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CHAPTER 15
ELECTRIC CHARGE, FORCES, AND FIELDS
Multiple Choice Questions:
1.
(c).
2.
(a).
3.
(c).
4.
(a). The fur is positively charged (see Exercise 15.5). When the positively charged fur is brought near an
electroscope, the leaves are charged by polarization. So the charges on the leaves are positive.
5.
(d). Water will be deflected towards the object regardless of its being positively or negatively charged. The water is
still neutral but polarized.
6.
(d). The balloon clings to the wall regardless of its being positively or negatively charged. The wall is still neutral but
polarized.
7.
(a).
8.
(d).
9.
(c). The electrical force between two point charges is inversely proportional to the square of the distance between
them. So if this distance is tripled, the force decreases by a factor of 1/3 2 = 1/9.
10.
(a). The electrical force between two charged particles is directly proportional to the product of the charges. To
increase the force by a factor of 9 (back to its original value), each charge should be 3 times its original value
because 3  3 = 9.
11.
(c).
12.
(b).
13.
(a), because the electron has a negative charge.
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259
14.
Chapter 15
Electric Charge, Forces, and Fields
(b). At the given point, the electric fields of the electrons point toward the electrons, so their y-components cancel,
leaving only the x-components pointing in the –x (left) direction.
15.
(d). At the given point, both fields point downward (–y direction) so the net field is also downward.
16.
(b), since there is no electric field inside a conductor in electrostatic equilibrium.
17.
(a), since there is no electric field inside a conductor in electrostatic equilibrium.
18.
(b).
19.
(b), since electric field lines point toward negative charge.
20.
(c).
21.
(c).
Conceptual Questions:
1.
2.
3.
There would be no effect since it is simply a sign convention.
No . Charges are simply moved from the object to another object.
If an object is positively charged, its mass decreases, because it loses electrons. If an object is negatively charged, its
mass increases, because it gains electrons.
4.
Since the metal chain is a conductor, it allows any excess electrical charge that may have been induced on the truck
(by friction of rubber on the road, for example) to leak off to ground instead of building up on the truck. If the excess
charge is not removed, it could result in a spark, causing a gasoline explosion.
5.
6.
No . The charges simply change location. There is no gain or loss of electrons.
Starting with an uncharged electroscope, touch the bulb with your finger, which grounds the electroscope by
providing a path by which electrons could escape from the bulb. Then bring a negatively charged rod near the bulb
without touching it. This repels electrons from the rod. Removing your finger while keeping the negatively charged
rod close would leave the electroscope with a net positive charge. If you bring a charged object near the top and the
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College Physics Seventh Edition: Instructor Solutions Manual
260
leaves collapse, the object must be negatively charged in order to push electrons to the leaf area and at least partially
cancel the positive charge there.
7.
The water would bend the same way as with the balloon. In this case, the positive charges in the rod would attract
the negative ends of the water molecules, causing the stream to bend toward the rod.
8.
The electrical force between two point charges is directly proportional to the product of the charges. If one charge is
doubled and the other tripled, the new force is proportional to (2q1)(3q2) = 6q1q2, which makes the force 6 times as
strong, or 6F .
9.
A positive charge can be placed midway between the two electrons so that each will experience a repulsive force
from the other and an attractive force from the positive charge. With the proper amount of positive charge, the two
forces on each electron will cancel.
10.
In the original case, Fe = kq1q2/d2, and in the new case Fe = k(27q1)(q2/3)/r2. Since the forces are equal, we can
equate them and solve for r, giving kq1q2/d2 = k(27q1)(q2/3)/r2, which gives r = 3d .
11.
(a) The object must be positively charged because the downward repulsion of the nearby positive charge of the
dipole is greater than the upward attraction of the more distant negative charge of the dipole. (b) The dipole would
accelerate downward because the downward attraction on the nearby negative end would be greater than the
upward repulsion of the more distant positive end.
12.
It is determined from the relative lengths of the electric field vectors.
13.
It is determined by the relative density or spacing of the field lines. The closer the lines are together, the greater
the field magnitude.
14.
The electric field is the force per unit charge at any location. The force on a charge can only point in one direction
and so therefore must the electric field. Thus the field lines never cross. If they did, that would indicate two different
force directions at the crossing point, something that is physically unreasonable since a charge placed and released
there will accelerate only in one direction, not two.
15.
If a positive charge is at center of the spherical shell, the electric field is not zero inside. The field lines run radially
outward to the inside surface of the shell where they stop at the induced negative charges on this surface. The field
lines reappear on the outside shell surface (which is positively charged) and continue radially outward as if
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261
Chapter 15
Electric Charge, Forces, and Fields
emanating from the point charge at the center. If the charge were negative, the field lines would point in the reverse
directions.
16.
The Earth’s surface must carry a negative charge, because the electric fields terminate at negative charges.
17.
(a) Yes, this is possible when the electric fields are equal in magnitude and opposite in direction at some location.
For example, at the midway point in between and along a line joining two charges of the same type and magnitude,
the electric field is zero. (b) No, this is not possible . Between the charges, the fields are in the same direction, so
they could not cancel. At other points, the fields could not cancel because they would either have different
magnitudes (for points along the line connecting the two charges) or would not have opposite directions (for all other
points).
18.
(a) Directly upward because the electric field due to a positive charge points away from the charge. (b)
Away from the plate in the same horizontal plane as the plate because the field points away from the positive plate.
(c) The field would have an upward component and a horizontal component horizontal component away from
the plate.
19.
At very large distances, the object looks very small—like a point. So the electric field pattern looks like the field due
to a point charge located at the object.
20.
Yes , because the car (a metal frame) keeps the electric field inside it zero.
21.
The surface must be spherical .
22.
This is because charges accumulate at sharp points , which makes the electric field strongest there. So lightning hits
the tall sharp rods first.
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College Physics Seventh Edition: Instructor Solutions Manual
23.
262
Note that electric field is zero inside the metallic slab. Since charges are mobile, the negative charges are attracted
toward the upper portion of the slab, while the positive charges move toward the lower portion. The amount of
charge induced on each side of the slab is the same in magnitude as that on each of the plates.
+
-
24.
The field lines start out perpendicular to the plates and end up curving so they are perpendicular to the surface of the
metal sphere.
+
F2
q1
Fnet
60
20 cm
25.
The net charges are equalF3 in magnitude but opposite in sign .
q2
26.
20 cm
q3
No . All it means is that there is more positive charge than negative charge. In other words, there is a net positive
charge.
Exercises:
1.
q = ne = (106)(1.6  1019 C) = 1.6  1013 C .
2.
q = ne,

n=
q
–50  106 C
=
= 3.1  1014 electrons .
e
1.60  1019 C
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263
3.
Chapter 15
Electric Charge, Forces, and Fields
(a) q = 2(2e) = 4e = 4(1.60  1019 C) = 6.40  10 19 C
(b) An alpha particle has two protons, so it needs two electrons to make it a neutral atom.
4.
(a) The charge on the silk must be (3) negative because of the conservation of charge. When one object becomes
positively charged, it loses electrons. These same electrons must be gained by another object, and therefore it is
negatively charged.
(b) 8.0  1010 C according to charge conservation.
q = ne,

n=
8.0  1010 C
q
=
= 5.0  109 electrons .
e
1.6  1019 C
(c) The mass is m = (9.11  1031 kg/electron)(5.0  109 electrons) = 4.6  1021 kg .
5.
(a) The charge on the fur must be (1) positive because of the conservation of charge. When one object becomes
negatively charged, it gains electrons. These same electrons must be lost by another object, and therefore it is
positively charged.
(b) +4.8  109 C according to charge conservation. From q = +ne,
n=
q
4.8  109 C
=
= 3.0  1010 electrons.
e
1.6  1019 C
The mass is (3.0  1010)(9.11  1031 kg) = 2.7  1020 kg .
(c) The electrons moved from fur to the rubber rod, so the mass is still 2.7  1020 kg .
6.
(2.50  1012 C)[(1 electron)/(1.60  1019 C)] = 1.56  107 electrons
7.
(3.22  108 electrons)[(1.60  1019 C)/(1 electron)] = 5.15  10 11 C = 51.5 pC
8.
(a) The answer is (4) 1/4 . Since Fe =
kq1q2
, F is inversely proportional to the square of the distance, r. If r
r2
doubles, Fe becomes 1/4 times the original force.
(b)
9.
Fe
ro2
12
= 2 =
= 9.
Feo
r
(1/3)2
(a) Since Fe =
So
Fe = 9F .
kq1q2
 q1q2, Fe is 2  12 = 1 , i.e., the same.
r2
(b) Fe = 12  12 = 1/4 .
(c) Fe = 12  1 = 1/2 .
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College Physics Seventh Edition: Instructor Solutions Manual
264
kq1q2
k(6e)(6e)
(9.0  109 N·m2/C2)(6)2 (1.6  1019 C)2
=
=
= 1.3  107 N .
2
2
r
r
(0.25  109 m)2
10.
Fe =
11.
(a) Fe =
kq1q2
(9.0  109 N·m2/C2)(1.6  1019 C)2
=
= 5.8  1011 N .
2
r
(2.0  109 m)2
(b) Zero , because they are internal forces of the system.
12.
kq1q2
, r is proportional to 1/ Fe . If Fe decreases by a factor of 10, r will increase but not by a
r2
(a) Since Fe =
factor of 10. So the answer is (1) less than 10 times the original distance.
(b) Fe =
kq1q2
,
r2
F2
r12
= 2,
F1
r2

F2
r12
= 2,
F1
r2
so
F1
r =
F2 1
r2 =
13.
Fe =
kq1q2
,
r2
14.
Fe =
kq1q2
(9.00  109 Nm2/C2)(1.60  1019 C)2
=
= 2.90  109 N .
2
r
(2.82  1010 m)2
15.
(a) If q1 and q2 are like charges, the third charge must be placed between q1 and q2 for it to be in electrostatic

so
r1 =
F2
r =
F1 2
10 (30 cm) = 95 cm .
5  (100 cm) = 2.24 m .
equilibrium. Also, q1 and q2 have the same magnitude; it must be at the center or at x = 0.25 m from symmetry.
(b) If q1 and q2 are unlike charges, the third charge could be placed
outside q1 and q3 (x < 0 and x > 0.50 m). However, since the magnitudes
+q3
F1
q1
q2
F2
d
of q1 and q2 are equal, there will be nowhere for it to happen according
–q3
F2
0
0.50 m
F1
to Coulomb’s law.
(c) Since q1 = +3.0 C and q2 = 7.0 C, a third charge of either type can be placed outside q1 (x < 0) for it to be in
electrostatic equilibrium. It cannot be placed to the right of q2 (x > 0.50 m), since the force by q2 will always be
larger due to the closer distance from it. Assume q3 is placed at d from q1 (or x = d).
From Coulomb’s law, we have
kq1q3
k|q2|q3
=
,
2
2
r13
r23
Taking the square root on both sides gives
Solving,
16.
d = 0.94 m.
Thus

q1
|q2|
=
.
d2
(d + 0.50 m)2
3.0
7.0
=
,
d
d + 0.50
so
1.73(d + 0.50) = 2.65d.
x = 0.94 m for q3 .
(a) Calling q the smaller charge, Coulomb’s law gives
Fe = kq1q2/r2 = kq(3q)/r2 = 3kq2/r2
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265
Chapter 15
Electric Charge, Forces, and Fields
3.15  10–6 N = (9.00  109 Nm2 /C2 )(3q2)/(0.10 m)2
q = 1.08  00–9 C = 1.08 nC
(b) Now each point charge has 2q of charge.
Fe = k(2q)(2q)/r2 = 4(kq2/r2) = 4/3 of the original force
Fe = (4/3)(3.15  10–6 N) = 4.20  10–6 N = 4.20 µN
17.
(a) The forces on the electron add numerically because they are in the same direction. Putting the negative charge on
the left and the positive charge on the right, we have
Fnet 
1 1
1 1
kqe kqe
 2  kqe  2  2   6.48  10 21  2  2 
2
r1
r2
 r1 r2 
 r1 r2 



1
1

At 5.0 cm: Fnet   6.48  10 21  
2
(0.05
m)
(0.25
m)2


18
  2.7  10 N

Likewise, at the other locations we get the forces
At 10.0 cm: 8.1  10 19 N
At 15.0 cm: 5.8  1019 N
At 20.0 cm: 8.1  10 19 N
At 25.0 cm: 2.7  1018 N
(b) According to the graph, the electron feels the least force midway between the two charges.
Fnet
|
x
15 cm
18.
F2 = F3 =
kq1q2
(9.0  109 N·m2/C2)(4.0  106 C)2
=
= 3.6 N.
2
r
(0.20 m)2
According to symmetry, the net force on q1 points in the +x direction.
Fnet = Fx = F2 cos 60 + F3 cos 60
= 2(3.6 N) cos 60 = 3.6 N in the positive x direction .
19.
(a) From Coulomb’s law: F1 =
So
kq1q2
(9.0  109 N·m2/C2)(10  106 C)(10  106 C)
=
= 90 N.
2
r
(0.10 m)2
F 1 = (90 N) ^x.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
College Physics Seventh Edition: Instructor Solutions Manual
F3 =
266
(9.0  109 N·m2/C2)(5.0  106 C)(10  106 C)
= 45 N.
(0.10 m)2
q1
F 3 = (45 N) ŷ .
So
F1
F4
(9.0  109 N·m2/C2)(5.0  106 C)(10  106 C)
F4 =
= 22.5 N.
(0.10 m)2 + (0.10 m)2
0.10 m
F 4 = F4 (cos 45) ^x + (sin 45) ŷ = (15.9 N) (^x + ŷ ).
So
q2
F3
q4
0.10 m
q3
Therefore the net force is F 2 = F 1 + F 3 + F 4 = (74.1 N) x + (60.9 N) ŷ .
Thus
(74.1 N)2 + (60.9 N)2 = 96 N ,
F2 =
. 
 609
 = tan1 
 = 39 below positive x-axis .
 741
.
(b) F1 =
(9.0  109 N·m2/C2)(10  106 C)(5.0  106 C)
= 45 N.
(0.10 m)2
So
F 1 = (45 N) ŷ .
q1
q2
0.10 m
F1
F2
(9.0  109 N·m2/C2)(10  106 C)(5.0  106 C)
F2 =
= 22.5 N.
(0.10 m)2 + (0.10 m)2
F3
q4
0.10 m
q3
F 2 = F2 (cos 45) ^x + (sin 45) ŷ = (15.9 N) (^x + ŷ ).
So
F3 =
(9.0  109 N·m2/C2)(5.0  106 C)(5.0  106 C)
= 22.5 N.
(0.10 m)2
So
F3 = (22.5 N) ^x.
Therefore the net force is F 4 = F 1 + F 2 + F 3 = (6.6 N) ^x + (60.9 N) ŷ .
Thus
20.
Fe =
 60.9 
 = tan1 
 = 84 above negative x-axis .
 6.6 
(6.6 N)2 + (60.9 N)2 = 61 N ,
F2 =
kq1q2
kq2
= 2 .
2
r
r
sin  =
In the vertical direction,
9.0 cm
,
30 cm
 = 17.5.

T cos = mg,

T=
F2
mg
.
cos
q1
Fnet
60
In the horizontal direction,
Therefore
Thus
21.
kq2
= mg tan,
r2
q=
(a) Since E =
F = T sin =

q=
mg
sin = mg tan.
cos
mg tan
r.
k
20 cm
F3
q2
20 cm
q3
(0.10  103 kg)(9.80 m/s2) tan 17.5
(0.18 m) = 3.3  108 C .
9.0  109 N·m2/C2
kq
, E is inversely proportional to the square of the distance.
r2
So if r doubled, E (2) decreased .
(b)
ro2
E
12
= 2 = 2 = 1/4.
Eo
r
2
Therefore
E=
Eo
1.0  104 N/C
=
= 2.5  105 N/C .
4
4
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267
Chapter 15
Electric Charge, Forces, and Fields
F
3.2  1014 N
=
= 2.0  105 N/C .
q
1.6  1019 C
22.
E=
23.
Use E = F / q to find the components of the electric field. Note that the direction of the field is opposite to the
direction of the force since the electron is negative.
Vertical component:
Ey = Fy/q = (2.7  1014 N)/(1.60  1019 C) = 1.69  105 N/C downward (y)
Horizontal component:
Ex = Fx/q = (3.8  1014 N)/(1.60  1019 C) = 2.38  105 N/C left (x)
The magnitude of the field is
E = (Ex2 + Ey2)1/2 = [(1.69  105 N/C)2 + (2.38  105 N/C)2]1/2 = 2.9  105 N/C
kq
(9.0  109 N·m2/C2)(2.0  1012 C)
= 3.2  102 N/C away from the charge .
2 =
r
(0.75  102 m)2
24.
E=
25.
Using E =
r=
26.
kq
.
r2
kq
=
E
(9.00  109 Nm2/C2)(1.60  1019 C )
= 1.2  107 m away from the charge .
1.0  105 N/C
(a) The net field is (3) directed toward the 5.0 C charge . The electric fields by the two charges are opposite in
direction, but the one by the –5.0-C charge has a greater magnitude.
(b) E4.0 =
E5.0 =
kq
(9.0  109 N·m2/C2)(4.0  106 C)
= 3.6  106 N/C,
2 =
r
(0.10 m)2
(9.0  109 N·m2/C2)(5.0  106 C)
= 4.5  106 N/C.
(0.10 m)2
So the net field is E = E5.0  E4.0 = 9.0  105 N/C .
27.
Electric force balances the weight of the particle, F = mg.
Proton: E =
F
mg
(1.67  1027 kg )(9.80 m/s2)
=
=
= 1.0  107 N/C upward .
q
q
1.6  1019 C
Electron: E =
28.
(9.11  1031 kg )(9.80 m/s2)
= 5.6  1011 N/C downward .
1.6  1019 C
(a) Since both charges are negative, a point between the two charges
could have zero electric field. The location is at
(1) left of the origin , because the charge closer to the point has the
–3.0 C
E3.0
–0.50 m
0
E4.0
–4.0 C
0.50 m
x
d
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College Physics Seventh Edition: Instructor Solutions Manual
268
smaller magnitude. The distance from the smaller charge has to be smaller to match the field by the bigger charge.
(b) Assume the point is d from the –3.0-C charge. For the electric field to be zero, E3.0 = E4.0.
So
E3.0 =
kq
k(3.0 C)
k(4.0 C)
=
= E4.0 =
.
r2
d2
(1.0  d)2
Taking the square root on both sides gives
Solving for
29.
d = 0.464 m.
3.0
4.0
=
.
d
1.0  d
Or
Therefore the coordinates of the point are (0.036 m‚ 0) .
 y
 = tan1  x  .
The electric field by the +2.5-C charge is
r=
x2 + y2 ,
E2.5 =
kq
(9.0  109 N·m2/C2)(2.5  106 C)
= 3.6  105 N/C,
2 =
r
(0.20 m)2 + (0.15 m)2
and
1.73(1.0  d) = 2.0d.
+2.5 C
(–0.20 m, 0.15 m)
2.5 = 36.9, so E 2.5 = (3.6  105 N/C)[(cos36.9) ^x  (sin 36.9) ŷ ]
–6.3 C
= (2.88  105 N/C) ^x + (2.16  105 N/C) ŷ .
–4.8 C
(–0.42 m, –0.32 m)
(0.50 m, –0.35 m)
The electric field by the 4.8-C charge is
E4.8 =
(9.0  109 N·m2/C2)(4.8  106 C)
= 1.16  105 N/C,
(0.50 m)2 + (0.35 m)2
4.8 = 35.0,
so E 4.8 = (1.16  105 N/C)[(cos 35.0) ^x  (sin 35.0) ŷ ] = (9.50  104 N/C) ^x + (6.65  104 N/C) ŷ .
The electric field by the –6.3-C charge is
E6.3 = c
(9.0  109 N·m2/C2)(6.3  106 C)
= 2.03  105 N/C,
(0.42 m)2 + (0.32 m)2
6.3 = 37.3,
so E 6.3 = [(2.03  105 N/C)(cos 37.3) ^x  (sin 37.3) ŷ ] = [(1.61  105 N/C) ŷ + (1.23  105 N/C) ŷ ].
ˆ .
Therefore the net electric field is E = E 2.5 + E 4.8 + E 6.3 = (2.2 105 N/C)xˆ + (  4.1  105 N/C) y.
30.
The point has to be between the two charges. Assume it is d from the +4.0 C charge.
For the electric field to be zero,
kq1
kq2
= 2,
r12
r2
Taking the square root on both sides gives
Solving,
31.
r=

4.0 C
9.0 C
=
.
d2
(0.30 m  d)2
2
3
=
.
d
0.30  d
Therefore
3d = 2(0.30  d).
d = 0.12 m = 12 cm from the charge of 4.0 C (between the charges) .
0.10 m
= 0.115 m.
cos 30
q1
kq
(9.0  109 N·m2/C2)(4.0  106 C)
E1 = E2 = E3 = 2 =
= 2.72  106 N/C.
r
(0.115 m)2
E1
E3
E2
r
Due to symmetry, the resultant of E 1 and E 2 will point toward q3.
30
The net field E = E3 + E1 cos 60 + E2 cos 60
q2
20 cm
q3
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269
Chapter 15
Electric Charge, Forces, and Fields
= 2.72  106 N/C + 2(2.72  106 N/C) cos 60 = 5.4  106 N/C toward the charge of –4.0 C .
q1
32.
Due to symmetry, E1 and E2 cancel out. The net field is the field by q3.
30
kq3
(9.0  109 N·m2/C2)(4.0  106 C)
E = E3 = 2 =
r
[(0.20 m) cos30]2
r
E2
E3
= 1.2  106 N/C toward the charge of –4.0 C .
E1
q2
33.
E1 = E2 =
kq
(9.0  109 N·m2/C2)(10  106 C)
= 1.8  107 N/C.
2 =
r
(0.05 m)2 + (0.05 m)2
E3 = E4 =
(9.0  109 N·m2/C2)(5.0  106 C)
= 9.0  106 N/C.
(0.05 m)2 + (0.05 m)2
20 cm
q1
q2
E1
0.10 m
Due to symmetry, the net electric field will be upward. It is equal to
q3
E2
E3
E4
r
q4
0.10 m
q3
E = (E1 + E3) cos 45 + (E2 + E3) cos 45
= 2(1.8  107 N/C + 9.0  106 N/C) cos 45 = 3.8  107 N/C in the +y direction .
34.
(a) First find the acceleration.
a=
F
qE
(2.0  106 C)(12 N/C)
=
=
= 1.2 m/s2.
m
m
2.0  105 kg
x = vot + 12 at2 = 12 (1.2 m/s2)(0.50 s)2 = 0.15 m .
(b) v = vo + at = (1.2 m/s2)(0.50 s) = 0.60 m/s .
(c) E =
4 kQ
,
A
so
Q=
EA
(12 N/C)(0.050 m)2
13
=
C.
9
2 2 = 2.7  10
4 k
4 (9.00  10 Nm /C )
The plates carry opposite charge, so they are 0.27 pC .
35.
(a) The field between two large parallel plates is E = 4πkQ/A, which gives Q/A = E/4πk
2 2
Q/A = (1.7  106 N/C)/[4π(9.00  109 Nm /C )] = 1.5  105 C/m 2 = 15 µC/m 2
(b) Q = A  (charge density) = (0.150 m)2(15 µC/m2) = 3.4  107 C = 0.34 µC
36.
(a) The field between two large parallel plates is
E = 4πkQ/A
2 2
E = 4π(9.00  Nm /C )(4.0  109 C)/(0.20 m)2 = 1.1 104 N/C
(b) F = qE = (1.60  1019 C)(1.13  104 N/C) = 1.8  10 15 N toward the positive plate.
(c) a = F/m = (1.18  1015 N)/(9.11  1031 kg) = 2.0  1015 m/s2
37.
E1 =
 4.0 
kq
(9.0  109 N·m2/C2)(10  106 C)
= 7.76  106 N/C, 1 = tan1  10  = 21.8.
2 =
2
2
r
(0.10 m) + (0.040 m)
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270
College Physics Seventh Edition: Instructor Solutions Manual
E 1 = (7.76  106 N/C)[(cos 21.8) ^x + (sin 21.8) ŷ ]
q1
= (7.20  106 N/C) ^x + (2.88  106 N/C) ŷ ;
1
0.10 m
(9.0  109 N·m2/C2)(10  106 C)
E2 =
= 5.63  107 N/C,
(0.040 m)2
E2
q2
4.0 cm
E1
E3
4
So
E 2 = (5.63  107 N/C) ^y;
E3 =
(9.0  109 N·m2/C2)(5.0  106 C)
= 1.25  107 N/C,
(0.060 m)2
so
E4 =
(9.0  109 N·m2/C2)(5.0  106 C)
= 3.31  106 N/C,
(0.10 m)2 + (0.060 m)2
 6.0 
4 = tan1  10  = 31.0,
q4
E4
0.10 m
6.0 m
q3
E 3 = (1.25  107 N/C) ŷ ;
Thus E 4 = (3.31  106 N/C)[(cos 31.0) ^x + (sin 31.0) ŷ ] = (2.84  106 N/C) ^x + (1.70  106 N/C) ŷ .
E = E 1 + E 2 + E 3 + E 4 = (4.4  106 N/C)xˆ + (7.3  107 N/C) yˆ .
Therefore
38.
(a) From symmetry, the net electric field is in the –y direction.
(b) r =
E– = E+ =
(d/2)2 + x2 ,
sin  =
d/2
=
r
+q
d/2
.
(d/2)2 + x2
r
d/2
x
kq
kq
=
.
r2
(d/2)2 + x2
d/2
E = Ey = E– sin  + E+ sin  = 2E+ sin 
2kq
=
(d/2)2 + x2

E–
–q
E+
E
d/2
kqd
=
.
2
2
2
[(d/2)
+ x2]3/2
(d/2) + x
(c) If x >> d, we can ignore the term (d/2)2 in (d/2)2 + x2.
So
E=
kqd
kqd
kqd
 2 3/2 = 3 .
[(d/2)2 + x2]3/2
(x )
x
(d) This is because the field is not created by a single point charge. We expect inverse square from single point
charges, but vector addition can cause other distance dependences as happens here.
39.
(a) The inner surface of the shell will have (1) negative charge due to induction.
(b) Zero since all excess charge resides on the surface of the conductor in electrostatic equilibrium.
(c) +Q since all excess charge resides on the surface of the conductor in electrostatic equilibrium.
(d) Q by induction and the conservation of charge.
(e) +Q by induction and the conservation of charge.
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271
40.
Chapter 15
Electric Charge, Forces, and Fields
(a) There is none , since the electric field is zero in the interior of the solid sphere.
(b) Outward from the center of the sphere, since the charge on the surface of the sphere is positive.
(c) There is none , since the electric field is zero. The field by the sphere cancels out the field by the inner surface
of the shell.
(d) Outward from the center of the sphere. The net excess charge is positive on the outer surface of the shell.
41.
(a) It is zero , since there is no electric field inside a conductor in electrostatic equilibrium.
(b) The charge on the surface of the sphere can be considered as if it were concentrated at the center.
So
E=
k(Q)
kQ
=  2 , the negative sign indicates it is toward the center.
r2
r
(c) Again, it is zero .
(d) All charges can be considered as if they were concentrated at the center.
So
E=
k(Q)
kQ
k(Q)
kQ
+ 2 +
=  2 , the negative sign indicates it is toward the center.
r2
r
r2
r
42.
More charges accumulate at sharper points, and the electric field lines are perpendicular to the surface.
43.
See the diagram below.
44.
(a) The electric field has moved negative charges in the slab to the top of the slab, so this field must point
downward .
(b) The net field in the metal must be zero. So the field due to the generated surface charges must be opposite to the
external field and of the same magnitude. Hence, E = 1250 N/C , and it points upward .
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College Physics Seventh Edition: Instructor Solutions Manual
272
(c) The top and bottom of the slab have equal but opposite charges Q. Inside the slab, these charges generate an
electric field of 1250 N/C. We can use the formula for the field between parallel plates, treating the surface charges
as the parallel plates.
E = 4πkQ/A
Q = EA/4πk = (1250 N/C)(0.0500 m)2/[4π(9.00  109 Nm2 /C2 )]
Q = 2.76  10 11 C = 27.6 pC
45.
The net number of field lines is +6  2  6 = 6 lines , or net of 6 lines entering it .
46.
(a) The magnitude of the unknown charge is (1) greater than 10.0 C , because the number of field lines is
proportional to the magnitude of the charge inside the Gaussian surface.
(b) q2 = 
47.
75
(+10.0 C) = 46.9 C .
16
(a) The other end of the dipole has the same magnitude charge as the positive end, except it is negative. Therefore
10 field lines will enter the Gaussian surface.
(b) The net charge of the dipole is zero, so the net number of field lines passing through the Gaussian surface would
be zero .
48.
(a) Fe =
kq1q2
(9.0  109 N·m2/C2)(1.6  1019 C)2
=
= 8.2  108 N .
r2
(5.3  1011 m)2
(b) The electric force provides the required centripetal force.
so
v=
(c) ac =
Fc r
=
m
Fe = Fc = m
v2
,
r
(8.2  108 N)(5.3  1011 m)
= 2.2  106 m/s .
9.11  1031 kg
v2
F
8.2  108 N
=
=
= 9.0  1022 m/s2 = 9.2  1021 g , where g = 9.80
r
m
9.11  1031 kg
m/s2.
T
49.
(a) Since the charge is negative, and the electric force has to be to the left to keep the ball
F
balanced, the direction of the electric field is to the right .
(b) In the vertical direction: T cos  = mg,

T=
mg
.
cos 
mg
In the horizontal direction: F = qE = T sin .
So E =
T sin 
mg sin 
mg tan 
(6.00  106 kg)(9.80 m/s2) tan 12.3
=
=
=
= 8.55  103 N/C .
q
q
q cos 
1.50  109 C
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273
50.
Chapter 15
Electric Charge, Forces, and Fields
(a) The bottom plate must be positively charged. In this configuration, the electric force (upward) can balance the
weight of the particle.
4 kQ
4 (9.00  10 Nm /C )(5.50  105 C)
=
= 5.14  108 N/C.
A
(0.110 m)2
9
(b) E =

F = qE = mg,
51.
m=
2
2
qE
(9.35  1012 C)(5.14  108 N/C)
=
= 4.90  104 kg .
g
9.80 m/s2
(a) The electron gains speed, so the force on it is to the right. Since electron carries negative charge, the charge is
positive on right plate and negative on left plate .
(b) The direction of the electric field is from right to left .
(c) First calculate the acceleration of the electron.
v = v + 2a(x  xo),
2
E=
So
52.
2
o
4 kQ
F
qE
4 kqQ
and a = m = m =
.
A
Am
Ama
(0.254 m)2 (9.11  1031 kg) (3.468  1010 m / s)
45
Q = 4 kq =
= 1.13  10 C .
4 (9.00  109 N  m 2 / C2 ) (1.60  1019 C)
(a) t =
x
0.10 m
=
= 1.626  109 s.
vx
6.15  107 m/s
d = 12 ay t2 ,
ay =

v2  v2o
(4.15  104 m/s)2  (1.63  104 m/s)2
a=
=
= 3.468  1010 m/s2.
2(0.0210 m)
2(x  xo)

F
qE
=
,
m
m
(b) Since E =
ay =

2d
2(4.70  103 m)
= 3.555  1015 m/s2.
2 =
t
(1.626  109 s)2
E=
ay m
ay m
(3.555  1015 m/s2)(9.11  1031 kg)
=
=
= 2.02  104 N/C .
q
q
(1.60  1019 C)
4 kQ
, the surface charge density is
A
Q
E
2.024  104 N/C
7
2
=
=
9
2 2 = 1.79  10 C/m .
A
4 k
4 (9.00  10 Nm /C )
53.
(a) The forces F and F (see figure) create a torque on the dipole which will turn it until the dipole moment p is
parallel to the electric field E .
+
p
F+
E
F-
-
(b) Add the torques due to the two forces, realizing that the torques are equal. Calling d the distance between the
charges and  the angle between p and E , we get
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College Physics Seventh Edition: Instructor Solutions Manual
274
 = 2F+(d/2) sin = qdE sin = pE sin 
(c) Fnet  0 because the two forces are of equal magnitude but opposite direction.
(d) The torque is a maximum when |sin| = 1, which is when  = 90° or 270°. When  = 0° or 180° , sin = 0 for the
minimum torque of zero.
54.
(a) First use kinematics to find the acceleration of the proton.
0 = (3.15  105 m/s)2 + 2a(0.0525 m)
a = 9.45  1011 m/s2
Newton’s second law gives
F = ma = qE
3
E = ma/q = (1.67  1027 kg)(9.45  1011 m/s2)/(1.60  1019 C) = 9.86  10 N/C
(b) Using kinematics, we have
v
v02  2a( x  x0 )  
3.15 10
5

2


m/s  2 9.45  1011 m/s2 (0.0350 m
5
v = 1.82  10 m/s
There are two answers because the proton first stops and then reverses direction.
55.
(a) The positive charge (+q) of the dipole is to the right of the origin with the negative charge (q) to the left of the
origin. The charge Q = 2.50 pC will attract the positive charge and repel the negative charge of the dipole. The ycomponents of the force will cancel, leaving only two equal x-components in the –x-direction. Calling r the distance
between the upper charge and each of the charges of the dipole and  the angle between the force and the x-axis, we
get
 kqQ 
Fx  2 
 cos 
 r2 




 9.00  109 N m2 /C2 4.55  1012 C 2.50  1012 C   3 
 
Fx = 2 

 5 
(0.0500 m)2


= 4.91  1011 N in the –x direction
Newton’s second law gives
ax = Fx/m = (4.91  1011 N)/(7.25  1012 kg) = 6.78 m/s 2 in the – x direction
(b) Since the torques are equal, the net torque is twice the torque on each charge of the dipole. Therefore
 = 2(F sin d/2) = kqQd/r2 sin
 = (9.00  109 Nm2 /C2 )(4.55  1012 C)(2.50  1012 C)(0.0600 m)(4/5)/(0.0500 m)2
= 1.97  1012 mN in the counterclockwise direction
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