 cm
36.
0.227 mm
38.
659 nm
68719_24_ch24_p256-281.indd 259
4.5 mm
1/7/11 2:53:01 PM
260
Chapter 24
40.
(a)
42.
(a) Three maxima (for m = −1, m = 0, and m = +1) will be observed.
(b)
(b)
2 complete orders
10.9°
for m = −1, q = − 45.2°; for m = 0, q = 0°; and for m = +1, q = + 45.2°.
44.
7.35°
46.
See Solution.
48.
(a)
2.381 × 10 −6 m
(b)
29.65°
(d)
26.69°, 1.140 m
(e)
2.000 × 10 −3 m
(c)
1.138 m
(f ) 1.537 × 10 −3 m; The two results agree to only 1 significant figure, showing that the
calculation is sensitive to rounding of intermediate answers.
50.
78.1 nm and 469 nm
52.
(a)
54.
36.9°
56.
60.5°
58.
0.336
(b)
0.164
(a)
45.0°
(b)
60.0°
60.
(a)
I = I0 2
(b)
54.7°
62.
632.8 nm
64.
(a)
(b)
m1 = 5, m2 = 6; 2.52 cm from the central maximum
66.
maxima at 0°, ± 29.1°, and ± 76.4°, minima at ±14.1° and ± 46.8°
68.
113 dark fringes (counting the m = 0 order at the line of contact)
70.
(a)
72.
See Solution.
74.
a = 20.0 × 10 −6 ( °C )
6m1 = 5m2
I f Ii = 0
(b)
(c)
65.9°
I f I i = 0.25
−1
PROBLEM SOLUTIONS
24.1
The location of the bright fringe of order m (measured from the position of the central maximum) is
(ybright )m = (lL d)m, m = 0, ± 1, ± 2, …. Thus, the spacing between successive bright fringes is
(
Δybright = ybright
)
m +1
(
− ybright
)
m
= ( lL d ) ( m + 1) − ( lL d ) m = lL d
The wavelength of the laser light must be
l=
68719_24_ch24_p256-281.indd 260
( Δy ) d = (1.58 × 10
bright
L
−2
m ) ( 0.200 × 10 −3 m )
5.00 m
= 6.32 × 10 −7 m = 632 nm
1/7/11 2:53:03 PM
Wave Optics
24.2
(a)
261
For a bright fringe of order m, the path difference is d = ml, where m = 0, 1, 2,…. At the
location of the third order bright fringe, m = 3 and
d = 3l = 3 ( 589 nm ) = 1.77 × 10 3 nm = 1.77 mm
(b)
1⎞
⎛
For a dark fringe, the path difference is d = ⎜ m + ⎟ l, where m = 0, 1, 2,….
⎝
2⎠
At the third dark fringe, m = 2 and
⎛
d = ⎜2+
⎝
24.3
(a)
1⎞
5
3
⎟⎠ l = ( 589 nm ) = 1.47 × 10 nm = 1.47 mm
2
2
The distance between the central maximum and the first order bright fringe is
Δy = ybright
Δy =
(b)
m=0
=
lL
,
d
or
The distance between the first and second dark bands is
m =1
− ydark
m=0
=
lL
= 2.62 mm as in (a) above.
d
In the double-slit interference pattern, the bright fringe of order m is found where d sinq = ml
with m = 0, ± 1, ± 2,…. Here, q is the angle between the line to the central maximum and the
line to the location of the bright fringe of interest. Thus, if q = 15.0° for the m = 1 bright fringe of
light having wavelength l = 620 nm, the spacing between the slits must be
d=
24.5
− ybright
−9
lL ( 546.1 × 10 m ) (1.20 m )
=
= 2.62 × 10 −3 m = 2.62 mm
d
0.250 × 10 −3 m
Δy = ydark
24.4
m =1
(a)
−9
ml (1)( 620 × 10 m )
=
= 2.40 × 10 − 6 m = 2.40 mm
sinq
sin15.0°
From d sinq = ml, the angle for the m = 1 maximum for the sound waves is
⎡
⎡m ⎛ v
⎞⎤
⎛ 354 m s ⎞ ⎤
1
⎛m ⎞
q = sin −1 ⎜ l ⎟ = sin −1 ⎢ ⎜ sound ⎟ ⎥ = sin −1 ⎢
⎟ ⎥ = 36.2°
⎜
⎝d ⎠
⎣ 0.300 m ⎝ 2 000 Hz ⎠ ⎦
⎣ d ⎝ f ⎠⎦
(b)
For 3.00-cm microwaves, the required slit spacing is
d=
(c)
The wavelength of the light is l = d sinq m , so the frequency is
f=
24.6
ml (1)( 3.00 cm )
=
= 5.08 cm
sinq
sin ( 36.2° )
(1)( 3.00 × 108 m s )
c
mc
=
=
= 5.08 × 1014 Hz
l d sinq (1.00 × 10 −6 m ) sin 36.2°
In a double-slit interference pattern, the screen position of the mth order maximum for
wavelength l is ym = (lL d)m. The separation between the maxima of orders m1 and m2 is
then Δy = (lL d)(m2 − m1 ).
continued on next page
68719_24_ch24_p256-281.indd 261
1/7/11 2:53:05 PM
262
Chapter 24
(a)
If l = 588 nm while m2 = 1 and m1 = 0 , the separation is
−9
lL ( 588 × 10 m )(1.50 m )
lL
=
= 1.52 × 10 −2 m = 1.52 cm
(1− 0 ) =
d
d
0.058 0 × 10 −3 m
Δy =
(b)
If l = 412 nm, m2 = 4, and m1 = 2, the spacing between these maxima is
Δy =
24.7
( 412 × 10
−9
m )(1.50 m )
0.058 0 × 10 −3 m
( 4 − 2 ) = 2.13 × 10 −2 m = 2.13 cm
As indicated in the sketch at the right, the path difference
between waves from the two antennas that travel toward
the car is given by d = d sinq . When d = ml, where
m = 0, 1, 2,… , constructive interference produces the
maximum of order m. Destructive interference produces the
minimum of order m when d = ( m + 12 ) l.
(a)
d
q
q
⎞ ( 300 m ) ⎡
d sinq d ⎛
y
⎢
= ⎜ 2
⎟=
⎢
m
2 ⎝ x + y2 ⎠
2
⎣
x
O
d =dsinq
At the m = 2 maximum, d = d sinq = 2 l, or
l=
(b)
y
400 m
(1000m ) + ( 400 m )
2
2
⎤
⎥ = 55.7 m
⎥
⎦
The next minimum encountered is the m = 2 minimum, and it occurs where d = 5l 2, or
⎡ 5 ( 55.7 m ) ⎤
⎛ 5l ⎞
q = sin −1 ⎜ ⎟ = sin −1 ⎢
⎥ = 27.7°
⎝ 2d ⎠
⎣ 2 ( 300 m ) ⎦
At this point, y = x tanq = (1 000 m ) tan 27.7° = 525 m, so the car must travel an additional
125 m .
24.8
The angular position of the bright fringe of order m is given by d sinq = ml . Thus, if the m = 1
bright fringe is located at q = 12° when l = 6.0 × 10 2 nm = 6.0 × 10 −7 m, the slit spacing is
d=
24.9
The screen position of the mth order bright fringe of wavelength l in a double-slit interference
pattern is ym = (lL d)m. Then, Δy = (lL d)(m2 − m1 )is the separation between the bright fringes of
orders m1 and m2 . If Δy = 0.552 mm for the first and second order bright fringes when L = 1.75 m,
and the slit separation is d = 0.552 mm, the wavelength incident upon the set of slits is
l=
24.10
(1) ( 6.0 × 10 −7 m )
ml
=
= 2.9 × 10 −6 m = 2.9 mm
sinq
sin12°
( Δy ) d
L ( m2 − m1 )
=
( 0.552 × 10
−3
m ) ( 2.10 × 10 −3 m )
(1.75 m )( 2 − 1)
= 6.62 × 10 −7 m = 662 nm
The angular deviation from the line of the central maximum is given by
⎛ 1.80 cm ⎞
⎛ y⎞
q = tan −1 ⎜ ⎟ = tan −1 ⎜
= 0.737°
⎝ 140 cm ⎟⎠
⎝ L⎠
(a)
The path difference is then
d = d sinq = ( 0.150 mm ) sin ( 0.737° ) = 1.93 × 10 −3 mm = 1.93 mm
continued on next page
68719_24_ch24_p256-281.indd 262
1/7/11 2:53:08 PM
Wave Optics
(b)
(c)
24.11
l
⎛
⎞
= 3.00 l
d = (1.93 × 10 −6 m ) ⎜
⎝ 643 × 10 −9 m ⎟⎠
Since the path difference for this position is a whole number of wavelengths (to three significant figures), the waves interfere constructively and produce a maximum at this spot.
The distance between the central maximum (position of A) and the first minimum is
y=
lL ⎛
⎜m +
d ⎝
Thus, d =
24.12
(a)
1⎞
lL
=
⎟⎠
2 m=0 2 d
lL ( 3.00 m ) (150 m )
=
= 11.3 m .
2y
2 ( 20.0 m )
The angular position of the bright fringe of order m is given by d sinq = ml, or
q m = sin −1 (ml d). If d = 25l, the first three bright fringes are found at
⎛ 1⎞
q1 = sin −1 ⎜ ⎟ = 2.3° ,
⎝ 25 ⎠
(b)
(c)
⎛ 2⎞
q 2 = sin −1 ⎜ ⎟ = 4.6° ,
⎝ 25 ⎠
and
⎛ 3⎞
q 3 = sin −1 ⎜ ⎟ = 6.9°
⎝ 25 ⎠
The angular position of the dark fringe of order m is given by d sinq = (m + 12 )l, or
q m = sin −1 [ (m + 12 )l d ] , m = 0, ± 1, ± 2,…. If d = 25l , the first three dark fringes are
found at
⎛ 12 ⎞
q 0 = sin −1 ⎜ ⎟ = 1.1° ,
⎝ 25 ⎠
24.13
263
⎛ 32 ⎞
q1 = sin −1 ⎜ ⎟ = 3.4° ,
⎝ 25 ⎠
and
⎛ 52 ⎞
q 2 = sin −1 ⎜ ⎟ = 5.7°
⎝ 25 ⎠
The answers are evenly spaced because the angles are small and q ≈ sinq . At larger
angles, the approximation breaks down and the spacing isn’t so regular.
As shown in the figure at the right, the path
difference in the waves reaching the telescope is
d = d 2 − d1 = d 2 (1 − sina ) . If the first minimum
(d = l 2) occurs when q = 25.0°, then
a = 180° − (q + 90.0° + q ) = 40.0° , and
d2 =
d
l 2
( 250 m 2) = 350 m
=
=
1− sina 1− sina 1− sin 40.0°
Thus, h = d 2 sin 25.0° = 148 m .
24.14
(a)
(b)
tanq1 =
y1 4.52 × 10 −3 m
=
= 2.51 × 10 −3
L
1.80 m
continued on next page
68719_24_ch24_p256-281.indd 263
1/7/11 2:53:13 PM
264
Chapter 24
(c)
q1 = tan −1 ( 2.51 × 10 −3 ) = 0.144° and sinq1 = sin ( 0.144° ) = 2.51 × 10 −3 .
The sine and the tangent are very nearly the same but only because the angle is small.
(d)
From d = d sinq = ml for the order m bright fringe,
l=
24.15
−4
d sinq1 ( 2.40 × 10 m ) sin ( 0.144° )
=
= 6.03 × 10 −7 m = 603 nm
1
1
(e)
⎡ 5 ( 6.03 × 10 −7 m ) ⎤
⎛ 5l ⎞
q 5 = sin −1 ⎜ ⎟ = sin −1 ⎢
⎥ = 0.720°
−4
⎝ d ⎠
⎣⎢ 2.40 × 10 m ⎥⎦
(f )
y5 = L tanq 5 = (1.80 m ) tan ( 0.720° ) = 2.26 × 10 −2 m = 2.26 cm
The path difference in the two waves received at the home is d = 2 d, where d is the distance
from the home to the mountain. Neglecting any phase change upon reflection, the condition for
destructive interference is
1⎞
⎛
d = ⎜ m + ⎟ l with m = 0, 1, 2,…
⎝
2⎠
so
24.16
(a)
d m in =
d m in ⎛
1 ⎞ l l 300 m
= ⎜0+ ⎟ = =
= 75.0 m
⎝
2⎠ 2 4
4
2
With a phase change in the reflection at the outer surface of the soap film and no change
on reflection from the inner surface, the condition for constructive interference in the light
reflected from the soap bubble is 2nfilm t = (m + 12 )l, where m = 0,1,2,…. For the lowest
order reflection, m = 0, so the wavelength is
l=
2nwater t
= 4 (1.333)(120 nm ) = 6.4 × 10 2 nm = 640 nm
( 0 + 1 2)
(b)
To strongly reflect the same wavelength light, a thicker film will need to make use of a
higher order constructive interference, i.e., use a larger value of m .
(c)
The next greater thickness of soap film that can strongly reflect 640 nm light corresponds to
m = 1, giving
t1 =
(1+ 1 2) l
2nfilm
=
3 ⎡ 640 nm ⎤
2
⎥ = 3.6 × 10 nm = 360 nm
⎢
2 ⎣ 2 (1.333) ⎦
and the third such thickness (corresponding to m = 2) is
t2 =
24.17
2nfilm
=
5 ⎡ 640 nm ⎤
2
⎥ = 6.0 × 10 nm = 600 nm
⎢
2 ⎣ 2 (1.333) ⎦
Light reflecting from the first (glass-iodine) interface suffers a phase change, but light reflecting
at the second (iodine-glass) interface does not have a phase change. Thus, the condition for constructive interference in the reflected light is 2nfilm t = (m + 12 )l, with m = 0,1,2,…. The smallest
film thickness capable of strongly reflecting the incident light is
tm in =
68719_24_ch24_p256-281.indd 264
( 2 + 1 2) l
(m
+ 1 2 ) l ( 0 + 1 2 ) λ 6.00 × 10 2 nm
=
=
= 85.4 nm
2nfilm
4 (1.756 )
2nfilm
m in
1/7/11 2:53:16 PM
Wave Optics
24.18
(a)
265
Phase changes are experienced by light reflecting at either surface of the oil film, a upper
air-oil interface and a lower oil-water interface. Under these conditions, the requirement
for constructive interference is 2nfilm t = m1 l , with m1 = 0,1,2,… , and the requirement for
destructive interference is 2nfilm t = (m2 + 12 )l , with m2 = 0,1,2,…. To have the thinnest
film that produces simultaneous constructive interference of l1 = 640 nm and destructive
interference of l2 = 512 nm, it is necessary that
1⎞
⎛
2nfilm t = m1 ( 640 nm ) = ⎜ m2 + ⎟ ( 512 nm )
⎝
2⎠
where both m1 and m2 are the smallest integers for which this is true. It is found that
m1 = m2 = 2 are the smallest integer values that will satisfy this condition, giving the
minimum acceptable film thickness as
tm in =
(b)
m1 l1 ( m2 + 12 ) l2 2 ( 640 nm ) ( 2.5)( 512 nm )
=
=
=
= 512 nm
2nfilm
2nfilm
2 (1.25)
2 (1.25)
From the discussion above, it is seen that in order to have a film thickness that produces
simultaneous constructive interference of 640-nm light and destructive interference of
512-nm light, it is necessary that
1⎞
⎛
m1 ( 640 nm ) = ⎜ m2 + ⎟ ( 512 nm )
⎝
2⎠
or
1
⎛ 640 nm ⎞
m1 ⎜
= m2 +
⎝ 512 nm ⎟⎠
2
1
This gives (1.25) m1 = m2 + , or 2.5 m1 = 2m2 + 1 .
2
24.19
With nfilm = 1.52, > nair , light reflecting from the upper surface (air to glass transition) experiences a phase change, while light reflecting from the lower surface (a glass to air transition) does
not experience such a change. Under these conditions, the requirement for constructive interference of normally incident waves reflecting from these two surfaces is 2nfilm t = (m + 12 )l, where m
is any integer value. Thus, if the thickness of the film is t = 0.420 mm = 420 nm, the wavelengths
that will strongly reflect from the film are
l=
yielding
2nfilm t
4 (1.52 )( 420 nm )
=
m +1 2
2m + 1
m = 0, 1, 2,…
m = 0 ⇒ l = 2.55 × 10 3 nm (infrared);
m = 1 ⇒ l = 851 nm (infrared);
m = 2 ⇒ l = 511 nm (visible);
m = 3 ⇒ l = 365 nm (ultraviolet),
with the wavelengths for all higher values of m being even shorter. Thus, the only visible light
wavelength that can strongly reflect from this film is 511 nm .
24.20
Since nair < noil < nwater , light reflected from both top and bottom surfaces of the oil film
experiences a phase change, resulting in zero net phase difference due to reflections.
Therefore, the condition for constructive interference in reflected light is
2 t = mln = m
l
nfilm
or
⎛ l ⎞
t = m⎜
where m = 0, 1, 2,…
⎝ 2 nfilm ⎟⎠
Assuming that m = 1, the thickness of the oil slick is t = (1)
68719_24_ch24_p256-281.indd 265
600 nm
l
=
= 233 nm .
2 nfilm 2 (1.29 )
1/7/11 2:53:19 PM
266
24.21
Chapter 24
There will be a phase change of the radar waves reflecting from both surfaces of the polymer,
giving zero net phase change due to reflections. The requirement for destructive interference in
the reflected waves is then
1⎞
⎛
2 t = ⎜ m + ⎟ ln
⎝
2⎠
or
t = ( 2m + 1)
l
where m = 0, 1, 2,…
4 nfilm
If the film is as thin as possible, then m = 0, and the needed thickness is
t=
l
3.00 cm
=
= 0.500 cm
4 nfilm
4 (1.50 )
This anti-reflectance coating could be easily countered by changing the wavelength of the
radar—to 1.50 cm—now creating maximum reflection!
24.22
(a)
With nair < nwater < noil , reflections at the air-oil interface experience a phase change, while
reflections at the oil-water interface experience no phase change. With one phase change at
the surfaces, the condition for constructive interference in the light reflected by the film is
2nfilm t = (m + 12 )l, where m is any positive integer. Thus,
lm =
2nfilm t 2 (1.45)( 280 nm ) 812 nm
=
=
m + 12
m + 12
m + 12
m = 0, 1, 2, 3,…
The possible wavelengths are: l0 = 1.62 × 10 3 nm, l1 = 541 nm, l2 = 325nm,….
Of these, only l1 = 541 nm (green) is in the visible portion of the spectrum.
(b)
The wavelengths that will be most strongly transmitted are those that suffer destructive
interference in the reflected light. With one phase change at the surfaces, the condition for
destructive interference in the light reflected by the film is 2nfilm t = ml, where m is any
positive, nonzero integer. The possible wavelengths are
lm =
or
2nfilm t 2 (1.45)( 280 nm ) 812 nm
=
=
m
m
m
m = 1, 2, 3,…
l1 = 812 nm, l2 = 406 nm, l3 = 271 nm,…
of which only l2 = 406 nm (violet) is in the visible spectrum.
24.23
(a)
For maximum transmission, we want destructive interference in the light reflected from the
front and back surfaces of the film.
If the surrounding glass has refractive index greater than 1.378, light reflected from the
front surface suffers no phase change, and light reflected from the back does undergo a
phase change. Under these conditions, the condition for destructive interference in the
reflected light is 2d = mln = m l nfilm , where m is a positive integer. Thus, for minimum
nonzero film thickness, we require that m = 1 and find
d=
(b)
l
656.3 nm
=
= 238 nm
2 nfilm
2 (1.378 )
The filter will expand. As d increases in 2nfilm d = l , so does l increase .
continued on next page
68719_24_ch24_p256-281.indd 266
1/7/11 2:53:22 PM
Wave Optics
(c)
267
Destructive interference of order 2 will occur for reflected light when 2 d = 2 l nfilm, or for
the wavelength
l = nfilm d = (1.378 )( 238 nm ) = 328 nm (near ultraviolet)
24.24
Light reflecting from the lower surface of the air
layer experiences a phase change, but light
reflecting from the upper surface of the layer does
not. The requirement for a dark fringe (destructive
interference) is then
2nfilm t = ml ,
where m = 0, 1, 2,…
At the thickest part of the film ( t = 2.00 mm ) , the order number is
m=
−6
2nfilm t 2 (1.00 )( 2.00 × 10 m )
=
= 7.32
l
546.1× 10 −9 m
Since m must be an integer, m = 7 is the order of the last dark fringe seen. Counting the m = 0
order along the edge of contact, a total of 8 dark fringes will be seen.
24.25
When the hair is inserted between one
B
end of the glass plates, which have
A
Δt
length L = 14.0 cm, a wedge-shaped
d
tm+1
air film is created as shown at the
tm
Δx
right. The maximum thickness of this
air wedge is equal to the diameter d
L = 14.0 cm
of the hair. If the bright interference
fringe of order m occurs at point A
(where the air wedge has thickness tm ) and the adjacent bright fringe of order m + 1 occurs at B
(where the thickness is tm +1 ), we may use the properties of similar triangles to write
d Δt tm +1 − tm
=
=
L Δx
Δx
or
⎛ Δt ⎞
d =⎜
L
⎝ Δ x ⎟⎠
[1]
Since nair < nglass , light waves reflecting at the upper surface of the air film do not undergo a phase
change, but those reflecting from the lower surface do experience such a change. Thus, the condition for constructive interference to produce a bright fringe is 2nair t = (m + 12 )l, or the thickness
of the film at the location of the bright fringe of order m is tm = (m + 12 )l 2nair . The change in
thickness between the locations of adjacent bright fringes is then
Δt = tm +1 − tm =
[( m + 1) + 1 2] l − [ m + 1 2] l =
2nair
2nair
l
l
=
2nair 2
If the observed spacing between adjacent bright fringes is Δx = 0.580 mm when using light of
wavelength l = 650 nm, Equation [1] gives the diameter of the hair as
(650 × 10−9 m ) (14.0 × 10 −2 m ) = 7.84 × 10 −5 m = 78.4 mm
lL
⎛ l 2⎞
d =⎜
=
L
=
⎝ Δx ⎟⎠
2 ( Δx )
2 ( 0.580 × 10 −3 m )
68719_24_ch24_p256-281.indd 267
1/7/11 2:53:24 PM
268
Chapter 24
24.26
From the geometry shown in the figure, R 2 = h 2 + r 2 = ( R − t ) + r 2 , or
2
t = R − R 2 − r 2 = 3.0 m −
( 3.0 m )2 − ( 9.8 × 10 −3 m ) = 1.6 × 10 −5 m
2
With a phase change upon reflection at the lower surface of the air layer but no change with
reflection from the upper surface, the condition for a bright fringe is
1⎞
1⎞ l ⎛
1⎞
⎛
⎛
2 t = ⎜ m + ⎟ ln = ⎜ m + ⎟
= ⎜ m + ⎟ l , where m = 0, 1, 2,…
⎝
⎠
⎝
⎠
⎝
2
2 nair
2⎠
At the 50th bright fringe, m = 49, and the wavelength is found to be
l=
24.27
2 (1.6 × 10 −5 m )
2t
=
= 6.5 × 10 −7 m = 6.5 × 10 2 nm
m +1 2
49.5
There is a phase change due to reflection at the bottom of the air film but not at the top of the
film. The requirement for a dark fringe is then
2nair t = ml
where
m = 0, 1, 2,…
At the 19th dark ring (in addition to the dark center spot), the order number is m = 19, and the
thickness of the film is
t=
24.28
−9
ml 19 ( 500 × 10 m )
=
= 4.75 × 10 −6 m = 4.75 mm
2nair
2 (1.00 )
With a phase change due to reflection at each surface of the magnesium fluoride layer, there is
zero net phase difference caused by reflections. The condition for destructive interference is then
1⎞
1⎞ l
⎛
⎛
2 t = ⎜ m + ⎟ ln = ⎜ m + ⎟
, where m = 0, 1, 2,…
⎝
⎝
2⎠
2 ⎠ nfilm
For minimum thickness, m = 0, and the thickness is
t = ( 2m + 1)
24.29
(550 × 10 m ) = 9.96 × 10 −8 m = 99.6 nm
l
= (1)
4 nfilm
4 (1.38 )
−9
Since the thin film has air on both sides of it, there is a phase change for light reflecting from the
first surface but no change for light reflecting from the second surface. Under these conditions,
the requirement to be met if waves reflecting from the two sides are to produce constructive
interference and a strong reflection is
1⎞
⎛
2nfilm t = ⎜ m + ⎟ l
⎝
2⎠
or
l=
4nfilm t
2m + 1
m = 0, 1, 2,…
continued on next page
68719_24_ch24_p256-281.indd 268
1/7/11 2:53:27 PM
Wave Optics
269
With nfilm = 1.473 and t = 542 nm, the wavelengths that produce strong reflections are given by
l = 4 (1.473)( 524 nm ) (2m + 1), which yields
m = 0 : l = 3.09 × 10 3 nm
m = 1: l = 1.03 × 10 3 nm
m = 2 : l = 617 nm
m = 3 : l = 441 nm
m = 4 : l = 343 nm
m = 5 : l = 281 nm
and many other shorter wavelengths. Of these, the only ones in the range 300 nm to 700 nm are
617 nm, 441 nm, and 343 nm .
24.30
With nair < nfilm < nglass , light undergoes a phase change when it reflects from either of the two
surfaces of the film. Therefore, the condition for destructive interference of normally incident
light reflecting from these two surfaces is
⎛
2nfilm t = ⎜ m +
⎝
(a)
24.31
⎛ l ⎞
t = ( 2m + 1) ⎜
⎝ 4nfilm ⎟⎠
or
where m = 0, 1, 2,…
[1]
The minimum thickness film that will minimize the light reflected from the lens is given by
m = 0 in Equation [1], and has the value
tm in =
(b)
1⎞
⎟l
2⎠
l
540 nm
=
= 97.8 nm
4nfilm
4 (1.38 )
Yes. From Equation [1], we see than any film thickness given by t = (2m + 1)tm in ,
where tm in = (l 4nfilm ) = 97.8 nm, will produce destructive interference and hence
minimum reflected light for wavelength l = 540 nm. For m = 1, 2, 3,… , we obtain
thicknesses of 293 nm, 489 nm, 685 nm, … .
In a single-slit diffraction pattern, with the slit having width a, the dark fringe of order m occurs
at angle q m , where sinq m = m(l a) and m = ±1, ± 2, ± 3,…. The location, on a screen located
distance L from the slit, of the dark fringe of order m (measured from y = 0 at the center of the
central maximum) is (ydark )m = L tanq m ≈ L sinq m = ml ( L a ) .
(a)
The central maximum extends from the m = −1 dark fringe to the m = +1 dark fringe, so the
width of this central maximum is
⎛ lL ⎞
⎛ lL ⎞ 2lL
− ( −1) ⎜
Central max. width = (ydark )1 − (ydark )−1 = 1⎜
=
⎟
⎝ a ⎠
⎝ a ⎟⎠
a
=
(b)
2 ( 5.40 × 10 −7 m )(1.50 m )
0.200 × 10 −3 m
= 8.10 × 10 −3 m = 8.10 mm
The first order bright fringe extends from the m = 1 dark fringe to the m = 2 dark fringe, or
( Δy ) = ( y ) − ( y )
bright
dark
1
=
2
(5.40 × 10
dark
−7
1
⎛ lL ⎞
⎛ lL ⎞ lL
− 1⎜
=
= 2⎜
⎝ a ⎟⎠
⎝ a ⎟⎠
a
m ) (1.50 m )
0.200 × 10 −3 m
= 4.05 × 10 −3 m = 4.05 mm
Note that the width of the first order bright fringe is exactly one-half the width of the
central maximum.
68719_24_ch24_p256-281.indd 269
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270
24.32
Chapter 24
(a)
Dark bands occur where sinq = m ( l a ) . At the first dark band, m = 1, and the distance
from the center of the central maximum is
⎛ 600 × 10 −9 m ⎞
⎛l⎞
= 2.3 mm
y1 = L tanq ≈ L sinq = L ⎜ ⎟ = (1.5 m ) ⎜
⎝ a⎠
⎝ 0.40 × 10 −3 m ⎟⎠
(b)
The central maximum extends from the m = 1 dark band to the m = −1 dark band. Its width
is (ydark )1 − (ydark )−1 = 1( lL a ) − ( −1) ( lL a ) = 2lL a , or
width =
24.33
(a)
2 ( 600 × 10 −9 m )(1.5 m )
0.40 × 10 −3 m
= 4.5 × 10 −3 m = 4.5 mm
Dark bands (minima) occur where sinq = m(l a). For the first minimum, m = 1, and the
distance from the center of the central maximum is y1 = L tanq ≈ L sinq = L ( l a ) . Thus,
the needed distance to the screen is
⎛ 0.75 × 10 −3 m ⎞
⎛ a⎞
= 1.1 m
L = y1 ⎜ ⎟ = ( 0.85 × 10 −3 m ) ⎜
⎝l⎠
⎝ 587.5 × 10 −9 m ⎟⎠
(b)
24.34
The width of the central maximum is 2 y1 = 2 ( 0.85 mm ) = 1.7 mm .
Note: The small angle approximation does not work
well in this situation. Rather, you should proceed as
follows.
At the first order minimum, sinq1 = (1) l a , or
⎛l⎞
⎛ 5.00 cm ⎞
q1 = sin −1 ⎜ ⎟ = sin −1 ⎜
= 7.98°
⎝ 36.0 cm ⎟⎠
⎝ a⎠
Then, y1 = L tanq1 = ( 6.50 m ) tan 7.98°
= 0.911 m = 91.1 cm
24.35
With the screen locations of the dark fringe of order m at
(ydark )m = L tanq m ≈ L sinq m = m(lL a)
the width of the central maximum is Δycentral
for
m = ±1, ± 2, ± 3,…
= (ydark )m = +1 − (ydark )m = −1 = 2(lL a), so
m axim um
⎛
⎞
a ⎜ Δycentral ⎟
−3
−3
⎝ m axim um ⎠ ( 0.600 × 10 m ) ( 2.00 × 10 m )
l=
=
= 4.62 × 10 −7 m = 462 nm
2L
2 (1.30 m )
24.36
At the positions of the minima, sinq m = m ( l a ) and ym = L tanq m ≈ L sinq m = m ⎡⎣ L ( l a ) ⎤⎦ .
Thus, y3 − y1 = ( 3 − 1) ⎡⎣ L ( l a ) ⎤⎦ = 2 ⎡⎣ L ( l a ) ⎤⎦ , and
a=
68719_24_ch24_p256-281.indd 270
2 ( 0.500 m )( 680 × 10 −9 m )
2Ll
=
= 2.27 × 10 −4 m = 0.227 mm
y3 − y1
3.00 × 10 −3 m
1/7/11 2:53:33 PM
Wave Optics
24.37
271
The locations of the dark fringes (minima) mark the edges of the maxima, and the widths of the
maxima equals the spacing between successive minima.
At the locations of the minima, sinq m = m(l a), and
⎡
⎛ 500 × 10 −9 m ⎞ ⎤
ym = L tanq m ≈ L sinq m = m ⎡⎣ L ( l a ) ⎤⎦ = m ⎢(1.20 m ) ⎜
⎥ = m (1.20 mm )
⎝ 0.500 × 10 −3 m ⎟⎠ ⎦
⎣
Then, Δy = Δm(1.20 mm), and for successive minima, Δm = 1. Therefore, the width of each
maximum, other than the central maximum, in this interference pattern is
width = Δy = (1)(1.20 mm ) = 1.20 mm
24.38
In a single-slit diffraction pattern, minima are found where
⎛l⎞
sinq dark = m ⎜ ⎟
⎝ a⎠
q
L
where m = ± 1, ± 2,…
y
or, using the small angle approximation y = L tanq ≈ L sinq , at screen
positions ym = m(lL a). Thus, if the m = 2 minimum is seen at y = 1.40 mm
on a screen located distance L = 85.0 cm from a single slit of width a = 0.800 mm,
the wavelength of the light passing through the slit must be
l=
24.39
−3
−3
y a (1.40 × 10 m ) ( 0.800 × 10 m )
=
= 6.59 × 10 −7 m = 659 nm
mL
2 (85.0 × 10 −2 m )
The grating spacing is d = (1 3 660 )cm = (1 3.66 × 10 5 ) m, and bright lines are found where
d sinq = ml .
(a)
The wavelength observed in the first-order spectrum is l = d sinq1 , or
9
⎛ 10 4 nm ⎞
⎛ 1 m ⎞ ⎛ 10 nm ⎞
sinq1 = ⎜
sinq1
l =⎜
5 ⎟ ⎜
⎟
⎝ 3.66 × 10 ⎠ ⎝ 1 m ⎠
⎝ 3.66 ⎟⎠
This yields: at q1 = 10.1°,
l = 698 nm .
(b)
l = 479 nm ; at q1 = 13.7°, l = 647 nm ; and at q1 = 14.8°,
In the second order, m = 2. The second order images for the above wavelengths will be
found at angles q 2 = sin −1 ( 2 l d ) = sin −1 [ 2sinq1 ] .
This yields: for l = 479 nm, q 2 = 20.5° ; for l = 647 nm, q 2 = 28.3° ; and for
l = 698 nm, q 2 = 30.7° .
24.40
(a)
The longest wavelength in the visible spectrum is 700 nm, and the grating spacing
is d = 1 mm 600 = 1.67 × 10 −3 mm = 1.67 × 10 −6 m. The maximum viewing angle is
q m ax = 90.0°, and maxima are found where ml = d sinq . Thus,
mm ax =
−6
d sin 90.0° (1.67 × 10 m ) sin 90.0°
=
= 2.38
lred
700 × 10 −9 m
so 2 complete orders will be observed.
continued on next page
68719_24_ch24_p256-281.indd 271
1/7/11 2:53:36 PM
272
Chapter 24
(b)
From l = d sinq , the angular separation of the red and violet edges in the first order will be
⎡l
Δq = sin −1 ⎢ red
⎣ d
or
24.41
⎤
−1 ⎡ lviolet
⎥ − sin ⎢ d
⎣
⎦
−9
−9
m⎤
m⎤
⎤
−1 ⎡ 700 × 10
−1 ⎡ 400 × 10
=
sin
−
sin
⎢
⎥
⎢
−6
−6
⎥
m
m ⎥⎦
1.67
×
10
1.67
×
10
⎦
⎣
⎦
⎣
Δq = 10.9°
The grating spacing is d =
1.00 cm ⎛ 1.00 m ⎞
1.00 m
.
⎜⎝ 2
⎟⎠ =
4 500 10 cm
4.50 × 10 5
From d sinq = ml, the angular separation between the given spectral lines will be
Δq = sin −1 ⎡⎣ m lred d ⎤⎦ − sin −1 ⎡⎣ m lviolet d ⎤⎦ , or
⎡ m ( 434 × 10 −9 m ) ( 4.50 × 10 5 ) ⎤
⎡ m ( 656 × 10 −9 m ) ( 4.50 × 10 5 ) ⎤
Δq = sin −1 ⎢
⎥
⎥ − sin −1 ⎢
1.00 m
1.00 m
⎢⎣
⎥⎦
⎢⎣
⎥⎦
The results obtained are: for m = 1, Δq = 5.91° ; for m = 2, Δq = 13.2° ; and for m = 3,
Δq = 26.5° . Complete orders for m ≥ 4 are not visible.
24.42
The array of wires will act as a diffraction grating for the ultrasound waves, with maxima
located at those angles given by the grating equation d sinq = ml, where m = 0, ± 1, ± 2,….
The spacing between adjacent slits in this grating is d = 1.30 cm, and the wavelength of these
ultrasound waves is
l=
(a)
vsound
343 m s
=
= 9.22 × 10 −3 m = 9.22 mm
37.2 × 10 3 Hz
f
The order number of the maximum found at angle q is m = (d l)sinq , so the maximum
order that can be found will be the largest integer satisfying the relation
1.30 × 10 −2 m
⎛d⎞
⎛d⎞
m ≤ ⎜ ⎟ sin q m ax = ⎜ ⎟ sin 90.0° =
= 1.41
⎝l⎠
⎝l⎠
9.22 × 10 −3 m
Thus, three maxima corresponding to m = −1, m = 0, and m = +1 can be found .
(b)
We use q = sin −1 (ml d) to find the direction for each maximum to be:
for m = −1, q = − 45.2° ; for m = 0, q = 0° ; and for m = +1, q = + 45.2° .
24.43
For diffraction by a grating, the angle at which the maximum of order m occurs is given by
d sinq m = ml, where d is the spacing between adjacent slits on the grating. Thus,
(1)( 632.8 × 10
ml
d=
=
sinq m
sin 20.5°
24.44
−9
m)
= 1.81× 10 −6 m = 1.81 mm
With 2 000 lines per centimeter, the grating spacing is
d=
1
cm = 5.00 × 10 −4 cm = 5.00 × 10 −6 m
2 000
Then, from d sinq = ml , the location of the first order for the red light is
⎛ 1( 640 × 10 −9 m ) ⎞
⎛ ml ⎞
−1
=
sin
q = sin −1 ⎜
⎜
⎟ = 7.35°
−6
⎝ d ⎟⎠
⎝ 5.00 × 10 m ⎠
68719_24_ch24_p256-281.indd 272
1/7/11 2:53:40 PM
273
Wave Optics
24.45
The spacing between adjacent slits on the grating is
d=
1 cm
10 −2 m
=
5 310 slits 5 310
The maximum of order m is located where d sinq m = ml, so
l=
y
d
d⎛
sinq m = ⎜ 2 m 2
m
m ⎜⎝ L + ym
⎞
10 −2 m ⎛
⎜
⎟=
⎟⎠ (1)( 5 310 ) ⎜⎝
0.488 m
(1.72 m )2 + ( 0.488 m )2
⎞
⎟
⎟⎠
= 5.14 × 10 −7 m = 514 nm
24.46
From d sinq m = ml, or sinq m = ml d , we see that
sinq v 2 =
2lviolet 800 nm
=
d
d
sinq v 3 =
3lviolet 1 200 nm
=
d
d
sinq r 2 =
2lred 1 400 nm
=
d
d
sinq r 3 =
3lred 2 100 nm
=
d
d
Since, for 0° ≤ q ≤ 90°, q increases as sinq increases,
we have that
q v 2 < q v3 < qr 2 < qr 3
so the second and third order spectra overlap in the range q v 3 ≤ q ≤ q r 2 .
24.47
1 cm
10 −2 m
=
= 3.64 × 10 −6 m. From d sinq = ml , or
2 750 2.75 × 10 3
q = sin −1 ( ml d ) , the angular positions of the red and violet edges of the second-order spectrum
are found to be
The grating spacing is d =
⎛ 2 ( 700 × 10 −9 m ) ⎞
⎛ 2l ⎞
q r = sin −1 ⎜ red ⎟ = sin −1 ⎜
⎟ = 22.6°
−6
⎝ d ⎠
⎝ 3.64 × 10 m ⎠
and
⎛ 2 ( 400 × 10 −9 m ) ⎞
⎞
⎛ 2l
q v = sin −1 ⎜ violet ⎟ = sin −1 ⎜
⎟ = 12.7°
−6
⎝ d ⎠
⎝ 3.64 × 10 m ⎠
Note from the sketch at the right that yr = L tanq r
and yv = L tanq v , so the width of the spectrum on the
screen is Δy = L ( tanq r − tanq v ) .
Screen
Grating
Since it is given that d = ( m + 1 2 ) l, the distance
from the grating to the screen must be
Δy
1.75 cm
L=
=
tanq r − tanq v tan ( 22.6° ) − tan (12.7° )
Δy
qv
qr
yv
yr
L
= 9.17 cm
24.48
(a)
d=
1 cm
= 2.381 × 10 −4 cm = 2.381 × 10 −6 m
4.200 × 10 3 slits
continued on next page
68719_24_ch24_p256-281.indd 273
1/7/11 2:53:43 PM
274
Chapter 24
⇒
⎡ 2 ( 589.0 × 10 −9 m ) ⎤
⎛ 2l ⎞
−1
=
sin
q 2 = sin −1 ⎜
⎢
⎥ = 29.65°
−6
⎝ d ⎟⎠
⎣⎢ 2.381 × 10 m ⎥⎦
(b)
d sinq m = ml
(c)
y2 = L tanq 2 = ( 2.000 m ) tan ( 29.65° ) = 1.138 m
(d)
⎡ 2 ( 589.6 × 10 −9 m ) ⎤
⎛ 2l′ ⎞
−1
=
sin
q ′2 = sin −1 ⎜
⎢
⎥ = 29.69°
−6
⎝ d ⎟⎠
⎣⎢ 2.381× 10 m ⎥⎦
y2′ = L tanq 2′ = ( 2.000 m ) tan ( 29.69° ) = 1.140 m
(e)
Δy = y2′ − y2 = 1.140 m − 1.138 m = 2.000 × 10 −3 m
(f )
⎡ ⎛
⎛ −1 ⎛ 2l ⎞ ⎞ ⎤
⎛ 2l ′ ⎞ ⎞
Δy = y2′ − y2 = L ⎢ tan ⎜ sin −1 ⎜
⎟⎠ ⎟ − tan ⎜ sin ⎜⎝
⎟ ⎥
⎝
d
d ⎠ ⎟⎠ ⎦
⎠
⎝
⎣ ⎝
⎡ ⎛
⎛
⎛ 2 ( 589.6 × 10 −9 m ) ⎞ ⎞
⎛ 2 ( 589.0 × 10 −9 m ) ⎞ ⎞ ⎤
−1
−
tan
sin
= ( 2.000 m ) ⎢ tan ⎜ sin −1 ⎜ −2
⎟
⎜
⎟
⎜ −2
⎟⎟ ⎥
⎜⎝
10 m 4.200 × 10 3 ⎠ ⎟⎠
10 m 4.200 × 10 3 ⎠ ⎟⎠ ⎥
⎢ ⎜⎝
⎝
⎝
⎣
⎦
= 1.537 × 10 −3 m
The two answers agree to only one significant figure. The calculation is sensitive to rounding at intermediate steps.
24.49
(a)
For a diffraction grating having distance d between adjacent slits, primary maxima for
wavelength l are produced at angles satisfying the grating equation, d sinq = ml. If the
m = 3 maximum for l = 500 nm is observed at q = 32.0°, the grating spacing is
d=
3 ( 500 × 10 −9 m )
ml
=
= 2.83 × 10 −6 m = 2.83 × 10 −4 cm
sinq
sin 32.0°
The number of rulings per centimeter on the grating is then
n=
(b)
1 cm
1 cm
=
= 3.53 × 10 3 grooves cm
d
2.83 × 10 −4 cm
The order number of a maximum appearing at angle q is m = d sinq l . The magnitude of
the largest order numbers visible is the largest integer satisfying the condition
d 2.83 × 10 −6 m
⎛d⎞
⎛d⎞
m ≤ ⎜ ⎟ sinq m ax = ⎜ ⎟ sin 90° = =
= 5.66
⎝l⎠
⎝l⎠
l 500 × 10 −9 m
Thus, 11 maxima for wavelength l = 500 nm will be observable with this grating. These
are the maxima corresponding to m = 0, ± 1, ± 2, ± 3, ± 4, and ± 5.
24.50
The grating spacing is d = 1 cm 1 200 = 8.33 × 10 −4 cm = 8.33 × 10 −6 m. Using sinq = ml d and
the small angle approximation, the distance from the central maximum to the maximum of order
m for wavelength l is ym = L tanq ≈ L sinq = (lL d)m. Therefore, the spacing between successive maxima is Δy = ym +1 − ym = lL d.
The longer wavelength in the light is found to be
l long =
( Δy ) d = (8.44 × 10
L
−3
m ) (8.33 × 10 −6 m )
0.150 m
= 4.69 × 10 −7 m = 469 nm
continued on next page
68719_24_ch24_p256-281.indd 274
1/7/11 2:53:47 PM
Wave Optics
275
Since the third order maximum of the shorter wavelength falls halfway between the central
maximum and the first order maximum of the longer wavelength, we have
3lshort L ⎛ 0 + 1 ⎞ llong L
=⎜
⎝ 2 ⎟⎠ d
d
24.51
(a)
or
⎛ 1⎞
lshort = ⎜ ⎟ ( 469 nm ) = 78.1 nm
⎝ 6⎠
From Brewster’s law, the index of refraction is
n2 = tanq p = tan ( 48.0° ) = 1.11
(b)
From Snell’s law, n2 sinq 2 = n1 sinq1 , the angle of refraction when q1 = q p is
⎛ n1 sinq p ⎞
⎛ (1.00 ) sin 48.0° ⎞
q 2 = sin −1 ⎜
= sin −1 ⎜
⎟⎠ = 42.0°
⎝
1.11
⎝ n2 ⎟⎠
Note that when q1 = q p , q 2 = 90.0° −q p as it should.
24.52
(a)
From Malus’s law, the fraction of the incident intensity of the unpolarized light that is
transmitted by the polarizer is
I ′ = I 0 ( cos 2 q )av = I 0 ( 0.500 )
The fraction of this intensity incident on the analyzer that will be transmitted is
I = I ′ cos 2 ( 35.0° ) = ⎡⎣ 0.500I 0 ⎤⎦ cos 2 ( 35.0° ) = 0.336I 0
Thus, the fraction of the incident unpolarized light transmitted is I I 0 = 0.336 .
(b)
The fraction of the original incident light absorbed by the analyzer is
I ′ − I 0.500I 0 − 0.336I 0
=
= 0.164
I0
I0
24.53
The more general expression for Brewster’s angle is given in problem P24.57 as
tanq p = n2 n1
24.54
(a)
⎛n ⎞
⎛ 1.52 ⎞
When n1 = 1.00 and n2 = 1.52, q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜
= 56.7° .
⎝ 1.00 ⎟⎠
⎝ n1 ⎠
(b)
⎛n ⎞
⎛ 1.52 ⎞
When n1 = 1.333 and n2 = 1.52, q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜
= 48.8° .
⎝ 1.333 ⎟⎠
⎝ n1 ⎠
The polarizing angle for light in air striking a water surface is
⎛n ⎞
⎛ 1.333 ⎞
q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜
= 53.1°
⎝ 1.00 ⎟⎠
⎝ n1 ⎠
This is the angle of incidence for the incoming sunlight (that is, the angle between the incident
light and the normal to the surface). The altitude of the Sun is the angle between the incident light
and the water surface. Thus, the altitude of the Sun is
a = 90.0° −q p = 90.0° − 53.1° = 36.9°
68719_24_ch24_p256-281.indd 275
1/7/11 2:53:50 PM
276
24.55
Chapter 24
(a)
Brewster’s angle (or the polarizing angle) is
⎛ nquartz ⎞
⎛n ⎞
⎛ 1.458 ⎞
= tan −1 ⎜
q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜
= 55.6°
⎝ 1.000 ⎟⎠
⎝ n1 ⎠
⎝ nair ⎟⎠
(b)
24.56
When the angle of incidence is the polarizing angle, q p , the angle of refraction of the
transmitted light is q 2 = 90.0° −q p . Hence, q 2 = 90.0° − 55.6° = 34.4° .
The critical angle for total internal reflection is q c = sin −1 ( n2 n1 ). Thus, if q c = 34.4° as light
attempts to go from sapphire into air, the index of refraction of sapphire is
nsapphire = n1 =
n2
1.00
=
= 1.77
sinq c sin 34.4°
Then, when light is incident on sapphire from air, the Brewster’s angle is
⎛n ⎞
⎛ 1.77 ⎞
q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜
= 60.5°
⎝ 1.00 ⎟⎠
n
⎝ 1⎠
24.57
From Snell’s law, the angles of incidence and refraction are related by n1 sinq1 = n2 sinq 2 .
If the angle of incidence is the polarizing angle (that is, q1 = q p), the refracted ray is perpendicular
to the reflected ray (see Figure 24.28 in the textbook), and the angles of incidence and refraction
are also related by q p + 90° +q 2 = 180°, or q 2 = 90° − q p .
Substitution into Snell’s law then gives
(
)
n1 sinq p = n2 sin 90° − q p = n2 cosq p
24.58
sinq p cosq p = tanq p = n2 n1
From Malus’s law, the intensity of the light transmitted by the polarizer is I = I 0 cos 2 q , where
I 0 is the intensity of the incident light, and q is the angle between the direction of the plane
of polarization of the incident light and the transmission axis of the polarizing disk. Thus,
q = cos −1 I I 0 .
(
24.59
or
)
(a)
1
I
=
I 0 2.00
⇒
⎛
1 ⎞
= 45.0°
q = cos −1 ⎜
⎝ 2.00 ⎟⎠
(b)
1
I
=
I 0 4.00
⇒
⎛
1 ⎞
q = cos −1 ⎜
⎟⎠ = 60.0°
4.00
⎝
(c)
1
I
=
I 0 6.00
⇒
⎛
1 ⎞
= 65.9°
q = cos −1 ⎜
⎝ 6.00 ⎟⎠
From Malus’s law, the intensity of the light transmitted by the first polarizer is I1 = I i cos 2 q1 .
The plane of polarization of this light is parallel to the axis of the first plate and is incident
on the second plate. Malus’s law gives the intensity transmitted by the second plate as
I 2 = I1 cos 2 (q 2 −q1 ) = I i cos 2 q1 cos 2 (q 2 −q1 ) . This light is polarized parallel to the axis of the
second plate and is incident upon the third plate. A final application of Malus’s law gives the
transmitted intensity as
I f = I 2 cos 2 (q 3 − q 2 ) = I i cos 2 q1 cos 2 (q 2 − q1 ) cos 2 (q 3 − q 2 )
continued on next page
68719_24_ch24_p256-281.indd 276
1/7/11 2:53:52 PM
Wave Optics
277
With q1 = 20.0°, q 2 = 40.0°, and q 3 = 60.0°, this result yields
I f = (10.0 units ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) = 6.89 units
24.60
(a)
Using Malus’s law, the intensity of the transmitted light is found to be
(
I = I 0 cos 2 ( 45° ) = I 0 1
(b)
(
(a)
)
3 = 54.7°
If light has wavelength l in vacuum, its wavelength in a medium of refractive index n is
ln = l n. Thus, the wavelengths of the two components in the specimen are
ln =
l 546.1 nm
=
= 413.7 nm
n1
1.320
ln =
l 546.1 nm
=
= 409.7 nm
n2
1.333
1
and
(b)
2
From Malus’s law, I I 0 = cos 2 q . Thus, if I I 0 = 1 3, we obtain cos 2 q = 1 3, or
q = cos −1 1
24.61
)
2 , or I = I 0 2 .
2
The numbers of cycles of vibration each component completes while passing through the
specimen of thickness t = 1.000 mm are
N1 =
t
1.000 × 10 −6 m
=
= 2.417
ln
413.7 × 10 −9 m
1
N2 =
and
t
1.000 × 10 −6 m
=
= 2.441
ln
409.7 × 10 −9 m
2
Thus, when they emerge, the two components are out of phase by N 2 − N1 = 0.024 cycles.
Since each cycle represents a phase angle of 360°, they emerge with a phase difference of
Δf = ( 0.024 cycles ) ( 360° cycle ) = 8.6°
24.62
In a single-slit diffraction pattern, the first dark fringe occurs where sin(q dark )1 = (1) l a. If no
diffraction minima are to be observed, the maximum width the slit can have is that which would
place the first dark fringe at the maximum viewable angle of 90.0°. That is, when a = am ax , we
will have sin(q dark )1 = (1) l am ax = sin 90.0° = 1.00, yielding
am ax = l = 632.8 nm
24.63
The light has passed through a single slit since the central
maximum is twice the width of other maxima (the space between
the centers of successive dark fringes). In a double-slit pattern, the
central maximum has the same width as all other maxima (compare
Active Figures 24.1(b) and 24.16(b) in the textbook).
Figure P24.63
In single-slit diffraction the width of the central maximum on the screen is given by
Δycentral
m axim um
⎛ (1) l ⎞
⎛ lL ⎞
= 2⎜
= 2L tan (q dark )m =1 ≈ 2L sin (q dark )m =1 = 2L ⎜
⎝ a ⎟⎠
⎝ a ⎟⎠
continued on next page
68719_24_ch24_p256-281.indd 277
1/7/11 2:53:55 PM
278
Chapter 24
The width of the slit is then
a=
2lL
Δycentral
=
2 ( 632.8 × 10 −9 m ) ( 2.60 m ) ⎛ 10 2 cm ⎞
−4
⎜⎝ 1 m ⎟⎠ = 1.2 × 10 m = 0.12 mm
(10.3 − 7.6 ) cm
m axim um
24.64
(a)
In a double-slit interference pattern, bright fringes on the screen occur where
(ybright )m = m(lL d). Thus, if bright fringes of the wavelengths l1 = 540 nm and
l2 = 450 nm are to coincide, it is necessary that
m1
( 540 nm ) L
d
= m2
( 450 nm ) L
d
or, dividing both sides by 90 nm, 6m1 = 5m2 , where both m1 and m2 are integers.
(b)
The condition found above may be written as m2 = (6 5)m1 . Trial and error reveals that the
smallest nonzero integer value of m1 that will yield an integer value for m2 is m1 = 5 ,
yielding m2 = 6 . Thus, the first overlap of bright fringes for the two given wavelengths
occurs at the screen position (measured from the central maximum)
ybright =
5 ( 540 × 10 −9 m )(1.40 m )
0.150 × 10 −3 m
=
6 ( 450 × 10 −9 m )(1.40 m )
0.150 × 10 −3 m
= 2.52 × 10 −2 m = 2.52 cm
24.65
Dark fringes (destructive interference) occur where d sinq = ( m + 1 2 ) l for m = 0, 1, 2,….
Thus, if the second dark fringe ( m = 1) occurs at q = (18.0 min ) (1.00° 60.0 min ) = 0.300°, the slit
spacing is
⎛
d = ⎜m +
⎝
24.66
−9
1⎞ l
⎛ 3 ⎞ ( 546 × 10 m )
= 1.56 × 10 −4 m = 0.156 mm
=
⎟
⎜ ⎟
2 ⎠ sinq ⎝ 2 ⎠ sin ( 0.300° )
The wavelength is l = vsound f = ( 340 m s ) ( 2 000 Hz ) = 0.170 m, and maxima occur
where d sinq = ml , or q = sin −1 ⎡⎣ m ( l d ) ⎤⎦ for m = 0, ± 1, ± 2,…. Since d = 0.350 m,
l d = 0.486, which gives q = sin −1 [ m ( 0.486 )]. For m = 0, ± 1, and ± 2, this yields
maxima at 0°, ± 29.1°, and ± 76.4° . No solutions exist for m ≥ 3, since that would
imply sinq > 1.
Minima occur where d sinq = ( m + 1 2 ) l, or q = sin −1 [( 2m + 1) l 2d ] for m = 0, ± 1, ± 2,….
With l d = 0.486 , this becomes q = sin −1 [( 2m + 1)( 0.243)]. For m = 0 and ± 1, we find
minima at ± 14.1° and ± 46.8° . No solutions exist for m ≥ 2, since that would imply sinq > 1.
24.67
The source and its image, located 1.00 cm below the mirror, act as a pair of coherent sources.
This situation may be treated as double-slit interference, with the slits separated by 2.00 cm, if
it is remembered that the light undergoes a phase change upon reflection from the mirror. The
existence of this phase change causes the conditions for constructive and destructive interference
to be reversed. Therefore, dark bands (destructive interference) occur where y = m ( lL d ) for
m = 0, 1, 2,….
The m = 0 dark band occurs at y = 0 (that is, at mirror level). The first dark band above the mirror
corresponds to m = 1, and is located at
−9
⎛ lL ⎞ ( 500 × 10 m ) (100 m )
=
y = (1) ⎜
= 2.50 × 10 −3 m = 2.50 mm
⎝ d ⎟⎠
2.00 × 10 −2 m
68719_24_ch24_p256-281.indd 278
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Wave Optics
24.68
279
Assuming the glass plates have refractive indices greater than that of both air and water, there
will be a phase change at the reflection from the lower surface of the film but no change from
reflection at the top of the film. Therefore, the condition for a dark fringe is
2 t = mln = m ( l nfilm ) for m = 0, 1, 2,…
If the highest order dark band observed is m = 84 (a total of 85 dark bands counting the m = 0
order at the edge of contact), the maximum thickness of the air film is
tm ax =
mm ax ⎛ l ⎞ 84 ⎛ λ ⎞
=
⎜
⎟ = 42 l
2 ⎜⎝ nfilm ⎟⎠ 2 ⎝ 1.00 ⎠
When the film consists of water, the highest order dark fringe appearing will be
⎛ 1.333 ⎞
⎛n ⎞
mm ax = 2 tm ax ⎜ film ⎟ = 2 ( 42 l ) ⎜
= 112
⎝ l ⎟⎠
⎝ l ⎠
Counting the m = 0 order, a total of 113 dark fringes are now observed.
24.69
In the figure at the right, observe that the path difference between
the direct and the indirect paths is
d = 2x − d = 2 h 2 + ( d 2 ) − d
2
With a phase change (equivalent to a half-wavelength shift)
occurring upon reflection at the ground, the condition for
constructive interference is d = ( m + 1 2 ) l, and the condition for
destructive interference is d = m l . In both cases, the possible
values of the order number are m = 0, 1, 2,….
(a)
d
The wavelengths that will interfere constructively are l =
. The longest of these is
m +1 2
for the m = 0 case and has a value of
l = 2d = 4 h 2 + ( d 2 ) − 2d = 4
2
(b)
The wavelengths that will interfere destructively are l = d m , and the largest finite one of
these is for the m = 1 case. That wavelength is
l = d = 2 h2 + ( d 2) − d = 2
2
24.70
( 50.0 m )2 + ( 300 m )2 − 2 ( 600 m ) = 16.6 m
( 50.0 m )2 + ( 300 m )2 − 600 m = 8.28 m
From Malus’s law, the intensity of the light transmitted by the first polarizer is I1 = I i cos 2 q1 .
The plane of polarization of this light is parallel to the axis of the first plate and is incident
on the second plate. Malus’s law gives the intensity transmitted by the second plate as
I 2 = I1 cos 2 (q 2 −q1 ) = I i cos 2 q1 cos 2 (q 2 −q1 ) . This light is polarized parallel to the axis of the
second plate and is incident upon the third plate. A final application of Malus’s law gives the
transmitted intensity as
I f = I 2 cos 2 (q 3 − q 2 ) = I i cos 2 q1 cos 2 (q 2 − q1 ) cos 2 (q 3 − q 2 )
(a)
If q1 = 45°, q 2 = 90°, and q 3 = 0°, then
I f I i = cos 2 45° cos 2 ( 90° − 45° ) cos 2 ( 0° − 90° ) = 0
continued on next page
68719_24_ch24_p256-281.indd 279
1/7/11 2:54:03 PM
280
Chapter 24
(b)
If q1 = 0°, q 2 = 45°, and q 3 = 90°, then
I f I i = cos 2 0° cos 2 ( 45° − 0° ) cos 2 ( 90° − 45° ) = 0.25
24.71
If the signal from the antenna to the
receiver station is to be completely
polarized by reflection from the water,
the angle of incidence where it strikes
the water must equal the polarizing angle
from Brewster’s law. This is given by
⎛n
⎞
q p = tan −1 ⎜ water ⎟ = tan −1 (1.333) = 53.1°
⎝ nair ⎠
From the triangle RST in the sketch, the
horizontal distance from the point of
refection, T, to shore is given by
x = ( 90.0 m ) tanq p = ( 90.0 m )(1.333) = 120 m
and from triangle ABT, the horizontal distance from the antenna to point T is
y = ( 5.00 m ) tanq p = ( 5.00 m )(1.333) = 6.67 m
The total horizontal distance from ship to shore is then x + y = 120 m + 6.67 m = 127 m .
24.72
There will be a phase change associated with the
reflection at one surface of the film but no change
at the other surface of the film. Therefore, the condition for a dark fringe (destructive interference) is
⎛ l ⎞
2 t = mln = m ⎜
where m = 0, 1, 2,…
⎝ nfilm ⎟⎠
From the figure, note that
2
R 2 = r 2 + ( R − t ) = r 2 + R 2 − 2Rt + t 2 , which
2
reduces to r = 2Rt − t 2 . Since t will be very
small in comparison to either r or R, we may
neglect the term t 2 , leaving r ≈ 2Rt .
For a dark fringe, t =
ml
, so the radii of the dark rings will be
2nfilm
⎛ ml ⎞
r ≈ 2R ⎜
=
⎝ 2nfilm ⎟⎠
24.73
mlR
nfilm
for m = 0, 1, 2,…
In the single-slit diffraction pattern, destructive interference (or minima) occur where
sinq = m ( l a ) for m = 0, ± 1, ± 2,…. The screen locations, measured from the center
of the central maximum, of these minima are at
ym = L tanq m ≈ L sinq m = m ( lL a )
continued on next page
68719_24_ch24_p256-281.indd 280
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Wave Optics
281
If we assume the first order maximum is halfway between the first and second order minima, then
its location is
y=
y1 + y2 (1+ 2 ) ( lL a ) 3lL
=
=
2
2
2a
and the slit width is
−9
3lL 3 ( 500 × 10 m )(1.40 m )
a=
=
= 3.50 × 10 −4 m = 0.350 mm
−3
2y
2 ( 3.00 × 10 m )
24.74
As light emerging from the glass reflects from the top of the air layer, there is no phase change
produced. However, the light reflecting from the end of the metal rod at the bottom of the air
layer does experience a phase change. Thus, the condition for constructive interference in the
reflected light is 2t = ( m + 12 ) l nair = ( m + 12 ) l.
As the metal rod expands, the thickness of the air layer decreases. The increase in the length of
the rod is given by
ΔL = Δt = ( mi + 12 )
l
l
l
− m f + 12
= Δm
2
2
2
(
)
The order number changes by one each time the film changes from bright to dark and back to
bright. Thus, during the expansion, Δm = 200, and the measured change in the length of the rod is
ΔL = ( 200 )
(500 × 10
l
= ( 200 )
2
2
−9
m)
= 5.00 × 10 −5 m
From ΔL = L0a ( ΔT ) , the coefficient of linear expansion of the rod is
a=
68719_24_ch24_p256-281.indd 281
ΔL
5.00 × 10 −5 m
−1
=
= 20.0 × 10 −6 ( °C )
L0 ( ΔT ) ( 0.100 m )( 25.0 °C )
1/7/11 2:54:08 PM
25
Optical Instruments
QUICK QUIZZES
1.
Choice (c). The corrective lens for a farsighted eye is a converging lens, while that for a
nearsighted eye is a diverging lens. Since a converging lens is required to form a real image of the
Sun on the paper to start a fire, the campers should use the glasses of the farsighted person.
2.
Choice (a). We would like to reduce the minimum angular separation for two objects below the
angle subtended by the two stars in the binary system. We can do that by reducing the wavelength
of the light—this in essence makes the aperture larger, relative to the light wavelength, increasing
the resolving power. Thus, we would choose a blue filter.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The amount of light focused on the film by a camera is proportional to the area of the aperture
through which the light enters the camera. Since the area of a circular opening varies as the
square of the diameter of the opening, the light reaching the film is proportional to the square of
the diameter of the aperture. Thus, increasing this diameter by a factor of 3 increases the amount
of light by a factor of 9, and (c) is the correct choice.
2.
The power of a lens in diopters equals the reciprocal of the focal length when that focal length is
expressed in meters. Hence, the power of a lens having a focal length of 25 cm is
P=
1
1
=
= 4.0 diopters
f 0.25 m
and choice (b) is the correct answer.
3.
Diffraction of light as it passes through, or reflects from, the objective element of a telescope can
cause the images of two sources having a small angular separation to overlap and fail to be seen
as separate images. The minimum angular separation two sources must have in order to be seen as
separate sources is inversely proportional to the diameter of the objective element. Thus, using a large
diameter objective element in a telescope increases its resolution, making (c) the correct choice.
4.
When the eye is longer than normal, the lens-cornea system tends to form images of distant
objects in front of the retina. Rays from near objects are more divergent and the lens-cornea
system brings them into focus farther from the lens, on the retina. This means that the eye can see
near objects clearly but is unable to focus on distant objects. Such an eye is nearsighted (myopia)
and needs a diverging corrective lens to make the rays from distant objects more divergent before
they enter the eye. Choice (b) is the correct answer.
5.
When the eye is shorter than normal, the lens-cornea system fails to bring light from near objects
into focus by the time it reaches the retina, resulting in a blurry image. Light rays entering the pupil
from distant objects are less divergent than those from near objects, and the lens-cornea system
can focus them on the retina. Such an eye is farsighted, or has hyperopia, and needs a converging
corrective lens to help bring rays from near objects to focus sooner. The correct choice is (c).
282
68719_25_ch25_p282-304.indd 282
1/7/11 2:55:10 PM
Optical Instruments
6.
283
The corrective lens must form an upright, virtual image located 55 cm in front of the lens
(q = −55 cm) when the object is 25 cm in front of the lens (p = +25 cm). The thin-lens equation
then gives the required focal length as
f =
pq
( 25 cm )( −55 cm )
=
= + 46 cm
p+q
25 cm − 55 cm
and the correct choice is (c).
7.
Stars are very distant, and the reciprocal of the object
distance p in the thin-lens or mirror equation
1 p +1 q = 1 f
is essentially zero. This means that q = f , or the images are formed at a distance equal to the focal
length from the objective element. Thus, the angular separation of the images, and hence the stars, is
a =
s 20.0 × 10 −3 m
⎛ 360° ⎞
= 1.00 × 10 −2 rad ⎜
=
= 0.573°
⎝ 2p rad ⎟⎠
2.00 m
f
and we see that choice (d) is the correct answer.
8.
When a compound microscope is adjusted for most relaxed viewing (i.e., the final image formed
by the eyepiece is at infinity), the approximate overall magnification produced by the microscope
is given by the expression
m=−
L ⎛ 25 cm ⎞
fo ⎜⎝ fe ⎟⎠
where L is the length of the microscope, fo is the focal length of the objective lens, and fe is the
focal length of the eyepiece. With the microscope described, the approximate magnification is
m=−
15 cm ⎛ 25 cm ⎞
2
⎜
⎟ = −1.2 × 10
0.80 cm ⎝ 4.0 cm ⎠
or
m ≈ 120
and the correct answer is choice (d).
9.
When using light of wavelength l, the resolving power needed to distinguish two closely spaced
spectral lines having a difference in wavelength of Δl is R = l Δl . Thus, if two lines in the visible spectrum differ in wavelength by Δl = 0.1 nm, the minimum resolving power of a diffraction
grating that might be used to separate them is
Rmin =
lmin 400 nm
=
= 4 × 10 3 = 4 000
Δl
0.1 nm
and the correct choice is (b).
10.
The angular separation of the two stars is q = s r = (10 −5 ly) (200 ly) = 5 × 10 −8 radians . The
limiting angle of resolution for a circular aperture is q min = 1.22(l D). Requiring that q min = q ,
and assuming a wavelength at the center of the visible spectrum (550 nm), the required diameter
of the aperture is found to be
D=
(
)
−9
1.22l 1.22 550 × 10 m
=
= 1 × 101 m = 10 m
q
5 × 10 −8 rad
so (c) is the best choice of the listed possible answers.
68719_25_ch25_p282-304.indd 283
1/7/11 2:55:10 PM
284
11.
Chapter 25
Diffraction of the light as it passes through the opening of the detector limits the ability of
that detector to resolve two closely spaced light sources. If the diameter of the opening of
the detector does not change, the intensity of the light is not a factor in the resolution of the
sources. Thus, if we neglect any contraction of the pupils of the eyes, your ability to resolve
the two headlights will not change when they are switched to high beam, and choice (c) is the
best response to the question. In reality, the natural response of the eyes to the increased light
intensity is to contract the pupils to some degree, thereby somewhat decreasing their ability to
resolve the two sources.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
Light nearly parallel to the horizontal ruler will
scatter from rule marks, distance d apart, to
produce a diffraction pattern on a vertical wall a
y
distance L away. At height y on the wall, where
the scattering angle (in radians) is q = y L , light
d
q
scattered from adjacent rule marks interferes
constructively to produce a bright spot if the path
d
L
difference is d = ml, where m = 1, 2, 3,…. The
path difference is given by d = d cosq ≈ d(1 − q 2 2),
where we have made use of a series expansion for cosq , valid for very small angles.
Combining these equations gives ml = d(1 − ym2 2 L2 ). Thus, the wavelength of the light may
be calculated by measurement of the heights ym of bright spots.
4.
There will be an effect on the interference pattern—it will be distorted. The high temperature
of the flame will change the index of refraction of air for the arm of the interferometer in which
the match is held. As the index of refraction varies randomly, the wavelength of the light in that
region will also vary randomly. As a result, the effective difference in length between the two
arms will fluctuate, resulting in a wildly varying interference pattern.
6.
The aperture of a camera is a close approximation to the iris of the eye. The retina of the eye
corresponds to the film of the camera, and a close approximation to the cornea of the eye is the
lens of the camera.
8.
Large lenses are difficult to manufacture and machine with accuracy. Also, their large weight
leads to sagging, which produces a distorted image. In reflecting telescopes, light does not
pass through glass; hence, problems associated with chromatic aberrations are eliminated.
Large-diameter reflecting telescopes are also technically easier to construct. Some designs
use a rotating pool of mercury as the reflecting surface.
10.
Under low ambient light conditions, a photoflash unit is used to ensure that light entering the
camera lens will deliver sufficient energy for a proper exposure to each area of the film. Thus,
the most important criterion is the additional energy per unit area (product of intensity and the
duration of the flash, assuming this duration is less than the shutter speed) provided by the
flash unit.
12.
The angular magnification produced by a simple magnifier is m = ( 25 cm ) f . Note that this
is proportional to the optical power of a lens, P = 1 f , where the focal length f is expressed in
meters. Thus, if the power of the lens is doubled, the angular magnification will also double.
14.
In a nearsighted person the image of a distant object focuses in front of the retina. The cornea
needs to be flattened so that its focal length is increased.
68719_25_ch25_p282-304.indd 284
1/7/11 2:55:13 PM
Optical Instruments
285
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
2.5 mm to 46 mm
4.
The image is 19 mm across and easily fits on the 35-mm slide.
6.
(a) See Solution.
8.
1.05 m to 6.30 m
10.
12.
(b) 1 100 s
(a) farsighted
(b)
18.0 cm
(c)
38.0 cm
(d) virtual image, negative
(e)
34.2 cm
(f )
+2.92 diopters
(c)
−3.70 diopters
(c)
+4.2 cm
(g)
p = 20.0 cm, q = − 40.0 cm, f = + 40.0 cm, P = +2.50 diopters
(a)
+50.8 diopters ≤ P ≤ + 60.0 diopters
(b) –0.800 diopters, diverging
14.
(a) Yes, by using a bifocal or progressive lens.
(b) +1.78 diopters
(c)
−1.18 diopters
16.
(a)
− 0.67 diopters
(b)
+0.67 diopters
18.
(a)
−25.0 cm
(b)
nearsighted
20.
(a) 4.17 cm in front of the lens
(b)
m = 6.00
22.
(a) +5.0
(b)
+6.0
24.
(a) mmax = 2.39
(b)
m = 1.39
26.
m = − 588
28.
0.809 mm
30.
(a)
(b)
2.00 cm toward the objective lens
32.
1.6 × 10 2 mi
34.
(a)
(b)
0.944 m
36.
(a) virtual image
L = fo [( m + 1) m ]
m = 7.50
(b) The final image is an infinite distance in front of the telescope.
fe = −5.00 cm, fo = 15.0 cm
(c)
38.
1.7 m
40.
(a)
68719_25_ch25_p282-304.indd 285
2.29 × 10 − 4 rad
(b)
43.7 m
1/7/11 2:55:14 PM
286
Chapter 25
42.
2.2 × 1011 m
44.
38 cm
46.
(a)
48.
(a) 449 nm
50.
39.6 mm
52.
1.000 5
54.
q min ≈ 2.0 × 10 −3 radians
56.
(a)
58.
(a) 1.96 cm
60.
5.07 mm
62.
m = 10.7
3.6 × 10 3 lines
− 4.3 diopters
(b) 1.8 × 10 3 lines
(b)
smaller, because the wavelength is longer
(b)
− 4.0 diopters, 44 cm
(b)
3.27
(c)
9.80
PROBLEM SOLUTIONS
25.1
The f-number (or focal ratio) of a lens is defined to be the ratio of focal length of the lens to its
diameter. Therefore, the f-number of the given lens is
f -number =
25.2
f
28 cm
=
= 7.0
D 4.0 cm
If a camera has a lens with focal length of 55 mm and can operate at f -numbers that range from
f 1.2 to f 22 , the aperture diameters for the camera must range from
Dmin =
f
55 mm
=
= 2.5 mm
22
( f -number )max
Dmax =
f
55 mm
=
= 46 mm
1.2
( f -number )min
to
25.3
The thin-lens equation, 1 + 1 = 1 , gives the image distance as
p q f
q=
pf
(100 m )( 52.0 mm )
=
= 52.0 mm
p − f 100 m − 52.0 × 10 −3 m
From the magnitude of the lateral magnification, M = h ′ h = −q p , where the height of the
image is h ′ = 0.092 0 m = 92.0 mm, the height of the object (the building) must be
h = h′ −
68719_25_ch25_p282-304.indd 286
p
100 m
= ( 92.0 mm ) −
= 177 m
q
52.0 mm
1/7/11 2:55:16 PM
Optical Instruments
25.4
287
Consider rays coming from
opposite edges of the object
and passing undeviated through
the center of the lens as shown
at the right. For a very distant
object, the image distance
equals the focal length of the
lens. If the angular width of the
object is q, the full image width
on the film is
⎛ 20° ⎞
h = 2 ⎡⎣ f tan (q 2 ) ⎤⎦ = 2 ( 55.0 mm ) tan ⎜
= 19 mm
⎝ 2 ⎟⎠
so the image easily fits within a 23.5 mm by 35.0 mm area.
25.5
The exposure time is being reduced by a factor of
t2 1 125 s 15
3
=
=
=
1 15 s 125 25
t1
Thus, to maintain correct exposure, the intensity of the light reaching the film should be increased
by a factor of 25 3. This is done by increasing the area of the aperture by a factor of 25 3, so in
terms of the diameter, p D22 4 = ( 25 3) p D12 4 , or D2 = D1 25 3.
(
)
The new f-number will be
( f -number )2 =
25.6
(a)
f
f
3
4.0 3
=
= ( f -number )1
=
= 1.4
D2 D1 25 3
25
5
or
f 1.4
The intensity is a measure of the
rate at which energy is received by
the film per unit area of the image,
or I ∝ 1 Aimage . Consider an object
with horizontal and vertical dimensions hx and hy as shown at the right.
If the vertical dimension intercepts
angle q , the vertical dimension of
the image is hy′ = qq , or hy′ ∝ q.
Similarly for the horizontal dimension, hx′ ∝ q, and the area of the image is Aimage = hx′ hy′
Assuming a very distant object, q ≈ f , so Aimage ∝ f 2, and we conclude that I ∝ 1 f 2 .
∝ q 2.
The intensity of the light reaching the film is also proportional to the cross-sectional area
of the lens and hence to the square of the diameter of that lens, or I ∝ D 2. Combining this
with our earlier conclusion gives
I
(b)
∝
D2
1
=
f 2 ( f D )2
or
I
∝
1
f
-number
(
)2
The total light energy delivered to the film is proportional to the product of intensity and exposure
time, It. Thus, to maintain correct exposure, this product must be kept constant, or I 2 t2 = I1t1 , giving
2
⎡ ( f -number )2 ⎤
⎛I ⎞
⎛ 4.0 ⎞ ⎛ 1 ⎞
t2 = ⎜ 1 ⎟ t1 = ⎢ 2
t
=
s ≈ 1 100 s
⎥
⎜
⎟
1
2
⎝ 1.8 ⎠ ⎜⎝ 500 ⎟⎠
⎝ I2 ⎠
⎢⎣ ( f1 -number ) ⎥⎦
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288
25.7
Chapter 25
Since the exposure time is unchanged, the intensity of the light reaching the film must be doubled
if the energy delivered is to be doubled. Using the result of Problem 25.6 (part a), we obtain
⎛
⎞
( f2 -number )2 = ⎜ II1 ⎟ ( f1 -number )2 = ⎛⎜⎝ 12 ⎞⎟⎠ (11)2 = 61,
⎝
2
⎠
or
f2 -number = 61 = 7.8
Thus, you should use the f 8.0 setting on the camera.
25.8
The image must always be focused on the film, so the image distance is the distance between
the lens and the film. From the thin-lens equation, 1 p + 1 q = 1 f , the object distance is
p = qf (q − f ), and the range of object distances this camera can work with is from
pmin =
qmax f
( 210 mm )(175 mm )
=
= 1.05 × 10 3 mm = 1.05 m
qmax − f
210 mm − 175 mm
pmax =
qmin f
(180 mm )(175 mm )
= 6.30 × 10 3 mm = 6.30 m
=
qmin − f
180 mm − 175 mm
to
25.9
The corrective lens must form an upright, virtual image at the near point of the eye (i.e.,
q = − 60.0 cm in this case) for objects located 25.0 cm in front of the eye (p = +25.0 cm). From
the thin-lens equation, 1 p + 1 q = 1 f , the required focal length of the corrective lens is
f =
pq
( 25.0 cm )( −60.0 cm )
=
= + 42.9 cm
p+q
25.0 cm − 60.0 cm
and the power (in diopters) of this lens will be
P=
25.10
1
1
=
= + 2.33 diopters
fin meters + 0.429 m
(a)
The person is farsighted , able to see distant objects but unable to focus on objects at the
normal near point for a human eye.
(b)
With the corrective lens 2.00 cm in front of the eye, the object distance for an object
20.0 cm in front of the eye is p = 20.0 cm − 2.00 cm = 18.0 cm .
(c)
The upright, virtual image formed by the corrective lens will serve as the object
for the eye, and this object must be 40.0 cm in front of the eye. With the lens
2.00 cm in front of the eye, the magnitude of the image distance for the lens will be
q = 40.0 cm − 2.00 cm = 38.0 cm .
(d)
The image must be located in front of the corrective lens, so it is a virtual image , and the
image distance is negative . Thus, q = −38.0 cm.
(e)
From the thin-lens equation, 1 p + 1 q = 1 f , the required focal length of the corrective lens is
f =
(f )
pq
(18.0 cm )( −38.0 cm )
= + 34.2 cm
=
18.0 cm − 38.0 cm
p+q
The power of the corrective lens is then
P=
1
fin meters
=
1
= + 2.92 diopters
+ 0.342 m
continued on next page
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Optical Instruments
(g)
289
With a contact lens, the lens-to-eye distance would be zero, so we would have
p = 20.0 cm and q = − 40.0 cm , giving a required focal length of
f =
( 20.0 cm ) ( − 40.0 cm )
pq
= + 40.0 cm
=
20.0 cm − 40.0 cm
p+q
and a power in diopters of
P=
25.11
1
fin meters
=
1
= + 2.50 diopters
+ 0.400 m
His lens must form an upright, virtual image of a very distant object (p ≈ ∞) at his far point,
80.0 cm in front of the eye. Therefore, the focal length is f = q = −80.0 cm.
If this lens is to form a virtual image at his near point (q = −18.0 cm), the object distance must be
p=
25.12
(a)
qf
( −18.0 cm ) ( − 80.0 cm ) = 23.2 cm
=
q − f −18.0 cm − ( − 80.0 cm )
When the child clearly sees objects at her far point ( pmax = 125 cm ) , the lens-cornea
combination has assumed a focal length suitable for forming the image on the retina
( q = 2.00 cm ). The thin-lens equation gives the optical power under these conditions as
Pfar =
1
1 1
1
1
= + =
+
= + 50.8 diopters
fin meters p q 1.25 m 0.020 0 m
When the eye is focused ( q = 2.00 cm ) on objects at her near point ( pmin = 10.0 cm ) , the
optical power of the lens-cornea combination is
Pnear =
1
1 1
1
1
= + =
+
= + 60.0 diopters
fin meters p q 0.100 m 0.020 0 m
Therefore, the range of the power of the lens-cornea combination is
+50.8 diopters ≤ P ≤ +60.0 diopters
(b)
If the child is to see very distant objects ( p → ∞ ) clearly, her eyeglass lens must form an
erect virtual image at the far point of her eye ( q = −125 cm ) . The optical power of the
required lens is
P=
1
1 1
1
= + = 0+
= − 0.800 diopters
fin meters p q
−1.25 m
Since the power, and hence the focal length, of this lens is negative, it is diverging .
25.13
(a)
The lens should form an upright, virtual image at the far point ( q = − 50.0 cm ) for very
distant objects ( p ≈ ∞ ) . Therefore, f = q = − 50.0 cm, and the required power is
P=
1
1
=
= − 2.00 diopters
f − 0.500 m
continued on next page
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290
Chapter 25
(b)
If this lens is to form an upright, virtual image at the near point of the unaided eye
( q = −13.0 cm ) , the object distance should be
p=
25.14
qf
( −13.0 cm ) ( − 50.0 cm ) = 17.6 cm
=
q − f −13.0 cm − ( − 50.0 cm )
(a)
Yes, a single lens can correct the patient's vision . The patient needs corrective action in
both the near vision (to allow clear viewing of objects between 45.0 cm and the normal
near point of 25 cm) and the distant vision (to allow clear viewing of objects more than
85.0 cm away). A single lens solution is for the patient to wear a bifocal or progressive
lens. Alternately, the patient must purchase two pairs of glasses, one for reading, and one
for distant vision.
(b)
To correct the near vision, the lens must form an upright, virtual image at the patient’s near
point (q = − 45.0 cm) when a real object is at the normal near point ( p = +25.0 cm). The
thin-lens equation gives the needed focal length as
f =
pq
( 25.0 cm )( −45.0 cm )
= + 56.3 cm
=
25.0 cm − 45.0 cm
p+q
so the required power in diopters is
P=
(c)
fin meters
1
= +1.78 diopters
0.563 m
1
fin meters
=
1
= −1.18 diopters
− 0.850 m
Considering the image formed by the cornea as a virtual object for the implanted lens, the object
distance for this lens is p = − 5.33 cm, and the image distance is q = + 2.80 cm. The thin-lens
equation then gives the focal length of the implanted lens as
f =
pq
( − 5.33 cm ) ( 2.80 cm ) = + 5.90 cm
=
p+q
− 5.33 cm + 2.80 cm
so the power is P =
25.16
=
To correct the distant vision, the lens must form an upright, virtual image at the patient’s far
point (q = − 85.0 cm) for the most distant objects (p → ∞). The thin-lens equation gives the
needed focal length as f = q = − 85.0 cm, so the needed power is
P=
25.15
1
(a)
1
1
=
= +17.0 diopters .
f + 0.059 0 m
The upper portion of the lens should form an upright, virtual image of very distant
objects ( p ≈ ∞ ) at the far point of the eye ( q = −1.5 m ). The thin-lens equation then
gives f = q = −1.5 m, so the needed power is
P=
1
1
=
= − 0.67 diopters
f −1.5 m
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Optical Instruments
(b)
The lower part of the lens should form an upright, virtual image at the near point of the eye
( q = − 30 cm ) when the object distance is p = 25 cm. From the thin-lens equation,
f =
( 25 cm ) ( − 30 cm )
pq
=
= +1.5 × 10 2 cm = +1.5 m
p+q
25 cm − 30 cm
Therefore, the power is P =
25.17
(a)
f =
1
fin meters
=
1
= − 2.50 diopters
− 0.400 m
1
1
=
= − 0.250 m = − 25.0 cm
P − 4.00 diopters
(b)
The corrective lens forms virtual images of very distant objects (p → ∞) at
q = f = −25.0 cm. Thus, the person must be very nearsighted , unable to see objects
clearly when they are more than (25.0 + 2.00) cm = 27.0 cm from the eye.
(c)
If contact lenses are to be worn, the far point of the eye will be 27.0 cm in front of the lens,
so the needed focal length will be f = q = −27.0 cm, and the power is
P=
25.19
1
1
=
= + 0.67 diopters .
f +1.5 m
The corrective lens should form an upright, virtual image at the woman’s far point
(q = − 40.0 cm) for a very distant object (p → ∞). The thin-lens equation gives the required
focal length as f = q = − 40.0 cm = − 0.400 m. Since f < 0, it is a diverging lens , and the
required power is
P=
25.18
291
(a)
1
fin meters
=
1
= − 3.70 diopters
− 0.270 m
The simple magnifier (a converging lens) is to form an upright, virtual image located 25 cm
in front of the lens ( q = −25 cm ) . The thin-lens equation then gives
p=
qf
( −25 cm )( 7.5 cm )
=
= +5.8 cm
q− f
−25 cm − 7.5 cm
so the stamp should be placed 5.8 cm in front of the lens .
(b)
When the image is at the near point of the eye, the angular magnification produced by the
simple magnifier is
m = mmax = 1 +
25.20
(a)
25 cm
25 cm
= 1+
= 4.3
f
7.5 cm
A simple magnifier produces maximum magnification when it forms an upright, virtual
image at the near point of the eye (25.0 cm for a normal eye). If the focal length of the
magnifier is f = +5.00 cm, the required object distance for maximum magnification with a
normal eye is
p=
qf
( −25.0 cm )( +5.00 cm )
=
= + 4.17 cm
q− f
−25.0 cm − 5.00 cm
or the object should be placed 4.17 cm in front of the lens .
continued on next page
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292
Chapter 25
(b)
Under the conditions described in part (a), the angular magnification produced is
m = mmax = 1 +
25.21
(a)
From the thin-lens equation,
f =
(b)
( 3.50 cm ) ( − 25.0 cm )
pq
= + 4.07 cm
=
3.50 cm − 25.0 cm
p+q
With the image at the normal near point, the angular magnification is
m = mmax = 1 +
25.22
(a)
25 cm 25 cm
=
= + 5.0
f
5.0 cm
When the object is positioned so the magnifier forms a virtual image at the near point of the
eye (q = −25 cm), maximum magnification is produced and this is
mmax = 1 +
(c)
(a)
25 cm
25 cm
= 1+
= + 6.0
f
5.0 cm
From the thin-lens equation, the object distance needed to yield the maximum magnification computed in part (b) above is
p=
25.23
25.0 cm
25.0 cm
= 1+
= + 7.14
f
4.07 cm
When the object is at the focal point of the magnifying lens, a virtual image is formed at
infinity and parallel rays emerge from the lens. Under these conditions, the eye is most
relaxed and the magnification produced is
m=
(b)
25.0 cm
25.0 cm
= 1+
= 6.00
f
5.00 cm
qf
( −25 cm )( 5.0 cm )
=
= + 4.2 cm
q− f
−25 cm − 5.0 cm
From the thin-lens equation, a real inverted image is formed at an image distance of
q=
pf
( 71.0 cm )( 39.0 cm )
=
= + 86.5 cm
p− f
71.0 cm − 39.0 cm
so the lateral magnification produced by the lens is
M=
h′
q
86.5 cm
=− =−
= −1.22
h
p
71.0 cm
and the magnitude is M = 1.22 .
(b)
If h is the actual length of the leaf, the small-angle approximation gives the angular width of the
leaf when viewed by the unaided eye from a distance of d = 126 cm + 71.0 cm = 197 cm as
q ≈
h
h
=
d 197 cm
continued on next page
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Optical Instruments
293
The length of the image formed by the lens is h ′ = M h = 1.22 h , and its angular width
when viewed from a distance of d ′ = 126 cm − q = 39.5 cm is
q′ ≈
h′
1.22 h
=
d ′ 39.5 cm
The angular magnification achieved by viewing the image instead of viewing the leaf
directly is
q ′ 1.22 h 39.5 cm 1.22 (197 cm )
≈
=
= 6.08
q
h 197 cm
39.5 cm
25.24
(a)
When a converging lens is used as a simple magnifier, maximum magnification is obtained
when the upright, virtual image is formed at the near point of the eye ( q = 25.0 cm for a
normal eye). For the given lens, this maximum magnification is
mmax = 1 +
(b)
When this simple magnifier is positioned for relaxed viewing (virtual image formed at
infinity), the magnification produced is
mrelaxed =
25.25
M1 ( 25 cm ) ( −12 ) ( 25 cm )
=
= 2.1 cm
m
−140
The approximate overall magnification of a compound microscope is given by
m = −(L fo )(25.0 cm fe ), where L is the distance between the objective and eyepiece lenses,
while fo and fe are the focal lengths of the objective and eyepiece lenses, respectively. Thus, the
described microscope should have an approximate overall magnification of
m=−
25.27
L
fo
⎛ 25.0 cm ⎞
⎛ 20.0 cm ⎞ ⎛ 25.0 cm ⎞
= −⎜
= −588
⎜⎝
⎝ 0.500 cm ⎟⎠ ⎜⎝ 1.70 cm ⎟⎠
fe ⎟⎠
The angular magnification produced by a telescope is m = fo fe , where fo is the focal length of
the objective element, and fe is that of the eyepiece lens. Also, the optical power of a lens is
P = 1 fmeters, where fmeters is the focal length of that lens, expressed in meters. Thus, fo = 1 Po and
fe = 1 Pe, so the angular magnification of this telescope is
m=
25.28
25.0 cm 25.0 cm
=
= 1.39
f
18.0 cm
The overall magnification is m = M1me = M1 ( 25 cm fe ) , where M1 is the lateral magnification
produced by the objective lens. Therefore, the required focal length for the eyepiece is
fe =
25.26
25.0 cm
25.0 cm
= 1+
= 2.39
f
18.0 cm
fo 1 Po Pe 35.0 diopters
=
=
=
= 12.7
fe 1 Pe Po 2.75 diopters
It is specified that the final image the microscope forms of the blood cell is 29.0 cm in front of
the eye and that the diameter of this image intercepts an angle of q = 1.43 mrad . The diameter of
this final image must then be
(
)(
)
he = rq = 29.0 × 10 −2 m 1.43 × 10 −3 rad = 4.15 × 10 −4 m
continued on next page
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294
Chapter 25
At this point, it is tempting to use Equation 25.7 from the textbook for the overall magnification of a
compound microscope and compute h = he m as the size of the blood cell serving as the object for the
microscope. However, the derivation of that equation is based on several assumptions, one of which
is that the eye is relaxed and viewing a final image located an infinite distance in front of the eyepiece.
This is clearly not true in this case, and the use of Equation 25.7 would introduce considerable error.
Instead, we shall return to basics and use the thin-lens equation to find the size of the original object.
The image formed by the objective lens is the object for the eyepiece, and we label the size of this
image as h ′. The lateral magnification of the objective lens is M1 = h ′ h = − q1 p1 and that of the
eyepiece is Me = he h ′ = − qe pe. The overall magnification produced by the microscope is
M total =
he ⎛ h ′ ⎞ ⎛ he ⎞
=⎜ ⎟⎜ ⎟
h ⎝ h ⎠ ⎝ h′ ⎠
which gives the size of the original object as h = he M total .
From the thin-lens equation, the required object distance for the eyepiece is
pe =
qe fe
( − 29.0 cm ) ( 0.950 cm ) = 0.920 cm
=
qe − fe
− 29.0 cm − 0.950 cm
and the magnification produced by the eyepiece is
Me = −
qe
( − 29.0 cm ) = + 31.5
=−
pe
0.920 cm
The image distance for the objective lens is then
q1 = L − pe = 29.0 cm − 0.920 cm = 28.1 cm
and the object distance for this lens is
p1 =
q1 fo
( 28.1 cm )(1.622 cm )
=
= 1.72 cm
q1 − fo
28.1 cm − 1.622 cm
The magnification by the objective lens is
M1 = −
q1
( 28.1 cm )
=−
= −16.3
p1
1.72 cm
and the overall lateral magnification is M total = M1 Me = ( −16.3) ( + 31.5) = − 513.
The actual diameter of the red blood cell serving as the original object is found to be
h=
25.29
he
4.15 × 10 −4 m
=
= 8.09 × 10 −7 m = 0.809 mm
513
M total
The sketch at the right shows an image
of the nebula formed by an objective
lens of diameter Do and focal length fo .
The diameter of the image of the nebula
formed on the film is given by
h ′ = fo ⋅q
Objective,
diameter = Do
nebula
h
image
q
q
h′
fo
continued on next page
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Optical Instruments
295
where q is the angular width of the nebula and its image as shown. If the intensity of the light
arriving at the objective lens from the nebula is I, the energy entering the lens during an exposure
time Δt is
⎛ p Do2 ⎞
E = IAobjective ( Δt ) = I ⎜
Δt
⎝ 4 ⎟⎠
The energy per unit area deposited on the film during the exposure time is then
2
4E
E
4 ⎛ p IDo2 Δt ⎞ ⎛ I ⎞ ⎛ Do ⎞
=
=
=
=⎜ ⎟
Δt
p (h ′)2 4 p ( fo ⋅q )2 pq 2 ⎜⎝ 4 fo2 ⎟⎠ ⎝ q 2 ⎠ ⎜⎝ fo ⎟⎠
E
Aimage
Telescope 1 has an objective diameter DO,1 = 200 mm, focal length fO,1 = 2 000 mm, and uses
an exposure time (Δt)1 = 1.50 min. If telescope 2, with DO,2 = 60.0 mm, and fO,2 = 900 mm, is to
deposit the same energy per unit area on the film as does telescope 1, it is necessary that
E
Aimage,2
⎛ I ⎞ ⎛ DO,2 ⎞
=⎜ 2⎟⎜
⎝ q ⎠ ⎝ fO,2 ⎟⎠
2
( Δt )2
⎛ I ⎞ ⎛ DO,1 ⎞
=⎜ 2⎟⎜
⎝ q ⎠ ⎝ fO,1 ⎟⎠
2
( Δt )1 =
E
Aimage,1
The required exposure time for the second telescope is therefore
( Δt )2 =
(I q 2 ) (DO ,1 fO ,1 )2
(I q 2 ) (DO , 2 fO , 2 )
2
⎛ DO ,1 ⎞ ⎛ fO , 2 ⎞
⎟ ⎜
⎟
⎝ fO ,1 ⎠ ⎝ DO , 2 ⎠
( Δt )1 = ⎜
2
2
( Δt )1
2
⎛ 200 mm ⎞ ⎛ 900 mm ⎞ 2
=⎜
⎜
⎟ (1.50 min ) = 3.38 min
⎝ 2 000 mm ⎟⎠ ⎝ 60.0 mm ⎠
25.30
(a)
For a refracting telescope, the overall length is L = fo + fe , and the magnification produced
is m = fo fe , where fo and fe are the focal lengths of the objective element and the eyepiece, respectively. Thus, we may write fe = fo m to obtain
L = fo +
(b)
fo
1⎞
⎛ m + 1⎞
⎛
= fo ⎜ 1 + ⎟ = fo ⎜
⎝ m ⎟⎠
⎝
m
m⎠
Using the result of part (a), the required change in the length of the telescope will be
⎛ m′ + 1 m + 1⎞
⎛ 101 51.0 ⎞
−2
ΔL = fo ⎜
−
−
⎟ = −2.00 × 10 m = −2.00 cm
⎟⎠ = ( 2.00 m ) ⎜⎝
⎝ m′
m
100 50.0 ⎠
or the telescope must be shortened by moving the eyepiece 2.00 cm forward toward the
objective lens.
25.31
(a)
The magnification of a telescope is m = fo fe , where fo and fe are the focal lengths of the
objective and eyepiece lenses, respectively. Thus, if m = 34.0, while fo = 86.0 cm, the focal
length of the eyepiece must be
fe =
(b)
fo 86.0 cm
=
= 2.53 cm
m
34.0
When the telescope is adjusted for relaxed-eye viewing, the distance between the objective
and eyepiece lenses is
L = fo + fe = 86.0 cm + 2.53 cm = 88.5 cm
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296
25.32
Chapter 25
The Moon may be considered an
infinitely distant object ( p → ∞ ) when
viewed with this lens, so the image
distance will be q = fo = 1 500 cm.
Considering the rays that pass
undeviated through the center of this
lens as shown in the sketch, observe
that the angular widths of the image and the object are equal. Thus, if w is the linear width of an
object forming a 1.00-cm-wide image, then
q =
or
25.33
(a)
w
1.0 cm
1.0 cm
=
=
3.8 × 108 m
fo
1 500 cm
⎛ 1.0 cm ⎞ ⎛ 1 mi ⎞
w = 3.8 × 108 m ⎜
= 1.6 × 10 2 mi
⎝ 1 500 cm ⎟⎠ ⎜⎝ 1 609 m ⎟⎠
(
)
From the thin-lens equation, q = p f ( p − f ), so the lateral magnification by the objective
lens is M = h ′ h = − q p = − f ( p − f ). Therefore, the image size will be
h′ = M h = −
fh
.
p
(b)
If p >> f , then f − p ≈ − p and h ′ ≈ −
(c)
Suppose the telescope observes the space station at the zenith. Then,
h′ ≈ −
25.34
fh
fh
=
p− f
f−p
fh
( 4.00 m )(108.6 m )
=−
= −1.07 × 10 −3 m = −1.07 mm
p
407 × 10 3 m
Use the larger focal length (lowest power) lens as the objective element and the shorter focal
length (largest power) lens for the eyepiece. The focal lengths are
fo =
(a)
1
1
= + 0.833 m, and fe =
= + 0.111 m
+1.20 diopters
+ 9.00 diopters
The angular magnification (or magnifying power) of the telescope is then
m=
(b)
fo + 0.833 m
=
= 7.50
fe + 0.111 m
The length of the telescope is
L = fo + fe = 0.833 m + 0.111 m = 0.944 m
25.35
The lens for the left eye forms an upright, virtual image at qL = − 50.0 cm when the object
distance is pL = 25.0 cm, so the thin-lens equation gives its focal length as
fL =
( 25.0 cm ) ( − 50.0 cm )
pL q L
= 50.0 cm
=
pL + qL
25.0 cm − 50.0 cm
Similarly for the other lens, qR = −100 cm when pR = 25.0 cm, and f R = 33.3 cm.
continued on next page
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Optical Instruments
(a)
Using the lens for the left eye as the objective,
fo
f
50.0 cm
= L =
= 1.50
fe
f R 33.3 cm
m=
(b)
297
Using the lens for the right eye as the eyepiece and, for maximum magnification, requiring
that the final image be formed at the normal near point ( qe = − 25.0 cm ) gives the object
distance for the eyepiece as
pe =
qe fe
( − 25.0 cm ) (33.3 cm ) = +14.3 cm
=
qe − fe
− 25.0 cm − 33.3 cm
The maximum magnification by the eyepiece is then
me = 1 +
25.0 cm
25.0 cm
= 1+
= +1.75
fe
33.3 cm
and the image distance for the objective is
q1 = L − pe = 10.0 cm − 14.3 cm = − 4.3 cm
The thin-lens equation then gives the object distance for the objective as
p1 =
q1 f1
( − 4.3 cm ) (50.0 cm ) = + 4.0 cm
=
q1 − f1
− 4.3 cm − 50.0 cm
The magnification by the objective is then
M1 = −
q1
( − 4.3 cm ) = +1.1
=−
p1
4.0 cm
and the overall magnification is m = M1me = ( +1.1) ( +1.75) = 1.9 .
25.36
Note: We solve part (b) before answering part (a) in this problem.
(b)
The objective forms a real, inverted
image, diminished in size, of a very
distant object at q1 = fo . This image
is a virtual object for the eyepiece at
pe = − fe , giving
1
1
1
1
1
=
− =
+
=0
qe pe fe − fe
fe
and
(a)
Fo
qo
qo
Fo
I
L1
qe → ∞
Parallel rays emerge from the
eyepiece, so the eye observes a
virtual image .
Fe
q
Fe
O
L2
continued on next page
68719_25_ch25_p282-304.indd 297
1/7/11 2:55:42 PM
298
Chapter 25
(c)
The angular magnification is m =
fo
= 3.00, giving fo = 3.00 fe . Also, the length of the
fe
telescope is L = fo + fe = 3.00 fe − fe = 10.0 cm, giving
fe = − fe = −
25.37
The angular separation of the lights is q = d h , where d = 1.00 m is their linear separation, and
h is the altitude of the satellite. If the lights are just resolved according to the Rayleigh criterion,
then q = q min = 1.22 (l D) , where l is the wavelength of the light, and D is the diameter of the
lens. Thus, the altitude of the satellite must be
h=
25.38
d
d
d⋅D
(1.00 m )( 0.300 m )
=
=
=
= 4.92 × 10 5 m = 492 km
q 1.22(l D) 1.22l 1.22 500 × 10 −9 m
(
)
The angular separation of two objects seen on the ground is q = d h , where d is their linear
separation and h is your altitude. If the objects are just resolved according to the Rayleigh
criterion, then q = q min = 1.22 (l D) , where l is the wavelength of the light, and D is the diameter
of the aperture (your pupil). Thus, the minimum separation of objects you can distinguish is
d = h ⋅q min =
25.39
10.0 cm
= − 5.00 cm and fo = 3.00 fe = 15.0 cm
2.00
(
)(
)
3
−9
1.22 ⋅ h ⋅ l 1.22 9.50 × 10 m 575 × 10 m
=
= 1.7 m
D
4.0 × 10 −3 m
The limit of resolution in air is q m in
air
= 1.22
l
= 0.60 mrad.
D
In oil, the limiting angle of resolution will be
q min oil = 1.22
or
25.40
(a)
q min oil =
( l noil ) = ⎛ 1.22 l ⎞ 1
loil
= 1.22
⎜⎝
⎟
D
D
D ⎠ noil
q min air
noil
(
)
500 × 10 −9 m
ln
l
= 1.22
= 1.22
= 2.29 × 10 −4 rad
D
nD
(1.33) 2.00 × 10 −3 m
(
)
From s = rq , the distance from the eye that two points separated by a distance s = 1.00 cm
will intercept this minimum angle of resolution is
r=
25.41
0.60 mrad
= 0.40 mrad
1.5
The wavelength of the light within the eye is ln = l n. Thus, the limiting angle of resolution
for light passing through the pupil (a circular aperture with diameter D = 2.00 mm) is
q min = 1.22
(b)
=
s
1.00 cm
=
= 4.37 × 10 3 cm = 43.7 m
q min 2.29 × 10 − 4 rad
The angular separation of the headlights when viewed from a distance of r = 10.0 km is
q =
s
2.00 m
=
= 2.00 × 10 −4 rad
r 10.0 × 10 3 m
continued on next page
68719_25_ch25_p282-304.indd 298
1/7/11 2:55:44 PM
Optical Instruments
299
If the headlights are to be just resolved, this separation must equal the limiting angle of resolution
for the circular aperture, q min = 1.22l D, so the diameter of the aperture is
D=
25.42
(
The angular separation of the two stars is q = d r , where d is their linear separation and r is
their distance from Earth. If the stars are just resolved according to the Rayleigh criterion, then
q = q min = 1.22 (l D) , where l is the wavelength of the light, and D is the diameter of the
aperture (telescope objective). Thus, the linear separation of the stars must be
d = r ⋅q min =
25.43
)
−9
1.22l 1.22l 1.22 885 × 10 m
=
=
= 5.40 × 10 −3 m = 5.40 mm
q min
q
2.00 × 10 −4 rad
(
)
(
)
15
−9
r ⋅1.22l ⎡⎣( 23 ly ) 9.461 × 10 m ly ⎤⎦ 1.22 575 × 10 m
= 2.2 × 1011 m
=
0.68 m
D
If just resolved, the angular separation of the objects is q = q min = 1.22
l
D
⎡
⎛ 500 × 10 −9 m ⎞ ⎤
and s = r q = 8.0 × 10 7 km ⎢1.22 ⎜
⎥ = 9.8 km
⎝ 5.00 m ⎟⎠ ⎦
⎣
(
25.44
)
If just resolved, the angular separation of the objects is q = q min = 1.22
λ
D
⎡
⎛ 550 × 10 −9 m ⎞ ⎤
and s = r q = 200 × 10 3 m ⎢1.22 ⎜
⎥ = 0.38 m = 38 cm
⎝ 0.35 m ⎟⎠ ⎦
⎣
(
25.45
)
The grating spacing is d = 1 cm 6 000 = 1.67 × 10 −4 cm = 1.67 × 10 −6 m, and the highest order of
600 nm light that can be observed is
mmax
(
)
1.67 × 10 −6 m (1)
d sin 90°
=
=
= 2.78 → 2 orders
l
600 × 10 −9 m
The total number of slits is N = (15.0 cm ) ( 6 000 slits cm ) = 9.00 × 10 4 , and the resolving power
of the grating in the second order is
(
)
Ravailable = Nm = 9.00 × 10 4 2 = 1.80 × 10 5
The resolving power required to separate the given spectral lines is
Rneeded =
l
600.000 nm
=
= 2.0 × 10 5
Δl
0.003 nm
These lines cannot be separated with this grating.
25.46
The resolving power of a diffraction grating is R = l Δl = Nm.
(a)
The number of lines the grating must have to resolve the Ha line in the first order is
N=
(b)
68719_25_ch25_p282-304.indd 299
R l Δl 656.2 nm
=
=
= 3.6 × 10 3 lines
m
0.18 nm
(1)
In the second order ( m = 2 ) , N =
R
656.2 nm
=
= 1.8 × 10 3 lines .
2 2 ( 0.18 nm )
1/7/11 2:55:46 PM
300
25.47
Chapter 25
A fringe shift occurs when the mirror moves distance l 4 . Thus, if the mirror moves distance
ΔL = 0.180 mm, the number of fringe shifts observed is
N shifts
25.48
(a)
(
)
−3
ΔL 4 ( ΔL ) 4 0.180 × 10 m
=
=
=
= 1.31 × 10 3
l 4
l
550 × 10 −9 m
When the central spot in the interferometer pattern goes through a full cycle from bright
to dark and back to bright, two fringe shifts have occurred, and the movable mirror
has moved a distance of 2(l 4) = l 2. Thus, if N cycles = 1 700 such cycles are observed
as the mirror moves distance d = 0.382 mm, it must be true that d = N cycles ( l 2 ) , or
l = 2d N cycles. The wavelength of the light illuminating the interferometer is therefore
l=
(
2 0.382 × 10 −3 m
1 700
) = 4.49 × 10
−7
m = 449 nm
which is in the blue region of the visible spectrum.
(b)
25.49
Red light has a longer wavelength than blue light, so fewer wavelengths would cover the
given displacement. Hence, N cycles would be smaller .
A fringe shift occurs when the mirror moves distance l 4 . Thus, the distance moved (length of
the bacterium) as 310 shifts occur is
⎛ 650 × 10 −9 m ⎞
⎛l⎞
−5
ΔL = N shifts ⎜ ⎟ = 310 ⎜
⎟⎠ = 5.04 × 10 m = 50.4 mm
⎝ 4⎠
4
⎝
25.50
A fringe shift occurs when the mirror moves distance l 4 . Thus, the distance the mirror
moves as 250 fringe shifts are counted is
⎛ 632.8 × 10 −9 m ⎞
⎛l⎞
−5
ΔL = N shifts ⎜ ⎟ = 250 ⎜
⎟⎠ = 3.96 × 10 m = 39.6 mm
⎝ 4⎠
4
⎝
25.51
When the optical path in one arm of a Michelson’s interferometer increases by one
wavelength, four fringe shifts will occur (one shift for every quarter-wavelength change
in path length).
The number of wavelengths (in a vacuum) that fit in a distance equal to a thickness t is
N vac = t l . The number of wavelengths that fit in this thickness while traveling through
the transparent material is N n = t ln = t ( l n ) = nt l . Thus, the change in the number
of wavelengths that fit in the path down this arm of the interferometer is
ΔN = N n − N vac = ( n − 1)
t
l
and the number of fringe shifts that will occur as the thin sheet is inserted will be
# fringe shifts = 4 ( ΔN ) = 4 ( n − 1)
68719_25_ch25_p282-304.indd 300
⎛ 15.0 × 10 −6 m ⎞
t
= 4 (1.40 − 1) ⎜
= 40
l
⎝ 600 × 10 −9 m ⎟⎠
1/7/11 2:55:49 PM
Optical Instruments
25.52
301
The wavelength of light within the tube decreases from l to ln = l n gas as the tube fills with
gas. Thus, the number of wavelengths that will fit in the length L of the tube increases from
L l to ngas L l . Four fringe shifts occur for each additional wavelength that fits within the tube,
so the number of fringes shifts to be seen as the tube fills with gas is
⎡ ngas L L ⎤ 4L
N shifts = 4 ⎢
− ⎥=
ngas − 1
l⎦ l
⎣ l
(
⎡ 600 × 10 −9 m
⎛ l ⎞
⎢
ngas = 1 + ⎜
=
1
+
N
−2
⎝ 4L ⎟⎠ shifts
⎢⎣ 4 5.00 × 10 m
Hence,
25.53
(a)
(
25.54
)
⎤
⎥ (160 ) = 1.000 5
⎥⎦
For a refracting telescope, the magnification is m = fo fe , where fo and fe are the focal
lengths of the objective lens and the eyepiece, respectively. Thus, when the Yerkes
telescope uses an eyepiece with fe = 2.50 cm, the magnification is
m=
(b)
)
fo
20.0 m
= 8.00 × 10 2 = 800
=
fe 2.50 × 10 −2 m
Standard astronomical telescopes form inverted images. Thus, the observer Martian polar
caps are upside down .
When viewed from a distance of 50 meters, the angular length of a mouse (assumed to have an
actual length of ≈ 10 cm) is
q =
s 0.10 m
=
= 2.0 × 10 −3 radians
r
50 m
Thus, the limiting angle of resolution of the eye of the hawk must be
q min ≈ 2.0 × 10 −3 radians
25.55
With 485 lines equally spaced in a height C, the distance separating adjacent lines is d = C 485.
When the screen is viewed from a distance L, the angular separation between adjacent lines
is q = d L . If the individual lines are not to be seen (i.e., the lines are to be unresolved), this
angular separation must be less than the minimum angle of resolution, q min = 1.22(l D) by the
Rayleigh criterion. That is, we must have
q =
or
25.56
(a)
d C 485
1.22 ⋅ l
=
< q min =
L
L
D
L
D
5.00 × 10 −3 m
>
=
= 15.4
C 485 (1.22 ⋅ l ) 485 (1.22 ) 550 × 10 −9 m
(
)
Since this eye can already focus on objects located at the near point of a normal eye
(25 cm), no correction is needed for near objects. To correct the distant vision, a corrective
lens (located 2.0 cm from the eye) should form virtual images of very distant objects
at 23 cm in front of the lens (or at the far point of the eye). Thus, we must require that
q = −23 cm when p → ∞. This gives
P=
1 1 1
1
= + = 0+
= − 4.3 diopters
f
p q
− 0.23 m
continued on next page
68719_25_ch25_p282-304.indd 301
1/7/11 2:55:51 PM
302
Chapter 25
(b)
A corrective lens in contact with the cornea should form virtual images of very distant objects
at the far point of the eye. Therefore, we require that q = −25 cm when p → ∞, giving
P=
1 1 1
1
= + = 0+
= − 4.0 diopters
f
p q
− 0.25 m
1
⎛
⎞
When the contact lens ⎜ f = = − 25 cm ⎟ is in place, the object distance which yields a
⎝
⎠
P
virtual image at the near point of the eye (that is, q = −16 cm) is given by
p=
25.57
(a)
qf
( −16 cm )( −25 cm )
=
= 44 cm
q − f −16 cm − ( −25 cm )
The lens should form an upright, virtual image at the near point of the eye, q = −75.0 cm,
when the object distance is p = 25.0 cm. The thin-lens equation then gives
f =
pq
( 25.0 cm )( −75.0 cm )
=
= 37.5 cm = 0.375 m
p+q
25.0 cm − 75.0 cm
so the needed power is P =
(b)
1
1
=
= + 2.67 diopters .
f 0.375 m
If the object distance must be p = 26.0 cm to position the image at q = −75.0 cm, the actual
focal length is
f =
and P =
pq
( 26.0 cm )( −75.0 cm )
=
= 0.398 m
p+q
26.0 cm − 75.0 cm
1
1
=
= + 2.51 diopters
f 0.398 m
The error in the power is
ΔP = ( 2.67 − 2.51) diopters = 0.16 diopters too low
25.58
(a)
The image must be formed on the back of the eye (retina), so we must have q = 2.00 cm
when p = 1.00 m = 100 cm. The thin-lens equation gives the required focal length as
f =
(b)
pq
(100 cm )( 2.00 cm )
=
= 1.96 cm
p + q 100 cm + 2.00 cm
The f-number of a lens aperture is the focal length of the lens divided by the diameter of the
aperture. Thus, the smallest f-number occurs with the largest diameter of the aperture. For
the typical eyeball focused on objects 1.00 m away, this is
( f -number )min =
(c)
Dmax
=
1.96 cm
= 3.27
0.600 cm
The largest f-number of the typical eyeball focused on a 1.00-m-distance object is
( f -number )max =
68719_25_ch25_p282-304.indd 302
f
f
1.96 cm
=
= 9.80
Dmin 0.200 cm
1/7/11 2:55:54 PM
Optical Instruments
25.59
(a)
The implanted lens should give an image distance of q = 22.4 mm for distant ( p → ∞ )
objects. The thin-lens equation then gives the focal length as f = q = 22.4 mm, so the
power of the implanted lens should be
Pimplant =
(b)
1
1
=
= + 44.6 diopters
f 22.4 × 10 −3 m
When the object distance is p = 33.0 cm, the corrective lens should produce parallel rays
( q → ∞ ) . Then the implanted lens will focus the final image on the retina. From the thinlens equation, the required focal length is f = p = 33.0 cm, and the power of this lens
should be
Pcorrective =
25.60
303
1
1
=
= + 3.03 diopters
f 0.330 m
We use n1 + n2 = n2 − n1 , with p → ∞ and q equal to the cornea to retina distance.
p q
R
⎛ n − n1 ⎞
⎛ 1.34 − 1.00 ⎞
Then, R = q ⎜ 2
= ( 2.00 cm ) ⎜
= 0.507 cm = 5.07 mm .
⎟
⎝ 1.34 ⎟⎠
⎝ n2 ⎠
25.61
When a converging lens forms a real image of a very distant object, the image distance equals the
focal length of the lens. Thus, if the scout started a fire by focusing sunlight on kindling 5.00 cm
from the lens, f = q = 5.00 cm.
(a)
When the lens is used as a simple magnifier, maximum magnification is produced when the
upright, virtual image is formed at the near point of the eye (q = −15 cm in this case). The
object distance required to form an image at this location is
p=
qf
( −15 cm )( 5.0 cm ) 15 cm
=
=
q− f
−15 cm − 5.0 cm
4.0
q
−15 cm
=−
= + 4.0 . Note
p
15 cm 4.0
that adapting Equation 25.5 for use with this “abnormal” eye would give an angular
magnification of mmax = 1 + q f = 1 + 15 cm 5.0 cm = + 4.0 .
and the lateral magnification produced is M = −
(b)
When the object is viewed directly while positioned at the near point of the eye, its angular
size is q 0 = h 15 cm . When the object is viewed by the relaxed eye while using the lens
as a simple magnifier (with the object at the focal point so parallel rays enter the eye), the
angular size of the upright, virtual image is q = h f . Thus, the angular magnification gained
by using the lens in this manner is
m=
25.62
q
h f
15 cm 15 cm
=
=
=
= + 3.0
q 0 h 15 cm
f
5.0 cm
The angular magnification is m = q q 0 , where q is the angle subtended by the final image, and q 0
is the angle subtended by the object as shown in the figure below. When the telescope is adjusted
for minimum eyestrain, the rays entering the eye are parallel. Thus, the objective lens must form
its image at the focal point of the eyepiece.
continued on next page
68719_25_ch25_p282-304.indd 303
1/7/11 2:55:56 PM
304
Chapter 25
From triangle ABC, q 0 ≈ tanq 0 = h ′ q1, and from triangle DEF, q ≈ tanq = h ′ fe . The angular
h ′ fe q1
q
magnification is then m =
=
= .
q 0 h ′ q1 fe
From the thin-lens equation, the image distance of the objective lens in this case is
q1 =
p1 f1
( 300 cm )( 20.0 cm )
=
= 21.4 cm
p1 − f1
300 cm − 20.0 cm
With an eyepiece of focal length fe = 2.00 cm, the angular magnification for this telescope is
m=
68719_25_ch25_p282-304.indd 304
q1 21.4 cm
=
= 10.7
fe 2.00 cm
1/7/11 2:55:59 PM
26
Relativity
QUICK QUIZZES
1.
False. One of Einstein’s two basic postulates in his special theory of relativity is the constancy
of the speed of light. This postulate states that the speed of light in a vacuum has the same value
in all inertial reference frames, regardless of the velocity of the observer or the velocity of the
source emitting the light.
2.
Choice (a). Less time will have passed for you in your frame of reference than for your employer
back on Earth. Thus, to maximize your paycheck, you should choose to have your pay calculated
according to the elapsed time on a clock on Earth.
3.
False. Clocks, including biological clocks, in the observer’s own reference frame appear to run
at normal speeds. It is only clocks in reference frames moving relative to the observer that the
observer will judge to be running slower than normal.
4.
No. From your perspective you are at rest with respect to the cabin, so you will measure yourself
as having your normal length and will require a normal-sized cabin.
5.
(i) Choices (a) and (e). The outgoing rocket will appear to have a shorter length and a slower
clock. (ii) Choices (a) and (e). The answers for the incoming rocket are the same as for the
outgoing rocket. Length contraction and time dilation depend only on the magnitude of the
relative velocity, not on the direction.
6.
False. According to the mass-energy equivalence equation, the particle’s total energy is
E = mc 2 1− v 2 c 2 , and it is seen that as v → c, the total energy becomes infinitely large. As the
total energy becomes infinitely large, so will the kinetic energy given by KE = E − mc 2 and the
relativistic momentum given by p = E 2 − (mc 2 )2 c.
7.
(a) False
(b) False
(c) True
(d) False
A reflected photon does exert a force on the surface. Although a photon has zero mass, a photon does
carry momentum. When it reflects from a surface, there is a change in the momentum, just like the
change in momentum of a ball bouncing off a wall. According to the momentum interpretation of
Newton’s second law, a change in momentum results in a force on the surface. This concept is used in
theoretical studies of space sailing. These studies propose building non-powered spacecraft with huge
reflective sails oriented perpendicularly to the rays from the Sun. The large number of photons from
the Sun reflecting from the surface of the sail will exert a force which, although small, will provide a
continuous acceleration. This would allow the spacecraft to travel to other planets without fuel.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
Einstein based his special theory of relativity on the two postulates included in choices (d) and (e).
The textbook refers to the postulate summarized in choice (d) as the principle of relativity and to
the postulate in choice (e) as the constancy of the speed of light.
305
68719_26_ch26_p305-323.indd 305
1/7/11 2:56:39 PM
306
Chapter 26
2.
The length of the ship is parallel to the relative velocity between the ship and the observer. Thus,
the observer at rest on Earth will measure the moving length of the ship to be shorter than its
length when it was at rest. Choice (b) is the correct answer.
3.
According to the second postulate of special relativity (the constancy of the speed of light), both
observers will measure the light speed to be c. Therefore, both choices (b) and (c) are true statements, and the other choices are false.
4.
The astronaut is moving with constant velocity and is therefore in an inertial reference frame.
According to the principle of relativity, all the laws of physics are the same in her reference frame
as in any other inertial reference frame. Thus, she should experience no effects due to her motion
through space, and choice (e) is the correct answer.
5.
The astronaut in the spaceship is at rest relative to the pendulum and measures the proper time
required for it to complete one oscillation (its period). The observer on Earth is in motion relative
to the pendulum and measures the time required for a full oscillation to be longer than does the
astronaut (i.e., T > P), and the correct choice is (c).
6.
If L p is the proper length of each edge of the cube, the volume measured by the observer at rest
relative to it is V = L3p . The observer moving relative to the cube sees the two dimensions that are
perpendicular to his velocity as unchanged, with each having value L p . The dimension parallel to
his velocity is measured to have the contracted length L = L p g . Thus, the observer moving
relative to the cube measures its volume to be V ′ = L p ⋅ L p ⋅ L = L2p ⋅ L p g = V 5, so the correct
answer is (c).
7.
The relativistic momentum of a particle is p = E 2 − ER2 c, where E is the total energy of the
particle and ER is its rest energy. In this case, each particle has the same total energy, E. Thus, the
particle with the smallest rest energy has the greatest momentum, while the particle with the greatest rest energy has the smallest momentum. This means that the correct ordering of the particles,
from smallest to greatest momentum, is proton, electron, photon, and (e) is the correct choice.
8.
The time interval between events appears to be dilated (or lengthened) to observers who are in
motion relative to those events. Thus, the intervals between successive “ticks” of the clock in orbit
will seem longer to observers on Earth than they do to an astronaut in orbit with the clock and at
rest relative to it. This means the Earth-based observers detect the orbiting clock to run slower than
the identical clock that was left at rest on Earth. The correct answer to this question is choice (c).
9.
When the ground observer is moving at speed v = 0.5c relative to the oscillating system, the time
he will measure for the period is T ′ = g T , where
g =
1
1− (v c)
2
=
1
1− (0.5)2
= 1.15
and the correct choice is (d).
10.
Choice (d) is the correct response. Each observer measures the intervals between “ticks” of the
clock in motion relative to him to be lengthened by the same factor g = 1 1− v 2 c 2 , where v is
the relative velocity between the two observers.
11.
According to the second postulate of special relativity, the speed of light is the same in all inertial
reference frames, regardless of the velocity of the observer or the velocity of the source emitting
the light. Thus, the speed of light from the rapidly moving quasar should be measured to be c,
the same as the speed of light emitted by a stationary source. This means that (b) is the correct
answer.
68719_26_ch26_p305-323.indd 306
1/7/11 2:56:40 PM
Relativity
307
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The two observers will agree on the speed of light and on the speed at which they move relative
to one another.
4.
Special relativity describes inertial reference frames—that is, reference frames that are not accelerating. General relativity describes all reference frames.
6.
You would see the same thing that you see when looking at a mirror when at rest. The theory of
relativity tells us that all experiments will give the same results in all inertial frames of reference.
8.
(a)
Attempts to apply special relativity to this situation would lead one to say that each
observer (one on Earth and one in orbit) would find the other’s clock to run slow relative
to their own clock. However, the observer in orbit is not in an inertial reference frame, and
special relativity does not apply here. According to general relativity, the clock in orbit and
undergoing an acceleration (the centripetal acceleration) will run slower.
(b)
When the moving clock returns to Earth, they are again in inertial reference frames and
subject to the same laws of physics. Thus, the identical clocks should tick at the same rate
once again. However, they will not be synchronized. The clock that underwent the accelerations (i.e., the orbiting clock) will have permanently lost time (or will have “aged less”) and
will be behind the clock that remained on Earth.
10.
The 8 light-years represents the proper length of a rod from Earth to Sirius, measured by an
observer seeing both the rod and Sirius nearly at rest. The astronaut sees Sirius coming toward
her at 0.8c but also sees the distance contracted to
d = (8 ly ) 1− ( v c ) = (8 ly ) 1− ( 0.8 ) = 5 ly
2
2
So the travel time measured on her clock is t =
d 5 ly ( 5 yr ) c
=
=
= 6 yr.
v 0.8c
0.8c
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a) 0.436 m
(b)
less than 0.436 m by an undetectable amount
4.
(a)
(b)
30 beats min
6.
(a) 65.0 beats min
(b)
10.6 beats min
8.
5.0 s
1.2 m, 2.0 m, 2.0 m
70 beats min
10.
26.0 min + 1.06 × 10 −11 s
12.
(a) rectangular box
(b)
14.
(a) 4.96 × 10 −19 kg ⋅ m s
(b) 3.52 × 10 −18 kg ⋅ m s
(c) No. Neglecting relativity in this case introduces an 86% error in the result.
16.
68719_26_ch26_p305-323.indd 307
0.94c toward the left
1/7/11 2:56:42 PM
308
Chapter 26
18.
+ 0.696c
20.
0.161 Hz
22.
(a) 939 MeV
24.
(a)
26.
(a) 0.582 MeV
28.
2.51× 10 −28 kg with speed 0.987c, and 8.84 × 10 −28 kg with speed 0.868c
30.
(a)
KE ≈ g ER = g m p c 2
KEi + mi c 2 = KE f + m f c 2
(b)
3.01 GeV
(c)
(b)
0.999 99c
(b)
2.45 MeV
2.07 GeV
(b)
236.052 588 u
(d) 0.190 976 u
(e)
177.893 MeV
32.
(a) 0.023 6c = 7.08 × 10 3 km s
(b)
(6.17 × 10 ) c = 185 km s
34.
(a) Yes. As the spring is compressed, positive work is done on it or energy is added to it. Since
mass and energy are equivalent, mass has been added to the spring.
−4
(b)
Δm = kx 2 2c 2
(c) 2.5 × 10 −17 kg
36.
(a)
2.50 MeV c
(b)
4.60 GeV c
38.
1.2 × 1013 m
40.
(a) t =
(b)
t′ =
42.
(a) ∼ 108 km
(b)
∼ 10 2 s
44.
(a) See Solution.
(b)
0.943c
46.
Rg = 1.47 km for the Sun
48.
(a) 21.0 yr
(b)
14.7 ly
(c) 10.5 ly
(d)
35.7 yr
(a)
(b)
0.160 ly
(d)
7.51× 10 22 J
50.
2d
c+v
0.946c
(c) 0.114 yr
(c) 235.861 612 u
2d
c
c−v
c+v
PROBLEM SOLUTIONS
26.1
(a)
Observers on Earth measure the time for the astronauts to reach Alpha Centauri as
Δt E = 4.42 yr. But these observers are moving relative to the astronaut’s internal biological
clock and hence experience a dilated version of the proper time interval Δt p measured on
that clock. From Δt E = g Δt p , we find
Δt p = Δt E g = Δt E 1− ( v c ) = ( 4.42 yr ) 1− ( 0.950 ) = 1.38 yr
2
2
continued on next page
68719_26_ch26_p305-323.indd 308
1/7/11 2:56:43 PM
Relativity
(b)
309
The astronauts are moving relative to the span of space separating Earth and Alpha Centauri.
Hence, they measure a length-contracted version of the proper distance, L p = 4.20 ly. The
distance measured by the astronauts is
L = L p g = L p 1− ( v c ) = ( 4.20 ly ) 1− ( 0.950 ) = 1.31 ly
2
26.2
(a)
2
The length of the meterstick measured by the observer moving at speed v = 0.900 c relative
to the meterstick is
L = L p g = L p 1− ( v c ) = (1.00 m ) 1− ( 0.900 ) = 0.436 m
2
(b)
26.3
2
If the observer moves relative to Earth in the direction toward the meterstick, the velocity
of the observer relative to the meterstick is greater than that in part (a). The measured
length of the meterstick will be less than 0.436 m under these conditions, but at the speed
the observer could run, the effect would be too small to detect.
The contracted length of the ship, L = L p − ΔL, as measured by the Earth-based observer is
2
2
L = L p − ΔL = L p 1− ( v c ) . This yields 1− ( v c ) = 1− ΔL L p , and solving for the speed v of
(
)
(
)
2
the ship, we find v = c 1− 1− ΔL L p . Thus, if the proper length of the ship is L p = 28.0 m, and
the observed contraction is ΔL = 0.150 m, the speed of the ship must be
2
0.150 m ⎞
⎛
v = c 1− ⎜ 1−
⎟ = 0.103c
⎝
28.0 m ⎠
26.4
(a)
The time for 70 beats, as measured by the astronaut and any observer at rest with respect
to the astronaut, is Δt p = 1.0 min. The observer in the ship then measures a rate of
70 beats min .
(b)
The observer on Earth moves at v = 0.90 c relative to the astronaut and measures the time
for 70 beats as
Δt = g Δt p =
Δt p
1− ( v c )
2
=
1.0 min
1− ( 0.90 )
2
= 2.3 min
This observer then measures a beat rate of 70 beats 2.3 min = 30 beats min .
26.5
26.6
Δt p
2.6 × 10 −8 s
(a)
Δt = g Δt p =
(b)
d = v ( Δt ) = ⎡⎣ 0.98 ( 3.0 × 108 m s ) ⎤⎦(1.3 × 10 −7 s ) = 38 m
(c)
d ′ = v Δt p = ⎡⎣ 0.98 ( 3.0 × 108 m s ) ⎤⎦( 2.6 × 10 −8 s ) = 7.6 m
(a)
1− ( v c )
2
=
1− ( 0.98 )
2
= 1.3 × 10 −7 s
( )
As measured by observers in the ship (that is, at rest relative to the astronaut), the time
required for 75.0 beats is Δt p = 1.00 min.
The time interval required for 75.0 beats as measured by the Earth observer is
Δt = g Δt p = (1.00 min) 1 − (0.500)2 , so the Earth observer measures a pulse rate of
75.0 75.0 1− ( 0.500 )
= 65.0 min
=
1.00 min
Δt
2
rate =
continued on next page
68719_26_ch26_p305-323.indd 309
1/7/11 2:56:46 PM
310
Chapter 26
(b)
1.00 min
If v = 0.990 c, then Δt = g Δt p =
1− ( 0.990 )
2
and the pulse rate observed on Earth is
75.0 75.0 1− ( 0.990 )
= 10.6 min
=
1.00 min
Δt
2
rate =
That is, the life span of the astronaut (reckoned by the total number of his heartbeats) is
much longer as measured by an Earth clock than by a clock aboard the space vehicle.
26.7
(a)
To the observer on Earth, the muon appears to have a lifetime of
Δt =
d
4.60 × 10 3 m
=
= 1.55 × 10 −5 s
v 0.990 ( 3.00 × 108 m s )
1
1
(b)
g =
(c)
To an observer at rest with respect to the muon, its proper lifetime is
1− ( v c )
Δt p =
(d)
2
=
= 7.09
1− ( 0.990 )
2
Δt 1.55 × 10 −5 s
=
= 2.19 × 10 −6 s
g
7.09
The muon is at rest relative to the observer traveling with the muon. Thus, the muon travels
zero distance as measured by this observer. However, during the observed lifetime of the
muon, this observer sees Earth move toward the muon a distance of
( )
d ′ = v Δt p = ⎡⎣ 0.990 ( 3.00 × 108 m s ) ⎤⎦ ( 2.19 × 10 − 6 s ) = 6.50 × 10 2 m = 650 m
(e)
As the third observer travels toward the incoming muon, his speed relative to the muon
is greater than that of the observer at rest on Earth. Thus, his observed gamma factor
(Δt = g Δt p) is higher, and he measures the muon’s lifetime as longer than that measured
by the observer at rest with respect to Earth.
Δt p
3.0 s
26.8
Δt = g Δt p =
26.9
The proper length of the faster ship is three times that of the slower ship L pf = 3L ps , yet they
both appear to have the same contracted length, L. Thus,
1− ( v c )
=
2
= 5.0 s
1− ( 0.80 )
2
(
(
L = L ps 1− ( vs c ) = 3L ps
2
)
(
)
(
1− v f c , or 1− ( vs c ) = 9 − 9 v f c
2
2
)
)
2
This gives
vf =
26.10
c 8 + ( vs c )
3
2
8 + ( 0.350 )
c = 0.950 c
3
2
=
The driver is the observer at rest with respect to the clock measuring the 26.0-min time interval.
Thus, this observer measures the proper time Δt p , and the Earth-based observer measures
−
the dilated time Δt = g Δt p = [1− (v c)2 ] ⋅ Δt p. In this case, v = 35.0 m s c, so we use the
1
2
continued on next page
68719_26_ch26_p305-323.indd 310
1/7/11 2:56:48 PM
Relativity
−1
binominal expansion [1− x]
fixed on Earth as
2
≈ 1+ 12 x when x
2
Δt = ( 26.0 min ) ⎡⎣1− ( v c ) ⎤⎦
or Δt ≈ 26.0 min +
26.11
( 26.0
− 12
311
1. This gives the time measured by the observer
1
2⎞
⎛
≈ ( 26.0 min ) ⎜ 1+ ( v c ) ⎟
⎝
⎠
2
)
2
min ⎛ 35.0 m s ⎞ ⎛ 60.0 s ⎞
−11
⎜⎝ 3.00 × 108 m s ⎟⎠ ⎜⎝ 1 min ⎟⎠ = 26.0 min + 1.06 × 10 s .
2
The trackside observer sees the supertrain length-contracted as
L = L p 1− ( v c ) = (100 m ) 1− ( 0.95) = 31 m
2
2
The supertrain appears to fit in the tunnel with 50 m − 31 m = 19 m to spare .
26.12
Length contraction occurs only in the dimension parallel to the
motion.
(a)
The sides labeled L2 and L3 in the figure at the right are
unaffected, but the side labeled L1 will appear contracted,
giving the box a rectangular shape, or more formally, the
shape of a rectangular parallelepiped.
(b)
The dimensions of the box, as measured by the observer
moving at v = 0.80 c relative to it, are
L2 = L2 p = 2.0 m , L3 = L3 p = 2.0 m , and
L1 = L1 p 1− ( v c ) = ( 2.0 m ) 1− ( 0.80 ) = 1.2 m
2
26.13
(a)
The classical expression for linear momentum is pclassical = mv, while the relativistic expression is p = g mv, where g = 1 1− (v c)2 . Thus, if p = 3pclassical, it is necessary that the
gamma factor have a value g = 3. Solving for the speed of the particle gives
v = c 1−
26.14
2
1
1 c 8
2c 2
= c 1− =
=
= 0.943c
2
g
9
3
3
(b)
Observe that the calculation above did not depend on the mass of the particle involved.
Thus, the result is the same for a proton or any other particle.
(a)
Classically,
p = mv = m ( 0.990c ) = (1.67 × 10 −27 kg )( 0.990 )( 3.00 × 108 m s ) = 4.96 × 10 −19 kg ⋅ m s
(b)
By relativistic calculations,
p=
mv
1− (v c)2
=
m ( 0.990c )
1− (0.990)2
=
(1.67 × 10
−27
kg )( 0.990 )( 3.00 × 108 m s )
1− (0.990)2
= 3.52 × 10 −18 kg ⋅ m s
(c)
68719_26_ch26_p305-323.indd 311
No , neglecting relativistic effects at such speeds would introduce an approximate 86%
error in the result.
1/7/11 2:56:51 PM
312
26.15
Chapter 26
Momentum must be conserved, so the momenta of the two fragments must add to zero. Thus,
their magnitudes must be equal, or g 2 m2 v2 = g 1 m1 v1 . This gives
(1.67 × 10
−27
kg ) v
1− ( v c )
2
=
( 2.50 × 10
−28
kg ) ( 0.893c )
1− ( 0.893)
2
⎡ (1.67 × 10 −27 kg ) 1− ( 0.893)2
and reduces to ⎢
⎢ ( 2.50 × 10 −28 kg )( 0.893)
⎣
2
⎤ v
⎥ ⎛⎜ ⎞⎟ = 1− ⎛⎜ v ⎞⎟ , or
⎝ c⎠
⎥⎝ c ⎠
⎦
12.3 ( v c ) = 1, and yields v = 0.285c .
2
26.16
We take to the right as the positive direction. Then, the velocities of the two ships relative to
Earth are vR E = + 0.70c and vLE = − 0.70c. The velocity of ship L relative to ship R is given by the
relativistic relative velocity relation (Equation 26.7 in the textbook) as
vLR =
26.17
Taking to the right as the positive direction, the velocity of the electron relative to the laboratory
is vEL = + 0.90c, and the velocity of the proton relative to the electron is vPE = − 0.70c. Thus, the
relativistic addition of velocities (Equation 26.8 in the textbook) gives the velocity of the proton
relative to the laboratory as
vPL =
26.18
vLE − vR E
−1.40c
( −0.70c ) − 0.70c
=
=
= − 0.94c = 0.94c toward the left
vLE vR E
−0.70c )( 0.70c ) 1+ 0.49
(
1−
1−
c2
c2
vPE + vEL
( − 0.70c ) + 0.90c = + 0.20c = + 0.54c = 0.54c toward the right
=
vPE vEL
( −0.70c )( 0.90c ) 1− 0.63
1+
1+
c2
c2
We choose the direction of the spaceship’s motion relative to Earth as the positive direction.
Then, the spaceship’s velocity relative to Earth is vSE = + 0.750c. It is desired to have the velocity
of the rocket relative to Earth be vR E = + 0.950c. The relativistic relative velocity relation
(Equation 26.7 in the textbook) then gives the required velocity of the rocket relative to
the ship as
vR S =
26.19
Taking away from Earth as the positive direction, the velocity of ship A relative to Earth is
vAE = + 0.800c, and the velocity of ship B relative to Earth is vB E = + 0.900c. The relativistic
relative velocity relation (Equation 26.7 in the textbook) gives the velocity of ship B relative
to ship A (and hence, the speed with which B is overtaking A) as
vB A =
26.20
vR E − vSE
0.950c − 0.750c
=
= + 0.696c
vR E vSE
( 0.950c )( 0.750c )
1−
1
−
c2
c2
vB E − vAE
0.900c − 0.800c
=
= + 0.357c
vB E vAE
( 0.900c )( 0.800c )
1−
1−
2
2
c
c
We first determine the velocity of the pulsar relative to the rocket. Taking toward the Earth as the
positive direction, the velocity of the pulsar relative to Earth is vPE = + 0.950c, and the velocity of the
rocket relative to Earth is vR E = − 0.995c. The relativistic relative velocity relation (Equation 26.7 in
the textbook) gives the velocity of the pulsar relative to the rocket as
continued on next page
68719_26_ch26_p305-323.indd 312
1/7/11 2:56:53 PM
Relativity
vPR =
313
vPE − vR E
0.950c − ( −0.995c )
=
= + 0.999 87c
vPE vR E
( 0.950c )( −0.995c )
1−
1−
2
2
c
c
The time interval between successive pulses in the pulsar’s own reference frame (i.e., on a clock
at rest with respect to the event being timed) is Δt p = 1 10.0 Hz = 0.100 s. The duration of this
interval in the rocket’s frame of reference is given by the time dilation relation as
Δt = g Δt p =
Δt p
1− ( vPR c )
2
=
0.100 s
1− ( 0.999 87 )
2
= 6.20 s
and the frequency of the pulses as observed in the rocket’s reference frame is
f=
26.21
1
1
=
= 0.161 Hz
Δt 6.20 s
Taking to the right as positive, it is given that the velocity of the rocket relative to observer A is
vR A = + 0.92c. If observer B observes the rocket to have a velocity vR B = − 0.95c, the velocity of
observer B relative to the rocket is vB R = + 0.95c. The relativistic velocity addition relation then
gives the velocity of B relative to the stationary observer A as
vB A =
26.22
0.998c toward the right
(a)
2 ⎛
1 MeV
ER = mc 2 = (1.67 × 10 −27 kg ) ( 3.00 × 108 m s ) ⎜
⎝ 1.60 × 10 −13
(b)
E = g mc 2 = g ER =
=
26.23
v B R + vR A
+ 0.95c + 0.92c
=
= + 0.998c or
v B R vR A
( 0.95c )( 0.92c )
1+
1+
c2
c2
939 MeV
1− ( 0.950 )
2
⎞
⎟ = 939 MeV
J⎠
ER
1− ( v c )
2
= 3.01× 10 3 MeV = 3.01 GeV
(c)
KE = E − ER = 3.01× 10 3 MeV − 939 MeV = 2.07 × 10 3 MeV = 2.07 GeV
(a)
The total energy is 400 times the rest energy, or E = g ER = 400ER , so it is necessary that
g = 400. But g = 1 1− (v c)2 , and solving for the speed gives
v = c 1−
(b)
1
1
= c 1−
= 0.999 997c
g2
( 400 )2
The kinetic energy is KE = E − ER = 400ER − ER = 399ER . For a proton, ER = 938.3 MeV.
Thus,
KE = 399 ( 938.3 MeV ) = 3.74 × 10 5 MeV
26.24
(a)
The rest energy of a proton is ER = 938.3 MeV. If the kinetic energy is KE = 175 GeV
we have KE = (g − 1)ER ER , or g − 1 1, and g − 1 ≈ g . Thus, we may write the
approximate relation KE = (g − 1) ER ≈ g EE = g m p c 2 .
ER ,
continued on next page
68719_26_ch26_p305-323.indd 313
1/7/11 2:56:56 PM
314
Chapter 26
The gamma factor is g = 1 1− (v c)2 , so solving for the speed gives v = c 1− g
Therefore, when KE ER and we may use the approximation KE ≈ g ER , we have
g ≈ ER KE . Then, for the proton having KE = 175 GeV,
(b)
26.25
−2
.
⎛E ⎞
⎛ 938.3 MeV ⎞
≈ c 1− ⎜ R ⎟ = c 1− ⎜
⎟ = 0.999 99c
⎝ KE ⎠
⎝ 175×10 3 MeV ⎠
2
v = c 1−g
−2
2
The nonrelativistic expression for kinetic energy is KE = 12 mv 2, while the relativistic expression is
KE = E − ER = (g − 1)ER = (g − 1)mc 2, where g = 1 1− ( v c ) . Thus, when the relativistic kinetic
energy is twice the predicted nonrelativistic value, we have
2
⎛
⎞
1
1
⎟ mc 2 = 2 ⎛⎜ mv 2 ⎞⎟
⎜
−
1
2
⎝2
⎠
⎜⎝ 1− ( v c )
⎟⎠
2
2
1 = ⎡⎣1+ ( v c ) ⎤⎦ 1− ( v c )
or
Squaring both sides of the last result and simplifying gives
( v c ) ⎡⎣( v c ) + ( v c )
2
4
2
− 1⎤⎦ = 0
Ignoring the trivial solution v c = 0, we must have ( v c ) + ( v c ) − 1 = 0. This is a quadratic
2
equation of the form x 2 + x − 1 = 0,with x = ( v c ) . Applying the quadratic formula gives
2
x = −1 ± 5 2. Since x = ( v c ) , we ignore the negative solution and find
4
(
2
)
−1+ 5
⎛ v⎞
x=⎜ ⎟ =
= 0.618
⎝ c⎠
2
2
which yields v = c 0.618 = 0.786 c .
26.26
The energy input to the electron will be W = E f − Ei = (g f − g i ) ER , or
⎛
1
W =⎜
⎜
⎜⎝ 1− v f c
(
(a)
)
2
−
1
1− ( vi c )
2
⎞
⎟E
⎟ R
⎟⎠
(a)
⎞
⎟ ( 0.511 Mev ) = 0.582 MeV
⎟⎠
When v f = 0.990 c and vi = 0.900 c, we have
⎛
1
1
W =⎜
−
2
⎜⎝ 1− ( 0.990 )2
1− ( 0.900 )
26.27
ER = 0.511 MeV
If v f = 0.900 c and vi = 0.500 c, then
⎛
1
1
W =⎜
−
2
⎜⎝ 1− ( 0.900 )2
1− ( 0.500 )
(b)
where
⎞
⎟ ( 0.511 Mev ) = 2.45 MeV
⎟⎠
From the work-energy theorem, Wnet = ΔKE. The ship starts from rest, so KEi = 0, and
ΔKE = KE f = (g − 1)ER = (g − 1)mc 2. Thus,
⎛
⎞
2
1
Wnet = ⎜
−
1
⎟ ( 2.40 × 10 6 kg ) ( 3.00 × 108 m s ) = 8.65 × 10 22 J
2
⎜⎝ 1− ( 0.700 )
⎟⎠
continued on next page
68719_26_ch26_p305-323.indd 314
1/7/11 2:56:59 PM
Relativity
(b)
If all the increase in the kinetic energy of the ship comes from the rest energy of the fuel
(i.e., ΔKE = ER, fuel = mfuel c 2), the mass of the fuel required is
ΔKE
8.65 × 10 22 J
=
= 9.61× 10 5 kg
2
2
8
c
(3.00 × 10 m s)
mfuel =
26.28
315
Let m1 be the mass of the fragment moving at v1 = 0.987c, and m2 be the mass having a speed of
v2 = 0.868c.
From conservation of mass-energy, E1 + E2 = ER, i , or g 1 m1 c 2 + g 2 m2 c 2 = mi c 2 , giving
m1
1− ( 0.987 )
2
m2
+
1− ( 0.868 )
2
= mi
and
6.22 m1 + 2.01m2 = 3.34 × 10 −27 kg
[1]
Since the original particle was at rest, the momenta of the two fragments after decay must add to
zero. Thus, the magnitudes must be equal, giving p2 = p1 , or g 2 m2 v2 = g 1 m1 v1, and yielding
2.01m2 ( 0.868c ) = 6.22 m1 ( 0.987c )
or
m2 = 3.52 m1
[2]
Substituting Equation [2] into [1] gives
( 6.22 + 7.08 ) m1 = 3.34 × 10 −27 kg , or m1 = 2.51× 10 −28 kg
Equation [2] then yields m2 = 3.52 ( 2.51× 10 −28 kg ) = 8.84 × 10 −28 kg .
26.29
The relativistic total energy is E = g ER = g mc 2, and the momentum is p = g mv, where
g = 1 1 − v 2 c 2 . Thus, E 2 c 2 = g 2 m 2 c 2, and p2 = g 2 m 2 v 2, so subtracting yields
⎛
⎞
E2
1
2 2
2
2
2 2
− p2 = g 2 m 2 ( c 2 − v 2 ) = g 2 m 2 c 2 (1− v 2 c 2 ) = ⎜
⎟ m c 1− v c = m c
2
2
2
c
⎝ 1− v c ⎠
(
)
Rearranging, this becomes
E2
= p2 + m 2 c 2
c2
26.30
(a)
E 2 = p2 c 2 + m 2 c 4
or
Since the total relativistic energy of a particle is E = KE + ER = KE + mc 2, requiring that
total energy be conserved (i.e., total energy before reaction = total energy after reaction)
gives KEi + mi c 2 = KE f + m f c 2 , where mi is the total mass of the particles present before
the reaction, KEi is the total kinetic energy before the reaction, m f is the total mass of the
particles present after the reaction, and KE f is the final total kinetic energy.
(b)
Using atomic masses from the tables in Appendix B of the textbook, the total mass of the
initial particles is found to be
mi = m
(c)
235
92 U
+m
1
0n
= 235.043 923 u + 1.008 665 u = 236.052 588 u
The total mass of the particles present after the reaction is
mf = m
148
57 La
+m
87
35 B r
+m
1
0n
= 147.932 236 u + 86.920 71119 u + 1.008 665 u = 235.861 612 u
continued on next page
68719_26_ch26_p305-323.indd 315
1/7/11 2:57:03 PM
316
Chapter 26
(d)
The mass that must have been converted to energy in the reaction is
Δm = mi − m f = ( 236.052 588 − 235.861 612 ) u = 0.190 976 u
(e)
Assuming that KEi = 0, the total kinetic energy of the product particles is then
⎛ 931.494 MeV c 2 ⎞
KE f = Δmc 2 + KEi = ( 0.190 976 u ) c 2 ⎜
⎟⎠ + 0 = 177.893 MeV
1u
⎝
26.31
26.32
E
20.0 GeV ⎛ 10 3 MeV ⎞
= 3.91 × 10 4
=
ER 0.511 MeV ⎜⎝ 1 GeV ⎟⎠
(a)
g =
(b)
L=
(a)
For an electron moving at ve = 0.750c, the gamma factor is
Lp
g
=
3.00 × 10 3 m
= 7.67 × 10 −2 m = 7.67 cm
3.91× 10 4
g e =1
1− (ve c)2 = 1
1− (0.750)2 = 1.51
The kinetic energy of a particle is KE = E − ER = (g − 1)ER, so if the kinetic energy of a
proton (ER, p = 938.3 MeV) equals that of the electron (ER, e = 0.511 MeV), we must have
(g p − 1)ER, p = (g e − 1)ER, e , or
g p = 1+ (g e − 1)
But g p = 1
1− (v p c)2 , so (v p c)2 = 1− 1 g p2 and the speed of the proton must be
v p = c 1−
(b)
ER, e
⎛ 0.511 MeV ⎞
= 1+ (1.51− 1) ⎜
= 1 + 2.78 × 10 − 4
⎝ 938.3 MeV ⎟⎠
ER, p
1
1
= c 1−
= 0.023 6c = 7.08 × 10 3 km s
−4 2
g p2
(1+ 2.78 × 10 )
As above, g e = 1.51 for an electron having speed ve = 0.750 c. The momentum of a particle is
p = g mv =
g mc 2 (v c) g ER (v c)
=
c
c
pc = g ER ( v c )
or
If the momentum of a proton equals that of the electron, then p p c = pe c, or
(
)
g p ER, p v p c = g e ER, e ( ve c )
( v c ) = g E ⎛ v ⎞ = (1.51)⎛ 0.511 MeV ⎞ ( 0.750 ) = 6.17 × 10
⎜⎝
⎟
⎜ ⎟
E ⎝ c⎠
938.3 MeV ⎠
1− ( v c )
Thus, ( v c ) = 6.17 × 10 1− ( v c ) , and
( v c ) = (6.17 × 10 ) − (6.17 × 10 ) ( v c ) , which yields
p
and
R, e
2
e
−4
e
R, p
p
2
−4
p
p
2
−4
2
−4
p
6.17 × 10
( v c ) = 1+( 6.17 × 10 )
(
)
2
p
2
2
p
−4
2
−4
2
= 3.81× 10 −7
and the speed of the proton must be
v p = c 3.81× 10 −7 = ( 3.00 × 108 m s ) ( 6.17 × 10 − 4 ) = 1.85 × 10 5 m s = 185 km s
68719_26_ch26_p305-323.indd 316
1/7/11 2:57:07 PM
Relativity
26.33
KE = E − ER = (g − 1) ER
so
(a)
g = 1+
KE
1
=
2
ER
1− ( v c )
v = c 1−
giving
2
1
(1+ KE E )
2
= c 1−
1
= 0.979c
(1+ 2.00 0.511)2
For a proton with KE = 2.00 MeV, the speed is
v p = c 1−
1
(1+ KE E )
2
= c 1−
R,p
1
= 0.065 2c
(1+ 2.00 938)2
(c)
ve − v p = 0.979c − 0.065 2c = 0.914 c
(a)
Yes. As the spring is compressed, positive work is done on it or energy is added to it.
Since mass and energy are equivalent, mass has been added to the spring.
(b)
Δm =
ΔER ΔPEs 12 kx 2
kx 2
= 2 = 2 =
2
c
c
c
2c 2
(c)
Δm =
( 2.0 × 10 N m )( 0.15 m )
2 ( 3.00 × 10 m s )
2
2
26.35
(1+ KE E )
R
R ,e
(b)
1
The speed of an electron having KE = 2.00 MeV will be
ve = c 1−
26.34
317
2
8
= 2.5 × 10 −17 kg
An observer who moves at speed v relative to an object (or span of space) having proper length
L p sees a contracted length given by L = L p g = L p 1− (v c)2 . Thus, if the proper distance to the
star is L p = 5.00 ly, and this length is to have a contracted value of L = 2.00 ly in the reference
frame of the spacecraft, the speed of the spacecraft relative to the star must be
2
2
⎛ L⎞
⎛ 2.00 ly ⎞
v = c 1− ⎜ ⎟ = c 1− ⎜
= 0.917c
⎝ 5.00 ly ⎟⎠
⎝ Lp ⎠
26.36
26.37
From E 2 = ( pc)2 + ER2 , with E = 5 ER , we find that p = ER 24 c.
( 0.511 MeV ) 24
= 2.50 MeV c .
(a)
For an electron, p =
(b)
For a proton, p =
(a)
Observers on Earth measure the distance to Andromeda to be
d = 2.00 × 10 6 ly = (2.00 × 10 6 yr)c. The time for the trip, in Earth’s frame
of reference, is Δt = g Δt p = 30.0 yr 1 − (v c)2 . The required speed is then
c
( 938.3 MeV ) 24
c
= 4.60 × 10 3
MeV
= 4.60 GeV c .
c
( )
v=
( 2.00 × 106 yr ) c
d
=
Δt 30.0 yr 1− ( v c )2
which gives (1.50 × 10 −5 ) ( v c ) = 1− ( v c ) . Squaring both sides of this equation and solv2
ing for v c yields v c = 1
1+ 2.25 × 10 −10 . Then, the approximation 1 1+ x = 1− x 2 gives
continued on next page
68719_26_ch26_p305-323.indd 317
1/7/11 2:57:10 PM
318
Chapter 26
2.25 × 10 −10
v
≈ 1−
= 1− 1.13 × 10 −10
c
2
(b)
KE = (g − 1) mc 2, and
g =
1
1− ( v c )
2
1
=
1− (1− 1.13 × 10 −10 )
2
=
1
2.26 × 10 −10
Thus,
⎛
⎞
2
1
KE = ⎜
− 1⎟ (1.00 × 10 6 kg ) ( 3.00 × 108 m s ) = 5.99 × 10 27 J
−10
⎝ 2.26 × 10
⎠
(c)
cost = KE × rate
⎡
⎛ 1 kWh ⎞ ⎤
= ⎢( 5.99 × 10 27 J ) ⎜
($0.13 kWh ) = $2.16 × 10 20
⎝ 3.60 × 10 6 J ⎟⎠ ⎥⎦
⎣
26.38
The clock, at rest in the ship’s frame of reference, will measure a proper time of Δt p = 10 h before
sounding. Observers on Earth move at v = 0.75c relative to the clock and measure an elapsed
time of
( )
Δt = g Δt p =
Δt p
1− ( v c )
2
=
10 h
1− ( 0.75)
2
= 15 h
The observers on Earth see the clock moving away at 0.75c and compute the distance traveled
before the alarm sounds as
⎛ 3600
d = v ( Δt ) = ⎡⎣ 0.75 ( 3.0 × 108 m s ) ⎤⎦ (15 h ) ⎜
⎝ 1h
26.39
s⎞
13
⎟⎠ = 1.2 × 10 m
(a)
Since Dina and Owen are at rest in the same frame of reference (S′), they both see the ball
traveling in the negative x ′ direction with speed vball
′ = 0.800c . Note that the velocity of
the ball relative to Dina is vB D = − 0.800c.
(b)
The distance between Dina and Owen, measured in their own rest frame, is L p = 1.80 × 1012 m.
Therefore, the time required for the ball to reach Dina, measured on her own clock, is
Δt p =
(c)
Lp
vball
′
=
1.80 × 1012 m
2.25 × 1012 m
= 7.50 × 10 3 s
=
0.800c
3.00 × 108 m s
Ed sees a contracted length for the distance separating Dina and Owen. According to him,
they are separated by a distance L = L p g = L p 1− (v c)2 , where v = 0.600c is the speed
of the S′ frame relative to Ed’s reference frame, S. Thus, according to Ed, the ball must
travel a distance
L = (1.80 × 1012 m ) 1− ( 0.600 ) = 1.44 × 1012 m
2
(d)
The ball has a velocity of vB D = − 0.800c relative to Dina, and Dina moves at velocity
vDE = + 0.600c relative to Ed. The relativistic velocity addition relation (Equation 26.8 from
the textbook) then gives the velocity of the ball relative to Ed as
continued on next page
68719_26_ch26_p305-323.indd 318
1/7/11 2:57:13 PM
Relativity
vB E =
319
vB D + vDE
− 0.800c + 0.600c
−0.200c
=
=
= − 0.385c
vB D vDE 1+ ( − 0.800 )( 0.600 )
0.520
1+
c2
Thus, the speed of the ball according to Ed is vball = vB E = 0.385c .
26.40
(a)
As seen by an observer at rest relative to the mirror in frame S, the light must travel
distance d before it strikes the mirror and then a distance d − d1 back to the ship after
reflection. Here, distance d1 = vt is the distance the ship moves toward the mirror in the
time t between when the pulse was emitted from the ship and when the reflected pulse
was received by the ship. Since all observers agree that light travels at speed c, the total
travel time for the light is
t=
d + ( d − d1 ) 2d − d1 2d ⎛ v ⎞
=
=
−⎜ ⎟t
c
c
c ⎝ c⎠
or
⎛
⎜⎝ 1+
v⎞
2d
⎟⎠ t =
c
c
Solving for the travel time t of the light gives t = 2d ( c + v ) .
(b)
From the viewpoint of observers in the spacecraft, the spacecraft is at rest, and the mirror
moves toward it at speed v. At the time the pulse starts toward the mirror, these observers
see a contracted initial distance d ′ = d 1− (v c)2 to the mirror. If the total time of travel for
the light is t ′, during the time t ′ 2 while the light is traveling toward the mirror, the mirror
moves a distance Δx = vt ′ 2 closer to the ship. Thus, the light must travel a distance d ′ − Δx
before reflection and the same distance d ′ − Δx back to the stationary ship after reflection. The total distance traveled by the light is then D = 2 ( d ′ − Δx ) , and the time required
(traveling at speed c) is
D 2d ′ − 2v(t ′ 2) 2d ′ ⎛ v ⎞
t′ = =
=
− ⎜ ⎟ t′
c
c
c ⎝ c⎠
or
⎛
⎜⎝ 1+
v⎞
2d ′ 2d 1− ( v c )
=
⎟⎠ t ′ =
c
c
c
2
Solving for the travel time t ′ measured by observers in the spacecraft then gives
c2 − v2
c2
( c + v ) t ′ = 2d
26.41
and
t′ =
2d
c
( c + v )( c − v ) 2d c − v
=
c c+v
( c + v )2
The length of the space ship, as measured by observers on Earth, is L = L p 1− ( v c ) . In Earth’s
frame of reference, the time required for the ship to pass overhead is
2
L L p 1− ( v c )
=
= Lp
v
v
2
Δt =
1
1
−
v2 c2
2
Thus,
2
⎛ 0.75 × 10 −6 s ⎞
1
1 ⎛ Δt ⎞
1
s2
−17
=
+
=
+
=
1.74
×
10
2
v 2 c 2 ⎜⎝ L p ⎟⎠
m2
(3.00 × 108 m s) ⎜⎝ 300 m ⎟⎠
or
v=
1
2
1.74 × 10 −17
26.42
s
m2
⎞
m⎞ ⎛
c
⎛
= 0.80 c
= ⎜ 2.4 × 108
⎟
⎝
s ⎠ ⎜⎝ 3.00 × 108 m s ⎟⎠
From KE = (g − 1) ER , we find
g = 1+
KE
1013 MeV
= 1.07 × 1010
= 1+
ER
938 MeV
continued on next page
68719_26_ch26_p305-323.indd 319
1/7/11 2:57:16 PM
320
Chapter 26
or g ~ 1010 . Thus, the speed of the proton (and hence, the speed of the galaxy as seen by the
proton) is v = c 1− 1 g 2 ~ c 1− 10 −20 ≈ c.
(a)
The diameter of the galaxy, as seen in the proton’s frame of reference, is
L = L p 1− ( v c ) =
2
or
(b)
g
10 5 ly
= 10 −5 ly
1010
~ 108 km
The proton sees the galaxy rushing by at v ≈ c. The time, in the proton’s frame of reference,
for the galaxy to pass is
1011 m
L 108 km
=
= 333 s
~
c
3.00 × 108 m s
v
~ 10 2 s
The difference between the relativistic momentum, p = g mv, and the classical momentum, mv, is
Δp = g mv − mv = (g − 1)mv.
(a)
(b)
26.44
~
⎛ 9.461× 1015 m ⎞ ⎛ 1 km ⎞
7
L ≈ 10 −5 ly ⎜
⎟⎠ ⎜⎝ 10 3 m ⎟⎠ = 9.461× 10 km
1 ly
⎝
Δt =
26.43
Lp
The error is 1.00% when Δp p = 0.010 0, or (g − 1)mv = 0.010 0g mv. This gives
2
2
g = 1 0.990 , or 1− ( v c ) = ( 0.990 ) , and yields v = 0.141c .
When the error is 10.0%, we have g = 1 0.900 , and 1− ( v c ) = ( 0.900 ) . In this case, the
speed of the particle is v = c 1− (0.900)2 = 0.436c .
2
2
The kinetic energy gained by the electron will equal the loss of potential energy, so
KE = q ( ΔV ) = e (1.02 MV ) = 1.02 MeV
(a)
If Newtonian mechanics remained valid, then KE = 12 mv 2, and the speed attained would be
v=
(b)
2 (1.02 MeV )(1.60 × 10 −13 J MeV )
2 KE
=
m
9.11× 10 −31 kg
KE = (g − 1) ER, so g = 1+
= 5.99 × 108 m s ≈ 2c
KE
1.02 MeV
= 1+
= 3.00
ER
0.511 MeV
The actual speed attained is
v = c 1− 1 g
26.45
(a)
= c 1− 1 ( 3.00 ) = 0.943c
2
When at rest, muons have a mean lifetime of Δt p = 2.2 ms. In a frame of reference where
they move at v = 0.95c, the dilated mean lifetime of the muons will be
( )
t = g Δt p =
(b)
2
Δt p
1− ( v c )
2
=
2.2 ms
1− ( 0.95)
2
= 7.0 ms
In a frame of reference where the muons travel at v = 0.95c, the time required to travel
3.0 km is
t=
d
3.0 × 10 3 m
=
= 1.05 × 10 −5 s = 10.5 ms
v 0.95 ( 3.00 × 108 m s )
continued on next page
68719_26_ch26_p305-323.indd 320
1/7/11 2:57:19 PM
Relativity
321
If N 0 = 5.0 × 10 4 muons started the 3.0 km trip, the number remaining at the end is
N = N 0 e − t t = ( 5.0 × 10 4 ) e − 10.5 m s 7.0 m s = 1.1× 10 4
26.46
The work required equals the increase in the gravitational potential energy, or W = GMSun m Rg .
If this is to equal the rest energy of the mass removed, then
mc 2 =
Rg =
26.47
GMSun m
Rg
(6.67 × 10
−11
GMSun
c2
Rg =
or
N ⋅ m 2 kg2 ) (1.99 × 10 30 kg )
(3.00 × 10
8
m s)
2
= 1.47 × 10 3 m = 1.47 km
According to Earth-based observers, the times required for the two trips are:
For Speedo:
ΔtS =
For Goslo:
ΔtG =
Lp
vS
Lp
vG
=
20.0 ly 20.0 yr
=
= 21.1 yr
0.950 c
0.950
=
20.0 ly 20.0 yr
=
= 26.7 yr
0.750 c
0.750
Thus, after Speedo lands, he must wait and age at the same rate as planet-based observers, for an
additional ΔTage, S = ΔtG − ΔtS = (26.7 − 21.1) yr = 5.6 yr before Goslo arrives.
The time required for the trip according to Speedo’s internal biological clock (which measures
the proper time for his aging process during the trip) is
Tage, S = Δt p, S =
ΔtS
2
2
= ΔtS 1− ( vS c ) = ( 21.1 yr ) 1− ( 0.950 ) = 6.59 yr
gS
When Goslo arrives, Speedo has aged a total of
TS = Tage, S + ΔTage, S = 6.59 yr + 5.6 = 12.2 yr
The time required for the trip according to Goslo’s internal biological clock (and hence the
amount he ages) is
TG = Δt p, G =
ΔtG
2
2
= ΔtG 1− ( vG c ) = ( 26.7 yr ) 1− ( 0.750 ) = 17.7 yr
gG
Thus, we see that when he arrives, Goslo is older than Speedo, having aged an additional
TG − TS = 17.7 yr − 12.2 yr = 5.5 yr
26.48
(a)
The proper lifetime is measured in the ship’s reference frame, and Earth-based observers
measure a dilated lifetime of
Δt = g Δt p =
(b)
Δt p
1− ( v c )
2
=
15.0 yr
1− ( 0.700 )
2
= 21.0 yr
d = v ( Δt ) = [ 0.700 c ]( 21.0 yr ) = ⎡⎣( 0.700 )(1.00 ly yr ) ⎤⎦ ( 21.0 yr ) = 14.7 ly
continued on next page
68719_26_ch26_p305-323.indd 321
1/7/11 2:57:23 PM
322
Chapter 26
(c)
Looking out the rear window, the astronauts see Earth recede at a rate of v = 0.700 c. The
distance it has receded, as measured by the astronauts, when the batteries fail is
( )
d ′ = v Δt p = ( 0.700 c ) (15.0 yr ) = ⎡⎣( 0.700 )(1.00 ly yr ) ⎤⎦ (15.0 yr ) = 10.5 ly
(d)
26.49
Mission control gets signals for 21.0 yr while the battery is operating and then for 14.7 yr
after the battery stops powering the transmitter, 14.7 ly away. The total time that signals are
received is 21.0 yr + 14.7 yr = 35.7 yr .
Note: Excess digits are retained in some steps given
below to more clearly illustrate the method of
solution.
We are given that L = 2.00 m and q = 30.0° (both
measured in the observer’s rest frame). The
components of the rod’s length as measured in the
observer’s rest frame are
Lx = L cosq = ( 2.00 m ) cos30.0° = 1.732 m
and
L y = L sinq = ( 2.00 m ) sin 30.0° = 1.00 m
The component of length parallel to the motion has been contracted, but the component perpendicular to the motion is unaltered. Thus, L py = L y = 1.00 m and
L px =
(a)
Lx
1− ( v c )
2
=
1.732 m
1− ( 0.995)
2
The proper length of the rod is then
(17.34 m )2 + (1.00 m )2 = 17.4 m
L p = L2px + L2py =
(b)
= 17.34 m
The orientation angle in the rod’s rest frame is
⎛ L py ⎞
−1 ⎛ 1.00 m ⎞
q p = tan −1 ⎜
⎟ = tan ⎜⎝ 17.34 m ⎟⎠ = 3.30°
L
⎝ px ⎠
26.50
(a)
Taking toward Earth as the positive direction, the velocity of the ship relative to Earth is
vSE = + 0.600c, and the velocity of the lander relative to the ship is v = + 0.800c. The relativistic velocity addition relation (Equation 26.8 in the textbook) then gives the velocity of
the lander relative to Earth as
LS
vLE =
(b)
vLS + vSE
0.800c + 0.600c
1.40c
=
=
= + 0.946c
vLS vSE 1+ ( 0.800 )( 0.600 ) 1.48
1+
c2
Observers at rest on Earth measure the proper length between the ship and Earth as
L p = 0.200 ly. The contracted distance measured by observers at rest relative to the
spaceship is
L = L p 1− ( vSE c ) = ( 0.200 ly ) 1− ( 0.600 ) = 0.160 ly
2
2
continued on next page
68719_26_ch26_p305-323.indd 322
1/7/11 2:57:25 PM
Relativity
(c)
Observers on the ship see the lander start toward Earth from an initial distance of
L = 0.160 ly. They see this distance diminish as the lander nibbles into it from one end at
vLS = 0.800c, and the Earth (as it appears to approach them) reducing it from the other end
at vSE = 0.600c. The time they compute it will take the lander and Earth to meet is
t=
(d)
323
L
0.160 ly 0.160 yr
=
=
= 0.114 yr
vLS + vSE
1.40c
1.40
The kinetic energy of the lander as observed in the Earth reference frame is
2
KE = E − ER = (g − 1) mc 2, where g = 1 1− ( vLE c ) . This gives
⎛
⎞
2
1
KE = ⎜
−
1
⎟ ( 4.00 × 10 5 kg ) ( 3.00 × 108 m s ) = 7.51× 10 22 J
2
⎜⎝ 1− ( 0.946 )
⎟⎠
68719_26_ch26_p305-323.indd 323
1/7/11 2:57:27 PM
27
Quantum Physics
QUICK QUIZZES
1.
True. When a photon scatters off an electron that was initially at rest, the electron must recoil to
conserve momentum. The recoiling electron possesses kinetic energy that was gained from the photon
in the scattering process. Thus, the scattered photon must have less energy than the incident photon.
2.
Choice (b). Some energy is transferred to the electron in the scattering process. Therefore, the
scattered photon must have less energy (and hence, lower frequency) than the incident photon.
3.
Choice (c). Conservation of energy requires the kinetic energy given to the electron be equal to
the difference between the energy of the incident photon and that of the scattered photon.
4.
False. The de Broglie wavelength of a particle of mass m and speed v is l = h p = h mv. Thus,
the wavelength decreases when the momentum increases.
5.
Choice (c). Two particles with the same de Broglie wavelength will have the same momentum p = mv = h l. If the electron and proton have the same momentum, they cannot have the
same speed because of the difference in their masses. For the same reason, remembering that
KE = p2 2m, they cannot have the same kinetic energy. Because the kinetic energy is the only
type of energy an isolated particle can have, and we have argued that the particles have different
energies, the equation f = E h tells us that the particles do not have the same frequency.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
According to Einstein’s photoelectric effect equation, e ΔVs = KEm ax = Ephoton − f , and the work
function of the metal is
f = Ephoton − e ΔVs = 3.56 eV − e (1.10 V ) = 2.46 eV
so the correct answer is choice (b).
2.
From Wien’s displacement law,
T=
0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K
= 6.10 × 10 3 K = 6 100 K
=
lm ax
475 × 10 −9 m
and choice (a) is correct.
3.
An electron accelerated from rest through a potential difference of 50.0 V will have a kinetic
energy of KE = 50.0 eV ER, e = 0.511 MeV. Thus, it is classical, and its momentum may be
expressed as p = 2me (KE). The de Broglie wavelength is then
l=
h
=
p
h
=
2me (KE)
h
=
2me (e ΔV )
6.63 × 10 −34 J ⋅s
2 ( 9.11× 10 −31 kg ) (1.60 × 10 −19 C )( 50.0 V )
= 0.174 nm
and the correct choice is (b).
324
68719_27_ch27_p324-340.indd 324
1/7/11 2:58:04 PM
Quantum Physics
4.
325
The maximum energy, or minimum wavelength, photon is produced when the electron loses all of
its kinetic energy in a single collision. Therefore, Em ax = hc lm in = KEi = e ΔV , or
lm in =
(6.63 × 10−34 J ⋅s) (3.00 × 108 m s) = 4.14 × 10−7 m = 414 nm
hc
=
e ΔV
(1.60 × 10−19 C) (3.00 J C)
and we see that (c) is the correct choice.
5.
One form of Heisenberg’s uncertainty relation is ΔxΔpx ≥ h 4p , which says that one cannot
determine both the position and momentum of a particle with arbitrary accuracy. Another form
of this relation is ΔEΔt ≥ h 4p , which sets a limit on how accurately the energy can be determined in a finite time interval. Thus, both (a) and (c) are true statements, and the other listed
choices are false.
6.
From the Compton shift formula, the change in the photon’s wavelength is
Δl = l − l0 =
h
6.63 × 10 −34 J ⋅s
(1− cosq ) =
(1− cos180°)
me c
(9.11× 10−31 kg) (3.00 × 108 m s)
= 4.85 × 10 −12 m = 4.85 × 10 −3 nm
and choice (d) is the correct answer.
7.
The comparative masses of the particles of interest are m p ≈ 1 840 me and mHe ≈ 4m p.
Assuming the particles are all classical, their momenta are p p = m p v, pHe = mHe v = 4m p v,
and pe = me ve = m p v 1 840. Their de Broglie wavelengths are l p = h p p = h m p v ,
lHe = h pHe = h 4m p v = l p 4 , and le = h pe = h (m p v 1 840) = 1 840l p . The ranking, from
longest to shortest, of these wavelengths is le > l p > lHe , and choice (e) gives the correct order.
8.
From Bragg’s law, the angle q at which constructive interference of order m will be observed
when x-rays reflect from crystalline planes separated by distance d is given by 2d sinq = ml.
Thus, the separation distance of the planes in the crystal is d = ml 2sinq . With m and l the
same for the two crystals, we see that d is inversely proportional to sinq . Since it is observed that
q A > q B (and hence, sinq A > sinq B), it must be true that d A < d B , and the correct answer is (c).
9.
Diffraction, polarization, interference, and refraction are all processes associated with waves.
However, to understand the photoelectric effect, we must think of the energy transmitted as light
coming in discrete packets, or quanta, called photons. Thus, the photoelectric effect most clearly
demonstrates the particle nature of light, and the correct choice is (b).
10.
Electron diffraction by crystals, first detected by the Davisson-Germer experiment in 1927,
confirmed de Broglie’s hypothesis and, of the listed choices, most clearly demonstrates the wave
nature of electrons. The correct answer is (e).
11.
During the scattering process, the photon will transfer some of its energy to the originally
stationary electron, and the electron will recoil following this process. Since the energy of a
photon is Ephoton = hf , where h is constant, the frequency f of the photon must decrease when the
photon gives up energy. Thus, the correct choice is (a).
12.
The magnitude of the charge is q = e for both the electron and proton, and since the two particles
are accelerated through the same potential difference, they are given identical kinetic energies
KEe = KE p = eΔV. The momentum of a non-relativistic particle having kinetic energy KE = eΔV
and mass m, is p = 2mKE = 2m(eΔV ). The de Broglie wavelength, l = h p, is then seen to be
inversely proportional to m. Since the mass of an electron is smaller than that of a proton, we see
that le > l p when the electron and proton have the same kinetic energy. Thus, the correct choice is (a).
68719_27_ch27_p324-340.indd 325
1/7/11 2:58:06 PM
326
Chapter 27
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
A microscope can see details no smaller than the wavelength of the waves it uses to produce
images. Electrons with kinetic energies of several electron volts have wavelengths of less than a
nanometer, which is much smaller than the wavelength of visible light (having wavelengths ranging from about 400 to 700 nm). Therefore, an electron microscope can resolve details of much
smaller size as compared to an optical microscope.
4.
Measuring the position of a particle implies having photons reflect from it. However, collisions
between photons and the particle will alter the velocity of the particle. Thus, the more accurately
you try to measure one quantity, the more uncertainty you create in your knowledge of the other.
6.
Light has both wave and particle characteristics. In Young’s double-slit experiment, light behaves
as a wave. In the photoelectric effect, it behaves like a particle. Light can be characterized as an
electromagnetic wave with a particular wavelength or frequency, yet at the same time, light can
be characterized as a stream of photons, each carrying a discrete energy, hf.
8.
Ultraviolet light has a shorter wavelength and higher photon energy, Ephoton = hf = hc l, than
visible light.
10.
Increasing the temperature of the substance increases the average kinetic energy of the electrons
inside the material. This makes it slightly easier for an electron to escape from the material when
it absorbs a photon.
12.
Most stars radiate nearly as blackbodies. Of the two stars, Vega has the higher surface temperature and, in agreement with Wien’s displacement law, radiates more intensely at shorter wavelengths and has a bluish appearance, while Arcturus has a more reddish appearance.
14.
No, the crystal cannot produce diffracted beams of visible light. The angles where one could
expect to observe diffraction maxima are given by Bragg’s law, 2d sinq = ml, where m is a nonzero
integer. Because sinq ≤ 1, this equation gives m = (2d l)sinq ≤ 2d l . Since the wavelengths of
visible light (l ∼ 10 2 nm) are much larger than the distances between atomic planes in crystals
(d ∼ 1 nm), there are no nonzero integer values for m that can satisfy Bragg’s law for visible light.
16.
Any object of macroscopic size—including a grain of dust—has an undetectably small wavelength,
so any diffraction effects it might exhibit are very small, effectively undetectable. Recall historically
how the diffraction of sound waves was at one time well known, but the diffraction of light was not.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a)
∼ 100 nm, ultraviolet
(b)
∼ 10 −1 nm, gamma rays
4.
(a)
5.78 × 10 3 K
(b)
501 nm
6.
(a)
2.90 × 10 −19 J photon
(b)
4.23 × 10 3 K
(c)
1.65 × 10 26 W
(d)
5.69 × 10 44 photon s
(c)
1.19 eV
8.
5.7 × 10 3 photons s
10.
(a)
288 nm
(b)
1.04 × 1015 Hz
12.
(a)
only lithium
(b)
0.81 eV
68719_27_ch27_p324-340.indd 326
1/7/11 2:58:09 PM
Quantum Physics
14.
(a) 1.90 eV
16.
(a)
18.
0.455 nm
20.
m = 2, q = 25.8°; m = 3, q = 40.8°; m = 4, q = 60.8°
8.29 × 10 −11 m
(b)
0.216 V
(b)
1.24 × 10 −11 m
(c)
327
lm in decreases
Higher orders cannot be found since sinq cannot exceed a value of 1.00.
22.
1.03 × 10 −3 nm
24.
4.85 × 10 −3 nm
26.
(a) 0.101 nm
28.
(a)
30.
(a) 0.709 nm
32.
(a)
1.99 × 10 −11 m
p = 2mqΔV
(b)
80.9°
(b)
1.98 × 10 −14 m
(b)
414 nm
(b)
l=h
2mqΔV
(c) The proton, with the larger mass, will have the shorter wavelength.
34.
23 m s
36.
(a)
38.
(a) See Solution.
40.
v=c 2 2
42.
(a)
44.
(a)
46.
0.14 keV
48.
0.785 eV
50.
(a) 148 days
(b)
(b)
2.25 m
(b)
5.3 MeV
2.49 × 10 −11 m
(b)
0.29 nm
l = 2.82 × 10 −37 m
(b)
ΔE ≥ 1.06 × 10 −32 J
0.250 m s
(c)
2.88 × 10 −35 %
This result is totally contrary to observations of the photoelectric effect.
PROBLEM SOLUTIONS
27.1
68719_27_ch27_p324-340.indd 327
From Wien’s displacement law,
(a)
T=
0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K
= 2.99 × 10 3 K, or ≈ 3 000 K
=
lm ax
970 × 10 −9 m
(b)
T=
0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K
=
= 2.00 × 10 4 K, or ≈ 20 000 K
lm ax
145 × 10 −9 m
1/7/11 2:58:11 PM
328
27.2
27.3
Chapter 27
Using Wien’s displacement law,
(a)
lm ax =
0.289 8 × 10 − 2 m ⋅K
= 2.898 × 10 −7 m ~100 nm
10 4 K
ultraviolet
(b)
lm ax =
0.289 8 × 10 − 2 m ⋅K
= 2.898 × 10 −10 m ~10 −1 nm
10 7 K
g -rays
From Wien’s displacement law,
lm ax =
27.4
(a)
0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K
= 9.47 × 10 −6 m = 9.47 mm (infrared)
=
T
306 K
The power radiated by an object with surface area A and absolute temperature T is given by
Stefan’s law (see Chapter 11 of the textbook) as P = s AeT 4 , where s is the Stefan-Boltzmann
constant equal to 5.669 6 × 10 −8 W m 2 ⋅K 4 . For an object that approximates a blackbody, the
emissivity is e = 1. Thus, the temperature of our Sun’s surface should be
1
1
1
⎤4
⎤4 ⎡
⎡ P ⎤4 ⎡
P
3.85 × 10 26 W
⎢
⎥
T=⎢
=
=
⎢
⎥
2
⎥
2
⎢ 4p ( 5.669 6 × 10 −8 W m 2 ⋅K 4 ) ( 6.96 × 108 m ) (1) ⎥
⎣ s Ae ⎦
⎢⎣ s ( 4p RSun ) e ⎥⎦
⎣
⎦
= 5.78 × 10 3 K
(b)
Wien’s displacement law gives the peak wavelength of the radiation from the Sun as
lm ax =
27.5
27.6
0.289 8 × 10 − 2 m ⋅K 0.289 8 × 10 − 2 m ⋅K
=
= 5.01× 10 −7 m = 501 nm
T
5.78 × 10 3 K
The energy of a photon is given by E = hf = hc l , where h = 6.63 × 10 −34 J ⋅s is Planck’s constant.
(a)
E=
− 34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV
=
⎜⎝
−2
l
5.00 × 10 m
1.60 × 10 −19
(b)
E=
− 34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV ⎞
=
⎜⎝
⎟ = 2.49 eV
l
500 × 10 −9 m
1.60 × 10 −19 J ⎠
(c)
E=
− 34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV ⎞
=
⎜⎝
⎟ = 249 eV
l
5.00 × 10 −9 m
1.60 × 10 −19 J ⎠
(a)
Epeak =
(b)
T=
(c)
The surface area of the spherical star is A = 4p r 2, and (considering it to be a blackbody) its
emissivity is e = 1. Stefan’s law (see Chapter 11 of the textbook) gives the radiated power as
⎞
−5
⎟ = 2.49 × 10 eV
J⎠
(6.63 × 10−34 J ⋅s) (3.00 × 108 m s) = 2.90 × 10−19 J photon
hc
=
lm ax
685 × 10 −9 m
0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K
=
= 4.23 × 10 3 K
lm ax
685 × 10 −9 m
P = s AeT 4 = s ( 4p r 2 ) eT 4
= 4p ( 5.669 6 × 10 −8 W m 2 ⋅K 4 ) (8.50 × 108 m ) (1)( 4.23 × 10 3 K ) = 1.65 × 10 26 W
2
4
continued on next page
68719_27_ch27_p324-340.indd 328
1/7/11 2:58:13 PM
Quantum Physics
(d)
329
Assuming the average energy of an emitted photon equals the energy of photons at the peak
of the radiation distribution, the number of photons emitted by the star each second is
ΔN
P
1.65 × 10 26 J s
=
=
= 5.69 × 10 44 photon s
Δt
Epeak 2.90 × 10 −19 J photon
27.7
27.8
The energy of a photon having frequency f is given by Ephoton = hf, where Planck’s constant has a
value of h = 6.63 × 10 −34 J ⋅s. This energy may be converted to units of electron volts by use of the
conversion factor 1 eV = 1.60 × 10 −19 J.
(a)
1 eV
⎛
Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 620 × 1012 Hz ) ⎜
⎝ 1.60 × 10 −19
(b)
1 eV
⎛
⎞
Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 3.10 × 10 9 Hz ) ⎜
= 1.28 × 10 −5 eV
⎝ 1.60 × 10 −19 J ⎟⎠
(c)
1 eV
⎛
Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 46.0 × 10 6 Hz ) ⎜
⎝ 1.60 × 10 −19
(a)
(b)
(c)
(a)
2
J s ⋅ m 2 ) ⎡⎢p (8.5×10 −3 m ) 4⎤⎥ (500 ×10 −9 m )
⎣
⎦
= 5.7 ×10 3 photons s
−34
8
(6.63×10 J ⋅ s) (3.00 ×10 m s)
(6.63 × 10
J ⋅s ) ( 3.00 × 108 m s ) ⎛
1 eV
⎜⎝
−9
350 × 10 m
1.60 × 10 − 19
− 34
⎞
⎟ − 1.31 eV = 2.24 eV
J⎠
− 34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛
1 eV
⎞
−7
lc =
=
⎜⎝
⎟ = 5.55 × 10 m = 555 nm
f
2.24 eV
1.60 × 10 − 19 J ⎠
fc =
c 3.00 × 108 m s
=
= 5.41× 1014 Hz
lc
555 × 10 −9 m
At the cutoff wavelength, KEm ax = 0, so the photoelectric effect equation (KEm ax = hc l − f)
gives the cutoff wavelength as
lc =
(b)
−11
From the photoelectric effect equation, the work function is f = hc l − KEm ax , or
f=
27.10
⎞
−7
⎟ = 1.91× 10 eV
J⎠
The energy entering the eye each second is P = I ⋅ A, where A is the area of the pupil. If the
light has wavelength l, the energy of a single photon is Ephoton = hc l . Hence, the number
of photons entering the eye in time Δt is given by N = ΔE Ephoton = P ⋅ ( Δt ) Ephoton , or
N = IA(Δt) (hc l) = IA(Δt)l hc. With I = 4.0 × 10 −11 W m 2 , l = 500 nm, Δt = 1.0 s, and the
opening of the pupil having an area of A = p d 2 4 , where d is the pupil diameter, we find
N ( 4.0 ×10
=
Δt
27.9
⎞
⎟ = 2.57 eV
J⎠
−34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛
1 ev
⎞
−7
=
⎜⎝
⎟ = 2.88 × 10 m = 288 nm
f
4.31 eV
1.60 × 10 −19 J ⎠
The lowest frequency of light that will free electrons from the material is fc = c lc , where
lc is the cutoff wavelength (calculated above). Thus,
fc =
c 3.00 × 108 m s
=
= 1.04 × 1015 Hz
lc
2.88 × 10 −7 m
continued on next page
68719_27_ch27_p324-340.indd 329
1/7/11 2:58:16 PM
330
Chapter 27
(c)
The photoelectric effect equation may be written as KEm ax = Ephoton − f. Therefore, if the
photons incident on this surface have energy Ephoton = 5.50 eV, the maximum kinetic energy
of the ejected electrons is
KEm ax = Ephoton − f = 5.50 eV − 4.31 eV = 1.19 eV
27.11
(a)
(b)
⎛ 1.60 × 10 −19 J ⎞
−18
f = 6.35 eV ⎜
⎟⎠ = 1.02 × 10 J
1 eV
⎝
The energy of a photon having the cutoff frequency or cutoff wavelength equals the work
function of the surface, or Ephoton = hfc = hc lc = f . Thus, the cutoff frequency of a surface
having a work function of f = 6.35 eV is
fc =
(c)
The cutoff wavelength is
lc =
27.12
⎛ 1.60 × 10 −19 J ⎞
f
6.35 eV
15
=
−34
⎟⎠ = 1.53 × 10 Hz
h 6.63 × 10 J ⋅s ⎜⎝
1 eV
hc c 3.00 × 108 m s
= =
= 1.96 × 10 −7 m = 196 nm
f
fc 1.53 × 1015 Hz
(d)
KEm ax = Ephoton − f = 8.50 eV − 6.35 eV = 2.15 eV
(e)
eVs = KEm ax , so the stopping potential is Vs = KEm ax e = 2.15 eV e = 2.15 V .
(a)
The energy of the incident photons is
Ephoton =
− 34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛
1 eV
⎞
=
⎜⎝
⎟ = 3.11 eV
−9
l
400 × 10 m
1.60 × 10 − 19 J ⎠
For photoelectric emission to occur, it is necessary that Ephoton ≥ f. Thus, of the three metals
given, only lithium will exhibit the photoelectric effect.
(b)
27.13
For lithium, KEm ax = Ephoton − f = 3.11 eV − 2.30 eV = 0.81 eV .
The two frequencies of the light allowed to strike the surface are
f1 =
and
c 3.00 × 108 m s
=
= 1.18 × 1015 Hz = 11.8 × 1014 Hz
l1
254 × 10 −9 m
f2 =
c 3.00 × 108 m s
=
= 6.88 × 1014 Hz
l2
436 × 10 −9 m
The graph you draw should look
somewhat like that given at the
right.
The desired quantities, read
from the axes intercepts of the
graph line, should agree within
their uncertainties with
fc = 4.8 × 1014 Hz and f = 2.0 eV
68719_27_ch27_p324-340.indd 330
1/7/11 2:58:19 PM
Quantum Physics
27.14
(a)
The maximum kinetic energy of the ejected electrons is related to the stopping potential
by the expression KEm ax = e ( ΔVs ) . Thus, if the stopping potential is Vs = 0.376 V when
the incident light has wavelength l = 546.1 nm, the photoelectric effect equation gives the
work function of this metal as
f=
=
(b)
331
hc
hc
− KEm ax =
− e ( ΔVs )
l
l
(6.63 × 10
J ⋅s ) ( 3.00 × 108 m s ) ⎛
1 eV
⎞
⎜⎝
⎟ − 0.376 eV = 1.90 eV
546.1× 10 −9 m
1.60 × 10 −19 J ⎠
−34
If light of wavelength l = 587.5 nm is incident on this metal, the maximum kinetic energy
of the ejected electrons is
e ( ΔVs ) = KEm ax =
=
hc
−f
l
(6.63 × 10
J ⋅s ) ( 3.00 × 108 m s ) ⎛
1 eV
⎞
⎜⎝
⎟ − 1.90 eV = 0.216 eV
587.5 × 10 −9 m
1.60 × 10 −19 J ⎠
−34
Thus, the stopping potential will be
ΔVs =
27.15
KEm ax 0.216 eV
=
= 0.216 V
e
e
Assuming the electron produces a single photon as it comes to rest, the energy of that photon is
Ephoton = ( KE )i = e ( ΔV ). The accelerating voltage is then
ΔV =
Ephoton
e
=
−34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) 1.24 × 10 −6 V ⋅ m
=
=
el
l
(1.60 × 10−19 C) l
For l = 1.0 × 10 −8 m, V =
1.24 × 10 −6 V ⋅ m
= 1.2 × 10 2 V
1.0 × 10 −8 m
and for l = 1.0 × 10 −13 m, V =
27.16
1.24 × 10 −6 V ⋅ m
= 1.2 × 10 7 V
1.0 × 10 −13 m
A photon of maximum energy and minimum wavelength is produced when the electron gives up
all its kinetic energy in a single collision, or lm in = hc KE = hc e ΔV .
(a)
If ΔV = 15.0 kV, lm in =
(6.63 × 10
(1.60 × 10
(b)
If ΔV = 100 kV, lm in =
(6.63 × 10 J ⋅s) (3.00 × 10 m s) = 1.24 × 10
(1.60 × 10 C) (100 × 10 V )
(c)
As seen above, lm in decreases when the potential difference, ΔV , increases.
−34
J ⋅s ) ( 3.00 × 108 m s )
−19
−34
27.17
= 8.29 × 10 −11 m .
8
−19
3
−11
m.
A photon of maximum energy or minimum wavelength is produced when the electron gives up
all of its kinetic energy in a single collision within the target. Thus, Em ax = hc lm in = KEe = e ΔV.
If lm in = 70.0 pm = 70.0 × 10 −12 m, the required accelerating voltage is
ΔV =
68719_27_ch27_p324-340.indd 331
C ) (15.0 × 10 3 V )
(6.63 × 10−34 J ⋅s) (3.00 × 108 m s) = 1.78 × 104 V = 17.8 kV
hc
=
elm in
(1.60 × 10−19 C) ( 70.0 × 10−12 m )
1/7/11 2:58:21 PM
332
27.18
Chapter 27
From the Bragg equation, 2d sinq = ml, the interplanar spacing when 0.129-nm x-rays produce
a first order diffraction maximum at q = 8.15° is
d=
27.19
Using Bragg’s law, the wavelength is found to be
l=
27.20
ml
(1)( 0.129 nm )
=
= 0.455 nm
2sinq
2sin8.15°
2 d sinq 2 ( 0.296 nm ) sin 7.6°
= 0.078 nm
=
1
m
From the Bragg equation, 2d sinq = ml, the angle at which the diffraction maximum of order m
will be found when the first order maximum is at q1 = 12.6°, is given by
sinq m = m ( l 2d ) = m sinq1 = m sin (12.6° )
Thus, additional orders will be found at
m = 2:
sinq 2 = 2sin12.6° = 0.436
and
q 2 = sin −1 ( 0.436 ) = 25.8°
m =3:
sinq 3 = 3sin12.6° = 0.654
and
q 3 = sin −1 ( 0.654 ) = 40.8°
m=4:
sinq 4 = 4 sin12.6° = 0.873
and
q 4 = sin −1 ( 0.873) = 60.8°
m =5:
sinq 5 = 5sin12.6° = 1.09
Impossible, since sinq ≤ 1 for all q .
No orders higher than m = 4 will be seen because it is mathematically impossible for the sinq to
be greater than one.
27.21
The interplanar spacing in the crystal is given by Bragg’s law as
d=
27.22
ml
(1)( 0.140 nm )
=
= 0.281 nm
2sinq
2sin14.4°
From the Compton shift equation, the wavelength shift of the scattered x-rays is
Δl =
h
6.63 × 10 −34 J ⋅s
(1− cosq ) =
(1− cos55.0°)
me c
(9.11× 10−31 kg) (3.00 × 108 m s)
⎛ 1 nm ⎞
= 1.03 × 10 −12 m ⎜ −9 ⎟ = 1.03 × 10 −3 nm
⎝ 10 m ⎠
27.23
If the scattered photon has energy equal to the kinetic energy of the recoiling electron, the energy
of the incident photon is divided equally between them. Thus,
Ephoton =
(E )
0
photon
2
⇒
hc
hc
=
, so l = 2 l0 and Δl = 2 l0 − l0 = l0 = 0.001 60 nm
l 2 l0
The Compton scattering formula, Δl = lC (1− cosq ) , then gives the scattering angle as
⎛ Δl ⎞
⎛ 0.001 60 nm ⎞
−1
q = cos−1 ⎜1−
⎟ = cos ⎜1−
⎟ = 70.0°
⎝ 0.002 43 nm ⎠
⎝ lC ⎠
68719_27_ch27_p324-340.indd 332
1/7/11 2:58:24 PM
Quantum Physics
27.24
At point A, the incident photon scatters at angle
q from the first electron. The shift in wavelength that occurs in this scattering process is
given by the Compton equation as
Incident
photon l
Electron 1
A
q
h
l′ − l =
(1− cosq )
me c
333
l′
l″
q
Reflected
photon
B
Electron 2
q ′ = 180° − q
At point B, this intermediate photon scatters from the second electron at angle q ′ = 180° −q .
Since cos(180° −q ) = cos180° cosq + sin180°sinq = − cosq , the Compton equation gives the
wavelength shift that occurs in the scattering at B as
l ′′ − l ′ =
h
(1+ cosq )
me c
Hence, the total wavelength shift that occurs in the two scattering processes is given by
l ′′ − l = ( l ′′ − l ′ ) + ( l ′ − l ) =
or
27.25
l ′′ − l =
h
h
2h
(1+ cosq ) +
(1− cosq ) =
me c
me c
me c
2 ( 6.63 × 10 −34 J ⋅s )
(9.11× 10−31 kg) (3.00 × 108 m s)
= 4.85 × 10 −12 m = 4.85 × 10 −3 nm
The figure at the right shows the situation before
and after the scattering process. Note that the scattering angle is q = 180°, so the Compton equation
gives
l′ − l =
Incident
photon
Before
+x
l, p, E
l′, p′, E′
h
2h
(1− cos180°) =
me c
me c
(a)
pe, KEe
After
Scattered
photon
l′ = l +
Electron
at rest
Recoiling
electron
+x
2 (6.63×10 −34 J ⋅ s)
2h
= 0.110 ×10 −9 m +
= 0.115×10 −9 m
−31
8
me c
9.11×10
kg
3.00
×10
m
s
(
)(
)
The momentum of the incident photon is p = h l, while that of the scattered photon is
p′ = − h l ′ (the negative sign is included since momentum is a vector quantity and the
scattered photon travels in the negative x-direction). Thus, conservation of momentum
gives pe − h l ′ = h l + 0, or the momentum of the recoiling electron is
pe =
⎛
⎞
h h
1
1
+ = (6.63×10 −34 J ⋅ s) ⎜
+
⎟
−9
−9
⎝
l l′
0.110 ×10 m 0.115×10 m ⎠
= 1.18 ×10 −23 kg ⋅ m s
(b)
Assuming the recoiling electron is nonrelativistic, its kinetic energy is given by
(1.18 ×10−23 kg ⋅ m s) = 7.64 ×10−17 J ⎛⎜ 1 eV ⎞⎟ = 478 eV
p2
KEe = e =
⎝ 1.60 ×10 −19 J ⎠
2me
2 (9.11×10 −31 kg)
2
continued on next page
68719_27_ch27_p324-340.indd 333
1/7/11 2:58:27 PM
334
Chapter 27
Note that this energy is very small in comparison to the rest energy of an electron. Thus,
our assumption that the recoiling electron would be non-relativistic is seen to be valid.
27.26
First, observe that v = 2.18 × 10 6 m s << c. Thus, the recoiling electron is nonrelativistic, and its
kinetic energy is KEe = 12 me v 2 , while its momentum is given by pe = me v.
(a)
The Compton equation gives the wavelength of the scattered photon in terms of that of the
incident photon as
l = l0 +
h
(1− cosq )
me c
[1]
Conservation of energy gives the kinetic energy of the recoiling electron as
2
1
2 me v = hc l0 − hc l = hc ( l − l0 ) l0 l. Substituting Equation [1] into this yields
hc ⎡⎣( h me c ) (1− cosq )⎤⎦
1
me v 2 =
2
l0 ⎡⎣l0 + ( h me c ) (1− cosq )⎤⎦
⎛ h ⎞
l0 ⎡⎣l0 + ( h me c ) (1− cosq )⎤⎦ = 2 ⎜
⎟ (1− cosq )
⎝ me v ⎠
2
or
Simplifying, this equation is seen to be of the form al02 + bl0 + c = 0, where a = 1, and the
constants b and c are as computed below:
b=
(6.63×10−34 J ⋅ s) (1− cos17.4°) = 1.11×10−13 m
h
(1− cosq ) =
me c
(9.11×10−31 kg) (3.00 ×108 m s)
⎛ h ⎞
− 2 (6.63×10 −34 J ⋅ s) (1− cos17.4°)
c = − 2⎜
= −1.02 ×10 −20 m 2
⎟ (1− cosq ) =
2
2
−31
6
m
v
⎝ e ⎠
(9.11×10 kg) (2.18 ×10 m s)
2
2
From the quadratic formula, l0 = ⎡ −b ± b 2 − 4ac ⎤ 2a , using only the upper sign since l0
⎣
⎦
must be positive, we find l0 = 1.01× 10 −10 m = 0.101 nm .
(b)
Choosing the x-axis to be the direction of the
incident photon’s motion, the initial momentum
in the y-direction is zero. Thus, conserving
momentum in the y-direction gives
+y
l0
q
where q = 17.4°, pe = me v, and p = h l . From
Equation [1] and the result of part (a), the
wavelength of the scattered photon is
l = l0 +
+x
f
pe sinf = psinq
l, p = h/l
me, pe, KEe
u
(6.63×10−34 J ⋅ s) (1− cos17.4°)
h
(1− cosq ) = 1.01×10−10 m +
me c
(9.11×10−31 kg) (3.00 ×108 m s)
or l = 1.011× 10 −10 m. Thus, the scattering angle for the electron is
f = sin −1 ( psinq pe ) = sin −1 ( h sinq lme v ) , which gives
⎡
⎤
(6.63 × 10−34 J ⋅s) sin17.4°
f = sin −1 ⎢
⎥ = 80.9°
−10
−31
6
⎢⎣ (1.011× 10 m ) ( 9.11× 10 kg ) ( 2.18 × 10 m s ) ⎥⎦
68719_27_ch27_p324-340.indd 334
1/7/11 2:58:29 PM
Quantum Physics
27.27
(a)
From l = h p = h mv, the speed is
v=
(b)
27.28
l=
h
6.63 × 10 −34 J ⋅s
=
= 1.46 × 10 3 m s = 1.46 km s
me l ( 9.11× 10 −31 kg ) ( 5.00 × 10 −7 m )
h
6.63 × 10 −34 J ⋅s
=
= 7.28 × 10 −11 m
me v ( 9.11× 10 −31 kg ) (1.00 × 10 7 m s )
The de Broglie wavelength of a particle of mass m is l = h p, where the momentum is given
2
by p = g mv = mv 1− ( v c ) . Note that when the particle is not relativistic, then g ≈ 1, and this
relativistic expression for momentum reverts back to the classical expression.
(a)
For a proton moving at speed v = 2.00 × 10 4 m s, v << c and g ≈ 1, so
l=
(b)
h
6.63 × 10 −34 J ⋅s
=
= 1.99 × 10 −11 m
m p v (1.67 × 10 −27 kg ) ( 2.00 × 10 4 m s )
For a proton moving at speed v = 2.00 × 10 7 m s,
l=
h
h
2
1− ( v c )
=
lm p v m p v
(6.63 × 10 J ⋅s)
(1.67 × 10 kg) ( 2.00 × 10
−34
=
27.29
335
−27
For relativistic particles, p =
E 2 − ER2
c
2
7
m s)
and l =
⎛ 2.00 × 10 7 m s ⎞
1− ⎜
= 1.98 × 10 −14 m
⎝ 3.00 × 108 m s ⎟⎠
h
=
p
hc
E − ER2
2
.
For 3.00 MeV electrons, E = KE + ER = 3.00 MeV + 0.511 MeV = 3.51 MeV, so
l=
27.30
(a)
(6.63 × 10
J ⋅s ) ( 3.00 × 108 m s ) ⎛ 1 MeV ⎞
−13
⎜⎝
⎟⎠ = 3.58 × 10 m
−13
2
2
1.60
×
10
J
( 3.51 MeV ) − ( 0.511 MeV )
−34
A 3.00 eV electron is nonrelativistic and its momentum is p = 2me KE , so its wavelength is
l=
h
h
6.63×10 −34 J ⋅ s
=
=
p
2me KE
2 (9.11×10 −31 kg) (3.00 eV ) (1.60 ×10 −19 J 1 eV )
= 7.09 ×10 −10 m = 0.709 nm
(b)
A photon has energy of Ephoton = hc l , so its wavelength is l = hc Ephoton. Because a photon
has zero rest energy, all of its energy is kinetic energy. In this case, Ephoton = 3.00 eV, and
the wavelength is
l=
27.31
(a)
hc
Ephoton
=
(6.63 × 10
−34
J ⋅s ) ( 3.00 × 108 m s )
( 3.00 eV )(1.60 × 10 −19 J 1 eV )
= 4.14 × 10 −7 m = 414 nm
The required electron momentum is
p=
1 keV
h 6.63 × 10 −34 J ⋅s ⎛
⎞
−7 keV ⋅s
=
⎜
⎟ = 4.1× 10
l
1.0 × 10 −11 m ⎝ 1.60 × 10 −16 J ⎠
m
continued on next page
68719_27_ch27_p324-340.indd 335
1/7/11 2:58:32 PM
336
Chapter 27
and the total energy is
E=
p2 c 2 + ER2
2
keV ⋅s ⎞
2
⎛
8
= ⎜ 4.1× 10 −7
⎟ ( 3.00 × 10 m s ) + ( 511 keV ) = 526 keV
⎝
m ⎠
2
The kinetic energy is then
KE = E − ER = 526 keV − 511 keV = 15 keV
27.32
27.33
−34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛
1 keV
⎞
2
=
⎜⎝
⎟ = 1.2 × 10 keV
l
1.0 × 10 −11 m
1.60 × 10 −16 J ⎠
(b)
Ephoton =
(a)
From conservation of energy, the increase in kinetic energy must equal the decrease in
potential energy, or KE − 0 = q ΔV. Since the particle is nonrelativistic, its kinetic energy
and momentum are related by the expression KE = p2 2m. Thus, the momentum
is p2 = 2m ( KE ) = 2mq ΔV , and p = 2mq ΔV .
(b)
The de Broglie wavelength is l = h p. Therefore, using the result of part (a) gives
l = h 2mq ΔV .
(c)
The electron and proton have the same magnitude charge. Thus, when both are
accelerated through the same magnitude potential difference, the resulting de Broglie
wavelengths are inversely proportional to the square root of the masses of the particles.
The proton (with the larger mass) will have the shorter wavelength .
From the uncertainty principle, the minimum uncertainty in the momentum of the electron is
Δpx =
h
6.63 × 10 −34 J ⋅s
=
= 5.3 × 10 −25 kg ⋅ m s
4p ( Δx ) 4p ( 0.10 × 10 −9 m )
so the uncertainty in the speed of the electron is
Δvx =
Δpx 5.3×10 −25 kg ⋅ m s
=
= 5.8 ×10 5 m s or ~ 10 6 m s
9.11×10 −31 kg
me
If the speed is on the order of the uncertainty in the speed, then v ~ 10 6 m s .
27.34
Assuming the electron is nonrelativistic, the uncertainty in its momentum is Δp = me Δv. From the
uncertainty principle, Δx Δpx ≥ h 4p , so if there is an uncertainty of Δx = 2.5 mm = 2.5 × 10 −6 m in
the position of the electron, the minimum uncertainty in its speed is
( Δv )
x
27.35
m in
=
( Δp )
x
me
m in
=
h
6.63×10 −34 J ⋅ s
=
= 23 m s
4p me ( Δx )m ax 4p (9.11×10 −31 kg) ( 2.5×10 −6 m )
The uncertainty in the magnitude of the velocity of each particle is
Δvx = vx ⋅ ( 0.010 0%) = ( 500 m s ) ( 0.010 0 × 10 −2 ) = 5.00 × 10 −2 m s
If the mass is known precisely, the uncertainty in momentum is Δpx = m(Δvx ), and the minimum
uncertainty in position is Δx m in = h 4p (Δpx ) = h 4p m(Δvx ).
continued on next page
68719_27_ch27_p324-340.indd 336
1/7/11 2:58:35 PM
Quantum Physics
337
For the electron:
Δx m in =
h
6.63×10 −34 J ⋅ s
=
4p me (Δvx ) 4p (9.11×10 −31 kg) (5.00 ×10 −2 m s)
= 1.16 ×10 −3 m = 1.16 mm
For the bullet:
Δx m in =
27.36
(a)
h
6.63 × 10 −34 J ⋅s
=
= 5.28 × 10 −32 m
4p m(Δvx ) 4p ( 0.020 0 kg ) ( 5.00 × 10 −2 m s )
With uncertainty Δx in position, the minimum uncertainty in the speed is
( Δv )
x
(b)
m in
=
( Δp )
x
m
m in
=
h
2p J ⋅s
=
= 0.250 m s
4p m ( Δx ) 4p ( 2.00 kg )(1.00 m )
If we knew Fuzzy’s initial position and velocity exactly, his final position after an elapsed
time t would be given by x f = xi + vx t. Since we have uncertainty in both the initial position
and the speed, the uncertainty in the final position is
Δx f = Δxi + ( Δvx ) t = 1.00 m + ( 0.250 m s )( 5.00 s ) = 2.25 m
27.37
The maximum time one can use in measuring the energy of the particle is equal to the lifetime
of the particle, or Δtm ax ≈ 2 ms. One form of the uncertainty principle is ΔE Δt ≥ h 4p . Thus, the
minimum uncertainty one can have in the measurement of a muon’s energy is
ΔEm in =
27.38
h
6.63 × 10 −34 J ⋅s
=
= 3 × 10 −29 J
4p Δtm ax
4p ( 2 × 10 −6 s )
(a)
For a nonrelativistic particle, KE = 12 mv 2 = ( mv ) 2 m = p2 2 m .
(b)
From the uncertainty principle,
2
Δpx ≥
h
6.63 × 10 −34 J ⋅s
=
= 5.3 × 10 −20 kg ⋅ m s
4p ( Δx ) 4p (1.0 × 10 −15 m )
Since the momentum must be at least as large as its own uncertainty, the minimum kinetic
energy is
(5.3 × 10−20 kg ⋅ m s) ⎛ 1 MeV
p2
= m in =
⎜
2m
2 (1.67 × 10 −27 kg ) ⎝ 1.60 × 10 −13
2
KEm in
27.39
(a)
p=
h 6.63 × 10 −34 J ⋅s
=
= 2.21× 10 −32 kg ⋅ m s
l
3.00 × 10 −2 m
(b)
f=
c 3.00 × 108 m s
= 1.00 × 1010 Hz
=
l
3.00 × 10 −2 m
(c)
E = hf = ( 6.63 × 10 −34 J ⋅s ) (1.00 × 1010 Hz )
⎞
⎟ = 5.3 MeV
J⎠
1 eV
⎛
⎞
= 6.63 × 10 −24 J ⎜
= 4.14 × 10 −5 eV
⎝ 1.60 × 10 −19 J ⎟⎠
68719_27_ch27_p324-340.indd 337
1/7/11 2:58:38 PM
338
27.40
Chapter 27
The de Broglie wavelength is l = h p, and the Compton wavelength is lc = h me c. Thus, if
l = lc , it is necessary that p = me c. The relativistic expression for the momentum of an electron
2
is p = g me v, so if p = me c, we must have g me v = me c, or v c = 1 g = 1− ( v c ) .
Squaring both sides of this result gives ( v c ) = 1− ( v c ) , or 2 ( v c ) = 1, and
2
v=
27.41
2
2
c
c 2
=
2
2
The total mechanical energy of this object is
Em ech = KE f = PEi = mgyi = ( 2.0 kg ) ( 9.80 m s 2 )( 5.0 m ) = 98 J
A photon of light having wavelength l = 5.0 × 10 −7 m has an energy of
Ephoton =
−34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s )
=
= 4.0 × 10 −19 J
l
5.0 × 10 −7 m
Thus, if all the initial gravitational potential energy of the object were converted to light with
l = 5.0 × 10 −7 m, the number of photons that would be produced is
n=
27.42
(a)
Em ech
98 J
=
= 2.5 × 10 20 photons
−19
Ephoton 4.0 × 10
J photon
Minimum wavelength photons are produced when an electron gives up all its kinetic energy
in a single collision. Then, Ephoton = 50 000 eV and
lm in =
(b)
hc
Ephoton
(6.63 × 10
=
(5.00 × 10
−34
4
J ⋅s ) ( 3.00 × 108 m s )
eV ) (1.60 × 10
−19
J eV )
= 2.49 × 10 −11 m
From Bragg’s law, the interplanar spacing is
(1)( 2.49 × 10 −11 m )
ml
d=
=
= 2.9 × 10 −10 m = 0.29 nm
2sinq
2sin ( 2.5° )
27.43
(a)
The peak radiation occurs at approximately 560 nm wavelength. From Wien’s
displacement law,
T=
27.44
0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K
=
≈ 5 200 K
lm ax
560 × 10 −9 m
(b)
Clearly, a firefly is not at this temperature, so this is not blackbody radiation .
(a)
From v 2 = v02 + 2a y ( Δy ) , Johnny’s speed just before impact is
v = 2 g Δy = 2 ( 9.80 m s )( 50.0 m ) = 31.3 m s
and his de Broglie wavelength is
l=
h
6.63 × 10 −34 J ⋅s
=
= 2.82 × 10 −37 m
mv ( 75.0 kg ) ( 31.3 m s )
continued on next page
68719_27_ch27_p324-340.indd 338
1/7/11 2:58:40 PM
Quantum Physics
(b)
The energy uncertainty is
ΔE ≥
(c)
27.45
339
% error =
h
6.63 × 10 −34 J ⋅s
=
= 1.06 × 10 −32 J
4p ( Δt ) 4p ( 5.00 × 10 −3 s )
(1.06 × 10−32 J)(100%) = 2.88 × 10−35 %
ΔE
(100%) ≥
mg Δy
( 75.0 kg) (9.80 m s2 )(50.0 m )
The magnetic force supplies the centripetal acceleration for the electrons, so m(v 2 r) = qvB, or
p = mv = qrB. The maximum kinetic energy is then KEm ax = p2 2m = q 2 r 2 B 2 2 m , or
KEm ax
(1.60 × 10
=
−19
J ) ( 0.200 m ) ( 2.00 × 10 −5 T )
2
2
2 ( 9.11× 10 −31 kg )
2
= 2.25 × 10 −19 J
The work function of the surface is given by f = Ephoton − KEm ax = hc l − KEm ax , or
f=
(6.63 × 10
−34
J ⋅s ) ( 3.00 × 108 m s )
450 × 10 −9 m
− 2.25 × 10 −19 J
1 eV
⎛
⎞
= 2.17 × 10 −19 J ⎜
= 1.36 eV
⎝ 1.60 × 10 −19 J ⎟⎠
27.46
The energy of the incident photon is E0 = 6.20 keV, and its wavelength is l0 = hc E0 .
In a head-on collision with an electron, the photon is back scattered, or the scattering angle
is q = 180°. The Compton equation then gives the wavelength of the scattered photon as
l = l0 + lc (1− cos180°) = l0 + 2lc , where lc = 2.43 × 10 −12 m is the Compton wavelength. Thus,
l=
(6.63 × 10−34 J ⋅s) (3.00 × 108 m s) + 2 2.43 × 10−12 m = 2.05 × 10−10 m
hc
+ 2lc =
(
)
E0
( 6.20 keV )(1.60 × 10 −16 J 1 keV )
The energy of the scattered photon is E = hc l , and from conservation of energy, the kinetic
energy of the recoiling electron is
KEe = E0 − E = 6.20 keV −
or
27.47
(6.63×10
J ⋅ s) (3.00 ×108 m s) ⎛ 1 keV
⎞
⎜
⎟
−10
−16
⎝ 1.60 ×10 J ⎠
2.05×10 m
−34
KEe = 6.20 keV − 6.06 keV = 0.14 keV
From the photoelectric effect equation, KEm ax = Ephoton − f =
For l = l0 , KEm ax = 1.00 eV,
For l =
l0
, KEm ax = 4.00 eV,
2
hc
− f.
l
so
1.00 eV =
hc
−f
l0
[1]
giving
4.00 eV =
2 hc
−f
l0
[2]
Multiplying Equation [1] by a factor of 2 and subtracting the result from Equation [2] gives the
work function as f = 2.00 eV .
68719_27_ch27_p324-340.indd 339
1/7/11 2:58:43 PM
340
27.48
Chapter 27
From the photoelectric effect equation, KEm ax = Ephoton − f = hc l − f.
For l = 670 nm, KEm ax = E1 ,
E1 =
so
For l = 520 nm, KEm ax = 1.50 E1 , giving
hc
−f
670 nm
1.50 E1 =
[1]
hc
−f
520 nm
[2]
Multiplying Equation [1] by a factor of −1.50 and adding the result to Equation [2] gives
1.00 ⎞
⎛ −1.50
0 = hc ⎜
+
+ (1.50 − 1.00 )f
⎝ 670 nm 520 nm ⎟⎠
The work function for the material is then
f=
or
27.49
(a)
(6.63 × 10
−34
J ⋅s ) ( 3.00 × 108 m s ) ⎛ 1.50
1.00 ⎞ ⎛ 1 nm ⎞ ⎛
1 eV
⎞
−
⎜⎝
⎟⎜
⎟⎜
⎟
0.50
670 nm 520 nm ⎠ ⎝ 10 −9 m ⎠ ⎝ 1.60 × 10 −19 J ⎠
f = 0.785 eV .
If the single photon produced has wavelength l = 1.00 × 10 −8 m, the kinetic energy of the
electron was
KEe =
−34
8
hc (6.63×10 J ⋅ s) (3.00 ×10 m s)
=
= 1.99 ×10 −17 J = 124 eV
l
1.00 ×10 −8 m
This electron is nonrelativistic and its speed is given by
v=
(b)
2 (1.99 ×10 −17 J) ⎛
⎞
2 ( KEe )
c
=
⎜
⎟ = 0.022 0 c
−31
8
me
9.11×10 kg ⎝ 3.00 ×10 m s ⎠
When the single photon produced has wavelength l = 1.00 × 10 −13 m,
−34
8
hc (6.63×10 J ⋅ s) (3.00 ×10 m s) ⎛ 1 MeV
KEe =
=
⎜
⎝ 1.60 ×10 −13
l
1.00 ×10 −13 m
⎞
⎟ = 12.4 MeV
J⎠
the electron is highly relativistic and KE = (g − 1) ER , giving
g = 1+
KE
12.4 MeV
= 1+
= 25.3
ER
0.511 MeV
Then, v = c 1− 1 g
27.50
(a)
2
= c 1− 1 ( 25.3) = 0.999 2c .
2
If light were a classical wave, the time required for a surface of area A = p r 2 to
absorb energy E when illuminated with radiation of intensity I would be given by
E = P ⋅(Δt) = IA ⋅(Δt) = I(p r 2 )(Δt). Using the given data values, we find the time to absorb
energy E = 1.00 eV to be
1.00 eV (1.60 × 10 −19 J 1 eV ) ⎛ 1 day ⎞
E
Δt =
=
⎟ = 148 days
2 ⎜
I (p r 2 ) ( 500 J s ) p ( 2.82 × 10 −15 m ) ⎝ 8.64 × 10 4 s ⎠
(b)
68719_27_ch27_p324-340.indd 340
This result is totally contrary to observations of the photoelectric effect. There, when a
surface is illuminated with light above the threshold frequency of the surface, photoelectrons
are seen to be ejected with a delay of less than 10 −9 s.
1/7/11 2:58:46 PM
28
Atomic Physics
QUICK QUIZZES
1.
Choice (b). The allowed energy levels in a one-electron atom may be expressed as
En = −Z 2 (13.6 eV ) n 2 , where Z is the atomic number. Thus, the ground state ( n = 1 level ) in
helium, with Z = 2, is lower than the ground state in hydrogen, with Z = 1.
2.
(a)
For n = 5, there are 5 allowed values of , namely = 0, 1, 2, 3, and 4.
(b)
Since m ranges from − to + in integer steps, the largest allowed value of ( = 4 in
this case) permits the greatest range of values for m . For n = 5, there are 9 possible values
for m : –4, –3, –2, –1, 0, +1, +2, +3, and +4.
(c)
For each value of , there are 2 + 1 possible values of m . Thus, there is 1 distinct pair
with = 0, 3 distinct pairs with = 1, 5 distinct pairs with = 2, 7 distinct pairs with = 3,
and 9 distinct pairs with = 4. This yields a total of 25 distinct pairs of and m that are
possible when n = 5.
3.
Choice (d). Krypton has a closed configuration consisting of filled n = 1, n = 2, and n = 3 shells
as well as filled 4s and 4 p subshells. The filled n = 3 shell (the next to outer shell in krypton)
has a total of 18 electrons, 2 in the 3s subshell, 6 in the 3p subshell and 10 in the 3d subshell.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
2
2
The energy levels in a single electron atom having atomic number Z are En = − Z (13.6 eV ) n .
For beryllium, Z = 4, and for the ground state, n = 1. Thus,
E1 =
− 4 2 (13.6 eV )
= −218 eV
12
and the correct answer is choice (b).
2.
Wavelengths of the hydrogen spectrum are given by 1 l = RH (1 n 2f − 1 ni2 ), where the Rydberg
constant is RH = 1.097 373 2 × 10 7 m −1 . Thus, with n f = 3 and ni = 5,
1
1⎞
⎛ 1
= 1.097 373 2 × 10 7 m −1 ⎜ 2 − 2 ⎟ = 7.80 × 10 5 m −1
⎝3
l
5 ⎠
and l = 1 (7.80 × 10 5 m −1 ) = 1.28 × 10 −6 m, so (a) is seen to be the correct choice.
3.
There are 6 distinct possible downward transitions with 4 energy levels. These transitions are:
4 → 1, 4 → 2, 4 → 3, 3 → 1, 3 → 2, and 2 → 1. Thus, assuming that each transition has a unique
photon energy, Ephoton = ΔE = Ei − E f , associated with it, there are 6 different wavelengths
l = hc Ephoton the atom could emit, and (e) is the correct choice.
341
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342
Chapter 28
4.
With a principal quantum number of n = 3, there are 3 possible values of the orbital quantum
number, = 0, 1, 2. There are a total of 2 ( 2 + 1) possible quantum states for each value of ;
2 + 1 possible values of the orbital magnetic quantum number m , and 2 possible spin orientations (ms = ± 12 ) for each value of m . Thus, there are 10 3d states (having n = 3, = 2), 6 3p
states (with n = 3, = 1), and 2 3s states (with n = 3, = 0), giving a grand total of 10 + 6 + 2 = 18
n = 3 states, and the correct choice is (e).
5.
The structure of the periodic table is the result of the Pauli exclusion principle, which states that
no two electrons in an atom can ever have the same set of values for the set of quantum numbers
n, , m , and ms . This principle is best summarized by choice (c).
6.
Of the electron configurations listed, (b) and (e) are not allowed. Choice (b) is not possible
because the Pauli exclusion principle limits the number of electrons in any p subshell to a
maximum of 6. Choice (e) is impossible because the selection rules of quantum mechanics
limit the maximum value of to n − 1. Thus, a 2d state (n = 2, = 2 ) cannot exist.
7.
All states associated with = 2 are referred to as d states. Thus, all 10 possible quantum states
having n = 3, = 2 are called 3d states (see Question 4 above), and the correct answer is choice (c).
8.
If it were possible for the spin quantum number to take on the four values ms = ± 32 and ± 12 ,
the first closed shell would occur for beryllium with 4 electrons in states of (1,0,0, 32 ), (1,0,0, 12 ),
(1,0,0,− 12 ), and (1,0,0,− 32 ). The correct answer is choice (c).
9.
Since the electron is in some bound quantum state of the atom, the atom is not ionized, and
choice (a) is false. The fact that the electron is in a d state means that its orbital quantum number
is = 2, so choice (b) is false. Also, since the maximum value of is n − 1, choice (e) is false.
Finally, the ground state of hydrogen is a 1s state, so choice (d) is false, leaving (c) as the only
true statement in the list of choices.
10.
According to de Broglie’s interpretation of Bohr’s quantization postulate, the circumference of
the n = 3 orbit would be exactly 3 wavelengths of an electron in that orbit, and the circumference
of the n = 1 orbit would equal the wavelength of an electron in this orbit. However, the de Broglie
wavelength of the electron is given by
ln =
⎛ 2p 2 ⎞
hr
hrn
2p rn 2p ⎛ n 2 2 ⎞
h
h
h
=
=
= n =
=
=
n
=
⎜⎝ m k e 2 ⎟⎠
pn me vn me vn rn rn
Ln
n ( h 2p )
n
n ⎜⎝ me ke e 2 ⎟⎠
e e
Thus, the wavelength of an electron in the n = 3 orbit is 3 times longer than the wavelength
of the electron when in the n = 1 orbit, and the circumference of the n = 3 orbit must be
C3 = 3l3 = 3 ( 3l1 ) = 9l1 = 9C1. Choice (c) is the correct answer for this question.
11.
An electron in the ground state of hydrogen must be given 10.2 eV of energy to be excited to an
n = 2 state in hydrogen. However, it would have to be given 13.6 eV of energy to remove it from
the atom. Thus, incident electrons having 10.5 eV of kinetic energy have sufficient energy to raise
the atom to an excited state but do not possess enough energy to ionize the atom. Choice (a) is the
correct answer.
12.
Choices (a), (b), (d), and (e) are basic postulates of the Bohr model of one-electron atoms.
However, Bohr made no assumptions regarding quantization of charge. Thus, the correct answer
for this question is choice (c).
13.
A photon emitted by an atom carries energy away from the atom. Thus, the atom emitting the
photon must move to a lower allowed energy state, and choice (c) is the correct answer.
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Atomic Physics
343
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
Neon signs do not emit a continuous spectrum, as can be determined by observing the light from
the sign through a spectrometer. The discrete wavelengths a neon gas can produce are determined
by the differences in the energies of the allowed states within the neon atom. The specific wavelengths and intensities account for the color of the sign.
4.
In a neutral helium atom, one electron can be modeled as moving in an electric field created by
the nucleus and the other electron. According to Gauss’s law, if the electron is above the ground
state it moves in the electric field of a net charge of +2e − 1e = +1e. We say the nuclear charge
is screened by the inner electron. The electron in a He + ion moves in the field of the unscreened
nuclear charge of 2 protons. Then, the potential energy function for the electron is about double
that of one electron in the neutral atom.
6.
Classically, the electron can occupy any energy state. That is, all energies would be allowed.
Therefore, if the electron obeyed classical mechanics, its spectrum, which originates from
transitions between states, would be continuous rather than discrete.
8.
Fundamentally, three quantum numbers describe an orbital wave function because we live in
three-dimensional space. They arise mathematically from boundary conditions on the wave
function, expressed as a product of a function of r, a function of q , and a function of f .
10.
In both cases the answer is yes. Recall that the ionization energy of hydrogen is 13.6 eV. The
electron can absorb a photon of energy less than 13.6 eV by making a transition to some intermediate state such as one with n = 2. It can also absorb a photon of energy greater than 13.6 eV,
but in doing so, the electron would be separated from the proton and have some residual kinetic
energy.
12.
It replaced the simple circular orbits in the Bohr theory with electron clouds. More important,
quantum mechanics is consistent with Heisenberg’s uncertainty principle, which tells us about the
limits of accuracy in making measurements. In quantum mechanics, we talk about the probabilistic nature of the outcome of a measurement on a system, a concept which is incompatible with
the Bohr theory. Finally, the Bohr theory of the atom contains only one quantum number n, while
quantum mechanics provides the basis for additional quantum numbers to explain the finer details
of atomic structure.
14.
Each of the given atoms has a single electron in an = 0 ( or s ) state outside a fully closedshell core, shielded from all but one unit of the nuclear charge. Since they reside in very similar
environments, one would expect these outer electrons to have nearly the same electrical potential energies and hence nearly the same ionization energies. This is in agreement with the given
data values. Also, since the distance of the outer electron from the nuclear charge should tend
to increase with Z (to allow for greater numbers of electrons in the core), one would expect the
ionization energy to decrease somewhat as atomic number increases. This is also in agreement
with the given data.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a)
1 875 nm, 1 281 nm, 1 094 nm
(b) All are in the infrared region of the spectrum.
4.
68719_28_ch28_p341-359.indd 343
1.94 mm
1/7/11 3:01:52 PM
344
Chapter 28
6.
45 fm
8.
(a)
2.19 × 10 6 m s
(b)
13.6 eV
(c)
− 27.2 eV
10.
(a)
1 281 nm
(b)
2.34 × 1014 Hz
(c)
0.970 eV
12.
(a) Transition II
(b) Transition I
14.
(a)
(b)
12.1 eV, 1.89 eV, 10.2 eV
16.
See Solution.
18.
(a) 6 distinct wavelengths
(b)
1.88 × 10 3 nm
(c)
Paschen series
20.
(a)
2.89 × 10 34 kg ⋅ m 2 s
(b)
2.75 × 10 68
(c)
7.27 × 10 −69
22.
(a)
KEn = − 12 PEn
(b)
ΔPE = 2E
(c)
ΔKE = −E
24.
(a)
En = −(122 eV) n 2
(b)
−7.63 eV
(c)
−30.5 eV
(d)
22.9 eV = 3.66 × 10 −18 J
(e)
5.52 × 1015 Hz, 54.3 nm
(f )
deep ultraviolet region
26.
(a)
See Solution.
(b)
See Solution.
28.
(a)
4:
(b)
7: m = −3, − 2, − 1, 0, +1, +2, and +3.
30.
(a)
1s 2 2s 2 2 p3
12.1 eV
= 0, 1, 2, and 3
(b) ( n = 1,
32.
= 0, m = 0, ms = ± 12 ); ( n = 2,
( n = 2,
= 1, m = −1, ms = ± 12 ); ( n = 2,
( n = 2,
= 1, m = 1, ms = ± 12 )
(c) Transitions II and III
= 0, m = 0, ms = ± 12 );
= 1, m = 0, ms = ± 12 );
(a) 30 allowed states – see Solution for tables
(b) 36 possible combinations – six possible states for each electron independently
34.
0.0311 nm
36.
germanium
38.
137
40.
(a)
4.20 mm
(c)
8.84 × 1016 photons mm 3
42.
(a) See Solution.
(d)
68719_28_ch28_p341-359.indd 344
(b)
1.05 × 1019 photons
(b)
2.54 × 10 74
(c)
1.18 × 10 −63 m
This number is much smaller than the radius of an atomic nucleus (∼ 10 −15 m), so the
distance between quantized orbits of the Earth is too small to observe.
1/7/11 3:01:53 PM
Atomic Physics
44.
(a) 135 eV
(b)
345
≈ 10 times the magnitude of the ground state energy
46.
PROBLEM SOLUTIONS
28.1
28.2
28.3
(a)
The wavelengths in the Lyman series of hydrogen are given by 1 l = RH (1 − 1 n 2 ), where
n = 2, 3, 4,… , and the Rydberg constant is RH = 1.097 373 2 × 10 7 m −1 . This can also be
written as l = (1 RH ) ⎡⎣ n 2 ( n 2 − 1) ⎤⎦ , so the first three wavelengths in this series are
l1 =
⎛ 22 ⎞
1
= 1.215 × 10 −7 m = 121.5 nm
7
−1 ⎜
1.097 373 2 × 10 m ⎝ 22 − 1 ⎟⎠
l2 =
⎛ 32 ⎞
1
= 1.025 × 10 −7 m = 102.5 nm
1.097 373 2 × 10 7 m −1 ⎜⎝ 32 − 1 ⎟⎠
l3 =
⎛ 42 ⎞
1
= 9.720 × 10 −8 m = 97.20 nm
7
−1 ⎜
1.097 373 2 × 10 m ⎝ 4 2 − 1 ⎟⎠
(b)
These wavelengths are all in the far ultraviolet region of the spectrum.
(a)
The wavelengths in the Paschen series of hydrogen are given by 1 l = RH (1 32 − 1 n 2 ),
where n = 4, 5, 6,… , and the Rydberg constant is RH = 1.097 373 2 × 10 7 m −1 . This can
also be written as l = (1 RH )[9n 2 (n 2 − 9)], so the first three wavelengths in this series
are
⎡ 9(4)2 ⎤
−6
⎢ 4 2 − 9 ⎥ = 1.875 × 10 m = 1 875 nm
⎣
⎦
l1 =
1
1.097 373 2 × 10 7 m −1
l2 =
⎡ 9(5)2 ⎤
1
= 1.281 × 10 −6 m = 1 281 nm
7
−1 ⎢ 2
1.097 373 2 × 10 m ⎣ 5 − 9 ⎥⎦
l1 =
1
1.097 373 2 × 10 7 m −1
⎡ 9(6)2 ⎤
−6
⎢ 62 − 9 ⎥ = 1.094 × 10 m = 1 094 nm
⎣
⎦
(b)
These wavelengths are all in the infrared region of the spectrum.
(a)
From Coulomb’s law,
k qq
(8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10−19 C)
F = e 21 2 =
2
r
(1.0 × 10−10 m )
2
= 2.3 × 10 −8 N
continued on next page
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346
Chapter 28
(b)
The electrical potential energy is
PE =
9
2
2
−19
−19
ke q1 q2 (8.99 × 10 N ⋅ m C ) ( −1.60 × 10 C ) (1.60 × 10 C )
=
r
1.0 × 10 −10 m
1 eV
⎛
= − 2.3 × 10 −18 J ⎜
⎝ 1.60 × 10 −19
28.4
If E2 , E5 , and E6 , are the energies of an electron in the second, fifth, and sixth excited states within
this atom, respectively, the energies of the photons produced are: Ephoton = hc l1 = E2 − E5 ,
1
Ephoton = hc l2 = E2 − E6 , and Ephoton = hc l3 = E5 − E6 . Thus, we have
2
3
hc l3 = E5 − E6 = ( E2 − E6 ) − ( E2 − E5 ) , or hc l3 = hc l2 − hc l1. This reduces to
(
)
(
l3 =
28.5
(a)
(c)
No.
(8.99 × 10 N ⋅ m C ) (1.60 × 10 C)
(9.11 × 10 kg) (1.0 × 10 m )
2
−19
2
−31
−10
2
= 1.6 × 10 6 m s
v 1.6 × 10 6 m s
=
= 5.3 × 10 −3 << 1, so the electron is not relativistic.
c 3.00 × 108 m s
The de Broglie wavelength for the electron is l = h p = h me v , or
l=
(d)
)
The electrical force supplies the centripetal acceleration of the electron, so
me v 2 r = ke e 2 r 2 or v = ke e 2 me r .
9
(b)
(
)
l1 l2
( 520 nm )( 410 nm )
=
= 1.94 × 10 3 nm = 1.94 mm
l1 − l2
520 nm − 410 nm
v=
28.6
⎞
⎟ = −14 eV
J⎠
6.63 × 10 −34 J ⋅ s
= 4.5 × 10 −10 m = 0.45 nm
(9.11 × 10−31 kg) (1.6 × 106 m s)
Yes. The wavelength and the atom are roughly the same size.
Assuming a head-on collision, the a-particle comes to rest momentarily at the point of closest
approach. From conservation of energy,
KE f + PE f = KEi + PEi
or
0+
ke ( 2e ) ( 79e )
k ( 2e ) ( 79e )
= KEi + e
rf
ri
With ri → ∞ , this gives the distance of closest approach as
rf =
9
2
2
−19
158ke e 2 158 (8.99 × 10 N ⋅ m C ) (1.60 × 10 C )
=
KEi
5.0 MeV (1.60 × 10 −13 J MeV )
2
= 4.5 × 10 −14 m = 45 fm
28.7
(a)
rn = n 2 a0 yields r2 = 4 ( 0.052 9 nm ) = 0.212 nm
(b)
With the electrical force supplying the centripetal acceleration, me vn2 rn = ke e 2 rn2, giving
vn = ke e 2 me rn , and pn = me vn = me ke e 2 rn .
continued on next page
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Atomic Physics
347
Thus,
me ke e 2
p2 =
=
r2
(9.11×10
−31
kg) (8.99 ×10 9 N ⋅ m 2 C2 ) (1.6 ×10 −19 C)
2
0.212 ×10 −9 m
= 9.95×10 −25 kg ⋅ m s
(c)
⎛ 6.63 × 10 −34 J ⋅ s ⎞
⎛ h ⎞
−34
Ln = n ⎜
→ L2 = 2 ⎜
⎟
⎟⎠ = 2.11 × 10 J ⋅ s
⎝ 2p ⎠
2p
⎝
(d)
(9.95 × 10−25 kg ⋅ m s) = 5.44 × 10−19
1
p2
KE2 = mv22 = 2 =
2me
2
2 ( 9.11 × 10 −31 kg )
(e)
(8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10−19 C)
k ( −e ) e
PE2 = e
=−
r2
( 0.212 × 10−9 m )
2
1 eV
⎛
⎞
= 3.40 eV
J⎜
⎝ 1.60 × 10 −19 J ⎟⎠
2
= −1.09 × 10 −18 J = − 6.80 eV
28.8
(f )
E2 = KE2 + PE2 = 3.40 eV − 6.80 eV = − 3.40 eV
(a)
With the electrical force supplying the centripetal acceleration, me vn2 rn = ke e 2 rn2 , giving
vn = ke e 2 me rn , where rn = n 2 a0 = n 2 (0.052 9 nm).
Thus,
k e2
v1 = e =
me r1
(b) KE1 =
(8.99 ×10 N ⋅ m C ) (1.60 ×10 C)
(9.11×10 kg) (0.052 9 ×10 m )
9
2
2
−31
−19
−9
2
= 2.19 ×10 6 m s
2
1
1
me v12 = (9.11×10 −31 kg) ( 2.19 ×10 6 m s)
2
2
⎛
⎞
1 eV
= 2.18 ×10 −18 J ⎜
⎟ = 13.6 eV
−19
⎝ 1.60 ×10 J ⎠
(c)
k (−e ) e
(8.99 ×109 N ⋅ m 2 C2 ) (1.60 ×10−19 C)
PE1 = e
=−
r1
(0.052 9 ×10−9 m )
2
= − 4.35×10 −18 J = − 27.2 eV
28.9
Since the electrical force supplies the centripetal acceleration,
me vn2 ke e 2
= 2
rn
rn
or
vn2 =
k2 e 2
me rn
From Ln = me rn vn = n , we have rn = n
vn2 =
me vn , so
k2 e 2 ⎛ me vn ⎞
⎜
⎟
me ⎝ n ⎠
which reduces to vn = ke e 2 n .
68719_28_ch28_p341-359.indd 347
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348
28.10
Chapter 28
(a)
(
)
The Rydberg equation is 1 l = RH 1 n 2f − 1 ni2 , or
l=
2 2
1 ⎛ ni n f ⎞
RH ⎜⎝ ni2 − n 2f ⎟⎠
With ni = 5 and n f = 3,
l=
28.11
1
⎡ ( 25) ( 9 ) ⎤
= 1.281 × 10 −6 m = 1 281 nm
7
−1 ⎢
1.097 373 2 × 10 m ⎣ 25 − 9 ⎥⎦
c 3.00 × 108 m s
=
= 2.34 × 1014 Hz
l 1 281 × 10 − 9 m
(b)
f =
(c)
Ephoton =
−34
8
hc ( 6.63 × 10 J ⋅ s ) ( 3.00 × 10 m s ) ⎛
1 eV
=
⎜⎝
−9
l
1 281 × 10 m
1.60 × 10 −19
⎞
⎟ = 0.970 eV
J⎠
The energy of the emitted photon is
Ephoton =
−34
8
hc ( 6.626 × 10 J ⋅s ) ( 2.998 × 10 m s ) ⎛
1 eV
=
⎜⎝
−9
l
656 × 10 m
1.60 × 10 −19
⎞
⎟ = 1.89 eV
J⎠
This photon energy is also the difference in the electron’s energy in its initial and final orbits. The
energies of the electron in the various allowed orbits within the hydrogen atom are
En = −
13.6 eV
n2
where
n = 1, 2, 3,…
giving E1 = −13.6 eV, E2 = −3.40 eV, E3 = −1.51 eV, E4 = −0.850 eV,….
Observe that Ephoton = E3 − E2 , so the transition was from the n = 3 orbit to the n = 2 orbit .
28.12
The change in the energy of the atom is
⎛ 1
1⎞
ΔE = E f − Ei = 13.6 eV ⎜ 2 − 2 ⎟
⎝ ni n f ⎠
Transition I:
⎛1 1 ⎞
ΔE = 13.6 eV ⎜ − ⎟ = 2.86 eV (absorption)
⎝ 4 25 ⎠
Transition II:
⎛ 1
ΔE = 13.6 eV ⎜
−
⎝ 25
Transition III:
1⎞
⎛ 1
ΔE = 13.6 eV ⎜
−
= − 0.572 eV (emission)
⎝ 49 16 ⎟⎠
Transition IV:
1⎞
⎛ 1
ΔE = 13.6 eV ⎜ − ⎟ = 0.572 eV (absorption)
⎝ 16 49 ⎠
(a)
Since l =
hc
hc
=
,
Ephoton −ΔE
1⎞
⎟ = −0.967 eV (emission)
9⎠
transition II emits the shortest wavelength photon.
continued on next page
68719_28_ch28_p341-359.indd 348
1/7/11 3:02:05 PM
Atomic Physics
28.13
(b)
The atom gains the most energy in transition I .
(c)
The atom loses energy in transitions II and III .
The energy absorbed by the atom is
Ephoton = E f − Ei =
28.14
⎛ 1
−13.6 eV (−13.6 eV)
1⎞
−
= +13.6 eV ⎜ 2 − 2 ⎟
2
2
nf
ni
⎝ ni n f ⎠
(a)
1⎞
⎛ 1
Ephoton = 13.6 eV ⎜ 2 − 2 ⎟ = 2.86 eV
⎝2
5 ⎠
(b)
1⎞
⎛ 1
Ephoton = 13.6 eV ⎜ 2 − 2 ⎟ = 0.472 eV
⎝4
6 ⎠
(a)
The energy absorbed is
ΔE = E f − Ei =
(b)
28.15
349
(a)
−13.6 eV (−13.6 eV)
⎛ 1 1⎞
−
= 13.6 eV ⎜ − ⎟ = 12.1 eV
⎝ 1 9⎠
n 2f
ni2
Three transitions are possible as the electron returns to the ground state. These transitions
and the emitted photon energies are
ni = 3 → n f = 1:
⎛1 1⎞
ΔE = 13.6 eV ⎜ 2 − 2 ⎟ = 12.1 eV
⎝1
3 ⎠
ni = 3 → n f = 2:
1⎞
⎛ 1
ΔE = 13.6 eV ⎜ 2 − 2 ⎟ = 1.89 eV
⎝2
3 ⎠
ni = 2 → n f = 1:
1⎞
⎛1
ΔE = 13.6 eV ⎜ 2 − 2 ⎟ = 10.2 eV
⎝1
2 ⎠
The photon of longest wavelength is produced in the transition for which the atom gives
up the smallest amount of energy. From Figure P28.15, this is seen to be the n = 3 to n = 2
transition, for which
Ephoton = E3 − E2 = −1.512 eV − ( −3.401 eV ) = 1.889 eV
(b)
l=
hc
Ephoton
=
(6.63 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s ) ⎛
1 eV
⎞
⎜⎝
⎟
1.889 eV
1.60 × 10 −19 J ⎠
= 6.58 × 10 −7 m = 658 nm
(c)
The photon of shortest wavelength is produced in the transition for which the atom gives
up the greatest amount of energy. From Figure P28.15, this is seen to be the n = 6 to n = 2
transition, for which
Ephoton = E6 − E2 = − 0.378 eV − (−3.401 eV ) = 3.02 eV
(d)
l=
hc
Ephoton
=
(6.63 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s ) ⎛
1 eV
⎞
⎜⎝
⎟
3.02 eV
1.60 × 10 −19 J ⎠
= 4.12 × 10 −7 m = 412 nm
continued on next page
68719_28_ch28_p341-359.indd 349
1/7/11 3:02:07 PM
350
Chapter 28
(e)
The shortest wavelength is produced for the n = ∞ to n = 2 transition. The photon energy
for this transition is Ephoton = 0 − (−3.401 eV ) = 3.401 eV, and its wavelength is
l=
hc
Ephoton
=
(6.63 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s ) ⎛
1 eV
⎜⎝
3.401 eV
1.60 × 10 −19
⎞
⎟
J⎠
= 3.66 × 10 −7 m = 366 nm
28.16
The magnetic force supplies the centripetal acceleration, so mv 2 r = qvB, or r = mv qB. If angular momentum is quantized according to
Ln = mvn rn = 2n , then mvn =
2n
rn
and the allowed radii of the path are given by
rn =
28.17
(a)
1 ⎛ 2n ⎞
qB ⎜⎝ rn ⎟⎠
or
rn =
2n
qB
The energy of the emitted photon is
1⎞
⎛ 1
Ephoton = E4 − E2 = −13.6 eV ⎜ 2 − 2 ⎟ = 2.55 eV
⎝4
2 ⎠
This photon has a wavelength of
l=
(b)
Ephoton
h
matom l
(6.63 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s )
( 2.55 eV ) (1.60 × 10
−19
J eV )
= 4.88 × 10 −7 m = 488 nm
=
6.63 × 10 −34 J ⋅ s
= 0.814 m s
(1.67 × 10−27 kg) ( 4.88 × 10−7 m )
(a)
Starting from the n = 4 state, there are 6 possible transitions as the electron returns to the
ground (n = 1) state. These transitions are: n = 4 → n = 1, n = 4 → n = 2, n = 4 → n = 3,
n = 3 → n = 1, n = 3 → n = 2, and n = 2 → n = 1. Since there is a different change in
energy associated with each of these transitions, there will be 6 distinct wavelengths
observed in the emission spectrum of these atoms.
(b)
The longest observed wavelength is produced by the transition involving the smallest
change in energy. This is the n = 4 → n = 3 transition, and the wavelength is
lm ax =
or
(c)
68719_28_ch28_p341-359.indd 350
=
Since momentum must be conserved, the photon and the atom go in opposite directions
with equal magnitude momenta. Thus, p = matom v = h l , or
v=
28.18
hc
(6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) ⎛ 1 eV ⎞ ⎛ 1 nm ⎞
hc
=
⎜⎝
⎟⎜
⎟
E4 − E3
1.60 × 10 −19 J ⎠ ⎝ 10 −9 m ⎠
−13.6 eV (1 4 2 − 1 32 )
lm ax = 1.88 × 10 3 nm .
Since the transition producing this wavelength terminates on the n = 3 level, this is part of
the Paschen series .
1/7/11 3:02:09 PM
Atomic Physics
28.19
351
(a)
For the absorption of the photon to ionize the hydrogen atom, the electron in this atom must
be in an excited state having an ionization energy less than or equal to the photon energy.
That is, we must have Eionization = −En ≤ Ephoton = 2.28 eV, or En ≥ −2.28 eV. The state with
the smallest value of n meeting this requirement is the n = 3 state, with E3 = −1.51 eV.
(b)
After the electron spends 1.51 eV of energy to escape from the atom, it will retain the
remaining absorbed photon energy (2.28 eV − 1.51 eV = 0.77 eV) as kinetic energy. Its
speed will then be
2KE
=
me
v=
2 ( 0.77 eV ) ⎛ 1.60 × 10 −19 J ⎞
5
⎟⎠ = 5.2 × 10 m s = 520 km s
9.11 × 10 −31 kg ⎜⎝
1 eV
(a)
( 7.36 × 1022 kg) 2p (3.84 × 108 m ) = 2.89 × 1034 kg ⋅ m 2 s
⎛ 2p r ⎞
L = M M vr = M M ⎜
r=
⎟
⎝ T ⎠
2.36 × 10 6 s
(b)
n=
(c)
The gravitational force supplies the centripetal acceleration, so M M v 2 r = GM E M M r 2, or
r v 2 = GM E . Then, from Ln = M M vn rn = n , or vn = n M M rn , we have
2
rn ( n M M rn ) = GM E , which gives rn = n 2 ( 2 GM E M M2 ) = n 2 r1 .
2
28.20
L
=
2p ( 2.89 × 10 34 kg ⋅ m 2 s )
1.05 × 10 −34 J ⋅ s
= 2.75 × 10 68
Therefore, when n increases by 1, the fractional change in the radius is
Δr rn +1 − rn ( n + 1) r1 − n 2 r1 2n + 1 2
=
=
=
≈ for n >> 1
r
rn
n 2 r1
n2
n
2
Δr
2
≈
= 7.27 × 10 −69
r
2.75 × 10 68
28.21
(a)
rn = n 2 a0 = n 2 (0.052 9 nm) ⇒ r3 = 32 (0.052 9 nm) = 0.476 nm
(b)
In the Bohr model, the circumference of an allowed orbits must be an integral multiple of
the de Broglie wavelength for the electron in that orbit, or 2p rn = nl. Thus, the wavelength
of the electron when in the n = 3 orbit in hydrogen is
l=
28.22
2p r3 2p ( 0.476 nm )
= 0.997 nm
=
3
3
(a)
The Coulomb force supplies the necessary centripetal force to hold the electron in orbit, so
me vn2 rn = ke e 2 rn2 or me vn2 = ke e 2 rn . But me vn2 = 2KEn , and ke e 2 rn = −PEn , where PEn is
the electrical potential energy of the electron-proton system when the electron is in
an orbit of radius rn . We then have 2KEn = −PEn , or KEn = − 12 PEn .
(b)
When the atom absorbs energy E, and the electron moves to a higher level, both the kinetic
and potential energies will change. Conservation of energy requires that E = ΔKE + ΔPE.
However, from the result of part (a), ΔKE = − 12 ΔPE, and we have
E=−
(c)
68719_28_ch28_p341-359.indd 351
ΔKE = −
1
1
ΔPE + ΔPE = + ΔPE
2
2
1
1
ΔPE = − ( 2E )
2
2
or
ΔPE = 2E
or
ΔKE = −E
1/7/11 3:02:12 PM
352
28.23
28.24
Chapter 28
The radii for atomic number Z are rn = n 2 ( 2 me ke e 2 ) Z = n 2 a0 Z , so r1 = a0 Z , where
a0 = 0.052 9 nm is the radius of the first Bohr orbit in hydrogen.
(a)
For He + , Z = 2 and r1 = 0.052 9 nm 2 = 0.026 5 nm .
(b)
For Li 2+ , Z = 3 and r1 = 0.052 9 nm 3 = 0.017 6 nm .
(c)
For Be 3+ , Z = 4 and r1 = 0.052 9 nm 4 = 0.013 2 nm .
(a)
The energy levels in a single electron atom with nuclear charge +Ze are
En = − Z 2 (13.6 eV) n 2. For doubly ionized lithium, Z = 3, giving En = −(122 eV) n 2 .
(b)
E4 =
−122 eV
= −7.63 eV
42
(c)
E2 =
−122 eV
= −30.5 eV
22
(d)
Ephoton = Ei − E f = −7.63 eV − ( −30.5 eV ) = 22.9 eV
⎛ 1.60 × 10 −19 J ⎞
−18
Ephoton = ( 22.9 eV ) ⎜
⎟⎠ = 3.66 × 10 J
1 eV
⎝
(e)
Ephoton
f =
h
l=
(f )
28.25
=
3.66 × 10 −18 J
= 5.52 × 1015 Hz
6.63 × 10 −34 J ⋅ s
c 3.00 × 108 m s
=
= 5.43 × 10 −8 m = 54.3 nm
f
5.52 × 1015 Hz
This wavelength is in the deep ultraviolet region of the spectrum.
The charge of the electron is q1 = −e, and that of the nucleus is q2 = Ze. The energy
levels for the one-electron atoms and ions can then be written as En = −kmZ 2 n 2 , where
k = ke2 e 4 2 2 = constant. The energy of the photon emitted in the n = 3 to n = 2 transition
becomes Ephoton = E3 − E2 = kmZ 2 (1 22 − 1 32 ) = mZ 2 ( 5k 36 ) , and the wavelength is
l=
hc
1 ⎛ 36hc ⎞
=
⎜
⎟
Ephoton mZ 2 ⎝ 5k ⎠
For hydrogen, Z H = 1 and the reduced mass is m H = m p me (m p + me ). Since the proton mass is
very large in comparison to the electron mass, m p + me ≈ m p and m H ≈ me .
(a)
For positronium, Z p = 1 and the reduced mass is m p = me me (me + me ) = me 2. Thus, the
ratio of the wavelength for the n = 3 to n = 2 transition in positronium to that for the same
transition in hydrogen is
lp
lH
=
(1 m
(1 m
p
H
Z p2 ) (36hc 5k )
Z H2 ) (36hc 5k )
=
me ⋅12
m H Z H2
=
=2
m p Z p2 (me 2) ⋅12
giving l p = 2lH = 2 ( 656.3 nm ) = 1313 nm = 1.313 mm (infrared region) .
continued on next page
68719_28_ch28_p341-359.indd 352
1/7/11 3:02:16 PM
Atomic Physics
(b)
353
For singly ionized helium, Z He = 2, and the reduced mass is m He ≈ 2m p me (2m p + me ).
Again, m p me and we find that m He ≈ me . The ratio of the wavelength for the n = 3 to
n = 2 transition in singly ionized helium to that for the same transition in hydrogen is then
lHe
m ⋅12 1
m Z2
= H H2 = e 2 =
lH
m He Z He me ⋅ 2
4
giving lHe =
28.26
(a)
For standing waves in a string fixed at both ends, L = nl 2 , or l = 2 L n. According to the
de Broglie hypothesis, p = h l. Combining these expressions gives p = mv = nh 2L .
(b)
Using E = 12 mv 2 = p2 2m, with p as found in (a) above:
En =
28.27
28.28
28.29
n2 h2
4 L2 ( 2m )
En = n 2 E0 where E0 =
or
In the 3d subshell, n = 3 and
h2
8mL2
= 2. The 10 possible quantum states are
n=3
=2
m = + 2 ms = + 12
n=3
=2
m = + 2 ms = − 12
n=3
=2
m = +1 ms = + 12
n=3
=2
m = +1 ms = − 12
n=3
=2
m = + 0 ms = + 12
n=3
=2
m = + 0 ms = − 12
n=3
=2
m = −1 ms = + 12
n=3
=2
m = −1 ms = − 12
n=3
=2
m = − 2 ms = + 12
n=3
=2
m = − 2 ms = − 12
(a)
For a given value of the principal quantum number n, the orbital quantum number varies
from 0 to n − 1 in integer steps. Thus, if n = 4, there are 4 possible values of
: = 0, 1, 2, and 3.
(b)
For each possible value of the orbital quantum number , the orbital magnetic quantum
number m ranges from − to + in integer steps. When the principal quantum number is
n = 4, and the largest allowed value of the orbital quantum number is = 3, and there are
7 distinct possible values for m . These values are: m = − 3, − 2, −1, 0, +1, + 2, and + 3.
The 3d subshell has n = 3 and = 2 . For r-mesons, we also have s = 1. Thus, there are
15 possible quantum states, as summarized in the table below.
n
68719_28_ch28_p341-359.indd 353
lH 656.3 nm
=
= 164.1 nm (ultraviolet region) .
4
4
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
m
+2
+2
+2
+1
+1
+1
0
0
0
–1
–1
–1
–2
–2
–2
ms
+1
0
–1
+1
0
–1
+1
0
–1
+1
0
–1
+1
0
–1
1/7/11 3:02:21 PM
354
Chapter 28
28.30
(a)
The electronic configuration for nitrogen (Z = 7) is 1s 2 2s 2 2 p3 .
(b)
The quantum numbers for the 7 electrons can be:
1s states
n =1
=0
m =0
2s states
n=2
=0
m =0
ms = + 12
ms = − 12
ms = + 12
ms = − 12
ms = + 12
m = −1
2 p states
n=2
ms = − 12
ms = + 12
m =0
=1
ms = − 12
ms = + 12
m =1
ms = − 12
28.31
In the table of electronic configurations (Table 28.4), or the periodic table on the inside back
cover of the text, look for the element whose last electron is in a 3p state and which has three
electrons outside a closed shell. Its electron configuration will then end in 3s 2 3p1. You should
find that the element is aluminum .
28.32
(a)
For Electron #1 and also for Electron #2, n = 3 and = 1. The other quantum numbers for
each of the 30 allowed states are listed in the tables below.
m
ms
m
ms
m
ms
m
ms
m
ms
m
ms
Electron #1
+1
+ 12
+1
+ 12
+1
+ 12
+1
− 12
+1
− 12
+1
− 12
Electron #2
+1
− 12
0
± 12
−1
± 12
+1
+ 12
0
± 12
−1
± 12
m
ms
m
ms
m
ms
m
ms
m
ms
m
ms
Electron #1
0
+ 12
0
+ 12
0
+ 12
0
− 12
0
− 12
0
− 12
Electron #2
+1
± 12
0
− 12
−1
± 12
+1
± 12
0
+ 12
−1
± 12
m
ms
m
ms
m
ms
m
ms
m
ms
m
ms
Electron #1
−1
+ 12
−1
+ 12
−1
+ 12
−1
− 12
−1
− 12
−1
− 12
Electron #2
+1
± 12
0
± 12
−1
− 12
+1
± 12
0
± 12
−1
+ 12
There are 30 allowed states , since Electron #1 can have any of three possible values of
m for both spin up and spin down, totaling six possible states. For each of these states,
Electron #2 can be in either of the remaining five states.
(b)
68719_28_ch28_p341-359.indd 354
Were it not for the exclusion principle, there would be 36 possible states, six for each
electron independently.
1/7/11 3:02:29 PM
Atomic Physics
28.33
355
(a)
Observe the electron configurations given in the periodic table on the inside back cover
of the textbook. Zirconium, with 40 electrons, has 4 electrons outside a closed krypton
core. The krypton core, with 36 electrons, has all states up through the 4 p subshell filled.
Normally, one would expect the next 4 electrons to go into the 4d subshell. However,
an exception to the rule occurs at this point, and the 5s subshell fills (with 2 electrons)
before the 4d subshell starts filling. The two remaining electrons in zirconium are in an
incomplete 4d subshell. Thus, n = 4 and = 2 for each of these electrons.
(b)
For electrons in the 4d subshell, with = 2, the possible values of m are
m = 0, ± 1, ± 2 , and those for ms are ms = ± 1 2 .
(c)
We have 40 electrons, so the electron configuration is:
1s 2 2s 2 2 p6 3s 2 3p6 3d 10 4s 2 4 p6 4d 2 5s 2 = [Kr]4d 2 5s 2
28.34
For electrons accelerated through a potential difference of 40.0 kV, starting from rest, the kinetic
energy is KE = e(ΔV ) = e(40.0 kV) = 40.0 keV. When these electrons strike the tungsten target,
the shortest wavelength possible is produced when the electron loses all its kinetic energy and
is brought to rest in a single collision, producing a single photon. The wavelength of the photon
produced is lm in = hc Ephoton
= hc KE = hc e ( ΔV ). This gives
(
(6.63 × 10
(1.60 × 10
)
−34
lm in =
28.35
(a)
m ax
J ⋅ s ) ( 3.00 × 108 m s )
C ) ( 40.0 × 10 3 V )
−19
= 3.11 × 10 −11 m = 0.0311 nm
Note that Z = 83 for bismuth. With a vacancy in the L-shell, an electron in the M-shell is
shielded from the nuclear charge by a total of 9 electrons, 2 electrons in the filled K-shell
and 7 electrons in the partially filled L-shell. Thus, our estimate for the energy of this
electron while it is in the M-shell is
− ( Z − 9 ) (13.6 eV ) − ( 74 ) (13.6 eV )
=
= −8.27 × 10 3 eV = −8.27 keV
32
9
2
EM ≈
2
When this electron drops down to fill the vacancy in the L-shell, it will continue to be
shielded from the nuclear charge by the 2 electrons in the filled K-shell. Our estimate for
the energy of this electron in the L-shell is then
− ( Z − 2 ) (13.6 eV ) − (81) (13.6 eV )
=
= −2.23 × 10 4 eV = −22.3 keV
22
4
2
EL ≈
2
Therefore, the estimate for the transitional energy, and hence the energy of the photon
produced, in a M- to L-shell transition in bismuth is
Ephoton = EM − EL ≈ −8.27 keV − ( −22.3 keV ) = 14 keV
(b)
The wavelength of the photon produced in the M- to L-shell transition should be approximately
l=
28.36
hc
Ephoton
≈
(6.63 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s ) ⎛
1 keV
⎜⎝
14 keV
1.60 × 10 −16
⎞
−11
⎟ = 8.9 × 10 m
J⎠
The energies in the K- and M-shells are
EK ≈ −
( Z − 1)2 (13.6 eV )
( Z − 9 )2 (13.6 eV )
and
E
≈
−
M
(1)2
( 3)2
continued on next page
68719_28_ch28_p341-359.indd 355
1/7/11 3:02:39 PM
356
Chapter 28
Thus,
2
2
Ephoton = EM − EK ≈ (13.6 eV ) ⎡⎣ − ( Z − 9 ) 9 + ( Z − 1) ⎤⎦ = (13.6 eV ) (8Z 2 9 − 8 )
and Ephoton = hc l gives Z 2 = 9 [8 + hc (13.6 eV ) l ] 8 , or
Z ≈ 9+
9 ( 6.63 × 10 −34 J ⋅ s ) ( 3.00 × 108 m s ) ⎛
1 eV
⎞
⎜⎝
⎟ = 32.0
−9
1.60 × 10 −19 J ⎠
8 (13.6 eV ) ( 0.101 × 10 m )
The element is germanium .
28.37
The transitions that produce the three longest
wavelengths in the K series are shown at the right.
The energy of the K-shell is EK = − 69.5 keV.
N-shell
M-shell
L-shell
Thus, the energy of the L-shell is
EL = EK +
or
hc
l3
EL = − 69.5 keV +
K-shell
(6.63 × 10
J ⋅ s ) ( 3.00 × 108 m s ) ⎛
1 keV
⎞
⎜⎝
⎟
−9
0.021 5 × 10 m
1.60 × 10 −16 J ⎠
−34
= − 69.5 keV + 57.8 keV = −11.7 keV
Similarly, the energies of the M- and N-shells are
and
EM = EK +
(6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) = −10.0 keV
hc
= − 69.5 keV +
l2
( 0.020 9 × 10−9 m ) (1.60 × 10−16 J keV )
EN = EK +
(6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) = − 2.30 keV
hc
= − 69.5 keV +
l1
( 0.018 5 × 10−9 m ) (1.60 × 10−16 J keV )
The ionization energies of the L-, M-, and N-shells are
11.7 keV, 10.0 keV, and 2.30 keV, respectively
28.38
According to the Bohr model, the radii of the electron orbits in hydrogen are given by
rn = n 2 a0, with a0 = 0.052 9 nm = 5.29 × 10 −11 m
Then, if rn ≈ 1.00 mm = 1.00 × 10 −6 m , the quantum number is
n=
28.39
(a)
(b)
rn
=
a0
ΔE = E2 − E1 = −13.6 eV (2)2 − ( −13.6 eV (1)2 ) = 10.2 eV
The average kinetic energy of the atoms must equal or exceed the needed excitation energy,
or 32 kB T ≥ ΔE, which gives
T≥
68719_28_ch28_p341-359.indd 356
1.00 × 10 −6 m
≈ 137
5.29 × 10 −11 m
−19
J eV )
2 ( ΔE ) 2 (10.2 eV ) (1.60 × 10
=
= 7.88 × 10 4 K
−23
3kB
3 (1.38 × 10
J K)
1/7/11 3:02:42 PM
Atomic Physics
28.40
(a)
L = c ( Δt ) = ( 3.00 × 108 m s ) (14.0 × 10 −12 s ) = 4.20 × 10 −3 m = 4.20 mm
(b)
N=
=
(c)
n=
Epulse
Ephoton
(a)
Epulse
hc l
=
Epulse l
hc
( 3.00 J ) ( 694.3 × 10 −9 m )
(6.63 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s )
= 1.05 × 1019 photons
N
N
4N
=
=
V L (p d 2 4 ) L (p d 2 )
=
28.41
=
357
4 (1.05 × 1019 photons )
( 4.20 mm ) p ( 6.00 mm )2
= 8.84 × 1016 photons mm 3
With one vacancy in the K-shell, an electron in the L-shell has one electron shielding it
from the nuclear charge, so Z eff = Z − 1 = 24 − 1 = 23. The estimated energy the atom gives
up during a transition from the L-shell to the K-shell is then
ΔE ≈ Ei − E f = −
2
⎡1
Z eff
1⎤
(13.6 eV ) ⎡ Z eff2 (13.6 eV ) ⎤ 2
− ⎢−
⎥ = Z eff (13.6 eV ) ⎢ 2 − 2 ⎥
2
2
ni
nf
⎢⎣
⎥⎦
⎢⎣ n f ni ⎥⎦
or
1
1
2
ΔE ≈ ( 23) (13.6 eV ) ⎡⎢ 2 − 2
2
⎣1
(b)
⎤ = 5.40 × 10 3 eV = 5.40 keV
⎥⎦
With a vacancy in the K-shell, we assume that Z − 2 = 24 − 2 = 22 electrons shield the
outermost electron (in a 4s state) from the nuclear charge. Thus, for this outer electron,
Z eff = 24 − 22 = 2, and the estimated energy required to remove this electron from the atom is
⎡ Z 2 (13.6 eV ) ⎤ 22 (13.6 eV )
Eionization = E f − Ei = 0 − Ei ≈ − ⎢ − eff
= 3.40 eV
⎥=
ni2
42
⎣
⎦
28.42
(c)
KE = ΔE − Eionization = 5.40 keV − 3.40 eV ≈ 5.40 keV
(a)
Requiring the angular momentum associated with the orbital motion of Earth to
satisfy Bohr’s postulate gives M E vr = n , or the speed of the Earth in the nth
allowed orbit would be vn = n M E rn . The gravitational force between Earth and the
Sun supplies the needed centripetal acceleration as Earth moves in its orbit. Thus,
M E v 2 r = GM S M E r 2 , or v 2 = GM S r. Substituting for the speed in the nth orbit from
2
above gives ( n M E rn ) = GM S rn , which reduces to rn = n 2 2 GM S M 2E .
(b)
From Table 7.3 and the back flysheet of the textbook, M S = 1.991 × 10 30 kg,
M E = 5.98 × 10 24 kg, r = 1.496 × 1011 m, G = 6.67 × 10 −11 N ⋅ m 2 kg2 , and
= 1.05 × 10 −34 J ⋅ s. We then find the quantum number for Earth’s orbit to be
n=
=
or
M E rn GM S
(5.98 × 10
24
kg )
(1.496 × 10
11
m ) ( 6.67 × 10 −11 N ⋅ m 2 kg2 ) (1.991 × 10 30 kg )
1.05 × 10 −34 J ⋅ s
n = 2.54 × 10 74
continued on next page
68719_28_ch28_p341-359.indd 357
1/7/11 3:02:44 PM
358
Chapter 28
(c)
The difference in the radii of the nth and the (n + 1)th allowed orbits of the Earth is
Δr = rn +1
⎡⎣( n + 1)2 − n 2 ⎤⎦
− rn =
GM S M 2E
2
=
[ 2n + 1]
2
GM S M
2
E
≈
2n 2
GM S M 2E
Using the values from part (b) above gives
2 ( 2.54 × 10 74 ) (1.05 × 10 −34 J ⋅ s )
2n 2
=
2
GM S M 2E ( 6.67 × 10 −11 N ⋅ m 2 kg2 ) (1.991 × 10 30 kg ) ( 5.98 × 10 24 kg )
2
Δr ≈
= 1.18 × 10 −63 m
28.43
28.44
(d)
The result computed in part (d) above is much smaller than the radius of an atomic nucleus
(∼ 10 −15 m), so the distance between quantized orbits of the Earth is too small to observe.
(a)
I=
(b)
2
W ⎞ ⎡p
⎛
E = I A ( Δt ) = ⎜ 4.24 × 1015
( 0.600 × 10−9 m ) ⎤⎥⎦ (1.00 × 10−9 s) = 1.20 × 10−12 J
⎟
⎝
m 2 ⎠ ⎢⎣ 4
(a)
−3
−9
P ( ΔE Δt ) 4 ( 3.00 × 10 J 1.00 × 10 s )
=
=
= 4.24 × 1015 W m 2
2
−6
A p d2 4
p ( 30.0 × 10 m )
Given that the de Broglie wavelength of the electron is l = 2 a0 , its momentum is
p = h l = h 2 a 0 . The kinetic energy of this nonrelativistic electron is
KE =
p2
h2
=
2 me 8 me a 20
(6.63 × 10
=
8 ( 9.11 × 10
−34
(b)
28.45
J ⋅ s ) (1 eV 1.60 × 10 −19 J )
2
−31
kg ) ( 0.052 9 × 10 −9 m )
2
= 135 eV
The kinetic energy of this electron is ≈ 10 times the magnitude of the ground state energy
of the hydrogen atom, which is –13.6 eV.
In the Bohr model,
2 4
1 ⎞ ⎤ 4p 2 me ke2 e 4 ⎡ 1
1⎤
En − En −1 1 ⎡ −me ke e ⎛ 1
=
−
− 2⎥
=
⎢
f =
⎢
2 ⎟⎥
2
2
2
3
⎜
h ⎣⎢ 2
n ⎦
2h
( n − 1) ⎠ ⎥⎦
⎝n
h
⎣ ( n − 1)
which reduces to
28.46
Ephoton =
f =
2p 2 me ke2 e 4 ⎛ 2n − 1 ⎞
⎜⎝ ( n − 1)2 n 2 ⎟⎠ .
h3
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 2.998 × 10 m s ) 1 240 eV ⋅ nm
=
=
= ΔE
l
l
l (1.602 × 10 −19 J eV ) (10 −9 m nm )
For l = 310.0 nm, ΔE = 4.000 eV; l = 400.0 nm, ΔE = 3.100 eV; and l = 1 378 nm,
ΔE = 0.900 0 eV.
continued on next page
68719_28_ch28_p341-359.indd 358
1/7/11 3:02:48 PM
Atomic Physics
The ionization energy is 4.100 eV. The energy level diagram having the smallest number of
levels, and consistent with these energy differences, is shown below.
359
29
Nuclear Physics
QUICK QUIZZES
1.
False. In a sample containing a very large number of identical radioactive atoms, 50% of the
original atoms remain after 1 half-life has elapsed. During the second half-life, half of the
remaining 50% of the atoms decay, leaving 25% of the original atoms still present. Thus, after
2 half-lives have elapsed, only 75% of the original radioactive atoms have decayed. An individual
radioactive atom may exist an indefinite time before decaying.
2.
Choice (c). At the end of the first half-life interval, half of the original sample has decayed and
half remains. During the second half-life interval, half of the remaining portion of the sample
decays, leaving one-quarter of the original sample. During the third half-life, half of that quarter
will decay. The total fraction of the sample that has decayed during the three half-lives is
1 1 ⎛ 1⎞ 1 ⎛ 1⎞ 7
+ ⎜ ⎟+ ⎜ ⎟=
2 2 ⎝ 2⎠ 2 ⎝ 4⎠ 8
3.
Choice (c). The half-life of a radioactive material is T1 2 = ln 2 l , where l is the decay constant
for that material. Thus, if l A = 2lB , we have
(T )
1 2
A
=
( )
T1 2
ln 2 ln 2
=
=
lA
2lB
2
B
=
4h
=2h
2
4.
Choices (a) and (b). Reactions (a) and (b) both conserve total charge and total mass number as
required. Reaction (c) violates conservation of mass number with the sum of the mass numbers
being 240 before the reaction and being only 223 after the reaction.
5.
Choice (b). The reactant nuclei in this endothermic reaction must supply Q = 2.17 MeV of energy
to be converted into mass during the reaction. However, the reactant nuclei must also supply the
emerging particles with sufficient kinetic energy to allow momentum to be conserved during the
reaction. Thus, the threshold kinetic energy for the reaction (minimum total kinetic energy of
the particles going into the reaction) must exceed the value Q = 2.17 MeV.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
Nuclei are approximately spherical with average radii of r = r0 A1 3, where r0 = constant. Thus, the
ratio of the volume of a 20 Ne nucleus to that of a 4 He is
(r0 A1Ne3 ) = ANe = 20 = 5
VNe 43 p rNe3
= 4 3 =
3
VHe
3 p rHe
(r0 A1He3 ) AHe 4
3
and (d) is the correct choice.
2.
⎛ 931.5 MeV c 2 ⎞
4
2
m = (15.999 u ) ⎜
⎟⎠ = 1.49 × 10 MeV c , and the correct answer is choice (e).
1u
⎝
360
68719_29_ch29_p360-376.indd 360
1/7/11 3:04:17 PM
Nuclear Physics
3.
361
⎛ 1.66 × 10 −27 kg ⎞
−26
m = (15.999 u ) ⎜
⎟⎠ = 2.66 × 10 kg, so choice (e) is the correct answer.
1u
⎝
4.
From the binding energy curve shown in Figure 29.4 of the textbook, the approximate binding energies per nucleon in the isotopes 35 Cl, 62 Ni, and 197 Au are seen to be
8.4 MeV, 8.8 MeV, and 7.9 MeV, respectively. Therefore, the correct ranking, from smallest to
largest, is gold, chlorine, nickel, which is choice (a).
5.
The nucleus 40
18 X contains A = 40 total nucleons, of which Z = 18 are protons. The remaining
A − Z = 40 − 18 = 22 are neutrons, and choice (c) is correct.
6.
In a large sample, one-half of the radioactive nuclei initially present remain in the sample after one
half-life has elapsed. Hence, the fraction of the original number of radioactive nuclei remaining after
n half-lives have elapsed is (1 2)n = 1 2n. In this case, the number of half-lives that have elapsed is
Δt T1 2 = 14 d 3.6 d ≈ 4. Therefore, the approximate fraction of the original sample that remains
undecayed is 1 24 = 1 16, and the correct answer is choice (d).
7.
The half-life of a substance is related to its decay constant by T1 2 = ln 2 l . The desired ratio is then
(T )
(T )
1 2
1 2
A
=
B
ln 2 l A lB 1
=
=
ln 2 lB l A 3
so (a) is the correct choice.
8.
9.
95
In the beta decay of 36
Kr, the emitted particles are an electron, −10 e, and an antineutrino, n e . The
emitted particles contain a total charge of −e and zero nucleons. Thus, to conserve both charge
95
and nucleon number, the daughter nucleus must be 37
Rb, which contains Z = 37 protons and
A − Z = 95 − 37 = 58 neutrons, making (a) the correct choice.
32
15
P decays to 32
16 S by means of beta decay, with the decay equation being
32
15
P→
32
16
S+
0
−1
e +n e
As will be discussed in Chapter 30, the antineutrino must be emitted along with the electron in
order to conserve electron-lepton number. The correct choices are (c) and (e).
10.
To conserve the total number of nucleons, it is necessary that A + 4 = 234, or A = 230. Conservation
of charge demands that Z + 2 = 90, or Z = 88. We then see that the correct answer is choice (c).
11.
In gamma decay, an unstable nucleus gives off excess energy by emitting a high-energy photon. The
daughter nucleus is the same as the parent, with both the charge and nucleon number unchanged,
simply in a lower energy state. Thus, the correct choice is (b).
12.
The Q-value for the reaction 49 Be + 42 He → 126 C + 01 n is
(
Q = ( Δm ) c 2 = m
9
4 Be
+m
4
2 He
−m
12
6C
)
− mn c 2
= [9.012 182 u + 4.002 603 u −12.000 000 u −1.008 665 u ] (931.5 Mev u )
= 5.70 MeV
so (d) is the correct choice.
68719_29_ch29_p360-376.indd 361
1/7/11 3:04:20 PM
362
Chapter 29
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The nucleus was at rest before decay, so the total linear momentum will be zero both before and
after decay. Thus, the alpha particle and the daughter nucleus carry equal amounts of momentum
in opposite directions. Since kinetic energy can be written as KE = p2 2m , the small-mass alpha
particle has much more of the decay energy than the recoiling daughter nucleus.
4.
Extra neutrons are required to overcome the increasing electrostatic repulsion of the protons. The
neutrons participate in the net attractive effect of the nuclear force but feel no Coulomb repulsion.
6.
An alpha particle is a doubly positive charged helium nucleus, is very massive, and does not penetrate very well. A beta particle is a singly negative charged electron, is very low mass, and only
slightly more difficult to shield from. A gamma ray is a high-energy photon or high frequency
electromagnetic wave and has high penetrating ability.
8.
After one half-life, one-half the radioactive atoms have decayed. After the second half-life,
one-half of the remaining atoms have decayed. Therefore, 12 + 12 ( 12 ) = 43 of the original radioactive
atoms have decayed after two half-lives.
10.
With a very small mass in comparison to alpha particles, beta particles have greater penetrating
ability than do alpha particles.
12.
The alpha particle, with a mass about 7 000 times that of the beta particle, is much more difficult
to deflect. Even under the influence of an electrical force whose magnitude is double that acting on
the beta particle, the alpha particle is deviated from its original path by a much smaller amount.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
(a) 1.9 fm
(b)
7.4 fm
4.
(a) 4.8 fm
(b)
4.6 × 10 −43 m 3
6.
(a)
r
12
(b)
8.
10.
= 7.89 cm , r
13
m
12
C
m
13
C
= 8.21 cm
r
12 u
= 0.961 and
13 u
r
(a) 1.11 MeV nucleon
12
C
13
C
=
7.89 cm
= 0.961
8.21 cm
(b)
7.07 MeV nucleon
(c) 8.79 MeV nucleon
7.57 MeV nucleon
56
59
For 55
25 Mn, Eb A = 8.765 MeV. For 26 Fe, Eb A = 8.790 MeV. For 27 Co, Eb A = 8.768 MeV. This
gives us finer detail than is shown in Figure 29.4.
14.
7.93 MeV
16.
8.7 × 10 3 Bq
18.
(a) 0.755
68719_29_ch29_p360-376.indd 362
=
C
2.3 × 1017 kg m 3
16 km
(d)
12.
C
(c)
(b)
0.570
(c)
9.77 × 10 − 4
1/7/11 3:04:23 PM
Nuclear Physics
363
(d) No. The decay model depends on large numbers of nuclei. After some long but finite time,
only one undecayed nucleus will remain. It is likely that the decay of this final nucleus will
occur before infinite time.
20.
1.72 × 10 4 yr
22.
1.7 × 10 3 yr
24.
(a)
26.
4.28 MeV
28.
(a)
30.
2.81× 10 4 yr
32.
1.00 MeV
34.
(a)
(d)
36.
12
6
66
28
(b)
C
Ni → 66
29 Cu +
0
−1
e +ne
4
2
He
(b)
186 keV
8.023 829 u
(b)
8.025 594 u
m p v = ( m B e + mn ) V
(e)
1
2
m p v2 =
1
2
(m
Be
+ mn ) V 2 − Q
14
6
(c)
−1.64 MeV
(f)
1.88 MeV
(c)
1.1× 10 6 yr
C
(a) 6.258 MeV, exothermic; 5.494 MeV, exothermic
(b)
the first reaction releases more energy
(c)
1.88 × 10 −15 m
38.
(a)
4
2
40.
The equivalent dose is 5.0 rad of heavy ions.
42.
(a)
2.00 J kg
(b)
4.78 × 10 − 4 °C
44.
(a)
2.5 × 10 −3 rem x-ray
(b)
≈ 38 times background levels
46.
(a) 10 h
(b)
3.2 m
48.
22.0 MeV
50.
156 keV
52.
12 mg
54.
(a)
60.6 Bq L
(b)
40.6 days
56.
(a)
N 0 = 2.5 × 10 24
(b)
R0 = 2.3 × 1012 Bq
68719_29_ch29_p360-376.indd 363
(c)
He + 147 N → 11 H + 178 O
(b)
7
3
Li + 11 H → 42 He + 42 He
1/7/11 3:04:25 PM
364
Chapter 29
PROBLEM SOLUTIONS
29.1
29.2
29.3
The average nuclear radii are r = r0 A1 3, where r0 = 1.2 × 10 −15 m = 1.2 fm and A is the mass
number.
r = (1.2 fm )( 2 ) = 1.5 fm
13
(a)
For 12 H,
(b)
For
60
27
(c)
For
197
79
Au,
r = (1.2 fm )(197 ) = 7.0 fm
(d)
For
239
94
Pu,
r = (1.2 fm )( 239 ) = 7.4 fm
(a)
For 42 He,
(b)
For
r = (1.2 fm )( 60 ) = 4.7 fm
13
Co,
238
92
13
13
r = r0 A1 3 = (1.2 × 10 −15 m )( 4 ) = 1.9 × 10 −15 m = 1.9 fm
13
r = r0 A1 3 = (1.2 × 10 −15 m )( 238 ) = 7.4 × 10 −15 m = 7.4 fm
13
U,
From M E = rnuclear V = rnuclear (4p r 3 3), we find
⎛ 3 ME
r=⎜
⎝ 4p r
nuclear
29.4
29.5
⎞
⎟⎠
13
⎡ 3 ( 5.98 × 10 24 kg ) ⎤
=⎢
⎥
17
3
⎢⎣ 4p ( 2.3 × 10 kg m ) ⎥⎦
13
= 1.8 × 10 2 m
(a)
r ≈ r0 A1 3 = (1.2 fm )( 65) = 4.8 fm
(b)
V=
3
4 3 4
p r ≈ p ( 4.8 × 10 −15 m ) = 4.6 × 10 −43 m 3
3
3
(c)
r=
−27
m 65 u 65 (1.66 × 10 kg )
≈
=
= 2.3 × 1017 kg m 3
V
V
4.6 × 10 −43 m 3
(a)
Fm ax
2
9
2
2 ⎡
−19
⎤
ke q1 q2 (8.99 × 10 N ⋅ m C ) ⎣( 2 )( 6 )(1.60 × 10 C ) ⎦
= 2 =
= 27.6 N
2
rm in
(1.00 × 10−14 m )
(b)
am ax =
(c)
13
Fm ax
27.6 N
=
= 4.16 × 10 27 m s 2
mα
6.64 × 10 −27 kg
2
9
2
2 ⎡
−19
⎤
ke q1 q2 (8.99 × 10 N ⋅ m C ) ⎣( 2 )( 6 )(1.60 × 10 C ) ⎦ ⎛ 1 MeV ⎞
=
=
⎜⎝
⎟,
rm in
1.00 × 10 −14 m
1.60 × 10 −13 J ⎠
PEm ax
yielding PEm ax = 1.73 MeV .
29.6
(a)
From conservation of energy, ΔKE = − ΔPE, or 12 mv 2 = q(ΔV ). Also, the centripetal acceleration is supplied by the magnetic force, so mv 2 r = qvB, or v = qBr m. The energy
equation then yields r = 2m ( ΔV ) qB 2 . Applying this to the two isotopes of carbon in this
case gives
r
12
C
=
2 ⎡⎣12 (1.66 × 10 −27 kg ) ⎤⎦ (1 000 V )
= 7.89 × 10 −2 m = 7.89 cm
2
−19
1.60
×
10
C
0.200
T
(
)
(
)
continued on next page
68719_29_ch29_p360-376.indd 364
1/7/11 3:04:28 PM
Nuclear Physics
365
and
r
13
(b)
C
=
2 ⎡⎣13 (1.66 × 10 −27 kg ) ⎤⎦ (1 000 V )
= 8.21× 10 −2 m = 8.21 cm
2
−19
1.60
×
10
C
0.200
T
(
)
(
)
From above, we see that the radii of the paths followed by singly ionized atoms of the two
isotopes should be r = 2m C ΔV eB 2 and r = 2m C ΔV eB 2 . Thus, the ratio of these
radii will be
12
12 C
r
12 C
r
13
C
=
2m C ΔV eB 2
12
2m
13
C
(ΔV ) eB
2
=
m
12
C
m
13
C
12
(a)
=
r
13
C
= 7.89 cm 8.21 cm = 0.961 , so they do
2ke ( 2e )( 79e )
ma rm in
316 (8.99 × 10 9 N ⋅ m 2 C2 ) (1.60 × 10 −19 C )
(6.64 × 10
−27
kg ) ( 3.2 × 10
−14
2
= 3.42 × 10 7 m s = 1.9 × 10 7 m s
m)
2
9
2
2 ⎡
−19
⎤
ke q1 q2 (8.99 × 10 N ⋅ m C ) ⎣( 2 )( 79 )(1.60 × 10 C ) ⎦
KEi = PE f =
=
rm in
3.2 × 10 −14 m
⎛ 1 MeV
= 1.14 × 10 −12 J ⎜
⎝ 1.60 × 10 −13
29.8
C
At the point of closest approach, PE f = KEi , so ke ( 2e ) ( 79e ) rm in = ma vi2 2, or
vi =
(b)
12 u
= 0.961
13 u
=
and the ratio of our computed radii is r
agree.
29.7
13
13 C
⎞
⎟ = 7.1 MeV
J⎠
If a star with a mass of two solar masses collapsed into a gigantic nucleus by converting all of its mass
into neutrons, the total number of nucleons (all neutrons), and hence the atomic number, would be
A=
30
m 2mSun 2 (1.99 × 10 kg )
=
=
= 2.38 × 10 57
mn
mn
1.67 × 10 −27 kg
and its approximate radius would be
r ≈ r0 A1 3 = (1.2 × 10 −15 m ) ( 2.38 × 10 57 )
13
29.9
(a)
= 1.6 × 10 4 m = 16 km
(
The total binding energy for 24
Mg is Eb = ( Δm ) c 2 = 12m
12
age binding energy per nucleon is
1
H
+ 12mn − m
) c , and the aver2
24
Mg
Eb ⎡⎣12 (1.007 825 u ) + 12 (1.008 665 u ) − 23.985 042 u ⎤⎦ ( 931.5 MeV u )
=
A
24
= 8.26 MeV nucleon
(b)
(
85
For 37
Rb, Eb = ( Δm ) c 2 = 37m
1
H
+ 48mn − m
) c , yielding
2
85
Rb
Eb ⎡⎣37 (1.007 825 u ) + 48 (1.008 665 u ) − 84.911 789 u ⎤⎦ ( 931.5 MeV u )
=
A
85
= 8.70 MeV nucleon
68719_29_ch29_p360-376.indd 365
1/7/11 3:04:31 PM
366
29.10
Chapter 29
For 12 H, Δm = 1(1.007 825 u ) + 1(1.008 665 u ) − ( 2.014 102 u ) = 0.002 388 u, and
(a)
Eb ( Δm ) c 2 ( 0.002 388 u ) ( 931.5 MeV u )
=
=
= 1.11 MeV nucleon
A
A
2
For 42 He, Δm = 2 (1.007 825 u ) + 2 (1.008 665 u ) − ( 4.002 603 u ) = 0.030 377 u, and
(b)
Eb ( Δm ) c 2 ( 0.030 377 u ) ( 931.5 MeV u )
=
=
= 7.07 MeV nucleon
A
A
4
For 56
26 Fe, Δm = 26 (1.007 825 u ) + 30 (1.008 665 u ) − ( 55.934 942 ) = 0.528 458 u, and
(c)
Eb ( Δm ) c 2 ( 0.528 458 u ) ( 931.5 MeV u )
=
=
= 8.79 MeV nucleon
A
A
56
For 238
, Δm = 92 (1.007 825 u ) + 146 (1.008 665 u ) − ( 238.050 783) = 1.934 207 u, and
92 U
(d)
Eb ( Δm ) c 2 (1.934 207 u ) ( 931.5 MeV u )
=
=
= 7.57 MeV nucleon
A
A
238
29.11
For 158 O, Δm = 8 (1.007 825 u ) + 7 (1.008 665 u ) − (15.003 065) = 0.120 190 u, and
Eb
15
O
= ( Δm ) c 2 = ( 0.120 190 u ) (931.5 MeV u ) = 112.0 MeV
For 157 N, Δm = 7 (1.007 825 u ) + 8 (1.008 665 u ) − (15.000 109 ) = 0.123 986 u, and
Eb
15
N
Therefore,
29.12
= ( Δm ) c 2 = ( 0.123 986 u ) ( 931.5 MeV u ) = 115.5 MeV
ΔEb = Eb
15
− Eb
N
15
O
= 3.5 MeV
Δm = Z m H + ( A − Z ) mn − m and Eb A = Δm ( 931.5 MeV u ) A
1
1
Nucleus
55
25
Mn
( in u )
Δm
( in u ) Eb A ( in MeV )
Z
(A – Z)
25
30
54.938 050
0.517 525
8.765
m
56
26
Fe
26
30
55.934 942
0.528 458
8.790
59
27
Co
27
32
58.933 200
0.555 355
8.768
Therefore, 56
26 Fe has a greater binding energy per nucleon than its neighbors. This gives us finer
detail than is shown in Figure 29.4.
29.13
(a)
For 1123 Na, Δm = 11(1.007 825 u ) + 12 (1.008 665 u ) − ( 22.989 770 u ) = 0.200 285 u, and
for 1223 Mg, Δm = 12 (1.007 825 u ) + 11(1.008 665 u ) − ( 22.994 127 u ) = 0.195 088 u. The
difference in the binding energy per nucleon for these two isobars is then
ΔEb ⎡⎣( Δm )
=
A
23
11 Na
− ( Δm )
23
12 M g
A
⎤ c 2 [ 0.200 285 u − 0.195 088 u ]( 931.5 MeV u )
⎦ =
23
= 0.210 MeV nucleon
(b)
68719_29_ch29_p360-376.indd 366
The binding energy per nucleon is greater by 0.210 MeV nucleon in 1123 Na. This is attributable
to less proton repulsion in 1123 Na than in 1223 Mg .
1/7/11 3:04:34 PM
Nuclear Physics
29.14
367
43
The sum of the mass of 42
20 Ca plus the mass of a neutron exceeds the mass of 20 Ca. This difference
in mass must represent the mass equivalence of the energy spent removing the last neutron from
43
42
20 Ca to produce 20 Ca plus a free neutron. Thus,
(
E= m
42
20 C a
+ mn − m
43
20 C a
)c
2
= ( 41.958 622 u + 1.008 665 u − 42.958 770 u ) c 2
or
E = ( 0.008 517 u ) ( 931.5 MeV u ) = 7.93 MeV
29.15
The mass of radon present at time t is equal to m = matom N = matom N 0 e − l t = m0 e − l t , where matom is
the mass of a single radon atom, N is the number or radon nuclei (and hence, atoms) present, and
N 0 is the number present at time t = 0, making m0 the mass of radon present at t = 0. The decay
constant for radon is l = ln 2 T1/ 2 = ln 2 (3.83 d), yielding
− t
m = m0 e − l t = m0 e (
29.16
)
T1 2 ln 2
= (1.1× 10 4 Bq ) e − ( 2.0 h 6.05 h ) ln 2 = 8.7 × 10 3 Bq
(a)
⎛ 8.64 × 10 4 s ⎞
5
T1 2 = 8.04 d ⎜
⎟⎠ = 6.95 × 10 s
1d
⎝
(b)
l=
(c)
⎛ 3.7 × 1010 Bq ⎞
4
R = 0.500 mCi = ( 0.500 × 10 −6 Ci ) ⎜
⎟⎠ = 1.9 × 10 Bq
1 Ci
⎝
(d)
From R = lN, the number of radioactive nuclei in a 0.500 mCi of 131 I is
ln 2
ln 2
=
= 9.97 × 10 −7 s −1
T1 2 6.95 × 10 5 s
N=
(e)
(a)
R
1.9 × 10 4 s −1
=
= 1.9 × 1010 nuclei
l 9.97 × 10 −7 s −1
The number of half-lives that have elapsed is n = t T1 2 = 40.2 d 8.04 d = 5.00 , so the
remaining activity of the sample is
R=
29.18
= ( 3.00 g ) e − (1.50 d 3.83 d ) ln 2 = 2.29 g
The activity is R = lN = lN 0 e − l t = R0 e − l t , where R0 is the activity at time t = 0, and the decay
constant is l = ln 2 T1 2 . Thus,
− t
R = R0 e (
29.17
)
T1 2 ln 2
R0
R0
6.40 mCi
= 5.00
=
= 0.200 mCi
n
2
2
32.0
From Equation (29.4a) in the textbook, the fraction remaining at t = 5.00 yr will be
N
− t T
ln 2
== e − l t = e ( ) = e − ( 5.00 yr 12.33 yr ) ln 2 = 0.755
N0
12
(b)
− t T
ln 2
At t = 10.0 yr, N N 0 = e − l t = e ( ) = e − (10.0 yr 12.33 yr ) ln 2 = 0.570 .
(c)
− t T
ln 2
At t = 123.3 yr, N N 0 = e − l t = e ( ) = e − (123.3 yr 12.33 yr ) ln 2 = e − (10.0 ) ln 2 = 9.77 × 10 − 4 .
(d)
68719_29_ch29_p360-376.indd 367
12
12
No. The decay model depends on large numbers of nuclei. After some long but finite time,
only one undecayed nucleus will remain. It is likely that the decay of this final nucleus will
occur before infinite time.
1/7/11 3:04:37 PM
368
29.19
Chapter 29
From R = lN = lN 0 e − l t = R0 e − l t , with R = ( 0.842 ) R0 , we find e − l t = R R0 , and lt = − ln ( R R0 ) .
Since the half-life may be expressed as T1 2 = ln 2 l, this yields
T1 2 = −
29.20
Using R = lN = lN 0 e − l t = R0 e − l t, with R R0 = 0.125, gives lt = − ln ( R R0 ) , or
t=−
29.21
t ln 2
( 2.00 d ) ln 2
=−
= 8.06 d
ln ( 0.842 )
ln ( R R0 )
(a)
ln ( R R0 )
l
⎡ ln ( R R0 ) ⎤
⎡ ln ( 0.125) ⎤
4
= − T1 2 ⎢
⎥ = − ( 5 730 yr ) ⎢
⎥ = 1.72 × 10 yr
⎣ ln 2 ⎦
⎢⎣ ln 2 ⎥⎦
The initial activity is R0 = 10.0 mCi, and at t = 4.00 h, R = 8.00 mCi. Then, from
R = lN = lN 0 e − l t = R0 e − l t, the decay constant is
ln ( R R0 )
l=−
t
=−
and the half-life is T1 2 =
29.22
ln ( 0.800 )
= 5.58 × 10 −2 h −1
4.00 h
ln 2
ln 2
=
= 12.4 h
l
5.58 × 10 −2 h −1
−3
10
−1
R0 (10.0 × 10 Ci ) ( 3.70 × 10 s 1 Ci )
=
= 2.39 × 1013 nuclei
l
(5.58 × 10−2 h−1 ) (1 h 3 600 s)
(b)
N0 =
(c)
− 5.58 ×10
R = R0 e − l t = (10.0 mCi ) e (
−2
h
−1
)( 30 h )
= 1.9 mCi
The number of 90
38 Sr nuclei initially present is
N0 =
total mass
5.0 kg
=
= 3.4 × 10 25
mass per nucleus (89.907 7 u ) (1.66 × 10 −27 kg u )
The half-life of 90
38 Sr is T1 2 = 29.1 yr (Appendix B), so the initial activity is
(3.4 × 1025 ) ln 2
N 0 ln 2
=
= 2.6 × 1016 counts s
T1 2
( 29.1 yr ) (3.156 × 10 7 s yr )
R0 = lN 0 =
From R = R0 e − l t, the time when the activity will be R = 10.0 counts min is
t=−
ln ( R R0 )
l
ln R R
( ) (ln 2 )
= − T1 2
0
⎡ 10.0 min −1 ⎛ 1 min ⎞ ⎤
ln ⎢
⎜
⎟
2.6 × 1016 s −1 ⎝ 60 s ⎠ ⎥⎦
= 1.7 × 10 3 yr
= − ( 29.1 yr ) ⎣
ln 2
29.23
We identify the missing nuclide in each case by requiring that both the total mass number and the
total charge number be the same on the two sides of the decay equation.
(a)
29.24
Bi →
Tl + 42 He
208
81
(b)
95
36
Kr →
95
37
Rb +
0
−1
e +n
(c)
144
60
Nd → 42 He + 140
Ce
58
We complete the decay formula in each case by requiring that both the total mass number and the
total charge number be the same on the two sides of the equation.
(a)
68719_29_ch29_p360-376.indd 368
212
83
12
5
B→
12
6
C +
0
−1
e +n
(b)
234
90
Th →
230
88
Ra + 42 He
(c)
14
6
C →
14
7
N+
0
−1
e +n
1/7/11 3:04:41 PM
Nuclear Physics
369
29.25
The more massive 56
Co decays into the less massive 56
Fe. To conserve charge, the charge of the
27
26
emitted particle must be +1e. Since the parent and the daughter have the same mass number, the
emitted particle must have essentially zero mass. Thus, the decay must be positron emission or
56
0
e + decay . The decay equation is 56
27 Co → 26 Fe + +1 e + n e .
29.26
The energy released in the decay 238
U → 42 He +
92
Q = ( Δm ) c 2 = ⎡⎣ m
238
U
(
− m
4
He
+m
234
Th
)⎤⎦ c
234
90
Th is
2
= ⎡⎣ 238.050 783 u − ( 4.002 603 u + 234.043 583 u ) ⎤⎦ ( 931.5 MeV u )
= 4.28 MeV
29.27
The Q-value of a decay, Q = ( Δm ) c 2, is the amount of energy released in the decay. Here, Δm is
the difference between the mass of the original nucleus and the total mass of the decay products.
If Q > 0, the decay may occur spontaneously.
(a)
40
+
For the decay 40
20 Ca → e + 19 K, the masses of the electrons do not automatically cancel.
Thus, we add 20 electrons to each side of the decay to yield neutral atoms and obtain
(
40
20
Ca + 20e − ) → e + + ( 1940 K + 19e − ) + e −
(
Then, Q = m
or
(b)
or
(b)
40
19 K atom
40
Ca atom → e + + 40 K atom + e −
)
− 2me c 2 = ⎡⎣39.962 591 u − 39.963 999 u − 2 ( 0.000 549 u ) ⎤⎦ c 2
so the decay cannot occur spontaneously .
4
140
In the decay 144
60 Nd →2 He + 58 Ce, we may add 60 electrons to each side, forming all neutral
atoms, and use masses from Appendix B to find
(
(a)
−m
Q = ( −0.002 506 u ) c 2 < 0
Q= m
29.28
40
20 C a atom
or
66
28
144
60
Nd
−m
−m
4
2 He
) c = (143.910 083 u − 4.002 603 u − 139.905 434 u) c
2
140
58 C e
2
Q = ( +0.002 046 u ) c 2 > 0 so the decay can occur spontaneously .
Ni →
66
29
Cu + −10 e +n e
Because of the mass differences, neglect the kinetic energy of the recoiling daughter nucleus
in comparison to that of the other decay products. Then, the maximum kinetic energy of the
beta particle occurs when the antineutrino is given zero energy. That maximum is
(
KEm ax = m
66
Ni
−m
66
Cu
)c
2
= ( 65.929 1 u − 65.928 9 u ) ( 931.5 MeV u )
= 0.186 MeV = 186 keV
29.29
In the decay 13 H → 32 He + −10 e +n e , the antineutrino is massless. Adding 1 electron to each side of
the decay gives ( 13 H + e − ) → ( 32 He +2 e − ) +n e , or 13 H atom → 32 Heatom +n e . Therefore, using neutral
atomic masses from Appendix B, the energy released is
(
E = ( Δm ) c 2 = m
3
H
−m
3
He
)c
2
= ( 3.016 049 u − 3.016 029 u ) ( 931.5 MeV u )
= 0.018 6 MeV = 18.6 keV
68719_29_ch29_p360-376.indd 369
1/7/11 3:04:45 PM
370
29.30
Chapter 29
The initial activity of the 1.00-kg carbon sample would have been
⎛ 15.0 counts min ⎞
4
−1
R0 = (1.00 × 10 3 g ) ⎜
⎟⎠ = 1.50 × 10 min
1.00 g
⎝
From R = lN = lN 0 e − l t = R0 e − l t, and T1 2 = 5 730 yr for 14 C (Appendix B), the age of the sample is
t=−
ln ( R R0 )
l
= − T1 2
= − ( 5 730 yr )
29.31
ln 2
ln ⎡⎣( 5.00 × 10 2 min −1 ) (1.50 × 10 4 min −1 ) ⎤⎦
ln 2
= 2.81 × 10 4 yr
From R = lN = lN 0 e − l t = R0 e − l t, and T1 2 = 5 730 yr for 14 C (Appendix B), the age of the
sample is
t=−
29.32
ln ( R R0 )
ln ( R R0 )
λ
= − T1 2
ln ( R R0 )
ln 2
= − ( 5 730 yr )
ln ( 0.600 )
= 4.22 × 10 3 yr
ln 2
The energy released in the reaction is given by
(
Q = ( Δm ) c 2 = m
1
1H
+m
27
13 Al
−m
27
14 Si
)
− mn c 2
= [1.007 825 u + 26.981 539 u − 26.986 721 u − 1.008 665 u ]( 931.5 MeV u )
= −5.61 MeV
If we require total energy to be conserved and ignore the kinetic energy of the recoiling product
nucleus, the kinetic energy of the emerging neutron must be
KE f = KEi + Q = 6.61 MeV − 5.61 MeV = 1.00 MeV
29.33
We identify the missing particles by requiring that both the total mass number and the total
charge number be the same on the two sides of the equation and by remembering that some form
of neutrino always accompanies the emission of a beta particle or positron.
(a)
29.34
21
10
Ne + 42 He → 1224 Mg + 10 n
U + 10 n → 90
Sr +
38
(b)
235
92
(c)
2 11 H → 12 H +
(a)
mi = m
(b)
mf = m
(c)
Q = mi − m f c 2 = [8.023 829 u − 8.025 594 u ]( 931.5 MeV u ) = −1.64 MeV
(d)
m p v = ( m B e + mn ) V
(e)
(
1
1H
+m
7
4 Be
0
+1
7
3 Li
144
54
Xe + 2 10 n
e + ne
= 1.007 825 u + 7.016 004 u = 8.023 829 u
+ mn = 7.016 929 u + 1.008 665 u = 8.025 594 u
)
From conservation of energy, the total kinetic energy before the reaction, plus the energy
released during the reaction, must equal the total kinetic energy after the reaction. That is,
KE p + Q = KEB e + KEn , or
1
2
m p v2 =
1
2
(m
Be
+ mn ) V 2 − Q .
continued on next page
68719_29_ch29_p360-376.indd 370
1/7/11 3:04:48 PM
Nuclear Physics
(f )
371
⎛
mp ⎞
⎛
1.007 825 u ⎞
m⎞
⎛
−1.64 MeV = 1.88 MeV
KEm in = ⎜ 1+ ⎟ Q = ⎜ 1+
⎟ Q = ⎜ 1+
⎝
⎠
⎟
⎜
7.016 004 u ⎟⎠
m Li ⎠
M
⎝
⎝
7
3
29.35
29.36
We determine the product nucleus by requiring that both the total mass number and the total
charge number be the same on the two sides of the reaction equation. The completed reaction
equations are given below:
B + 42 He → 11 H +
(a)
10
5
(a)
For the first reaction:
(
Q1 = mn + m
13
6
2
1H
C
(b)
−m
3
1H
)c
1
1H
+m
2
1H
−m
10
5
B
= [1.008 665 u + 2.014 102 u − 3.016 049 u ]( 931.5 MeV u )
exothermic
H + 21 H → 23 He
3
2 He
= 5.494 Mev > 0
2
⇒
1
1
For the second reaction:
(
C + 11 H → 42 He +
n + 21 H → 31 H
= 6.258 Mev > 0
Q2 = m
13
6
)c
2
⇒
= [1.007 825 u + 2.014 102 u − 3.016 029 u ]( 931.5 MeV u )
exothermic
(b)
Since Q1 > Q1 , the first reaction released more energy . One reason for this is that the
product nucleus in the second reaction contains 2 protons. Some energy had to be left
stored as electrical potential energy of this system, leaving less energy to be released as
kinetic energy of the product nucleus.
(c)
Assuming the electrical potential energy of the 2 protons in 23 He fully accounts for the
difference in the Q-values of the two reactions, we have ΔQ = PEe = ke e 2 r, where r is the
distance separating the 2 protons. Thus,
9
2
2
−19
ke e 2 (8.99 × 10 N ⋅ m C ) (1.60 × 10 C ) ⎛ 1 MeV ⎞
−15
r=
=
⎜⎝
⎟ = 1.88 × 10 m
ΔQ
1.60 × 10 −13 J ⎠
( 6.258 − 5.494 ) MeV
2
29.37
(a)
Requiring that both charge and the number of nucleons (atomic mass number) be conserved,
1
198
0
the reaction is found to be 197
79 Au + 0 n → 80 Hg + −1 e +n e . Note that the antineutrino has been
included to conserve electron-lepton number, which will be discussed in the next chapter.
(b)
We add 79 electrons to both sides of the reaction equation given above to produce neutral atoms
1
198
so we may use mass values from Appendix B. This gives 197
79 Au atom + 0 n → 80 Hgatom +n e , and,
remembering that the antineutrino is massless, the Q-value is found to be
(
Q = ( Δm ) c 2 = m
197
79 Au
+ mn − m
198
80 Hg
)c
2
= (196.966 552 u + 1.008 665 u − 197.966 750 u ) ( 931.5 MeV u ) = 7.89 MeV
The kinetic energy carried away by the daughter nucleus is negligible. Thus, the energy
released may be split in any manner between the electron and antineutrino, with the
maximum kinetic energy of the electron being 7.89 MeV .
29.38
We complete the reaction equation in each case by requiring that both the total mass number and
the total charge number be the same on the two sides of the equation.
(a)
68719_29_ch29_p360-376.indd 371
4
2
He + 147 N → 11 H + 178 O
(b)
7
3
Li + 11 H → 42 He + 42 He
1/7/11 3:04:51 PM
372
29.39
Chapter 29
(a)
(b)
Determine the product of the reaction by requiring that both the total mass number and the
total charge number be the same on the two sides of the equation. The completed reaction
equation is 73 Li + 42 He → 105 B + 01 n.
(
Q = Δmc 2 = ⎡⎢ m
⎣
7
3 Li
+m
4
2 He
) − (m
10
5B
+m
1
0n
)⎤⎦⎥ c
2
= ⎡⎣( 7.016 004 u + 4.002 603 u ) − (10.012 937 u + 1.008 665 u ) ⎤⎦ ( 931.5 MeV u )
= −2.79 MeV
29.40
For equal amounts of biological damage, the two doses (in rem units) must be equal, or
( heavy ion dose in rad ) × RBE
heavy ions
= ( x-ray dose in rad ) × RBE x-rays
or
ion dose in rad =
29.41
( x-ray dose in rad ) × RBE
x-rays
RBE heavy ions
=
(100 rad )(1.0 )
20
= 5.0 rad
For each rad of radiation, 10 − 2 J of energy is delivered to each kilogram of absorbing material.
Thus, the total energy delivered in this whole body dose to a 75.0-kg person is
J kg ⎞
⎛
E = ( 25.0 rad ) ⎜ 10 − 2
⎟ ( 75.0 kg ) = 18.8 J
⎝
rad ⎠
29.42
(a)
Each rad of radiation delivers 10 − 2 J of energy to each kilogram of absorbing material.
Thus, the energy delivered per unit mass with this dose is
E
J kg ⎞
⎛
= ( 200 rad ) ⎜ 10 − 2
⎟ = 2.00 J kg
⎝
m
rad ⎠
(b)
From E = Q = mc ( ΔT ) , the expected temperature rise with this dosage is
ΔT =
29.43
E m
2.00 J kg
=
= 4.78 × 10 − 4 °C
c
4 186 J kg ⋅°C
The rate of delivering energy to each kilogram of absorbing material is
J kg
⎛ E m⎞
⎛ − 2 J kg ⎞
⎜⎝
⎟⎠ = (10 rad s ) ⎜⎝ 10
⎟⎠ = 0.10
Δt
rad
s
The total energy needed per unit mass is
⎛
J ⎞
E m = c ( ΔT ) = ⎜ 4 186
( 50°C ) = 2.1× 105 J kg
kg ⋅°C ⎟⎠
⎝
so the required time will be
Δt =
68719_29_ch29_p360-376.indd 372
energy needed 2.1× 10 5 J kg
1d
⎛
⎞
=
= 2.1× 10 6 s ⎜
= 24 d
⎝ 8.64 × 10 4 s ⎟⎠
delivery rate
0.10 J kg ⋅s
1/7/11 3:04:53 PM
Nuclear Physics
29.44
(a)
373
The number of x-rays taken per year is
production = (8 x-ray d ) ( 5 d week ) ( 50 weeks yr ) = 2.0 × 10 3 x-ray yr
so the exposure per x-ray taken is
exposure rate =
(b)
exposure
5.0 rem yr
= 2.5 × 10 −3 rem x-ray
=
production 2.0 × 10 3 x-ray yr
The exposure due to background radiation is 0.13 rem yr . Thus, the work-related exposure
of 5.0 rem yr is
5.0 rem yr
≈ 38 times background levels
0.13 rem yr
29.45
(a)
(
)
From N = R l = R0 e − l t l = T1 2 R0 ln 2 e
the 10-day period is
⎛ T1 2 R0 ⎞
− t ln 2 T
1− e
ΔN = N 0 − N = ⎜
⎟
⎝ ln 2 ⎠
(
12
− t ln 2 T1 2
, the number of decays occurring during
)
⎡ (14.3 d )(1.31× 10 6 decay s ) ⎛ 8.64 × 10 4 s ⎞
− (10.0 d ) ln 2 14.3 d
=⎢
⎜⎝
⎟⎠ 1− e
ln 2
1d
⎢⎣
(
⎤
)⎥⎥
⎦
= 8.97 × 1011 decays , and one electron is emitted per decay
(b)
The total energy deposited is found to be
⎛
⎛ 1.60 × 10 −16 J ⎞
keV ⎞
8.97 × 1011 decays ) ⎜
E = ⎜ 700
(
⎟⎠ = 0.100 J
⎟
decay ⎠
1 keV
⎝
⎝
(c)
The total absorbed dose (measured in rad) is given by
dose =
29.46
(a)
energy deposited per unit mass ( 0.100 J 0.100 kg )
=
= 100 rad
energy deposition per rad
⎛ −2 J kg ⎞
10
⎜⎝
⎟
rad ⎠
The dose (in rem) received in time Δt is given by
rad ⎞ ⎤
rem ⎞
⎡⎛
⎛
dose = ( dose in rad ) × RBE = ⎢⎜ 100 × 10 − 3
⎟⎠ Δt ⎥ × (1.00 ) = ⎜⎝ 0.100
⎟ Δt
⎝
h
h ⎠
⎣
⎦
If this dose is to be 1.0 rem, the required time is
Δt =
(b)
1.0 rem
= 10 h
0.100 rem h
Assuming the radiation is emitted uniformly in all directions, the intensity of the radiation
is given by I = I 0 4p r 2 . Therefore,
I 0 4p r 2
Ir
(1.0 m )
=
=
2
I1 I 0 4p (1.0 m )
r2
2
continued on next page
68719_29_ch29_p360-376.indd 373
1/7/11 3:04:54 PM
374
Chapter 29
r = (1.0 m )
and
29.47
(a)
I1
100 mrad h
= 3.2 m
= (1.0 m )
Ir
10 mrad h
The mass of a 116 C atom is
matom = (11.011 u)(1.661× 10 −27 kg u) = 1.829 × 10 −26 kg = 1.829 × 10 −23 g
and the mass of 1 mole of 116 C is
M = matom NA = (1.829 × 10 −23 g atom ) ( 6.022 × 10 23 atoms mol ) = 11.01 g mol
The number of moles in a 3.50 mg sample is then
n=
(b)
msam ple
M
=
3.50 × 10 −6 g
= 3.18 × 10 −7 mol
11.01 g mol
The number of nuclei in the original sample is
N0 = nNA = ( 3.18 × 10 −7 mol ) ( 6.022 × 10 23 atoms mol ) = 1.91× 1017 atoms
or alternatively, N0 =
(c)
msam ple
matom
=
3.50 × 10 −6 g
= 1.91× 1017 atoms .
1.829 × 10 −23 g atom
The initial activity is
⎛ ln 2 ⎞
ln 2
R0 = lN 0 = ⎜
N0 =
(1.91× 1017 ) = 1.08 × 1014 Bq
⎟
( 20.4 min )( 60.0 s min )
⎝ T1 2 ⎠
(d)
The activity after an elapsed time of t = 8.00 h = 480 min will be
R = R0 e − l t = R0 e
29.48
− t ln 2 T1 2
= (1.08 × 1014 Bq ) e − ( 480 m in ) ln 2 ( 20.4 m in ) = 8.92 × 10 6 Bq
(
We must first calculate the Q value, given by Q = (Δm)c 2 = ⎡⎢ m n + m
⎣
1
0
4
2 He
) − (m
2
1H
+m
)⎤⎥⎦ c .
2
3
1H
Q = ⎡⎣(1.008 665 u + 4.002 603 u ) − ( 2.014 102 u + 3.016 049 u ) ⎤⎦ ( 931.5 MeV u )
= −17.6 MeV
The threshold kinetic energy for the incident neutron is then
⎛
mn ⎞
⎛ 1.008 665 u ⎞
KEm in = ⎜ 1+
(17.6 MeV ) = 22.0 MeV
⎟ Q = ⎜ 1+
⎜⎝
4.002 603 u ⎟⎠
M He ⎟⎠
⎝
1
0
4
2
29.49
From R = R0 e − l t, the elapsed time is
t=−
29.50
ln ( R R0 )
l
= − T1 2
ln ( R R0 )
ln 2
= − (14.0 d )
ln ( 20.0 mCi 200 mCi )
= 46.5 d
ln 2
To compute the Q value for the reaction 146 C → 147 N + e − +n , we add 6 electrons to each side
of the equation to obtain ( 146 C + 6e − ) → ( 147 N + 7e − ) +n , or 146 Catom → 147 N atom +n . Now, we use
the atomic masses from Appendix B of the textbook. Since the neutrino is a massless particle,
Q = (m C − m N )c 2, giving
14
6
atom
14
7
atom
Q = (14.003 242 u − 14.003 074 u ) ( 931.5 MeV u ) = 0.156 MeV = 156 keV
68719_29_ch29_p360-376.indd 374
1/7/11 3:04:56 PM
Nuclear Physics
29.51
(a)
If we assume all the 87 Sr nuclei in a gram of material came from the decay of 87 Rb nuclei,
the original number of 87 Rb nuclei was N0 = 1.82 × 1010 + 1.07 × 10 9 = 1.93 × 1010. Then,
from N = N 0 e − l t, the elapsed time is
t =−
29.52
375
T ln ( N N 0 )
ln ( N N 0 )
=− 12
=−
l
ln 2
(4.8 ×10
10
⎛ 1.82 ×1010
yr ) ln ⎜
10
⎝ 1.93×10
ln 2
⎞
⎟
⎠
= 4.1×10 9 yr
(b)
It could be no older. It could be younger if some 87 Sr were initially present.
(c)
We have assumed that all the
87
Sr present came from the decay of
87
Rb.
From R = lN = lN0 e − l t, if we have an activity of R after time t has passed, the number
of unstable nuclei that must have been present initially is given by N0 = Re l t l . With
R = 10 Ci = 10(3.7 × 1010 Bq) = 3.7 × 1011 decay s, l = ln 2 T1 2 = ln 2 5.2 yr , and
t = 30 months = 2.5 yr, this yields
⎡ ( 5.2 yr ) ( 3.156 × 10 7 s 1 yr ) ⎤ ( ln 2 5.2 yr )( 2.5 yr )
N0 = ( 3.7 × 1011 s −1 ) ⎢
= 1.2 × 10 20
⎥e
ln 2
⎢⎣
⎥⎦
Thus, the initial mass of 60 C required is found to be (using Appendix B from the text)
m = N 0 matom = (1.2 × 10 20 ) ( 59.933 822 u ) = 7.2 × 10 21 u
or
29.53
⎛ 1.66 × 10 −27 kg ⎞ ⎛ 10 6 mg ⎞
m = 7.2 × 10 21 u ⎜
⎟⎠ ⎜⎝ 1 kg ⎟⎠ = 12 mg
1u
⎝
The total activity of the working solution at t = 0 was
(R )
0
total
= ( 2.5 mCi mL )(10 mL ) = 25 mCi
Therefore, the initial activity of the 5.0-mL sample, drawn from the 250-mL working solution, was
(R )
0
sam ple
⎛ 5.0 mL ⎞
⎛ 5.0 mL ⎞
= ( R0 )total ⎜
= ( 25 mCi ) ⎜
= 0.50 mCi = 5.0 × 10 − 4 Ci
⎝ 250 mL ⎟⎠
⎝ 250 mL ⎟⎠
With a half-life of 14.96 h for 24 Na (Appendix B), the activity of the sample after 48 h is
R = R0 e − l t = R0 e
− t ln 2 T1 2
= ( 5.0 × 10 − 4 Ci ) e − ( 48 h ) ln 2 (14.96 h )
= 5.4 × 10 − 5 Ci = 54 mCi
29.54
(a)
The mass of a single
40
K atom is
m = ( 39.964 u )(1.66 × 10 −27 kg u ) = 6.63 × 10 −26 kg = 6.63 × 10 −23 g
Therefore, the number of 40 K nuclei in a liter of milk is
N=
total mass of 40 K present ( 2.00 g L ) ( 0.011 7 100 )
= 3.53 × 1018 L
=
6.63 × 10 −23 g
mass per atom
continued on next page
68719_29_ch29_p360-376.indd 375
1/7/11 3:04:59 PM
376
Chapter 29
and the activity due to potassium is
R = lN =
(b)
Using R = R0 e − l t, the time required for the 131 I activity to decrease to the level of the potassium
is given by
t =−
29.55
(3.53 × 1018 L ) ln 2
N ln 2
=
= 60.6 Bq L
T1 2
(1.28 × 109 yr ) (3.156 × 107 s yr )
ln ( R R0 )
l
=−
T1 2 ln ( R R0 )
ln 2
=−
(8.04 d ) ln ( 60.6 2 000 )
ln 2
= 40.6 d
The decay constant for 235 U is l235 = ln 2 T1 2 = ln 2 0.70 × 10 9 yr = 9.9 × 10 −10 yr −1 , while
that for 238 U is l238 = ln 2 4.47 × 10 9 yr = 1.55 × 10 −10 yr −1 . Assuming there were N 0 nuclei of
each isotope present initially, the number of each type still present should be N 235 = N 0 e − l t
and N 238 = N 0 e − l t . With the currently observed ratio of 235 U to 238 U being 0.007, we have
− l − l )t
N 235 N 238 = e (
= 0.007, or our estimate of the elapsed time since the release of the elements
forming our Earth is
235
238
235
t=−
29.56
(a)
ln ( 0.007 )
ln ( 0.007 )
=−
= 5.9 × 10 9 yr
l235 − l236
9.9 × 10 −10 yr −1 − 1.55 × 10 −10 yr −1
Obtaining the mass of a single 239
94 Pu atom from the table of Appendix B in the text, we find
N0 =
(b)
238
msam ple
matom
=
1.0 kg
( 239.052 156 u ) (1.66 × 10
−27
kg u )
= 2.5 × 10 24
The initial activity of this sample is
⎡
⎤
⎛ ln 2 ⎞
ln 2
N0 = ⎢
R0 = lN 0 = ⎜
⎥ ( 2.5 × 10 24 ) = 2.3 × 1012 Bq
⎟
7
⎝ T1 2 ⎠
⎢⎣ ( 24 000 yr ) ( 3.156 × 10 s yr ) ⎥⎦
(c)
From R = R0 e − l t , the time that must elapse before this sample will have a “safe” activity
level of 0.10 Bq is
t=−
68719_29_ch29_p360-376.indd 376
ln ( R R0 )
l
=−
T1 2 ln ( R R0 )
ln 2
=−
( 24 000 yr ) ln ( 0.10
ln 2
2.3 × 1012 )
= 1.1× 10 6 yr
1/7/11 3:05:02 PM
30
Nuclear Physics and
Elementary Particles
QUICK QUIZZES
1.
Choice (a). This reaction fails to conserve charge and also fails to conserve baryon number. For
each of these reasons, it cannot occur.
2.
Choice (b). This reaction fails to conserve charge and cannot occur.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The total energy released was E = (17 × 10 3 ton) ( 4.0 × 10 9 J 1 ton ) = 6.8 × 1013 J, and according to
the mass-energy equivalence (E = mc 2), the mass converted was
m=
E
6.8 × 1013 J
=
= 7.6 × 10 −4 kg = 0.76 g
2
c 2 ( 3.00 × 108 m s )
or m ∼ 1 g, and the correct choice is seen to be (d).
2.
The energy released in the decay n → p + e − +n e is Q = (Δm)c 2 = (mn − m p − me )c 2, or combining
the proton and electron to form a neutral hydrogen atom, Q = (mn − m H )c 2. We may then use
the atomic masses from Appendix B in the textbook to obtain
2
1
atom
Q = (1.008 665 u − 1.007 825 u ) ( 931.5 MeV u ) = 0.782 MeV
Alternately, we may use the particle masses (in energy units) from Table 30.2 in the textbook
to obtain
Q = (mn − m p − me )c 2 = (939.6 MeV c 2 − 938.3 MeV c 2 − 0.511 MeV c 2 )c 2 = 0.789 MeV
From either approach, we see that the best choice is (a).
3.
Both the charge and mass of a particle are independent of its spin, so both choices (c) and (d) are
false. A spin- 12 particle could be among the decay products, provided it is possible for the spins of
all the decay products to couple to 32 and conserve angular momentum. Also, in a magnetic field,
a spin- 32 particle could have spin states of
ms = 3 2, 1 2, − 1 2, and − 3 2
so choices (a) and (e) are false, while choice (b) is true.
4.
The decay p → +10 e +n e would conserve charge (+1 → +1+ 0), electron lepton number (0 → −1+ 1),
and strangeness (0 → 0 + 0), and can conserve energy if the total kinetic energy of the decay
377
68719_30_ch30_p377-408.indd 377
1/7/11 3:06:01 PM
378
Chapter 30
products equals the energy equivalent of the mass loss. However, it does not conserve baryon
number (+1 → 0 + 0), and the decay cannot occur. The correct choice is then (b).
5.
Positively charged particles, such as protons and alpha particles, have difficulty approaching the
target nuclei because of Coulomb repulsion. Fast-moving particles may not stay in close proximity
with a uranium nucleus long enough to have a good probability of producing a reaction. The best
particles to trigger a fission reaction of the uranium nuclei are slow-moving neutrons, so choice (d)
is the correct answer.
6.
Slow neutrons have a much higher probability of causing fission in a collision with a nucleus in
the fuel elements than do fast or high energy neutrons. The purpose of the moderator is to slow
the neutrons down and without it the chain reaction would quickly die out. The correct choice
is (c).
7.
The annihilation −10 e + +10 e → g can conserve energy [2(0.511 MeV) = 1.02 MeV], does conserve
charge [−1+ 1 = 0], conserves baryon number [0 + 0 = 0], and conserves lepton number [+1− 1 = 0].
However, the total momentum is zero before annihilation, and the momentum of the single
photon afterward is p = 1.02 MeV c ≠ 0. Thus, it cannot occur, and the correct choice is (b).
8.
1
137
96
1
In the fission reaction 235
92 U + 0 n → 53 I + 39 Y + n( 0 n), where n is some unknown number of
neutrons, we see that charge is conserved (92 + 0 = 53 + 39 + 0) regardless of the value of n. The
reaction must also conserve baryon number, so it is necessary that
235 + 1 = 137 + 96 + n
or
n=3
and (c) is seen to be the correct choice.
9.
The reaction of choice (c) fails to conserve charge [0 + 1 ≠ 0 + 0], while the reaction of choice
(d) fails to conserve baryon number [+1 ≠ 2(+1) + 0 + 0], so neither of these reactions can occur.
The reactions of choices (a), (b), and (e) satisfy all conservation laws and may occur. The correct
answers for this question are choices (c) and (d).
10.
The reaction of choice (a) fails to conserve baryon number [1+ 1 ≠ 1+ 1− 1], while the reaction of
choice (e) fails to conserve tau-lepton number [+1 ≠ 0 − 1+ 0], so neither of these reactions can
occur. The reactions of choices (b), (c), and (d) satisfy all conservation laws and may occur. The
correct answers for this question are choices (a) and (e).
11.
In a fusion reactor, the plasma must have a very high temperature so the nuclei have sufficient
energy to overcome Coulomb repulsion and collide. Also, the plasma must be sufficiently
dense to yield a good probability of collisions between nuclei and must be contained long
enough to allow a large number of collisions to occur, so the reactions can be self-sustained.
However, there are no requirements that the nuclei in the plasma be radioactive or that it
consist only of hydrogen. Fusion of elements other than hydrogen does occur inside stars, but
much higher temperatures are required for this to occur. The correct choices for this question
are (a), (c), and (d).
12.
When a particle of mass m and charge q enters a magnetic field perpendicular to the direction of
the field, it is deviated into a circular path of radius r = mv qB = 2m(KE) qB. In this case, the
kinetic energy, KE, and the magnetic field strength, B, are the same for the electron and the alpha
particle. However, the ratio m q is much smaller for the electron than for the alpha particle, so
the path of the electron has a smaller radius. This means the electron is deflected more than the
alpha particle, and the correct choice is (b).
68719_30_ch30_p377-408.indd 378
1/7/11 3:06:03 PM
Nuclear Physics and Elementary Particles
379
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The two factors presenting the most technical difficulties are the requirements of a high plasma
density and a high plasma temperature. These two conditions must be met simultaneously and
make containment of the plasma very difficult.
4.
Notice in the fusion reactions discussed in the text that the most commonly formed by-product of
the reactions is helium, which is inert and not radioactive.
6.
They are hadrons. Such particles decay into other strongly interacting particles such as p, n, and
p with very short lifetimes. In fact, they decay so quickly that they cannot be detected directly.
Decays which occur via the weak force have lifetimes of 10 −13 s or longer; particles that decay via
the electromagnetic force have lifetimes in the range of 10 −16 s to 10 −19 s.
8.
Each flavor of quark can have three colors, designated as red, green, and blue. Antiquarks are
colored antired, antigreen, and antiblue. Baryons consist of three quarks, each having a different
color. Mesons consist of a quark of one color and an antiquark with a corresponding anticolor.
Thus, baryons and mesons are colorless or white.
10.
The decays of the neutral pion, eta, and neutral sigma occur by the electromagnetic interaction.
These are the three shortest lifetimes in the table. All produce photons, which are the quanta of
the electromagnetic force, and all conserve strangeness.
12.
A neutron inside a nucleus is stable because it is in a lower energy state than a free neutron and
lower in energy than it would be if it decayed into a proton (plus electron and antineutrino). The
nuclear force gives it this lower energy by binding it inside the nucleus and by favoring pairing
between neutrons and protons.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.
192 MeV
4.
9.4 × 10 2 g
6.
2.63 kg
8.
(a) 3.1× 1010 g
(b)
1.3 × 108 mol; 7.8 × 10 31 atoms
(c)
2.6 × 10 21 J
(d)
5.5 yr
(e)
Fission alone cannot meet the world’s energy needs.
(a)
4 × 1015 g
10.
(b)
5 × 10 3 yr
(c) The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the
oceans, but the Earth’s supply of uranium is not renewable. However, if breeder reactors are
used, the current ocean supply can last about a half-million years.
12.
5.49 MeV
14.
(a)
68719_30_ch30_p377-408.indd 379
13
7
N
(b)
13
6
C
(c)
14
7
N
1/7/11 3:06:05 PM
380
16.
Chapter 30
(d)
15
8
(a)
8.68 MeV
O
(e)
15
7
N
(f)
12
6
C
(b) The proton and the boron nucleus must each have enough kinetic energy to overcome the
repulsive electrical force one exerts on the other.
18.
118 MeV
20.
(a) may occur via the weak interaction
(b) violates conservation of baryon number – cannot occur by any interaction
22.
24.
(a)
cannot occur, violates baryon number
(b)
can occur, no laws violated
(c)
can occur, no laws violated
(d)
can occur, no laws violated
(e)
cannot occur, violates baryon number and muon-lepton number
(a)
See Solution.
(b)
The second reaction violates conservation of strangeness and cannot occur via the strong or
electromagnetic interactions.
(c)
If a neutral kaon, K 0, was produced in addition to those shown, the second reaction could
occur without violating any conservation laws. Because of the greater total mass of the
product particles, the total kinetic energy of the incident particles in the this reaction would
have to be greater than in the first reaction.
26.
See Solution.
28.
See Solution.
30.
(a) Q = −e, antiproton
32.
(a) conservation of charge
(b)
Q = 0, antineutron
(b) conservation of electron-lepton number and conservation of muon-lepton number
(c) conservation of baryon number
34.
n m and ne
36.
a neutron
38.
(a)
7.85 × 1013 J
(b)
$1.74 × 10 6
(c)
$1.50 × 10 3
(d) At a 10% profit margin, the profit is $1.74 × 10 5 on each kilogram of deuterium.
68719_30_ch30_p377-408.indd 380
1/7/11 3:06:06 PM
Nuclear Physics and Elementary Particles
3.61× 10 30 J
381
1.63 × 108 yr
40.
(a)
(b)
42.
The first reaction has a net 1u and 2d quarks both before and after the reaction. The second
reaction has a net 1u and 2d quarks before the reaction and 1u, 3d, and 1s quark present
afterwards.
44.
0.827 9c
PROBLEM SOLUTIONS
30.1
The energy consumed by a 100-W lightbulb in a 1.0-h time period is
⎛ 3 600 s ⎞
= 3.6 × 10 5 J
E = P ⋅ Δt = (100 J s )(1.0 h ) ⎜
⎝ 1 h ⎟⎠
The number of fission events, yielding an average of 208 MeV each, required to produce this
quantity of energy is
n=
30.2
E
3.6 × 10 5 J ⎛ 1 MeV
=
⎜
208 MeV 208 MeV ⎝ 1.60 × 10 −13
The energy released in the reaction 10 n +
Q = ( Δm ) c 2 = ⎡ m
⎣
235
92 U
− 2 mn − m
98
40 Zr
235
92
⎞
16
⎟ = 1.1× 10
J⎠
U →
−m
135
52 Te
98
40
1
Zr + 135
52 Te + 3 0 n is
⎤ c2
⎦
= ⎡⎣ 235.043 923 u − 2 (1.008 665 u ) − 97.912 0 u − 134.908 7 u ⎤⎦ ( 931.5 MeV u )
= 192 MeV
30.3
The energy released in the reaction 10 n +
Q = ( Δm ) c 2 = ⎡ m
⎣
235
92 U
− 11mn − m
88
38 Sr
235
92
U →
−m
136
54 Xe
88
38
1
Sr + 136
54 Xe + 12 0 n is
⎤ c2
⎦
= ⎡⎣ 235.043 923 u − 11(1.008 665 u ) − 87.905 614 u − 135.907 220 u ⎤⎦ ( 931.5 MeV u )
= 126 MeV
30.4
The total energy released was
⎛ 4.0 × 10 9 J ⎞ ⎛
⎞
1 MeV
= 5.0 × 10 26 MeV
E = 20 × 10 3 ton TNT ⎜
−13
⎜
⎟
J ⎟⎠
⎝ 1 ton TNT ⎠ ⎝ 1.60 × 10
(
)
The number of 235 U nuclei that must have undergone fission to yield this energy is
n=
E
5.0 × 10 26 MeV
=
= 2.4 × 10 24 nuclei
208 MeV nucleus 208 MeV nucleus
The mass of 235 U which will contain this number of atoms is
⎛ n ⎞
⎛ 2.4 × 10 24 atoms ⎞ ⎛
g ⎞
2
m=⎜
M
=
⎜⎝ 235
⎟ = 9.4 × 10 g
m
ol
23
⎜
⎟
⎟
mol ⎠
⎝ 6.02 × 10 atoms mol ⎠
⎝ NA ⎠
68719_30_ch30_p377-408.indd 381
1/7/11 3:06:08 PM
382
30.5
Chapter 30
(a)
With a specific gravity of 4.00, the density of soil is r = 4.00 × 10 3 kg m 3. Thus, the mass
of the top 1.00 m of soil is
2
1m ⎞ ⎤
kg ⎞ ⎡
⎛
2 ⎛
m = rV = ⎜ 4.00 × 10 3
1.00
m
43
560
ft
= 1.62 × 10 7 kg
(
)
(
)
⎜
⎟
⎟
⎢
⎝ 3.281 ft ⎠ ⎥⎦
⎝
m3 ⎠ ⎣
At a rate of 1 part per million, the mass of uranium in this soil is then
mU =
(b)
235
Since 0.720% of naturally occurring uranium is 235
92 U, the mass of 92 U in the soil of part (a) is
m
30.6
m 1.62 × 10 7 kg
=
= 16.2 kg
10 6
10 6
235
92 U
= ( 7.20 × 10 −3 ) mU = ( 7.20 × 10 −3 ) (16.2 kg ) = 0.117 kg = 117 g
With a power output of Pout = 1.00 × 10 9 W and efficiency of e = 0.400, the total energy input
required each day is
Einput =
(
)(
)
1.00 × 10 9 J s 1.00 d ⎛
⎞ ⎛ 8.64 × 10 4 s ⎞
Pout ⋅ Δt
1 MeV
27
=
⎜⎝ 1.60 × 10 −13 J ⎟⎠ ⎜⎝
⎟⎠ = 1.35 × 10 MeV
0.400
1 d
e
At 200 MeV per fission event, the number of 235 U atoms consumed each day is
n=
Einput
200 MeV atom
=
1.35 × 10 27 MeV
= 6.75 × 10 24 atoms
200 MeV atom
The mass of 235 U which will contain this number of atoms is
⎛ n ⎞
⎛ 6.75 × 10 24 atoms ⎞ ⎛
g ⎞
3
m=⎜
M
=
⎜⎝ 235
⎟ = 2.63 × 10 g = 2.63 kg
m
ol
23
⎜
⎟
⎟
mol ⎠
⎝ 6.02 × 10 atoms mol ⎠
⎝ NA ⎠
30.7
(a)
For a sphere of radius a, the surface area is A = 4p a 2, and the volume is V = 4p a 3 3. Thus,
the surface-to-volume ratio is
ratiosphere =
(b)
A
4p a 2
=
= 3a
V 4p a 3 3
For a cube of side s, A = 6 ⋅ Aface = 6s 2 and V = s ⋅ s ⋅ s = s 3. The surface-to-volume ratio for a
cube is then
ratiocube =
A 6s 2 6
= 3 =
V
s
s
If the cube has the same volume as the sphere, then s 3 = 4p a 3 3, or s = ( 4p 3) a, and the
surface-to-volume ratio for the cube becomes
1
3
ratiocube =
(c)
68719_30_ch30_p377-408.indd 382
A 6s 2
6
= 3 =
= 3.72 a
s
V
( 4p 3) a
1
3
The sphere has the better shape for minimum leakage since it has the smaller surface-tovolume ratio.
1/7/11 3:06:09 PM
Nuclear Physics and Elementary Particles
30.8
(a)
The mass of 235 U in the reserve is
m
(b)
383
235
U
kg ⎞ ⎛ 3 g ⎞
⎛ 0.70 ⎞
⎛
4.4 × 10 6 metric ton ) ⎜ 10 3
=⎜
= 3.1× 1010 g
⎟ 10
⎝ 100 ⎟⎠ (
⎝
ton ⎠ ⎜⎝
kg ⎟⎠
The number of moles is n = m M m ol = ( 3.1× 1010 g ) ( 235g mol ) = 1.3 × 108 mol , and the
number of 235 U atoms in this reserve is
N = nN A = (1.3 × 108 mol ) ( 6.02 × 10 23 atoms mol ) = 7.8 × 10 31 atoms
(c)
Assuming all atoms undergo fission and all released energy captured, the total energy
available is
MeV ⎞
MeV ⎞ ⎛ 1.60 × 10 −13 J ⎞
⎛
⎛
= ( 7.8 × 10 31 atoms ) ⎜ 208
E = N ⎜ 208
= 2.6 × 10 21 J
⎟
⎟
⎝
⎝
atom ⎠
atom ⎠ ⎜⎝ 1 MeV ⎟⎠
(d)
At a consumption rate of 1.5 × 1013 J s, the maximum time this energy supply could last is
t=
(e)
30.9
1 yr
E
2.6 × 10 21 J ⎛
⎞
=
⎜
⎟ = 5.5 yr
P 1.5 × 1013 J s ⎝ 3.156 × 10 7 s ⎠
Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram
of uranium.
The total energy required for one year is
E = ( 2 000 kWh month ) ( 3.60 × 10 6 J kWh )(12.0 months ) = 8.64 × 1010 J
The number of fission events needed will be
N=
E
Eevent
=
8.64 × 1010 J
= 2.60 × 10 21
( 208 MeV ) (1.60 × 10 −13 J MeV )
and the mass of this number of 235 U atoms is
⎛ N ⎞
⎛ 2.60 × 10 21 atoms ⎞
m=⎜
=
M
m ol
⎜⎝ 6.02 × 10 23 atoms mol ⎟⎠ ( 235 g mol ) = 1.01 g
⎝ N A ⎟⎠
30.10
(a)
At a concentration of c = 3 mg m 3 = 3 × 10 −3 g m 3, the mass of uranium dissolved in
the oceans covering two-thirds of Earth’s surface to an average depth of hav = 4 km is
mU = cV = c( 23 A) ⋅ hav = c[ 23 (4p RE2 )]⋅ hav , or
2
g ⎞ ⎛ 2⎞
⎛
mU = ⎜ 3 × 10 −3 3 ⎟ ⎜ ⎟ 4p ( 6.38 × 10 6 m ) ( 4 × 10 3 m ) = 4 × 1015 g
⎝
⎠
⎝
⎠
m
3
(b)
Fissionable 235 U makes up 0.7% of all uranium, so m U = 0.70mU 100. If we assume all of
the 235 U is collected and caused to undergo fission, with the release of about 200 MeV per
event, the potential energy supply is
235
E = ( number of
235
⎤⎛
MeV ⎞
MeV ⎞ ⎡⎛ 0.7mU 100 ⎞
⎛
= ⎢⎜
N A ⎥ ⎜ 200
U atoms ) ⎜ 200
⎟
⎟
⎟
⎝
⎝
M m ol
atom ⎠
atom ⎠ ⎢⎣⎝
⎠
⎥⎦
continued on next page
68719_30_ch30_p377-408.indd 383
1/7/11 3:06:12 PM
384
Chapter 30
and at a consumption rate of P = 1.5 × 1013 J s, the time this could supply the world’s
energy needs is t = E P, or
⎡ 0.7 ⎛ mU ⎞ ⎤ ( 200 MeV atom )
t=⎢
⎜
⎟⎥
P
⎢⎣ 100 ⎝ M m ol ⎠ ⎥⎦
⎡ 0.7 ⎛ 4 × 1015 g ⎞ ⎛
⎤ 200 MeV atom ⎛ 1.6 × 10 −13 J ⎞ ⎛
⎞
1 yr
23 atoms ⎞
=⎢
6.02
×
10
⎜
⎟
⎥
13
7
⎜
⎟
⎜
⎟
⎜
⎝
⎠
mol ⎦ 1.5 × 10 J s ⎝ 1 MeV ⎠ ⎝ 3.156 × 10 s ⎟⎠
⎣ 100 ⎝ 235 g mol ⎠
= 5 × 10 3 yr
(c)
30.11
The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the
oceans, but the Earth’s supply of uranium is not renewable. However, if breeder reactors are
used, the current ocean supply can last about a half-million years.
(a)
4
2
He + 42 He →
8
4
(b)
8
4
Be + 42 He →
12
6
(c)
Be + g
C +g
The total energy released in this pair of fusion reactions is
Q = ( Δm ) c 2 = ⎡⎣3m
4
He
−m
12
C
⎤⎦ c 2
= ⎡⎣3 ( 4.002 603 u ) − 12.000 000 u ⎤⎦ ( 931.5 MeV u ) = 7.27 MeV
30.12
The energy released in the reaction 11 H + 12 H → 32 He + g is
Q = ( Δm ) c 2 = ⎡ m
⎣
1
1H
+m
2
1H
−m
3
2 He
⎤ c2
⎦
= [1.007 825 u + 2.014 102 u − 3.016 029 u ]( 931.5 MeV u ) = 5.49 MeV
30.13
The energy released in the reaction 21 H + 21 H → 31 H + 11 H is Q = ⎡ 2m
⎣
2
1H
−m
3
1H
− m H ⎤ c 2, or
⎦
1
1
Q = ⎡⎣ 2 ( 2.014 102 u ) − 3.016 049 u − 1.007 825 u ⎤⎦ ( 931.5 MeV u ) = 4.03 MeV
30.14
30.15
H + 126 C →
13
7
N +g
(b)
13
7
N→
13
6
C + 11 H →
14
7
N +g
(d)
14
7
N + 11 H →
15
8
O +g
15
8
O→
(f )
15
7
N + 11 H →
12
6
C + 42 He
(a)
1
1
(c)
(e)
15
7
N +
0
+1
e +n
13
6
C+
0
+1
e +n
With the deuteron and triton at rest, the total momentum before the reaction is zero. To
conserve momentum, the neutron and the alpha particle must move in opposite directions with
equal magnitude momenta after the reaction, or pa = pn . Neglecting relativistic effects, we
use the classical relationship between momentum and kinetic energy, KE = p2 2m, and write
2ma KEa = 2mn KEn , or KEa = (mn ma )KEn .
To conserve energy, it is necessary that the kinetic energies of the reaction products satisfy
the relation KEn + KEa = Q = 17.6 MeV. Then, using the result from above, we have
KEn + (mn ma )KEn = 17.6 MeV, or the kinetic energy of the emerging neutron must be
KEn =
68719_30_ch30_p377-408.indd 384
17.6 MeV
= 14.1 MeV
1+ (1.008 665 u ) ( 4.002 603 u )
1/7/11 3:06:15 PM
Nuclear Physics and Elementary Particles
30.16
(a)
385
The energy released in the reaction 11 H + 115 B → 3( 42 He) is
Q = ( Δm ) c 2 = ⎡ m
⎣
1
1H
+m
11
5B
− 3m
4
2 He
⎤ c2
⎦
= ⎡⎣1.007 825 u + 11.009 306 u − 3 ( 4.002 603 u ) ⎤⎦ ( 931.5 MeV u )
= 8.68 MeV
(b)
30.17
The proton and the boron nucleus both have positive charges but must come very close to
one another in order for fusion to occur. Thus, they must have sufficient kinetic energy to
overcome the repulsive Coulomb force one exerts on the other.
Note that pair production cannot occur in a vacuum. It must occur in the presence of a massive
particle which can absorb at least some of the momentum of the photon and allow total linear
momentum to be conserved.
When a particle-antiparticle pair is produced by a photon having the minimum possible frequency,
and hence minimum possible energy, the nearby massive particle absorbs all the momentum of
the photon, allowing both components of the particle-antiparticle pair to be left at rest. In such an
event, the total kinetic energy afterwards is essentially zero, and the photon need only supply the
total rest energy of the pair produced.
The minimum photon energy required to produce a proton-antiproton pair is Ephoton = 2 ( ER )proton.
Thus, the minimum photon frequency is
f=
and l =
30.18
Ephoton
h
=
2 ( ER )proton
h
=
2 ( 938.3 MeV )(1.60 × 10 −13 J MeV )
6.63 × 10 −34 J ⋅s
= 4.53 × 10 23 Hz
c 3.00 × 108 m s
=
= 6.62 × 10 −16 m = 0.662 fm
f 4.53 × 10 23 Hz
The total kinetic energy after the pair production is
KEtotal = Ephoton − 2 ( ER )proton = 2.09 × 10 3 MeV − 2 ( 938.3 MeV ) = 213 MeV
The kinetic energy of the antiproton is then
KE p = KEtotal − KE p = 213 MeV − 95.0 MeV = 118 MeV
30.19
The total rest energy of the p 0 is converted into energy of the photons. Since the total momentum
was zero before the decay, the two photons must go in opposite directions with equal magnitude
momenta (and hence equal energies). Thus, the rest energy of the p 0 is split equally between the
two photons, giving for each photon
Ephoton =
pphoton =
and
68719_30_ch30_p377-408.indd 385
f=
ER ,p
0
2
Ephoton
c
Ephoton
h
=
135 MeV
= 67.5 Mev
2
= 67.5 MeV c
=
67.5 MeV ⎛ 1.60 × 10 −13 J ⎞
= 1.63 × 10 22 Hz
6.63 × 10 −34 J ⋅s ⎜⎝ 1 MeV ⎟⎠
1/7/11 3:06:18 PM
386
30.20
Chapter 30
Observe that the given reactions involve only mesons and baryons. With no leptons before or after
the reactions, we do not have to consider the conservation laws concerning the various lepton
numbers. All interactions always conserve both charge and baryon numbers. The strong interaction
also conserves strangeness. Conservation of strangeness may be violated by the weak interaction
but never by more than one unit. With these facts in mind consider the given interactions:
K0 → p + + p −
K 0 total before p + p − total after
Charge
0
0
+1 −1
0
Baryon number
0
0
0
0
0
Strangeness
+1
+1
0
0
0
This reaction conserves both charge and baryon number but does violate strangeness by one unit.
Thus, it can occur via the weak interaction but not other interactions.
Λ0 → p + + p −
+
total before p
Λ0
Charge
p−
total after
0
0
+1
–1
0
Baryon number +1
+1
0
0
0
–1
0
0
0
Strangeness
–1
This reaction fails to conserve baryon number and cannot occur via any interaction.
30.21
(a)
Reaction
p + p → m + + e−
(b)
p +p→ p+p
(c)
p+p→ p+p +
−
Conservation Law Violated
Le : ( 0 + 0 → 0 + 1); and L m : ( 0 + 0 → −1+ 0 )
+
Charge, Q:
Baryon Number, B:
(d) p + p → p + p + n
(e)
( −1+ 1 → + 1+ 1)
Baryon Number, B:
g +p→ n+p 0
Charge, Q:
(1+ 1 → 1+ 0 )
(1+ 1 → 1+ 1 + 1)
(0 + 1 → 0 + 0)
30.22
(a)
Reaction
p→p + +p 0
(b)
p+p→ p+p+p
(c)
p → m +n m
No violations − The reaction can occur.
(d)
n → p + e − +n e
No violations − The reaction can occur.
(e)
p + → m+ + n
+
+
Conservation Law Violated
Baryon Number, B:
0
No violations − The reaction can occur.
Baryon Number, B:
68719_30_ch30_p377-408.indd 386
(1 → 0 + 0 )
( 0 → 0 + 1)
Muon-Lepton Number, L m :
( 0 → −1+ 0 )
1/7/11 3:06:19 PM
Nuclear Physics and Elementary Particles
387
30.23
Reaction
Conclusion
Baryon Number - violated: ( 0 + 1 → 0 )
Cannot Occur
p + p → 2h
(b)
K − + n → Λ0 + p −
All conservation laws observed.
May occur via the Strong
Interaction
(c)
K →p +p
0
Strangeness violated by 1 unit
( −1 → 0 + 0 ). All other conservation
laws observed.
Can occur via Weak
Interaction, but not
by Electromagnetic or
Strong Interactions.
(d)
Ω → Ξ +p
0
Strangeness violated by 1 unit
( −3 → −2 + 0 ). All other conservation
laws observed.
Can occur via Weak
Interaction, but not
by Electromagnetic or
Strong Interactions.
All conservation laws observed.
Allowed via all interactions, but photons are the
mediator of the electromagnetic interaction and
the lifetime of the h 0 is
consistent with decay by
that interaction.
(e)
30.24
Conservation Law
(a)
(a)
−
−
−
−
−
0
h 0 → 2g
p + +p→K+ + Σ+
Baryon number, B:
0 +1→ 0 +1
ΔB = 0
Charge, Q:
1+ 1 → 1+ 1
ΔQ = 0
Baryon number, B:
0 +1→ 0 +1
ΔB = 0
Charge, Q:
1+ 1 → 1+ 1
ΔQ = 0
p + +p→p + + Σ+
(b)
Strangeness is conserved in the first reaction:
Strangeness, S:
0 + 0 → 1− 1
ΔS = 0
The second reaction does not conserve strangeness:
Strangeness, S:
0 + 0 → 0 −1
ΔS = −1
The second reaction cannot occur via the strong or electromagnetic interactions .
(c)
If one of the neutral kaons were also produced in the second reaction, giving
p + + p → p + + Σ + + K 0, then strangeness would no longer be violated:
Strangeness, S:
0 + 0 → 0 − 1 +1
ΔS = 0
Because the total mass of the product particles in this reaction would be greater than
that in the first reaction [see part (a)], the total incident energy of the reacting particles
would have to be greater for this reaction than for the first reaction .
30.25
(a)
Λ0 → p + p −
Strangeness, S:
−1 → 0 + 0
⇒ ΔS ≠ 0
Not Conserved
continued on next page
68719_30_ch30_p377-408.indd 387
1/7/11 3:06:23 PM
388
Chapter 30
(b)
p − + p → Λ0 + K 0
Strangeness, S:
(c)
0 + 0 → +1− 1
⇒ ΔS = 0
Conserved
0 + 0 → 0 −1
⇒ ΔS ≠ 0
Not Conserved
−2 → −1+ 0
⇒ ΔS ≠ 0
Not Conserved
−2 → 0 + 0
⇒ ΔS ≠ 0
Not Conserved
Ξ− → Λ 0 + p −
Strangeness, S:
(f)
Conserved
p − + p → p − + Σ+
Strangeness, S:
(e)
⇒ ΔS = 0
p + p → Λ0 + Λ0
Strangeness, S:
(d)
0 + 0 → −1+ 1
Ξ0 → p + p −
Strangeness, S:
30.26
30.27
proton
u
u
d
total
strangeness
0
0
0
0
0
baryon
number
1
1
3
1
3
1
charge
e
1
3
2e 3
2e 3
−e 3
e
neutron
u
d
d
total
strangeness
0
0
0
0
baryon
number
1
1
3
0
1
3
1
3
1
charge
0
2e 3
−e 3
−e 3
0
The number of water molecules in one liter (mass = 1 000 g) of water is
⎛ 1 000 g ⎞
N =⎜
(6.02 × 1023 molecules mol ) = 3.34 × 1025 molecules
⎝ 18.0 g mol ⎟⎠
Each molecule contains 10 protons, 10 electrons, and 8 neutrons. Thus, there are
N e = 10 N = 3.34 × 10 26 electrons , N p = 10 N = 3.344 ×10 26 protons,
and
N n = 8 N = 2.68 × 10 26 neutrons
Each proton contains 2 up quarks and 1 down quark, while each neutron has 1 up quark and 2
down quarks. Therefore, there are
N u = 2 N p + N n = 9.36 × 10 26 up quarks , and N d = N p + 2 N n = 8.70 × 10 26 down quarks .
68719_30_ch30_p377-408.indd 388
1/7/11 3:06:26 PM
Nuclear Physics and Elementary Particles
389
30.28
K 0 Particle
K0
d
s
total
strangeness
1
0
1
1
baryon number
0
13
−1 3
0
charge
0
−e 3
e3
0
Λ0 Particle
30.29
30.30
30.31
L0
u
d
s
total
strangeness
−1
0
0
−1
−1
baryon
number
1
13
13
13
1
charge
0
2e 3
−e 3
−e 3
0
Compare the given quark states to the entries in Table 30.4:
(a)
suu = Σ +
(b) ud = p
(c)
sd = K 0
(d) ssd = Ξ −
(a)
⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞
uud: charge = ⎜ − e ⎟ + ⎜ − e ⎟ + ⎜ + e ⎟ = − e . This is the antiproton .
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠
(b)
⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
udd: charge = ⎜ − e ⎟ + ⎜ + e ⎟ + ⎜ + e ⎟ = 0 . This is the antineutron .
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠
−
The reaction is Σ 0 + p → Σ + + g + X, or on the quark level, uds + uud → uus + 0 + x.
The left side has a net 3 u, 2 d, and 1s. The right side has 2 u, 0 d, and 1s, plus the quark composition
of the unknown particle. To conserve the total number of each type of quark, the composition of the
unknown particle must be x = udd . Therefore, the unknown particle must be a neutron .
30.32
30.33
(a)
p − + p → Σ + + p 0 is forbidden by conservation of charge .
(b)
m − → p − +n e is forbidden by both conservation of electron-lepton number , and
conservation of muon-lepton number .
(c)
p → p + + p + + p − is forbidden by conservation of baryon number .
To the reaction for nuclei, 11 H + 32 He → 42 He + +01 e + n e , we add three electrons to both sides to
obtain 11 H atom + 32 Heatom → 42 Heatom + −01 e + +01 e + ne. Then we use the masses of the neutral atoms
from Appendix B of the textbook to compute
Q = ( Δm ) c 2 = ⎡ m
⎣
1
1H
+m
3
2
He
−m
4
2
He
− 2 me ⎤ c 2
⎦
= ⎡⎣1.007 825 u + 3.016 029 u − 4.002 603 u − 2 ( 0.000 549 u ) ⎤⎦ ( 931.5 MeV u )
= 18.8 MeV
68719_30_ch30_p377-408.indd 389
1/7/11 3:06:30 PM
390
Chapter 30
30.34
For the particle reaction, m + + e − → 2n, the lepton numbers before the event are L m = −1 and
Le = +1. These values must be conserved by the reaction so one of the emerging neutrinos must
have L m = −1 while the other has Le = +1. The emerging particles are n m and n e .
30.35
(a)
K+ + p → X + p
Since this occurs via the strong interaction, all conservation rules must be observed.
Baryon Number:
0 +1→ B +1
ΔB = 0 ⇒ B = 0
so X is not a baryon.
Strangeness:
+1+ 0 → S + 0
ΔS = 0 ⇒ S = +1
or X has S = +1
Charge:
+e + e → Q + e
ΔQ = 0 ⇒ Q = +e
or X has Q = +e
Lepton Numbers:
0+0→ L+0
ΔL = 0 ⇒ L = 0
or X has Le = L m = Lt = 0
Of the particles in Table 30.2, the only non-baryon with S = +1 and Q = +e is the positive kaon,
K + . Thus, this is an elastic scattering process.
The weak interaction observes all conservation rules except strangeness, and ΔS = ±1.
(b)
Ω− → X + p −
Baryon Number:
1→ B + 0
ΔB = 0 ⇒ B = +1
or X has B = +1
Strangeness:
−3→ S + 0
ΔS = +1 ⇒ S = −2
or X has S = −2
Charge:
−e → Q − e
ΔQ = 0 ⇒ Q = 0
or X has Q = 0
Lepton Numbers:
0→ L+0
ΔL = 0 ⇒ L = 0
or X has Le = L m = Lt = 0
The particle must be a neutral baryon with strangeness S = −2. Thus, it is the Ξ0 .
(c)
K + → X + m + +n m
Baryon Number:
0→ B+0+0
ΔB = 0 ⇒ B = 0
so X is not a baryon.
Strangeness:
+1 → S + 0 + 0
ΔS = ±1 ⇒ S = 0, + 2
or X has S = 0 or + 2
Charge:
+e → Q + e + 0
ΔQ = 0 ⇒ Q = 0
or X has Q = 0
Lepton Numbers:
0 → Le + 0 + 0
ΔLe = 0 ⇒ Le = 0
or X has Le = 0
0 → L m − 1+ 1
ΔL m = 0 ⇒ L m = + 0
or X has L m = 0
0 → Lt + 0 + 0
ΔLt = 0 ⇒ Lt = + 0
or X has Lt = 0
The particle must be a neutral non-baryon with strangeness S = 0 or S = +2, and
Le = L m = Lt = 0. This is the neutral pion, p 0 .
30.36
Assuming a head-on collision, the total momentum is zero both before and after the reaction
p + p → p + p + + X. Therefore, since the proton and the pion are at rest after the reaction, particle
X must also be left at rest.
continued on next page
68719_30_ch30_p377-408.indd 390
1/7/11 3:06:34 PM
Nuclear Physics and Elementary Particles
391
Particle X must be a neutral baryon in order to conserve charge and baryon number in the reaction.
Conservation of energy requires that the rest energy this particle be
ER, X = 2 ( ER, p + 70.4 MeV ) − ER, p − ER, π = ER, p − ER, π +140.8 MeV
+
or
+
ER, X = 938.3 MeV −139.6 MeV +140.8 MeV = 939.5 MeV
Particle X is a neutron .
30.37
If a neutron starts with kinetic energy KEi = 2.0 MeV and loses one-half of its kinetic energy in
each collision with a moderator atom, its kinetic energy after n collisions will be KE f = KEi 2n.
The average kinetic energy associated with particles in a gas at temperature T = 20.0°C = 293 K
(see Chapter 10 of the textbook) is
KE f =
3
3
1 eV
⎛
⎞
kB T = (1.38 × 10 −23 J K )( 293 K ) ⎜
= 0.0379 eV
⎝ 1.60 × 10 −19 J ⎟⎠
2
2
Thus, the number of collisions the neutron must make before it reaches the energy associated
with a room temperature gas is n ln 2 = ln(KEi KE f ), or
6
⎛ 1 ⎞ ⎛ 2.0 × 10 eV ⎞
n=⎜
= 26
ln
⎝ ln 2 ⎟⎠ ⎜⎝ 0.0379 eV ⎟⎠
30.38
(a)
The number of deuterons in one kilogram of deuterium is
N=
m
1 kg
=
= 3.00 × 10 26
matom ( 2.01 u )(1.66 × 10 −27 kg u )
Each occurrence of the reaction 21 D + 21 D → 23 He + 01 n consumes two deuterons and releases
Q = 3.27 MeV of energy. The total energy available from the one kilogram of deuterium is
then
⎛ 3.00 × 10 26 ⎞
⎛ 1.60 × 10 −13 J ⎞
⎛ N⎞
13
E =⎜ ⎟Q =⎜
3.27
Mev
(
)
⎟⎠
⎜⎝ 1 MeV ⎟⎠ = 7.85 × 10 J
⎝ 2⎠
2
⎝
(b)
At a rate of eight cents per kilowatt-hour, the value of this energy is
⎛ $0.08 ⎞ ⎛ 1 kWh
value = E × rate = ( 7.85 × 1013 J ) ⎜
⎝ kWh ⎟⎠ ⎜⎝ 3.60 × 10 6
30.39
⎞
6
⎟ = $1.74 × 10 = $1 740 000
J⎠
(c)
Deuterium makes up four-twentieths or one-fifth of the mass of a heavy water molecule.
Thus, five kilograms of heavy water are necessary to obtain one kilogram of deuterium. The
cost for this water (5 kg)($300 kg) = $1 500 .
(d)
Whether it would be cost-effective depends on how much it cost to fuse the deuterium and
how much net energy was produced. If the cost is nine-tenths of the value of the energy
produced, each kilogram of deuterium would still yield a profit of $174 000.
(a)
The number of moles in 1.0 gal of water is
⎛ 3.786 L ⎞ ⎛ 10 3 cm 3 ⎞
g ⎞
⎛
1.0 gal ) ⎜
⎜⎝ 1.0
3 ⎟(
cm ⎠
⎝ 1 gal ⎟⎠ ⎜⎝ 1 L ⎟⎠
m rV
n=
=
=
= 2.1× 10 2 mol
M
M
18 g mol
continued on next page
68719_30_ch30_p377-408.indd 391
1/7/11 3:06:40 PM
392
Chapter 30
so the number of hydrogen atoms (2 per water molecule) will be
N H = 2 ( nN A ) = 2 ( 2.1× 10 2 mol ) ( 6.02 × 10 23 mol −1 ) = 2.5 × 10 26
and one of every 6 500 of these contains a deuteron. Thus, the number of deuterons contained in 1.0 gal of water is
N D = N H 6 500 = 2.5 × 10 26 6 500 = 3.8 × 10 22 deuterons
and the available energy is
E = ( 3.8 × 10 22 deuterons ) (1.64 MeV deuteron ) (1.6 × 10 −13 J MeV ) = 1.0 × 1010 J
(b)
At a consumption rate of P = 1.0 × 10 4 J s, the time that this could supply a person’s
energy needs is
Δt =
30.40
(a)
E
1.0 × 1010 J ⎛ 1 d ⎞
=
= 12 d
P 1.0 × 10 4 J s ⎜⎝ 86 400 s ⎟⎠
The number of water molecules in the oceans is
NH
2O
⎛ mocean ⎞
⎛ 1.32 × 10 21 kg ⎞
water
⎟ NA = ⎜
=⎜
6.02 × 10 23 molecules mol )
(
−3
⎟
M
18.0
×
10
kg
mol
⎝
⎠
⎜⎝ m ol ⎟⎠
= 4.41× 10 46 molecules
Since there are 2 hydrogen atoms per water molecule, and the fraction of all hydrogen
atoms which contain deuterons is 1.56 × 10 −4, the number of deuterons in the oceans is
N D = ( 2N H
2O
) ×1.56 ×10
−4
= 2 ( 4.41×10 46 ) (1.56 ×10 −4 ) = 1.38 ×10 43
Each fusion event consumes 2 deuterons, so the number of fusion events possible is
N events = 12 ND . The reaction 21 D+ 21 D → 23 He + 01 n releases 3.27 MeV per event, so the total
energy that could be released is
E = (3.27 MeV ) N events = (3.27 MeV ) ( 12 N D )
⎛ 1.38 ×10 43 ⎞ ⎛ 1.60 ×10 −13
= (3.27 MeV ) ⎜
⎟⎜
2
⎝
⎠ ⎝ 1 MeV
(b)
J⎞
30
⎟ = 3.61×10 J
⎠
One hundred times the current world electric power consumption is
P = 100 ( 7.00 × 1012 W ) = 7.00 × 1014 J s
At this rate, the time the energy computed in part (a) would last is
t=
30.41
⎞
E
3.61× 10 30 J ⎛
1 yr
=
= 1.63 × 108 yr
14
7
⎜
P 7.00 × 10 J s ⎝ 3.156 × 10 s ⎟⎠
Conserving energy in the decay p − → m − +n m , with the p − initially at rest gives E m + En = E R ,p , or
−
E m + En = 139.6 MeV
[1]
continued on next page
68719_30_ch30_p377-408.indd 392
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Nuclear Physics and Elementary Particles
393
Since the total momentum was zero before the decay, conservation of momentum requires
the muon and antineutrino to recoil in opposite directions with equal magnitude momenta,
or pm = pn . The relativistic relation between total energy and momentum of a particle gives
for the antineutrino:
En = pn c, or pn = En c. The same relation applied to the muon gives
2
2
E m2 = pm c + ER2 , m . Since we must have pm = pn , this may be rewritten as E m2 = ( pn c ) + E R2 , m , and
using the fact that
pn c = En , we have E m2 = En 2 + E R2 , m . Rearranging and factoring then gives
( )
(
E m2 − En2 = E m + En
)(E
m
)
− En = E R2 , m = (105.7 MeV )
2
and
E m − En =
(105.7 MeV )2
[2]
E m + En
Substituting Equation [1] into [2] gives E m − En = (105.7 MeV)2 139.6 MeV, or
E m − En = 80.0 MeV
[3]
Subtracting Equation [3] from Equation [1] yields 2En = 59.6 MeV, and
En = 29.8 MeV
30.42
The reaction p − + p → K 0 + Λ 0 is ud + uud → d s + uds at the quark level. There is
a net 1u and 2 d quarks both before and after the reaction . This reaction conserves the net
number of each type quark.
For the reaction p − + p → K 0 + n, or ud + uud → d s + udd, there is a net
1u and 2 d before the reaction and 1u, 3d, and 1 s quark afterwards . This reaction does not
conserve the net number of each type quark.
30.43
To compute the energy released in each occurrence of the reaction
4p + 2e − → a + 2n e + 6g
we add two electrons to each side to produce neutral atoms and obtain
4( 11 H atom ) → 42 Heatom + 2n e + 6g . Then, recalling that the neutrino and the photon both have zero
rest mass, and using the neutral atomic masses from Appendix B in the textbook gives
Q = ( Δm ) c 2 = ⎡ 4m
⎣
1
1 H atom
−m
4
2 He atom
⎤ c2
⎦
= ⎡⎣ 4 (1.007 825 u ) − 4.002 603 u ⎤⎦ ( 931.5 MeV u )
= 26.7 Mev
Each occurrence of this reaction consumes four protons. Thus, the energy released per proton
consumed is E1 = 26.4 MeV 4 protons = 6.68 MeV proton.
Therefore, the rate at which the Sun must be fusing protons to provide its power output is
rate =
68719_30_ch30_p377-408.indd 393
P
E1
=
3.85 × 10 26 J s ⎛ 1 MeV ⎞
= 3.60 × 10 38 proton s
6.68 MeV proton ⎜⎝ 1.60 × 10 −13 J ⎟⎠
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394
30.44
Chapter 30
With the kaon initially at rest, the total momentum was zero before the decay. Thus, to conserve
momentum, the two pions must go in opposite directions and have equal magnitudes of momentum after the decay. Since the pions have equal mass, this means they must have equal speeds and
hence, equal energies. The rest energy of the kaon is then split equally between the two pions, and
the energy of each is
Ep = ER ,K 2 = 497.7 Mev 2 = 248.9 MeV
0
The total energy of one of the pions is related to its rest energy by Ep = g ER ,p , where
2
g = 1 1− ( v c ) . Therefore, the speed of each of the pions after the decay will be
2
2
⎛E ⎞
⎛ 139.6 MeV ⎞
v = c 1− ⎜ R ,p ⎟ = c 1− ⎜
= 0.827 9c
⎝ 248.9 MeV ⎟⎠
⎝ Ep ⎠
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