17
ELECTRIC CHARGE AND ELECTRIC FIELD
Answers to Multiple-Choice Problems
1. D 2. C
3. C
4. B, C
5. E 6. B
7. C
8. A 9. B
10. B
11. D 12. A 13. B
14. C
15. A
Solutions to Problems
17.1. Set Up: Unlike charges attract and like charges repel. In a conductor some of the negative charge is free to
move. In an insulator the charge can shift position only slightly.
Solve: (a) Aluminum is a conductor and negative charge in the sphere moves away from the rod. The distribution of
charge is sketched in Figure 17.1a.
(b) The charges in the nonconducting sphere displace slightly, with negative charge moving away from the rod. The
distribution of charge is sketched in Figure 17.1b.
⫺
⫺
⫺
⫺
⫺
⫹
⫹
⫹
⫹
⫹
⫺⫹ ⫺⫹
⫺⫺⫺
⫺⫹
⫺⫹
⫺⫺⫺
⫺⫹ ⫺⫹
(b)
(a)
Figure 17.1
17.2. Set Up: Copper is a conductor, so some of the electrons are free to move. The positive charge on the rod
attracts the negative charge in the ball. When the ball is connected to the earth by a conducting wire, charge can flow
between the ball and the earth.
Solve: (a) Electrons move toward the rod. The distribution of charge is sketched in Figure 17.2a.
(b) Electrons from the earth are attracted by the region of positive charge on the ball and flow onto the ball, giving it a
net negative charge. When the rod is removed, this net charge distributes uniformly over the surface of the ball, as
sketched in Figure 17.2b.
⫹⫹ ⫺⫺
⫹
⫺
⫺
⫹
⫺
⫹
⫹⫹ ⫺⫺
⫺
⫹⫹⫹⫹⫹
(a)
⫺
⫺
⫺
⫺
⫺
⫺
⫺
(b)
Figure 17.2
17.3. Set Up: For an isolated sphere, the excess charge is uniformly distributed over the surface of the conductor.
Unlike charges attract and like charges repel, and in a conductor the excess charge is free to move.
Solve: (a) The uniform distribution of charge over the surface of each sphere is sketched in Figure 17.3a.
(b) When the spheres are close to each other, the negative and positive excess charges are drawn toward each other, as
shown in Figure 17.3b.
17-1
17-2
Chapter 17
(c) When the spheres are close to each other, the excess negative charges on each sphere repel, as shown in Figure 17.3c.
⫹
⫹
⫹
⫹
⫹
⫹
⫺
⫺
⫹
⫺
(a)
⫺
⫺
⫺
⫺
⫺
⫺
⫺
⫺
(b)
⫺
⫺
⫺
⫺
⫺
⫹
(c)
Figure 17.3
Reflect: We will learn later in the chapter that the excess charge on a conductor is on the surface of the conductor.
17.4. Set Up: Unlike charges attract and in a conductor the excess charge is free to move.
Solve: Negative charge in the earth is pulled to the surface beneath the charged cloud, as sketched in Figure 17.4.
1
2
2
1
1 1
2
2
Figure 17.4
17.5. Set Up: The charge of one electron is 2e 5 21.60 3 10219 C. 1 mC 5 1026 C; 1 nC 5 1029 C.
0Q0
2.50 3 1026 C
13
Solve: (a) N 5
5
5 1.56 3 10 electrons
e
1.60 3 10219 C
0Q0
2.50 3 1029 C
(b) N 5
5
5 1.56 3 1010 electrons
e
1.60 3 10219 C
17.6. Set Up: The total charge is the number if ions times the charge of each. e 5 1.60 3 10219 C.
Solve: N 5 1 5.6 3 1011 / m 2 1 1.5 3 1022 m 2 5 8.4 3 10 9 ions
Q 5 Ne 5 1 8.4 3 109 2 1 1.60 3 10219 C 2 5 1.3 3 1029 C 5 1.3 nC
17.7. Set Up: Charge conservation requires that the total charge Q of the reactants equals the total charge of the
1
products. qn 5 0, qp 5 1e and qe 5 2e. He2 has charge 12e. X21 and Y21 each have charge 2e.
Solve: (a) reactants: Q 5 1e; products Q 5 2e. Could not occur.
(b) reactants: Q 5 0; products Q 5 1e 1 1 2e 2 5 0. Could occur.
(c) reactants: Q 5 12e 1 2e 5 14e; products Q 5 12e 1 e 1 0 5 13e. Could not occur.
(d) reactants: Q 5 1e 1 2e 5 13e; products Q 5 12e 1 1 2e 2 5 1e. Could not occur.
(e) reactants: Q 5 1e 1 e 1 1 2e 2 5 1e; products Q 5 1e 1 e 1 e 1 1 2e 2 1 1 2e 2 5 1e. Could occur.
Reflect: Just because the reaction obeys charge conservation doesn’t mean it actually occurs. There could be other
reasons why the reaction does not occur.
17.8. Set Up: Like charges repel and unlike charges attract. F 5 k
0 q1q2 0
r2
.
Solve: q1 5 q2 5 13.00 3 1026 C. The force is repulsive. F 5 1 8.99 3 109 N # m2/ C2 2
17.9. Set Up: F 5 k
k 0 q1q2 0
0 q1q2 0
r2
1 3.00 3 1026 C 2 2
5 2.02 N
1 0.200 m 2 2
. An electron has charge 2e and a proton has charge 1e, where e 5 1.60 3 10219 C.
1 8.99 3 109 N # m2 / C2 2 1 1.60 3 10219 C 2 2
5 1.07 3 10214 m. Since an electron and a
Å F
Å
2.00 N
proton have the same magnitude of charge the distance is the same for two protons as for two electrons.
Solve: r 5
5
Electric Charge and Electric Field
17.10. Set Up: F 5 k
17-3
0 q1q2 0
. Like charges repel and unlike charges attract.
r2
Solve: (a) The force is attractive so the unknown charge q2 is positive.
0 q2 0 5
1 0.200 N 2 1 0.300 m 2 2
Fr 2
5
5 3.64 3 1026 C 5 3.64 mC
1 8.99 3 109 N # m2 / C2 2 1 0.550 3 1026 C 2
k 0 q1 0
(b) The unknown charge exerts a downward force of 0.200 N on the other charge.
17.11. Set Up: The electrical force is given by Coulomb’s law, with k 5 8.99 3 109 N # m2 / C2. A proton has
charge 1e and an electron has charge 2e.
0 q1q2 0
1 8.99 3 109 N # m2 / C2 2 1 1.60 3 10219 C 2 2
Solve: (a) F 5 k 2 5
5 230 N. Yes, this force is about 52 lbs.
1 1.00 3 10215 m 2 2
r
1
1.00 3 10215 m
(b) The force is smaller than in part (a) by a factor of
1.00 3 10210 m
very small.
2
2
so it is 2.30 3 1028 N. No, this force is
17.12. Set Up: One mole of carbon contains NA 5 6.02 3 1023 atoms. Each electron has charge 2e 5
21.60 3 10219 C. If charge 2q is removed from the sphere, the sphere is left with a charge of 1q.
1.00 g
5 8.33 3 1022 mol. The number of atoms is N 5 nNA 5
Solve: (a) The number of moles is n 5
12.0 g / mol
1 8.33 3 1022 mol 2 1 6.02 3 1023 atoms / mol 2 5 5.01 3 1022 atoms. Each atom has 6 electrons, so the number of
electrons is Ne 5 6N 5 3.01 3 1023 electrons. Each electron has charge 2e so the total negative charge of the electrons is qe 5 2eNe 5 2 1 1.60 3 10219 C 2 1 3.01 3 1023 2 5 24.82 3 104 C.
(b) The sphere would be left with positive charge qsphere 5 14.82 3 104 C. The force would have magnitude
0 qeqsphere 0
1 4.82 3 104 C 2 2
5 9.28 3 1018 N. This is an immense force. The
1 1.50 m 2 2
r2
charges are opposite sign so the force is attractive.
F5k
5 1 8.99 3 109 N # m2 / C2 2
17.13. Set Up: One mole of Ca contains NA 5 6.02 3 1023 atoms. Each proton has charge e 5 1.60 3 10219 C.
Solve: (a) The mass of one hand is 1 0.010 2 1 75 kg 2 5 0.75 kg 5 750 g. The number of moles of Ca is
n5
The number of atoms is
750 g
40.18 g / mol
5 18.7 mol.
N 5 nNA 5 1 18.7 mol 2 1 6.02 3 1023 atoms / mol 2 5 1.12 3 1025 atoms.
(b) Each Ca atom contains positive charge 20e. The total positive charge in each hand is
Ne 5 1 1.12 3 1025 2 1 20 2 1 1.60 3 10219 C 2 5 3.58 3 107 C.
If 1.0% is unbalanced by negative charge, the net positive charge of each hand is
q 5 1 0.010 2 1 3.58 3 107 C 2 5 3.6 3 105 C.
(c) The repulsive force each hand exerts on the other would be
F5k
q2
r2
5 1 8.99 3 109 N # m2 / C2 2
1 3.6 3 105 C 2 2
5 4.0 3 1020 N.
1 1.7 m 2 2
This is an immense force; our hands would fly off.
Reflect: Ordinary objects contain a very large amount of charge. But negative and positive charge is present in
almost equal amounts and the net charge of a charged object is always a very small fraction of the total magnitude of
charge that the object contains.
17-4
Chapter 17
17.14. Set Up: F 5 k
0 q1q2 0
r2
.
Solve: (a) q1 5 q2 5 q. F 5 k
(b) q2 5 4q1 . F 5
k1
4q12
2
r
2
q2
r
2
and q 5 r
and q1 5 12 r
F
0.220 N
5 1 0.150 m 2
5 7.42 3 1027 C,
Åk
Å 8.99 3 109 N # m2 / C2
F
5 3.71 3 1027 C. q2 5 1.48 3 1026 C.
Åk
17.15. Set Up: The mass of a proton is 1.67 3 10227 kg and the charge of a proton is 1.60 3 10219 C. The
0 q1q2 0
8
distance from the earth to the moon is 3.84 3 10 m. The electrical force has magnitude Fe 5 k 2 , with
r
m 1m 2
k 5 8.99 3 109 N # m2 / C2. The gravitational force has magnitude Fgrav 5 G 2 , with
r
G 5 6.67 3 10211 N # m2 / kg2.
Solve: (a) The number of protons in each box is N 5
1.0 3 1023 kg
1.67 3 10227 kg
23
219
box is q 5 Ne 5 1 5.99 3 10 2 1 1.60 3 10 C 2 5 9.58 3 104 C.
5 5.99 3 1023. The total charge of each
1 9.58 3 104 C 2 2
5 560 N 5 130 lb.
1 3.84 3 108 m 2 2
r2
The tension in the string must equal this repulsive electrical force. The weight of the box on earth is w 5 mg 5
9.8 3 1023 N and the weight of the box on the moon is even less, since g is less on the moon. The gravitational
forces exerted on the boxes by the earth and by the moon are much less than the electrical force and can be neglected.
1 1.0 3 1023 kg 2 2
m 1m 2
5 4.5 3 10234 N
(b) Fgrav 5 G 2 5 1 6.67 3 10211 N # m2 / kg2 2
1 3.84 3 108 m 2 2
r
Fe 5 k
q2
5 1 8.99 3 109 N # m2 / C2 2
Reflect: Both the electrical force and the gravitational force are proportional to 1 / r 2. But in SI units the coefficient k
in the electrical force is much greater than the coefficient G in the gravitational force. And a small mass of protons
contains a large amount of charge. It would be impossible to put 1.0 g of protons into a small box, because of the
very large repulsive electrical forces the protons would exert on each other.
17.16. Set Up: F 5 k
Solve: F 5 k
q2
r
2
0 q1q2 0
. 0q0 5 r
r2
. The charge of an electron is 2e 5 21.60 3 10219 C.
F
4.57 3 10221 N
5 1 0.200 m 2
5 1.426 3 10216 C.
Åk
Å 8.99 3 109 N # m2 / C2
The number of excess electrons is
0q0
e
5
1.426 3 10216 C
5 891.
1.60 3 10219 C
17.17. Set Up: An electron has mass 9.11 3 10231 kg and charge q 5 2e 5 21.60 3 10219 C. The nucleus of a
0 q1q2 0
hydrogen atom is a proton, with charge q 5 1e. F 5 k
Solve: mg 5 k
e2
ke 2
gives r 5
5
r
Å
Å mg
.
r2
1 8.99 3 109 N # m2 / C2 2 1 1.60 3 10219 C 2 2
1 9.11 3 10231 kg 2 1 9.80 m / s2 2
5 5.08 m.
Reflect: The electrical force is much stronger than the gravitational force. The electrical binding force within the
atom is much larger than the weight of the electron.
17.18. Set Up: A proton has charge 1e and an electron has charge 2e, with e 5 1.60 3 10219 C. The force
0 q1q2 0
e2
between them has magnitude F 5 k
5 k 2 and is attractive since the charges have opposite sign. A proton has
r2
r
mass mp 5 1.67 3 10227 kg and an electron has mass 9.11 3 10231 kg. The acceleration is related to the net force
S
S
S
F by F 5 ma .
Electric Charge and Electric Field
Solve: F 5 k
1 1.60 3 10219 C 2 2
e2
9
2
2
#
1
2
5
8.99
3
10
N
m
C
/ 1 2.0 3 10210 m 2 2 5 5.75 3 1029 N
r2
proton: ap 5
F
5.75 3 1029 N
5
5 3.4 3 1018 m / s2
mp
1.67 3 10227 kg
electron: ae 5
17-5
F
5.75 3 1029 N
5
5 6.3 3 1021 m / s2
me
9.11 3 10231 kg
The proton has an initial acceleration of 3.4 3 1018 m / s2 toward the electron and the electron has an initial acceleration of 6.3 3 1021 m / s2 toward the proton. Note that the force the electron exerts on the proton is equal in magnitude
to the force the proton exerts on the electron, but the accelerations produced by this force are different because the
masses are different.
17.19. Set Up: a 5 25.0g 5 245 m / s2. F 5 ma, with F 5 k
0 q1q2 0
r2
. An electron has charge 2e 5 21.60 3 10219 C.
Solve: F 5 ma 5 1 8.55 3 1023 kg 2 1 245 m / s2 2 5 2.09 N. The spheres have equal charges q, so F 5 k
0q0
0q0 5 r
q2
r2
and
F
2.09 N
5 1 0.150 m 2
5 2.29 3 1026 C.
Åk
Å 8.99 3 109 N # m2 / C2
2.29 3 1026 C
5 1.43 3 1013 electrons. The charges on the spheres have the same sign so the electrie
1.60 3 10219 C
cal force is repulsive and the spheres accelerate away from each other.
Reflect: As the spheres move apart the repulsive force they exert on each other decreases and their acceleration
decreases.
N5
5
17.20. Set Up: qe 5 21.60 3 10219 C. qp 5 11.60 3 10219 C. The net force is the vector sum of the forces
exerted by each electron. Each force is attractive so is directed toward the electron that exerts it.
Solve: Each force has magnitude
0 q1q2 0
1 8.988 3 109 N # m2 / C2 2 1 1.60 3 10219 C 2 2
e2
5
5 1.023 3 1028 N.
1 1.50 3 10210 m 2 2
r2
r2
The vector force diagram is shown in Figure 17.20.
F1 5 F2 5 k
5k
y
F2
65.0°
1
Figure 17.20
F1
x
17-6
Chapter 17
F1x 5 1.023 3 1028 N; F1y 5 0.
F2x 5 F2 cos 65.0° 5 4.32 3 1029 N; F2y 5 F2 sin 65.0° 5 9.27 3 1029 N.
Fx 5 F1x 1 F2x 5 1.46 3 1028 N; Fy 5 F1y 1 F2y 5 9.27 3 1029 N. F 5 "Fx2 1 Fy2 5 1.73 3 1028 N.
tan u 5
Fy
Fx
5
9.27 3 1029 N
5 0.6349 and u 5 32.4°. The net force is 1.73 3 1028 N and is directed toward a
1.46 3 1028 N
point midway between the two electrons.
17.21. Set Up: F 5 k
0 qqr 0
r2
. Like charges repel and unlike charges attract. The charges and their forces on q3 are
shown in Figure 17.21.
y
x
q2
0.400 m
q3
0.200 m
F1
q1
Solve: F1 5 k
0 q1q3 0
F2 5 k
S
r12
F2
Figure 17.21
5 1 8.99 3 109 N # m2 / C2 2
0 q2q3 0
r22
1 1.50 3 1029 C 2 1 5.00 3 1029 C 2
5 1.69 3 1026 N.
1 0.200 m 2 2
5 1 8.99 3 109 N # m2 / C2 2
1 3.20 3 1029 C 2 1 5.00 3 1029 C 2
5 8.99 3 1027 N.
1 0.400 m 2 2
S
F1 and F2 are in the same direction so F 5 F1 1 F2 5 2.59 3 1026 N and the net force is in the 2y direction.
Reflect: The forces are vectors and must be added as vectors. We add the forces by adding their magnitudes only
because the two force vectors are in the same direction.
17.22. Set Up: F 5 k
0 qqr 0
r2
. Like charges repel and unlike charges attract. The charges and their forces on q3 are
shown in Figure 17.22.
y
0.200 m
F1
0.300 m
q2
Figure 17.22
F2
q3
q1
x
Electric Charge and Electric Field
Solve: F1 5 k
F2 5 k
0 q1q3 0
r12
0 q2q3 0
r22
5 1 8.99 3 109 N # m2 / C2 2
5 1 8.99 3 109 N # m2 / C2 2
17-7
1 4.00 3 1029 C 2 1 0.600 3 1029 C 2
5 5.394 3 1027 N.
1 0.200 m 2 2
1 5.00 3 1029 C 2 1 0.600 3 1029 C 2
5 2.997 3 1027 N.
1 0.300 m 2 2
Fx 5 F1x 1 F2x 5 1F1 2 F2 5 2.40 3 1027 N. The net force has magnitude 2.40 3 1027 N and is in the 1x
direction.
17.23. Set Up: F 5 k
0 qqr 0
. Like charges repel and unlike charges attract. The charges and the forces on the
r2
1.50 cm
charges q1 and q2 in the dipole are shown in Figure 17.23. Use the coordinates shown. sin u 5
and u 5 48.6°.
2.00 cm
y
F1x
q1
1.50 cm
F1
F1y
u
axis
x
1.50 cm
F2x
q3
u
2.00 cm
q2
F2y
F2
Figure 17.23
Solve: F1 5 F2 5 1 8.99 3 109 N # m2 / C2 2
1 5.00 3 1026 C 2 1 10.0 3 1026 C 2
5 1.124 3 103 N.
1 0.0200 m 2 2
Fx 5 F1x 1 F2x 5 0. Fy 5 F1y 1 F2y 5 22F1 sin u 5 22 1 1.124 3 103 N 2
1
2
1.50 cm
5 21.69 3 103 N.
2.00 cm
The net force has magnitude 1.69 3 103 N and is in the direction from the 15.00 mC to the 25.00 mC charge.
(b) F1x and F2x each produce clockwise torques and each have a moment arm of 1.50 cm. F1y and F2y each have zero
moment arm and produce no torque.
t 5 2F1x 1 0.0150 m 2 5 2F1 cos u 1 0.0150 m 2 5 2 1 1.124 3 103 N 2 1 cos 48.6° 2 1 0.0150 m 2 5 22.3 N # m
The torque is clockwise.
Reflect: The x components of the two forces are in opposite directions so they cancel and the net force has no x
component. But the torques of F1x and F2x are in the same direction and therefore produce a net torque.
17-8
Chapter 17
17.24. Set Up: In the O-H-N combination the O2 is 0.170 nm from the H1 and 0.280 nm from the N2. In the
N-H-N combination the N2 is 0.190 nm from the H1 and 0.300 nm from the other N2. Like charges repel and unlike
charges attract. The net force is the vector sum of the individual forces.
0 q1q2 0
e2
Solve: (a) F 5 k 2 5 k 2
r
r
O-H-N
1 1.60 3 10219 C 2 2
O2-H 1 F 5 1 8.99 3 109 N # m2 / C2 2
5 7.96 3 1029 N, attractive
1 0.170 3 1029 m 2 2
O2-N2 F 5 1 8.99 3 109 N # m2 / C2 2
1 1.60 3 10219 C 2 2
5 2.94 3 1029 N, repulsive
1 0.280 3 1029 m 2 2
N2-H 1 F 5 1 8.99 3 109 N # m2 / C2 2
1 1.60 3 10219 C 2 2
5 6.38 3 1029 N, attractive
1 0.190 3 1029 m 2 2
N-H-N
N2-N2 F 5 1 8.99 3 109 N # m2 / C2 2
1 1.60 3 10219 C 2 2
5 2.56 3 1029 N, repulsive
1 0.300 3 1029 m 2 2
The total attractive force is 1.43 3 1028 N and the total repulsive force is 5.50 3 1029 N. The net force is attractive
and has magnitude 1.43 3 1028 N 2 5.50 3 1029 N 5 8.80 3 1029 N.
1 1.60 3 10219 C 2 2
e2
5 8.22 3 1028 N
(b) F 5 k 2 5 1 8.99 3 109 N # m2 / C2 2
1 0.0529 3 1029 m 2 2
r
The bonding force of the electron in the hydrogen atom is a factor of 10 larger than the bonding force of the adeninethymine molecules.
17.25. Set Up: In the O-H-O combination the O2 is 0.180 nm from the H 1 and 0.290 nm from the other O2. In the
N-H-N combination the N2 is 0.190 nm from the H 1 and 0.300 nm from the other N2. In the O-H-N combination
the O2 is 0.180 nm from the H 1 and 0.290 nm from the other N2. Like charges repel and unlike charges attract. The
net force is the vector sum of the individual forces.
0 q1q2 0
e2
Solve: F 5 k 2 5 k 2 . The attractive forces are: O2-H 1 , 7.10 3 1029 N; N2-H 1 , 6.37 3 1029 N; O2-H 1 ,
r
r
7.10 3 1029 N. The total attractive force is 2.06 3 1028 N. The repulsive forces are: O2-O2, 2.74 3 1029 N;
N2-N2, 2.56 3 1029 N; O2-N2, 2.74 3 1029 N. The total repulsive force is 8.04 3 1029 N. The net force is attractive and has magnitude 1.26 3 1028 N.
17.26. Set Up: Like charges repel and unlike charges attract. The force increases as the distance between the
charges decreases.
Solve: The forces on the dipole that is between the slanted dipoles are sketched in Figure 17.26a. The forces are
attractive because the 1 and 2 charges of the two dipoles are closest. The forces are toward the slanted dipoles so
have a net upward component. In Figure 17.26b, in adjacent dipoles charges of opposite sign are closer than charges
of the same sign so the attractive forces are larger than the repulsive forces and the dipoles attract.
F1y
F1
F1x
F2y
F2
F2x
2 1
2 1
(a)
Figure 17.26
(b)
2 1
2 1
Electric Charge and Electric Field
17-9
17.27. Set Up: The central charge will be 1.85 nm from the charge on the left and 1.15 nm from the charge on the
right. Like charges repel. The force diagram for the central charge is given in Figure 17.27.
1.85 nm
q1
1.15 nm
F1
F2
q2
Figure 17.27
Solve: F1 5 k
1 2e 2 2
4 1 1.60 3 10219 C 2 2
5 1 8.99 3 109 N # m2 / C2 2
5 2.690 3 10210 N
1
1 1.85 3 1029 m 2 2
r2
F2 5 k
1 2e 2 2
4 1 1.60 3 10219 C 2 2
5 1 8.99 3 109 N # m2 / C2 2
5 6.961 3 10210 N
2
1 1.15 3 1029 m 2 2
r2
The net force is 4.27 3 10210 N, to the left.
Reflect: The electrical forces are vectors and must be added as vectors. At the initial position of the central charge
the net force on it is zero. If the central charge is displaced in either direction, the net force on it is in the direction
that pushes it back toward the equilibrium position.
0 q1q2 0
17.28. Set Up: F 5 k
r2
. The new charges are q1r 5 2q1 and q2r 5 2q2 .
Solve: The new force is Fr 5 k
17.29. Set Up: F 5 k
Fr 5 3F.
Solve: Fr 5 k
0 q1q2 0
1 rr 2
2
0 q1q2 0
r
5 3k
2
0 q1rq2r 0
r
2
5k
0 2q12q2 0
r
2
5 4k
0 q1q2 0
r2
. Initially, r 5 D and F 5 k
0 q1q2 0
D
2
, so
5 4F.
0 q1q2 0
D2
. The new separation is rr. The new force is
1
3
5 2 and rr 5 D / "3 .
2
1 rr 2
D
Reflect: The force is proportional to 1 / r 2, so a decrease in r increases the force.
0 q1q2 0
17.30. Set Up: If the new acceleration ar equals a / 5 then the new force Fr equals F / 5. F 5 k 2 . Initially,
r
0 q1q2 0
r 5 d and F 5 k
. The new separation is rr.
d2
0 q1q2 0
1 0 q1q2 0
Solve: Fr 5 k
5 k 2 , so rr 5 d "5 .
2
1 rr 2
5 d
S
S
17.31. Set Up: F 5 qE. A proton has charge q 5 1e 5 11.60 3 10219 C.
Solve: (a) F 5 0 q 0 E. E 5
20.0 3 1029 N
F
5
5 2.50 N / C. Since the charge is negative, the force and electric
0q0
8.00 3 1029 C
field are in opposite directions and the electric field is upward.
(b) F 5 0 q 0 E 5 eE 5 1 1.60 3 10219 C 2 1 2.50 N 2 5 4.00 3 10219 N. The charge is positive so the force is in the
same direction as the electric field; the force is upward.
17-10
Chapter 17
S
S
17.32. Set Up: F 5 qE. A proton has charge q 5 1e 5 11.60 3 10219 C and mass m 5 1.67 3 10227 kg.
Solve: (a) The gravity force is downward so the electrical force must be upward. An upward force from a downward
electric field requires a negative charge. mg 5 0 q 0 E.
0q0 5
1 1.45 3 1023 kg 2 1 9.80 m / s2 2
mg
5 2.19 3 1025 C.
5
E
650 N / C
The charge is q 5 22.19 3 1025 C.
1 1.67 3 10227 kg 2 1 9.80 m / s2 2
mg
5
5 1.02 3 1027 N / C
(b) mg 5 eE. E 5
e
1.60 3 10219 C
17.33. Set Up: Use the coordinates shown in Figure 17.33. Since the electric field is uniform, the force is constant
S
S
S
and the acceleration is constant. F 5 qE and F 5 ma . v0y 5 0. For an electron, q 5 2e 5 21.60 3 10219 C and
m 5 9.11 3 10231 kg.
S
y
1
Positive plate
1
1
1
1
2
2
2
2
3.20 cm
2
Negative plate
x
Figure 17.33
Solve: (a) y 5 v0y t 1 12 ay t 2. ay 5
Fy
2y
t2
5
2 1 3.20 3 1022 m 2
5 2.84 3 1014 m / s2.
1 1.5 3 1028 s 2 2
Fy 5 may 5 1 9.11 3 10231 kg 2 1 2.84 3 1014 m / s2 2 5 2.59 3 10216 N.
2.59 3 10216 N
5 1.62 3 103 N / C and E 5 1.62 3 103 N / C.
0q0
1.60 3 10219 C
(b) vy 5 v0y 1 ayt 5 0 1 1 2.84 3 1014 m / s2 2 1 1.5 3 1028 s 2 5 4.26 3 106 m / s
Reflect: We could also use the work-energy theorem and set the work done by the force equal to the gain in kinetic
energy of the electron.
Ey 5
5
17.34. Set Up: For a point charge, E 5 k
0q0
S
. E is toward a negative charge and away from a positive charge.
r2
Solve: (a) The field is toward the negative charge so is downward.
(b) r 5
k0q0
Å E
E 5 1 8.99 3 109 N # m2 / C2 2
5
Å
1 8.99 3 109 N # m2 / C2 2 1 3.00 3 1029 C 2
12.0 N / C
17.35. Set Up: For a point charge, E 5 k
Solve: 0 q 0 5
Er 2
5
k
3.00 3 1029 C
5 432 N / C.
1 0.250 m 2 2
5 1.50 m
0q0
.
r2
1 6.50 3 103 N / C 2 1 0.100 m 2 2
8.99 3 109 N # m2 / C2
5 7.23 3 1029 C
Electric Charge and Electric Field
17.36. Set Up: For a point charge, E 5 k
Solve: r 5
k0q0
Å E
Å
5
17-11
0q0
.
r2
1 8.99 3 109 N # m2 / C2 2 1 5.00 3 1029 C 2
4.00 N / C
5 3.35 m
17.37. Set Up: The electric field of a point charge has magnitude E 5 k
e 5 1.60 3 10219 C.
0q0
r2
. A proton has charge q 5
Solve: (a) E 5 1 8.99 3 109 N # m2 / C2 2
1.60 3 10219 C
5 5.8 3 1019 N / C
1 5.0 3 10215 m 2 2
1.60 3 10219 C
(b) E 5 1 8.99 3 109 N # m2 / C2 2
5 5.8 3 109 N / C
1 5.0 3 10210 m 2 2
Reflect: The electric fields inside atoms are very large.
17.38. Set Up: The weight of an electron is w 5 mg, with m 5 9.11 3 10231 kg. The magnitude of the force on
the electron due to the electric field is Fe 5 0 q 0 E 5 eE. In part (b), the electric field of a proton at a distance r from
the proton is E 5 k
e
.
r2
1 9.11 3 10231 kg 2 1 9.80 m / s2 2
mg
5
5 5.58 3 10211 N / C.
e
1.60 3 10219 C
1 8.99 3 109 N # m2 / C2 2 1 1.60 3 10219 C 2
e
ke
(b) E 5 k 2 so r 5
5
5 5.08 m
Å
ÅE
r
5.58 3 10211 N / C
Solve: (a) w 5 Fe gives mg 5 Ee and E 5
17.39. Set Up: If the axon is modeled as a point charge, its electric field is E 5 k
q
. The electric field of a point
charge is directed away from the charge if it is positive.
Solve: (a) 5.6 3 1011 Na1 ions enter per meter so in a 0.10 mm 5 1.0 3 1024 m section, 5.6 3 107 Na1 ions enter.
This number of ions has charge q 5 1 5.6 3 107 2 1 1.60 3 10219 C 2 5 9.0 3 10212 C.
(b) E 5 k
(c) r 5
0q0
r
2
5 1 8.99 3 109 N # m2 / C2 2
k0q0
Å E
5
Å
r2
9.0 3 10212 C
5 32 N / C, directed away from the axon.
1 5.00 3 1022 m 2 2
1 8.99 3 109 N # m2 / C2 2 1 9.0 3 10212 C 2
1.0 3 1026 N / C
5 280 m
17.40. Set Up: The electric field of a negative charge is directed toward the charge. E 5 k
0q0
. The net field is the
r2
vector sum of the fields produced by each charge. Point A is 0.100 m from q2 and 0.150 m from q1 . Point B is
S
S
S
0.100 m from q1 and 0.350 m from q2 . A charge q in an electric field E experiences a force F 5 qE.
0.150 m
0.100 m
0.100 m
2
2
q1
q2
E1
(a)
Figure 17.40
A
E2
E1
B
E2
(b)
0.250 m
2
2
q1
q2
17-12
Chapter 17
Solve: (a) The electric fields due to the charges at point A are shown in Figure 17.40a.
E1 5 k
E2 5 k
0 q1 0
rA12
5 1 8.99 3 109 N # m2 / C2 2
rA22
5 1 8.99 3 109 N # m2 / C2 2
0 q2 0
6.25 3 1029 C
5 2.50 3 103 N / C
1 0.150 m 2 2
12.5 3 1029 C
5 1.124 3 104 N / C
1 0.100 m 2 2
Since the two fields are in opposite directions, we subtract their magnitudes to find the net field. E 5
E2 2 E1 5 8.74 3 103 N / C, to the right.
(b) The electric fields at points B are shown in Figure 17.40b.
E1 5 k
E2 5 k
0 q1 0
rB12
5 1 8.99 3 109 N # m2 / C2 2
rB22
5 1 8.99 3 109 N # m2 / C2 2
0 q2 0
6.25 3 1029 C
5 5.619 3 103 N / C
1 0.100 m 2 2
12.5 3 1029 C
5 9.17 3 102 N / C
1 0.350 m 2 2
Since the fields are in the same direction, we add their magnitudes to find the net field. E 5 E1 1 E2 5
6.54 3 103 N / C, to the right.
(c) At A, E 5 8.74 3 103 N / C, to the right. The force on a proton placed at this point would be F 5 qE 5
1 1.60 3 10219 C 2 1 8.74 3 103 N / C 2 5 1.40 3 10215 N, to the right.
17.41. Set Up: For a point charge, E 5 k
0q0
S
. E is directed toward a negative charge and away from a positive
r2
charge. Let the points a, b and c be the locations where the field is calculated in parts (a), (b) and (c). The three points
and the electric fields produced at those points by each of the two charges are shown in Figure 17.41.
y
0.20 m
c
E2
0.20 m
q1 E1
E1
q2
a
2
1
E2
b
E2
x
E1
0.80 m
1.20 m
(a) E1 5 k
0 q1 0
r12
Figure 17.41
5
1 8.99 3 109 N # m2 / C2 2 1 4.00 3 1029 C 2
5 899 N / C
1 0.200 m 2 2
E2 5 k
0 q2 0
r22
5
1 8.99 3 109 N # m2 / C2 2 1 6.00 3 1029 C 2
5 150 N / C
1 0.600 m 2 2
Ex 5 E1x 1 E2x 5 2899 N / C 1 12150 N / C 2 5 21049 N / C. The field is 1050 N/C, in the 2x direction.
(b) E1 5 k
0 q1 0
r12
5
1 8.99 3 109 N # m2 / C2 2 1 4.00 3 1029 C 2
5 25.0 N / C
1 1.20 m 2 2
E2 5 k
0 q2 0
r22
5
1 8.99 3 109 N # m2 / C2 2 1 6.00 3 1029 C 2
5 337 N / C
1 0.400 m 2 2
Ex 5 E1x 1 E2x 5 225.0 N / C 1 337 N / C 5 312 N / C. The field is 312 N/C, in the 1x direction.
(c) E1 5 k
0 q1 0
r12
5
1 8.99 3 109 N # m2 / C2 2 1 4.00 3 1029 C 2
5 899 N / C
1 0.200 m 2 2
0 q2 0
1 8.99 3 109 N # m2 / C2 2 1 6.00 3 1029 C 2
5 53.9 N / C
1 1.00 m 2 2
r22
Ex 5 E1x 1 E2x 5 1899 N / C 1 1 253.9 N / C 2 5 845 N / C. The field is 845 N/C, in the 1x direction.
Reflect: In each case the two electric fields must be added as vectors.
E2 5 k
5
Electric Charge and Electric Field
17.42. Set Up: For a point charge, E 5 k
0q0
17-13
S
. E is toward a negative charge and away from a positive charge. The
r2
two charges and their fields at each point are shown in Figures 17.42a-d.
y
y
E1
q2
E2 q1
0.15 m
0.15 m
x
0.15 m
q2
E1
0.15 m
x
q1
E2
0.15 m
(a)
(b)
y
y
E1y
E1
q2 0.15 m 0.15 m q1
E2y
E2
x
u
0.50 m
0.40 m
u
E1x
E1x
E2x
u
u
E1
E2y
u
E2
q2 0.15 m
0.25 m
0.20 m
u
x
0.15 m q1
(d)
(c)
Figure 17.42
S
S
Solve: (a) E1 5 E2 and E 1 5 E 2 are in opposite directions, so the resultant electric field is zero. Ex 5 Ey 5 E 5 0.
6.00 3 1029 C
(b) E1 5 1 8.99 3 109 N # m2 / C2 2
5 2397 N / C
1 0.150 m 2 2
E2 5 1 8.99 3 109 N # m2 / C2 2
6.00 3 1029 C
5 266 N / C
1 0.450 m 2 2
Ex 5 E1x 1 E2x 5 E1 1 E2 5 2660 N / C. Ey 5 0. The resultant electric field has magnitude 2660 N/C and is in the
1x direction.
6.00 3 1029 C
(c) E1 5 1 8.99 3 109 N # m2 / C2 2
5 337 N / C. E1x 5 0, E1y 5 2337 N / C.
1 0.400 m 2 2
E2 5 1 8.99 3 109 N # m2 / C2 2
6.00 3 1029 C
5 216 N / C. u 5 53.1°. E2x 5 E2 cos u 5 130 N / C.
1 0.500 m 2 2
E2y 5 2E2 sin u 5 2173 N / C.
Ex 5 E1x 1 E2x 5 0 1 130 N / C 5 1130 N / C. Ey 5 E1y 1 E2y 5 2337 N / C 2 173 N / C 5 2510 N / C.
tan f 5
PE P
Ey
S
5 3.92. f 5 75.7° and E makes an angle 360° 2 75.7° 5 284° counterclockwise from the 1x axis.
x
E 5 "Ex2 1 Ey2 5 526 N / C.
(d) u 5 53.1°. E1 5 E2 5 1 8.99 3 109 N # m2 / C2 2
6.00 3 1029 C
5 863 N / C.
1 0.250 m 2 2
Ex 5 E1x 1 E2x 5 0. Ey 5 E1y 1 E2y 5 2E1 sin u 5 2 1 863 N / C 2 sin 53.1° 5 1380 N / C. E has magnitude 1380 N/C
and is in the 1y direction.
S
17-14
Chapter 17
17.43. Set Up: For a point charge, E 5 k
0q0
S
. E is toward a negative charge and away from a positive charge. Let
r2
S
S
q1 5 10.500 nC and q2 5 18.00 nC. For the net electric field to be zero, E 1 and E 2 must have equal magnitudes
and opposite directions.
III
II
I
E1
q1 E2
E2
1
E1
q2
E1
1
E2
x
E2
E1
q1
E1
q2
1
E2
2
x
1.2 m
III
II
I
E1
E2
1.2 m
(b)
(a)
Figure 17.43
Solve: The two charges and the directions of their electric fields in three regions are shown in Figure 17.43a. Only in
region II are the two electric fields in opposite directions. Consider a point a distance x from q1 so a distance
0.500 nC
8.00 nC
1.20 m 2 x from q2 . E1 5 E2 gives k
5k
. 16x 2 5 1 1.20 m 2 x 2 2. 4x 5 6 1 1.20 m 2 x 2
1 1.20 2 x 2 2
x2
and x 5 0.24 m is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from the
0.500 nC charge.
(b) Let q2 5 28.00 nC be the negative charge. The two charges and the directions of their electric fields in three
S
S
regions are shown in Figure 17.43b. E 1 and E 2 are in opposite directions in regions I and III. But for the magnitudes of the fields to be equal the point must be closer to the charge q1 that has smaller magnitude, and that
occurs only for region I. Consider a point a distance x to the left of q1 so 1.20 m 1 x from q2 . E1 5 E2 gives
0.500 nC
8.00 nC
k
5k
. 16x 2 5 1 1.20 m 1 x 2 2. 4x 5 6 1 1.20 m 1 x 2 and x 5 0.40 m is the positive solu2
1 1.20 1 x 2 2
x
tion. The electric field is zero at a point 0.40 m from q1 and 1.60 m from q2 .
Reflect: In each case there is only one point along the line connecting the two charges where the net electric field is
zero.
17.44. Set Up: The electric field of a negative charge is directed toward the charge. E 5 k
0q0
. The net field is the
r2
vector sum of the fields due to each charge. Label the charges q1 , q2 and q3 , as shown in Figure 17.44a. This figure
also shows additional distances and angles. The electric fields at point P are shown in Figure 17.44b. This figure also
S
S
S
shows the xy coordinates we will use and the x and y components of the fields E 1 , E 2 and E 3 .
y
E1
q1
8.0 cm
E1y
10.0 cm
53.1°
P
6.0 cm
53.1°
8.0 cm
10.0 cm
q2
x
q3
(a)
Figure 17.44
E2
E1x
53.1°
E3x
53.1°
E3
(b)
E3y
Electric Charge and Electric Field
Solve: E1 5 E3 5 1 8.99 3 109 N # m2 / C2 2
17-15
5.00 3 1026 C
5 4.49 3 106 N / C
1 0.100 m 2 2
E2 5 1 8.99 3 109 N # m2 / C2 2
2.00 3 1026 C
5 4.99 3 106 N / C
1 0.0600 m 2 2
Ey 5 E1y 1 E2y 1 E3y 5 0 and Ex 5 E1x 1 E2x 1 E3x 5 E2 1 2E1 cos 53.1° 5 1.04 3 107 N / C
E 5 1.04 3 107 N / C, toward the 22.00 mC charge.
S
S
S
17.45. Set Up: The force on a charge q that is in an electric field E is F 5 qE.
Solve: (a) The force on 1q is F1 5 qE, to the right. The force on 2q is F2 5 qE, to the left. The net force on the
dipole is zero.
(b) The axis is at the midpoint of the line connecting the charges, and perpendicular to the plane of the figure. The
torque is zero when each force, F1 and F2 , has zero moment arm. This is the case for u 5 0° and u 5 180°, as
shown in Figure 17.45a and b.
E
E
F2
F1
2q
1q
F2
F1
2q
1q
(b)
(a)
1q
E
F1
F1
E
1q
F2
(c)
2q
F2
2q
(d)
Figure 17.45
(c) For u slightly greater than zero, as shown in Figure 17.45c, the torque on the dipole is clockwise and is directed
so as to return the dipole to the equilibrium position. For u slightly less than zero the torque is counterclockwise
and again tends to rotate the dipole back to its equilibrium position. The u 5 0° position of Figure 17.45a is a stable orientation.
For u slightly less than 180° (Figure 17.45d), the net torque on the dipole rotates it away from the u 5 180°
equilibrium position. The same is true for u slightly greater than 180°. The u 5 180° position of Figure 17.45b is an
unstable orientation.
(d) The electric field of the dipole is directed from the positive charge and toward the negative charge. Thus, in Figure 17.45a the electric field of the dipole is to the left, opposite to the direction of the external field.
Reflect: For any orientation of the dipole the net force on the dipole is zero. But only for u 5 0° and u 5 180° is the
net torque zero.
17-16
Chapter 17
S
S
17.46. Set Up: F 5 qE. t 5 Fl, where l is the moment arm. The forces and moment arms are shown in Figure 17.46. The axis is at the midpoint of the line connecting the charges, and perpendicular to the page.
E
F1
1.75 nm
axis
l1
30°
30°
l2
1.75 nm
F2
Figure 17.46
Solve: F1 5 F2 5 0 q 0 E 5 1 2.50 3 1026 C 2 1 7800 N / C 2 5 1.95 3 1022 N
l1 5 l2 5 1 1.75 nm 2 sin 30° 5 8.75 3 10210 m
Each force produces a clockwise torque, so the net torque is
t 5 t1 1 t2 5 2t1 5 2 1 1.95 3 1022 N 2 1 8.75 3 10210 m 2 5 3.4 3 10211 N # m
The net torque is 3.4 3 10211 N # m, clockwise.
17.47. Set Up: E 5 k
0q0
r2
0q0
1 0q0
. r2 5 2r1 . E2 5 k
5 k
5 E/4
1 2r1 2 2 4 r12
r12
0q0
1 0q0
5 k
5 E/9
(b) Now r3 5 3r1 . E3 5 k
1 3r1 2 2 9 r12
Solve: (a) For r1 5 1.0 cm, E1 5 E 5 k
17.48. Set Up: E 5 k
0q0
0Q0
r2
0Q0
0Q0
k0Q0
k0Q0
k0Q0
5
5 "10
5 "10R
Solve: For r 5 R, E 5 k 2 . E2 5 k 2 . E2 5 E / 10, so r2 5
Å E2
Å E / 10
Å k 0 Q 0 /R2
R
r2
17.49. Set Up: Electric fields come out of positive charge and go into negative charge. The electric field is stronger
where the electric field lines are closer together.
Solve: (a) The field lines go from B to A, so B must be positive and A must be negative.
(b) The field lines are closer together near A, so the electric field has greater magnitude near A.
Electric Charge and Electric Field
17-17
17.50. Set Up: The proton has charge 1e and the electron has charge 2e. The electric field of a point charge is
directed away from the charge if it is positive and toward the point charge if it is negative. Use coordinates where the
x axis is along line ABC and the 1x direction is toward C and where the 1y direction is toward the proton. The
electric fields due to each charge at A, B and C are sketched in Figure 17.50.
1
A
B
F1
C
F2
F2
F1
2
F1
F2
Figure 17.50
Solve: (a) At point A the x components of the fields cancel and the net field is in the 2y direction.
(b) At B both fields are in the 2y direction and the net field is in the 2y direction.
(c) At C, as at A, the x component of the fields cancel and the net field is in the 2y direction.
17.51. Set Up: Electric field is away from positive charge and toward negative charge.
E
E1
E1
E2
B
B
E1
E
B
E2
E
A
E2
q1
A
E2
A
E2
E1
q2
q1
E2
q1
E1
q2
E1
q2
E2
E
C
E2
E1
E
E2
C
E1
E
E1
C
E
(a)
(b)
(c)
Figure 17.51
S
S
S
Solve: (a) The electric fields E 1 and E 2 and their vector sum, the net field E, are shown for each point in Figure 17.51a. The electric field is toward A at points B and C and the field is zero at A.
(b) The electric fields are shown in Figure 17.51b. The electric field is away from A at B and C. The field is zero at A.
(c) The electric fields are shown in Figure 17.51c. The field is horizontal and to the right at points A, B and C.
Reflect: Compare your results to the field lines shown in Figure17.24a and b in the textbook.
17-18
Chapter 17
17.52. Set Up: Electric field points toward negative charge and away from positive charge.
E1
E
EQ
P
E1q
E2
E2q
P
(b)
(a)
Figure 17.52
S
S
S
Solve: (a) Figure 17.52a shows E Q and E 1q at point P. E Q must have the direction shown, to produce a resultant
S
field in the specified direction. E Q is toward Q, so Q is negative. In order for the horizontal components of the two
fields to cancel, Q and q must have the same magnitude.
(b) No. If the lower charge were negative, its field would be in the direction shown in Figure 17.52b. The two possiS
S
ble directions for the field of the upper charge, when it is positive ( E 1) or negative (E 2), are shown. In neither case is
the resultant field in the direction shown in the figure in the problem.
17.53. Set Up: Electric field is directed away from positive charge and toward negative charge. By symmetry, far
from the edges of the sheets the field lines are perpendicular to the sheets; there is no reason to prefer to the left or to
the right for a component of electric field.
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
2
2
2
(a)
(b)
(c)
(d)
(e)
Figure 17.53
Solve: (a) The fields of each sheet are sketched in Figure 17.53a. The solid lines are the field due to the upper positive sheet and the dashed lines are the field due to the lower negative sheet. Between the two sheets the fields are in the
same direction and add. Outside the two sheets the fields are in opposite directions and cancel. The net electric field is
sketched in Figure 17.53b. This result is consistent with the field in the similar case described in Example 17.5.
(b) The fields of each sheet are sketched in Figure 17.53c. The solid lines are the field due to the upper sheet and the
dashed lines are the field due to the lower sheet. Since both sheets are positive, the field of each sheet is directed
away from that sheet. Between the two sheets the fields are in opposite directions and cancel. Outside the two sheets
the fields are in the same direction and add. The net field is sketched in Figure 17.53d.
(c) The fields are the same as in part (b), except they are toward each sheet. The net field of the two sheets is sketched
in Figure 17.53e.
Reflect: More advanced treatments show that the electric field of an infinite sheet is uniform and does not depend on
the distance from the sheet.
Electric Charge and Electric Field
17.54. Set Up: Gauss’s law says FE 5
17-19
Qencl
. P0 5 8.854 3 10212 C2 / 1 N # m2 2 .
P0
2.50 3 1026 C
5 2.82 3 105 N # m2 / C
8.854 3 10212 C2 / 1 N # m2 2
(b) Qencl 5 P0FE 5 1 8.854 3 10212 C2 / 1 N # m2 2 2 1 1.40 N # m2 / C 2 5 1.24 3 10211 C
Solve: (a) FE 5
17.55. Set Up: Gauss’s law says that the net electric flux FE through any closed surface equals Qencl / P0 , where
Qencl is the net charge enclosed by the surface.
Solve: S1: Qencl 5 0 so FE 5 0
S2: Qencl 5 15.0 mC 1 9.0 mC 1 1 27.0 mC 2 5 17.0 mC and FE 5 17.0 mC / P0 5 7.9 3 105 N # m2 / C.
S3: Qencl 5 19.0 mC 1 1.0 mC 1 1 210.0 mC 2 5 0 and FE 5 0.
S4: Qencl 5 18.0 mC 1 1 27.0 mC 2 5 11.0 mC and FE 5 11.0 mC / P0 5 1.1 3 105 N # m2 / C.
S5: Qencl 5 8.0 mC 1 1 27.0 mC 2 1 5.0 mC 1 9.0 mC 1 1.0 mC 1 1 210.0 mC 2 5 16.0 mC and
FE 5 16.0 mC / P0 5 6.8 3 105 N # m2 / C.
Reflect: The electric field varies in a complicated way, in both magnitude and direction, for each of the surfaces and
direct calculation of the flux as a E'DA would be very difficult. Gauss’s law gives us a very simple way of calculating the net flux.
17.56. Set Up: The cube is a closed surface so we can apply Gauss’s law to it: FE 5
is the same through each of the 6 faces of the cube.
Qencl
. By symmetry the flux
P0
Qencl
8.00 3 1029 C
5
5 904 N # m2 / C.
P0
8.854 3 10212 C2 / 1 N # m2 2
904 N # m2 / C
FE
5
5 151 N # m2 / C.
(b) The flux through each face is
6
6
Solve: (a) The total flux through the cube is FE 5
17.57. Set Up: Example 17.10 shows that E 5 0 inside a uniform spherical shell and that E 5 k
the shell.
Solve: (a) E 5 0
0q0
r2
outside
(b) r 5 0.060 m and E 5 1 8.99 3 109 N # m2 / C2 2
15.0 3 1026 C
5 3.75 3 107 N / C
1 0.060 m 2 2
15.0 3 1026 C
5 1.11 3 107 N / C
(c) r 5 0.110 m and E 5 1 8.99 3 109 N # m2 / C2 2
1 0.110 m 2 2
Reflect: Outside the shell the electric field is the same as if all the charge were concentrated at the center of the
shell. But inside the shell the field is not the same as for a point charge at the center of the shell; inside the shell the
electric field is zero.
17.58. Set Up: A spherically symmetric uniform sphere of charge can be modeled as a series of concentric shells,
0q0
so E 5 k
r2
outside the sphere. A proton has charge 1e.
Solve: (a) E 5 k
0q0
r2
5 1 8.99 3 109 N # m2 / C2 2
92 1 1.60 3 10219 C 2
5 2.4 3 1021 N / C
1 7.4 3 10215 m 2 2
(b) For r 5 1.0 3 10210 m, E 5 1 2.4 3 1021 N / C 2
(c) E 5 0, inside a spherical shell.
1
7.4 3 10215 m
1.0 3 10210 m
2
2
5 1.3 3 1013 N / C
17-20
Chapter 17
17.59. Set Up: The method of Example 17.10 shows that the electric field outside the sphere is the same as for a
point charge of the same charge located at the center of the sphere. The charge of an electron has magnitude
e 5 1.60 3 10219 C.
0q0
Solve: (a) E 5 k 2 . For r 5 R 5 0.150 m, E 5 1150 N / C so
r
1 1150 N / C 2 1 0.150 m 2 2
Er 2
0q0 5
5 2.88 3 1029 C.
5
k
8.99 3 109 N # m2 / C2
The number of excess electrons is
2.88 3 1029 C
5 1.80 3 1010 electrons.
1.60 3 10219 C / electron
(b) r 5 R 1 0.100 m 5 0.250 m. E 5 k
0q0
r
2
5 1 8.99 3 109 N # m2 / C2 2
2.88 3 1029 C
5 414 N / C.
1 0.250 m 2 2
17.60. Set Up: The charge distribution has spherical symmetry, so the electric field is radial and depends only on
the distance from the center of the charged sphere. The surface area of a sphere is A 5 4pr 2.
Solve: (a) Apply Gauss’s law to a spherical surface of radius r $ R, concentric with the sphere of charge. The electric field is constant over the Gaussian surface and perpendicular to it, so FE 5 EA 5 E 1 4pr 2 2 . The charge
Q
Q
. This is the same E as for a point charge Q at the
enclosed is Q, so Gauss’s law gives E 1 4pr 2 2 5 and E 5
P0
4pP0r 2
center of the sphere.
(b) No direction is preferred, so E 5 0.
17.61. Set Up: The charge distribution has spherical symmetry, so the electric field, if nonzero, is radial and
depends only on the distance from the center of the shell.
Solve: (a) Apply Gauss’s law to a sphere of radius r , a and concentric with the shell. The electric field, if nonzero,
is constant over the Gaussian surface and perpendicular to it, so FE 5 E 1 4pr 2 2 . But no charge is enclosed by the
Gaussian surface, so Qencl 5 0. Gauss’s law gives E 1 4pr 2 2 5 0 and E 5 0.
Q
(b) Apply Gauss’s law to a sphere of radius r . b. FE 5 E 1 4pr 2 2 . Qencl 5 Q. Gauss’s law gives E 1 4pr 2 2 5 and
P0
Q
E5
.
4pP0r 2
(c) The thick shell can be constructed from a series of concentric thin shells. E 5 0 inside each of these thin shells,
so E 5 0 inside the thick shell.
Reflect: In using Gauss’s law it is very helpful to select a Gaussian surface of appropriate symmetry, so it is simple
to express the flux in terms of the electric field at the surface.
17.62. Set Up: Section 17.9 shows that the net charge of a charged conductor is entirely on its outer surface.
Solve: (a) q 5 0 on the surface of the inner cavity of the conducting car.
(b) All the net charge is on the outer surface, 2850 mC.
17.63. Set Up: E 5 0 everywhere within the conductor. Any net charge must be on the inner and outer surfaces of
the conductor.
Gaussian surface
Figure 17.63
Electric Charge and Electric Field
17-21
Solve: (a) and (b) Apply Gauss’s law to a surface that is within the conductor, just outside the cavity, as shown in
Figure 17.63. E 5 0 everywhere on the Gaussian surface so FE 5 0 for that surface. Gauss’s law then says that Qencl
for this surface is zero. The conductor is neutral, so if the outer surface has charge 212 mC the inner surface must
have charge 112 mC. To make Qencl 5 0 there must be 212 mC within the hole.
Reflect: The charge in the hole creates the charge separation in the conductor. It pulls 112 mC to the inner surface
and that leaves 212 mC on the outer surface.
17.64. Set Up: E 5 0 everywhere within the conductor. Any net charge must be on the inner and outer surfaces of
the conductor.
Solve: Apply Gauss’s law to a surface that is within the conductor, just outside the cavity, as shown in Figure 17.64.
E 5 0 everywhere on the Gaussian surface so FE 5 0 for that surface.
Gaussian surface
Figure 17.64
(a) Qencl 5 0 for the Gaussian surface means that q 5 0 on the inner surface, so the 116 mC of net charge is on the
outer surface of the conductor.
(b) Qencl 5 0 for the Gaussian surface means that there must be 111 nC on the inner surface of the conductor. Since
the total net charge of the conductor is 116 nC, if there is 111 nC on the inner surface there must be 15 nC on the
outer surface.
17.65. Set Up: F 5 k
0 qqr 0
. Like charges repel and unlike charges attract. Charges q1 and q2 and the forces they
r2
S
S
exert on q3 at the origin are sketched in Figure 17.65a. For the net force on q3 to be zero, F1 and F2 from q1 and q2
must be equal in magnitude and opposite in direction.
y
q2
q3
1
1
0.30 m
F1
q1
F2
x
2
0.20 m
(a)
y
I
F2
1
II
F1
q3
(b)
Figure 17.65
q2
1
1
q3
III
F1
F2
q1 F1
1
2
q3
F2
x
17-22
Chapter 17
S
S
1
S
Solve: (a) Since F1 , F2 and Fnet are all in the 1x direction, F 5 F1 1 F2 . This gives
4.00 3 1026 N 5 k
0 q1q3 0
r12
1k
0 q2q3 0
r22
.
2.50 3 1029 C
4.50 3 1029 C
4.00 3 1026 N
5 q3
1
2
9
2
2
3 0.200 m 4
3 0.300 m 4 2
8.99 3 10 N # m / C
2
and q3 5 3.17 3 1029 C 5 3.17 nC.
S
S
S
S
S
(b) Both F1 and F2 are in the 1x direction, so Fnet 5 F1 1 F2 is in the 1x direction.
S
S
(c) The forces F1 and F2 on q3 in each of the three regions are sketched in Figure 17.65b. Only in regions I (to the left
S
S
of q2 ) and III (to the right of q1 ) are F1 and F2 in opposite directions. But since 0 q2 0 , 0 q1 0 , q3 must be closer to q2
than to q1 in order for F1 5 F2 , and this is the case only in region I. Let q3 be a distance d to the left of q2 , so it is
a distance d 1 0.500 m from q1 . F1 5 F2 gives
k
4.50 nC
2.50 nC
5k
. 1.80d 2 5 1 d 1 0.500 m 2 2. "1.80d 5 6 1 d 1 0.500 m 2 .
1 d 1 0.500 m 2 2
d2
The positive solution is d 5 1.46 m. This point is at x 5 20.300 m 2 1.46 m 5 21.76 m.
Reflect: At the point found in part (c) the electric field is zero. The force on any charge placed at this point will be
zero.
17.66. Set Up: Let q1 and q2 be the charges on the spheres. F 5 k
r2
and q1 1 q2 5 60.0 nC.
1 0.270 3 10 N 2 1 0.100 m 2
Fr 2
5
5 3.00 3 10216 C2. q2 5 60.0 3 1029 C 2 q1 so
k
8.99 3 109 N # m2 / C2
23
Solve: q1q2 5
0 q1q2 0
2
q1 1 60.0 3 1029 C 2 q1 2 5 3.00 3 10216 C2. q12 2 1 60.0 3 1029 C 2 q1 1 3.00 3 10216 C2 5 0.
The quadratic formula gives q1 5 12 1 6.00 3 1028 6 " 1 6.00 3 1028 2 2 2 4 1 3.00 3 10216 2 2 C 5 0.55 3 1028 C
or 5.45 3 1028 C. For q1 5 0.55 3 1028 C, q2 5 5.45 3 1028 C and for q1 5 5.45 3 1028 C, q2 5 0.55 3 1028 C.
One sphere has charge 0.55 3 1028 C and the other has charge 5.45 3 1028 C.
17.67. Set Up: F 5 k
0 qqr 0
r2
are shown in Figure 17.67.
. Like charges repel and unlike charges attract. The three charges and the forces on q3
y
F1
F1y
u
1
F1x
q3
50
m
0.0
0.030 m
F2
u
1 q1
Figure 17.67
0.040 m
q2
2
x
Electric Charge and Electric Field
Solve: (a)F1 5 k
0 q1q3 0
r12
17-23
1 5.00 3 1029 C 2 1 6.00 3 1029 C 2
5 1.079 3 1024 C.
1 0.0500 m 2 2
5 1 8.99 3 109 N # m2 / C2 2
u 5 36.9°. F1x 5 1F1 cos u 5 8.63 3 1025 N. F1y 5 1F1 sin u 5 6.48 3 1025 N.
F2 5 k
0 q2q3 0
r22
5 1 8.99 3 109 N # m2 / C2 2
1 2.00 3 1029 C 2 1 6.00 3 1029 C 2
5 1.20 3 1024 C.
1 0.0300 m 2 2
F2x 5 0, F2y 5 2F2 5 21.20 3 1024 N. Fx 5 F1x 1 F2x 5 8.63 3 1025 N.
Fy 5 F1y 1 F2y 5 6.48 3 1025 N 1 1 21.20 3 1024 N 2 5 25.52 3 1025 N.
Fy
(b)F 5 "Fx2 1 Fy2 5 1.02 3 1024 N. tanf 5 P P 5 0.640. f 5 32.6°, below the 1x axis.
Fx
Reflect: The individual forces on q3 are computed from Coulomb’s law and then added as vectors, using components.
17.68. Set Up: F 5 k
0 q1q2 0
. Like charges repel and unlike charges attract. The positions of the three charges are
r2
sketched in Figure 17.68a and each force on q3 is shown. The distance between q1 and q3 is 5.00 cm.
y
q3
F1x
1
q2
F2
1
u
4.0 cm
F1y
F1
u
x
q1 2
5.0 cm
(a)
y
Fx
x
u
Fy
F
(b)
0 q1q3 0
Figure 17.68
1 3.00 3 1029 C 2 1 5.00 3 1029 C 2
5 5.394 3 1025 N
1 5.00 3 1022 m 2 2
r12
F1x 5 2F1 cos u 5 2 1 5.394 3 1025 N 2 1 0.600 2 5 23.236 3 1025 N;
F1y 5 2F1 sin u 5 2 1 5.394 3 1025 N 2 1 0.800 2 5 24.315 3 1025 N
0 q2q3 0
1 2.00 3 1029 C 2 1 5.00 3 1029 C 2
F2 5 k 2 5 1 8.99 3 109 N # m2 / C2 2
5 9.989 3 1025 N
1 3.00 3 1022 m 2 2
r2
F2x 5 9.989 3 1025 N; F2y 5 0
Fx 5 F1x 1 F2x 5 9.989 3 1025 N 1 1 23.236 3 1025 N 2 5 6.75 3 1025 N; Fy 5 F1y 1 F2y 5 24.32 3 1025 N
Solve: (a) F1 5 k
5 1 8.99 3 109 N # m2 / C2 2
17-24
Chapter 17
Fy
S
(b) F and its components are shown in Figure 17.68b. F 5 "Fx2 1 Fy2 5 8.01 3 1025 N. tanu 5 P P 5 0.640 and
S
Fx
u 5 32.6°. F is 327° counterclockwise from the 1x axis.
17.69. Set Up: F 5 k
0 qqr 0
r2
. Like charges repel and unlike charges attract.
q1
q1
q3
0.20 m
F3
F1
0.20 m
q3
F1
F3
y
0.20 m
0.20 m
x
q4
F2
q2
q2
q4
F2
(b)
(a)
Figure 17.69
Solve: (a) The charges and the forces on the 21.00 mC charge are shown in Figure 17.69a. The distance r14 between
S
S
q1 and q4 is r 5 1 12 2 "2 1 0.200 m 2 5 0.1414 m. F2 and F3 are equal in magnitude and opposite in direction, so
S
S
S
S
F2 1 F3 5 0 and Fnet 5 F1 .
F1 5 k
0 q1q4 0
r142
5 1 8.99 3 109 N # m2 / C2 2
1 3.00 3 1029 C 2 1 1.00 3 1026 C 2
5 1.35 3 1023 N.
1 0.1414 m 2 2
The resultant force has magnitude 1.35 3 1023 N and is directed away from the vacant corner.
(b) The charges and the forces on the 21.00 mC charge are shown in Figure 17.69b. The distance r14 between q1
and q4 is "2 1 0.200 m 2 5 0.2828 m. Use coordinates as shown in the figure. F2x 5 2F3x and F1x 5 0, so Fx 5 0.
Fy 5 F1y 1 F2y 1 F3y 5 F1 1 2F2 cos 45°.
F1 5 k
0 q1q4 0
F2 5 k
0 q2q4 0
r142
5 1 8.99 3 109 N # m2 / C2 2
r242
5 1 8.99 3 109 N # m2 / C2 2
1 3.00 3 1029 C 2 1 1.00 3 1026 C 2
5 3.372 3 1024 N.
1 0.2828 m 2 2
1 3.00 3 1029 C 2 1 1.00 3 1026 C 2
5 6.742 3 1024 N
1 0.200 m 2 2
Fy 5 3.372 3 1024 N 1 2 1 6.742 3 1024 N 2 cos 45° 5 1.29 3 1023 N.
The resultant force has magnitude 1.29 3 1023 N and is directed toward the center of the square.
17.70. Set Up: The force on the negatively charged electron is opposite to the direction of the field. Since the field
is uniform, the force and acceleration are constant. Use coordinates with 1y upward and 1x to the right. The
electron is at x 5 10.0200 m when y 5 10.00500 m. ax 5 0. v0x 5 15.00 3 106 m / s, v0y 5 0. For an electron,
q 5 2e 5 21.60 3 10219 C and m 5 9.11 3 10231 kg. x 5 v0xt 1 12 axt 2.
t5
x
0.0200 m
5
5 4.00 3 1029 s.
v0x
5.00 3 106 m / s
2 1 0.00500 m 2
5
5 6.25 3 1014 m / s2.
1 4.00 3 1029 s 2 2
t2
may
Fy
1 9.11 3 10231 kg 2 1 6.25 3 1014 m / s2 2
52
Fy 5 may and Ey 5 2 5 2
5 23.56 3 103 N / C.
e
e
1.60 3 10219 C
The electric field has magnitude 3.56 3 103 N / C.
y 5 v0yt 1 12 ayt 2 gives ay 5
2y
Electric Charge and Electric Field
17-25
17.71. Set Up: The ball is in equilibrium, so for it a Fx 5 0 and a Fy 5 0. The force diagram for the ball is
S
S
S
given in Figure 17.71. FE is the force exerted by the electric field. F 5 qE. Since the electric field is horizontal, FE is
horizontal. Use the coordinates shown in the figure. The tension in the string has been replaced by its x and y
components.
y
T
Ty
u
FE
x
Tx
mg
Figure 17.71
mg
Solve: a Fy 5 0 gives Ty 2 mg 5 0. T cos u 2 mg 5 0 and T 5
.
cos u
1 2
a Fx 5 0 gives FE 2 Tx 5 0. FE 2 T sin u 5 0.
mg
FE 5
sin u 5 mg tan u 5 1 12.3 3 1023 kg 2 1 9.80 m / s2 2 tan17.4° 5 3.78 3 1022 N
cos u
FE
3.78 3 1022 N
5 3.41 3 104 N / C
FE 5 0 q 0 E so E 5
5
0q0
1.11 3 1026 C
S
S
q is negative and FE is to the right, so E is to the left in the figure.
Reflect: The larger the electric field E the greater the angle the string makes with the wall.
17.72. Set Up: E 5 k
0q0
2
S
S
. E is toward a negative charge and away from a positive charge. At the origin, E 1 due to
r
the 25.00 nC charge is in the 1x direction, toward the charge.
1 5.00 3 1029 C 2
5 31.2 N / C. E1x 5 131.2 N / C.
Solve: (a) E1 5 1 8.99 3 109 N # m2 / C2 2
1 1.20 m 2 2
Ex 5 E1x 1 E2x . Ex 5 145.0 N / C, so E2x 5 Ex 2 E1x 5 145.0 N / C 2 31.2 N / C 5 13.8 N / C.
0Q0
1 13.8 N / C 2 1 0.600 m 2 2
E2r 2
S
E is away from Q so Q is positive. E2 5 k 2 gives 0 Q 0 5
5
5 5.53 3 10210 C.
k
r
8.99 3 109 N # m2 / C2
(b) Ex 5 245.0 N / C, so E2x 5 Ex 2 E1x 5 245.0 N / C 2 31.2 N / C 5 276.2 N / C.
1 76.2 N / C 2 1 0.600 m 2 2
E2r 2
S
E is toward Q so Q is negative. 0 Q 0 5
5
5 3.05 3 1029 C.
k
8.99 3 109 N # m2 / C2
S
S
17.73. Set Up: The gravity force (weight) has magnitude w 5 mg and is downward. The electric force is F 5 qE.
Solve: (a) To balance the weight the electric force must be upward. The electric field is downward, so for an upward
force the charge q of the person must be negative. w 5 F gives mg 5 0 q 0 E and
(b) F 5 k
0 qqr 0
0q0 5
1 60 kg 2 1 9.80 m / s2 2
mg
5
5 3.9 C.
E
150 N / C
5 1 8.99 3 109 N # m2 / C2 2
r2
not a feasible means of flight.
1 3.9 C 2 2
5 1.4 3 107 N. The repulsive force is immense and this is
1 100 m 2 2
17-26
Chapter 17
17.74. Set Up: The force diagram for the 26.50 mC charge is given in Figure 17.74. FE is the force exerted on the
charge by the uniform electric field. The charge is negative and the field is to the right, so the force exerted by the
field is to the left. Fq is the force exerted by the other point charge. The two charges have opposite signs, so the force
is attractive. Take the 1x axis to be to the right, as shown in the figure.
y
E
FE
T
x
Fq
Figure 17.74
Solve: (a) F 5 0 q 0 E 5 1 6.50 3 1026 C 2 1 1.85 3 108 N / C 2 5 1.20 3 103 N
0 q1q2 0
1 6.50 3 1026 C 2 1 8.75 3 1026 C 2
5 8.18 3 102 N
1 0.0250 m 2 2
r2
a Fx 5 0 gives T 1 Fq 2 FE 5 0 and T 5 FE 2 Fq 5 382 N.
Fq 5 k
5 1 8.99 3 109 N # m2 / C2 2
(b) Now Fq is to the left, since like charges repel.
3
a Fx 5 0 gives T 2 Fq 2 FE 5 0 and T 5 FE 1 Fq 5 2.02 3 10 N.
17.75. Set Up: The force on the electron is upward in the figure so the electric field must be downward. To produce
a net electric field that is downward, it must be that q1 is positive, q2 is negative and 0 q1 0 5 0 q2 0 . The field due to q1
and q2 at the location of the electron are sketched in Figure 17.75. The electron is r 5 3.61 cm from each charge.
3.00 cm
tanu 5
and u 5 56.3°.
2.00 cm
q1
u
r
2.0 cm
E1x
3.0 cm
x
E2x
u
u
E1y
E2y
E1
E2
q2
y
Figure 17.75
Solve: The net electric field is E 5 2E1y 5 2k
q1
r2
cos u. F 5 eE 5
2keq1
r2
cos u. F 5 ma so
2keq1
r2
cos u 5 ma and
1 3.61 3 1022 m 2 2 1 9.11 3 10231 kg 2 1 8.25 3 1018 m / s2 2
r 2ma
5
5 6.14 3 1026 C.
2ke cos u
2 1 8.99 3 109 N # m2 / C2 2 1 1.60 3 10219 C 2 cos 56.3°
q2 5 26.14 3 1026 C.
Reflect: The force that an electric field exerts on a negative charge is opposite to the direction of the electric field.
We could also do this problem by considering the force that each charge exerts on the electron.
q1 5
Electric Charge and Electric Field
17-27
17.76. Set Up: When the forces balance, a 5 0 and the molecule moves with constant velocity.
q
Kv
5
R
E
Eq
Eq
T5
(b) v 5
and is constant. x 5 vt 5
KR
KR
q
ET q
ET
, where
52
(c) x 5
is constant.
K R
K
R 2
1 2 1 2
12 12 12 12
1 21 2 1 21 2
1 21 2 1 21 2
Solve: (a) F 5 FD so qE 5 KRv and
1 2
x2 5
ET q
K R
52
2
ET q
K R
ET q
K R
q
q
and
R 1
R
53
3
5 2x1; x3 5
1
q
R
1
ET q
K R
53
3
ET q
K R
5 3x1
1
m1m2
v2
2pr
.
a
5
. The period T is
. The mass of the sun is mS 5
For
a
circular
orbit,
2
r
v
r
1.99 3 1030 kg and the orbit radius of the earth is 1.50 3 1011 m.
Gm S
m SmE
v2
.
Solve: F 5 ma gives G 2 5 mE , so v 5
r
Å r
r
1 1.50 3 1011 m 2 3/2
r
r 3/2
T 5 2pr
5 2p
5 2p
5 2.73 3 1023 s
Å GmS
"Gm S
" 1 9.0 3 109 N # m2 / kg2 2 1 1.99 3 1030 kg 2
17.77. Set Up: Fgrav 5 G
A present year is 3.146 3 107 s, so the year in the parallel universe would be 8.7 3 10211 of a present year.
1 9.0 3 109 N # m2 / kg2 2 1 1.99 3 1030 kg 2
Gm S
Reflect: v 5
5
5 3.5 3 1014 m / s. This exceeds by a large
Å
Å r
1.50 3 1011 m
factor the speed of light in our universe, so the parallel universe must have a much larger speed of light than ours, in
order for this speed to be possible.