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Practice Problems
A
1.
C
Charge transfer can only occur when there is contact.
The positive rod attracts negative charge on A; leaving
a net positive charge on B.
C
When the rod is removed, the charges return to normal
(equal on both spheres).
Like charges repel and unlike charges attract
5.
a.
conduction
induction
b.
Place a grounding wire on the right side.
6.
– 1 C of excess charge on the piece of silk.
positively charged
conduction
negatively charged
induction
8.
A
Fe = kQAQB/r2; since QB is doubled, while all else is the
same, then Fe is doubled (2 x 1 N).
10.
C
The force is the same for both spheres (Newton's third
law).
11.
D
Fe = kQAQB/r2; since QA and QB are both doubled, while
all else is the same, then Fe is 2 x 2 x 1 N.
12.
A
Fe = kQAQB/r2; since r is doubled, while all else is the
same, then Fe is 1/22 x 1 N.
13.
D
Fe = kQAQB/r2; since r is halved, while all else is the
same, then Fe is 1/½2 x 1 N.
14.
B
Sphere A is attracted to the wall and opposite charges
attract  the wall must be negative.
15.
B
F3 = F1sin53 = 3 N
18.
B
The electric force is the same for both the electron and
proton (Newton's third law).
19.
B
Opposite charges attract.
20.
A
Fe = kQAQB/r2; since r is decreasing the force will
increase.
21.
A
B
The electric field is away from +Q () and toward –Q
( )   .
28.
The electric field is toward -Q () and away from +Q
( )   .
29.
D
The electric field equals zero closest to |smaller|
charge and outside when charges have different signs.
C
The electric field equals zero closest to |smaller|
charge and between when charges have same sign.
31.
B
The electric field equals zero closest to |smaller|
charge and between when charges have same sign.
32.
A
The electric field equals zero closest to |smaller|
charge and outside when charges have different signs.
33.
D
The electric field is away from the positive charges.
34.
B
The electric field is away from the positive charge and
toward the negative charge.
35.
C
The electric field is toward the negative charges.
36.
A
The electric field is away from the positive charges and
toward the negative charges.
D
The electric fields cancel each other out.
38.
F1 = F3/cos53 = 5 N (This is a 3-4-5 triangle)
17.
B
27.
The two electric fields cancel each other out.
37.
F3 = Fg = mg = (0.4 kg)(10 m/s2) = 4 N
16.
C
C
30.
Like charges repel, which for B is away from A.
9.
C
e = kQ/r2; since r is doubled, while all else is the same,
then E is 1/22 x E.
26.
A
7.
E = kQ/r2; since Q is doubled, while all else is the
same, then E is doubled (2 x E).
25.
A
4.
The direction of the electric field is away from the
positive charge.
24.
C
3.
Since the electron has greater acceleration, it will
travel a greater distance before collision.
23.
A
2.
B
22.
F = ma; since the electron is lighter it will have greater
acceleration.
D
The electric fields cancel each other out.
39. a.
Fe = kQ1Q2/r2
Fe = (9.0 x 10-9 N•m2/C2)(1.60 x 10-19 C)2/(5.3 x 10-11 m)2
Fe = 8.2 x 10-8 N
b.
Fg = GM1M2/r2
Fg = (6.67 x 10-11)(1.67 x 10-27)(9.11 x 10-31)/(5.3 x 10-11 m)2
Fg = 3.6 x 10-47 N
c.
Fe + Fg = Fc = mv2/r
8.2 x 10-8 N = (9.11 x 10-31 kg)v2/(5.3 x 10-11 m)
v = 2.2 x 106 m/s
40. a.
Fe = kQ1Q2/r2 = (9 x 109)(10 x 10-6)(8 x 10-6)/(0.4)2
Fe = 4.5 N (attraction)
b.
54.
(-10 C + 8 C)/2 = -1 C per sphere
c.
Fe = kQ1Q2/r2 = (9 x 109)(1 x 10-6)2/(0.4)2
Fe1`` = 0.056 N (repulsion)
41. a.
Fg
Fg = Fe  mg = qE  m = qE/g
m = (150)(1.6 x 10-19 C)(300 N/C)/(10 m/s2) = 7.2 x 10-16 kg
55. a.
Fe = qE = (1 x 10-6 C)(1000 N/C) = 1 x 10-3 N
b.
Fe = ma
a = FE/m = 1 x 10-3 N/1 x 10-6 kg = 1000 m/s2
c.
v2 = vo2 + 2ad
v = [(2)(1000 m/s2)(0.01 m)]½ = 4.5 m/s
Ft
FE
b.
Fe/Fg = tan 30
tan30 = kQ2/r2/mg  Q = (mgr2tan30/k)½
Q = [(0.030 kg)(10 m/s2)(0.50 m)2/(9 x 109)]½ = 2.2 x 10-6 C
56.
–1 C of excess charge on the piece of silk.
57.
42.
The strength of E inside a good conductor is zero.
43.
conduction
positive
induction
negative
59.
E = kQ/r2
E = (9 x 109 N•m2/C2)(10 x 10-6C)/(5 m)2 = 3600 N/C
The net charge is located on the surface.
44.
The electric field is perpendicular to the surface.
45. a.
b.
+
     
+ –

     
–
46. a.
The 50 C generates the stronger field.
b.
The charges exert equal force.
47.
E = kQ/r2
E = (9.0 x 109 N•m2/ C2)(10 x 10-6 C)/(3 m)2 = 1 x 104 N/C
48.
E1 = kQ/r2 = (9.0 x 109 N•m2/ C2)(5 x 10-6 C)/(1 m)2
E1 = 4.5 x 104 N/C 
E2 = kQ/r2 = (9.0 x 109 N•m2/ C2)(10 x 10-6 C)/(3 m)2
E2 = 1 x 104 N/C 
E = 4.5 x 104 + 1 x 104 = 5.5 x 104 N/C 
49.
E1 = kQ/r2 = (9.0 x 109 N•m2/ C2)(5 x 10-6 C)/(1 m)2
E1 = 4.5 x 104 N/C 
E2 = kQ/r2 = (9.0 x 109 N•m2/ C2)(10 x 10-6 C)/(3 m)2
E2 = 1 x 104 N/C 
E = [(4.5 x 104)2 + (1 x 104)2]½ = 4.6 x 104 N/C
tan = 4.5 x 104/1 x 104 = 4.5
 = 77o  (measured from – x axis)
50.
E = kQ/r2 = (9.0 x 109 N•m2/ C2)(8 x 10-6 C)/(2½ m)2
E1 = 3.6 x 104 N/C (toward the open corner)
51.
Fe = qE
Fe = (3 x 10-7 C)(500 N/C) = 1.5 x 10-4 N east
52.
E1 = kQ1/r2 = (9 x 109)(10 x 10-6)/(5)2 = 3,600 N/C 
E3 = kQ3/r2 = (9 x 109)(10 x 10-6)/(10)2 = 900 N/C 
Etot = E 1 + E3 = 3,600 + 900 = 4,500 N/C 
Fe = qEtot = (2 x 10-6 C)(4,500 N/C) = 9 x 10-3 N 
53. a.
E40 = kQ/r2 = (9 x 109)(40 x 10-6)/(5)2 = 14,400 N/C ()
E-10 = kQ/r2 = (9 x 109)(10 x 10-6)/(5)2 = 3600 N/C ()
Etot = 3600 N/C + 14,400 N/C = 18,000 N/C 
b.
E40 = E-10
kQ40/x2 = kQ-10/(x – 10 m)2
60.
Fe = kQ2/r2
Fe = (9.0 x 109)(1.6 x 10-19)2/(1.0 x 10-10)2 = 2.3 x 10-8 N
61.
Fe = qE = (5 x 10-7 C)(200 N/C) = 1 x 10-4 N west
62. a.
E = kQ/r2 = (9.0 x 109 N•m2/C2)(2 x 10-6 C)/(2m)2
E = 4.5 x 103 N/C
b.
Fe = qE = (8 x 10-6 C)(4.5 x 103 N/C) = 0.036 N
c.
Fe = qE = (4 x 10-6 C)(4.5 x 103 N/C) = 0.018 N
63.
tan 30 = Fe/Fg = qE/mg
tan30 = q(2500 N/C)/(0.010 kg)(10 m/s2)  q = -2.3 x 10-5 C
64. a.
Fe = kQ1Q2/r2
Fe = (9 x 109)(32 x 10-6)(8 x 10-6)/(2)2 = 0.576 N
b.
The same as the +8 C charge; 0.576 N
c.
The 32 C generates the stronger field.
d.
E32 = kQ/r2 = (9 x 109)(32 x 10-6)/(1)2 = 288,000 N/C 
E8 = kQ/r2 = (9 x 109)(8 x 10-6)/(1)2 = 72,000 N/C 
Etotal = 288,000 N/C – 72,000 N/C = 216,000 N/C 
e.
Fe = qE = (1 x 10-6 C)(216,000 N/C) = 0.216 N 
f.
E32 = E8
kQ32/x2 = kQ8/(2 – x)2 x = 4/3 m
65. a.
Fe = qE = (10 x 10-6 C)(30,000 N/C) = 0.3 N
b.
F = ma
a = Fe/m = 0.3 N/0.005 kg = 60 m/s2
c.
v2 = 2ad = 2(60 m/s2)(2 m) = 240  v = 15.5 m/s
66. a.
Fe = qE = -ma
(1.6 x 10-19 C)(3 x 105 N/C) = -(1.67 x 10-27 kg)a
a = -2.9 x 1013 m/s2
b.
vt2 = vo2 + 2ad
0 = (3 x 106 m/s)2 + 2(-2.9 x 1013 m/s2)d  d = 0.16 m
c.
vt = vo + at
0 = (3 x 106 m/s) + (-2.9 x 1013 m/s2)t  t = 1 x 10-7 s
67. a.
E1 = E2 
kQ1/x2 = kQ2/(6 m – x)2 x = 4.5 m
b.
E1 = E2 
kQ1/x2 = kQ2/(x – 6 m)2 x = 9 m
68. a. B
b. A
c. D
d. C
Practice Multiple Choice
1.
A
Fe attracts (unlike charges) or repels (like charges),
but Fg only attracts.
2.
D
Charge is on the surface, E = 0, without contact no
charge, two spheres share charge until equal potential
|/r2
Fe = k|Q1Q2
If r is ½  (½)2 = ¼ in the denominator, makes Fe 4 x.
4.
D
GMm/r2
kQq/r2
r2
Fg =
and Fe =
have in the
denominator, but spring force does not (Fs = kx).
5.
B
When the charges touch they share the total charge
in equally (they are the same size): -2 + -4 = -6/2 = -3.
6.
A
Fe = k|Q1Q2|/r2
Fe= (9 x 109)(1 x 10-6)2/(0.3)2 = 0.1 N
7.
C
E is away from positive charges (EA  and EB ). The
horizontal components cancel  .
8.
B
Electric field, E, is toward negative charges  .
All action and reaction forces are equal and in the
opposite direction (Newton's third law).
A
C
D
D
D
13.
D
Fe = |q|E
8.0 x 10-15 N = (1.6 x 10-19 C)E  E = 5 x 104 N/C
14.
C
The electric field is strongest closest to the surface,
but outside the sphere (E inside the sphere = 0).
15.
B
E = k|Q|/r2 Since B is twice as far from +Q as A, the
electric field at B is ¼ as strong as it is at A.
The electric field is away from the positive charge 
and toward the negative charge   .
21.
A
E = k|Q|/r2
E = (9 x 109 N•m2/C2)(5 x 10-6 C)/(0.5 m)2 = 1.8 x 105 N/C
22.
B
Fe = |q|E
Fe = (1 x 10-6 C)(1.8 x 105 N/C) = 1.8 x 10-1 N = 0.18 N
23.
E = kQA/rA2 = kQB/rB2
4/x2 = 1/(x – 6)2 x = 12
24.
C
Fe = kQAQB/r2 = (9 x 109)(1 x 10-6)2/(0.3)2= 0.1 N
Fe = ma  a = Fe/m = 0.1 N/⅓ kg = 0.3 m/s2
25.
D
26.
B
F1 = k|Q1Q2|/r2 = k(+2Q)(1Q)/d2 = 2kQ2/d2
touch: Q1 = Q2 = ½Q  F2 = k|½Q|2/d2 = ¼kQ2/d2 (1/8)
E = kQ/d2  Etot = [(kQ/d2)2 + (kQ/d2)]½ = 2(kQ/d2)
Fe = qE = q(2kQ/d2) = 2kqQ/d2
27.
A
E1 = k|Q1|/r12 = (9 x 109)(-16 x 10-6 C)/4 m2 = 9 x 103 N/C
E2 = k|Q2|/r22 = (9 x 109)(+9 x 10-6 C)/3 m2 = 9 x 103 N/C
28.
B
Etot = (E12 + E22)½ = (2E12)½ = 2(E1)
Etot = 2(9 x 103 N/C) = 1.4 x 9 x 103 = 13,000 N/C
29.
F1 is  (like charges repel)
F2 is  (opposite charges attract)  
30.
A
Fe = |q|E
Fe = (4 x 10-6 C)(1.3 x 104 N/C) = 5.2 x 10-2 N
Practice Free Response
1.
Fe = |q|E
Fe = (1.6 x 10-19 C)(3 x 103 N/C) = 4.8 x 10-16 N
Electric forces exist only between objects that have a
net charge  no charge, no force.
With opposite charges, E is zero outside the two
charges and closest to the smaller charge.
20.
12.
A
Fe = |q|E: qe = qp  Fe is the same, but opposite
directions. v and d are different (ap  ae).
19.
11.
A
E = k|Q|/r2
E = (9 x 109 N•m2/C2)(4 x 10-6 C)/(2m)2 = 9 x 103 N/C
18.
10.
Fe on –q is opposite E. Since the force is up the field
must be down; toward plate B.
The electric field inside a conductor, whether
charged, uncharged, hollow or solid, is always zero.
17.
C
9.
C
A
D
3.
D
16.
2.
a.
tan = Fe/Fg  tan37 = Fe/mg
0.75 = Fe/(0.04 kg)(10 m/s2)  Fe = 0.30 N
b.
Fe = qE
0.30 N = (30 x 10-6 C)E  E = 10,000 N/C 
a.
Fe = k|QAQB|/r2
Fe = (9 x 109)(9 x 10-6)(1 x 10-6)/(3)2 = 0.009 N
b.
Fe = ma
0.009 N = (0.01 kg)a  a = 0.9 m/s2 
c.
EA = k|Q|/r2 = (9 x 109)(9 x 10-6)/(1.5)2 = 36,000 N/C 
EB = k|Q|/r2 = (9 x 109)(1 x 10-6)/(1.5)2 = 4,000 N/C 
Etot = 36,000 N/C + 4,000 N/C = 40,000 N/C 
d.
EA = EB to the right of B
k|QA|/x2 = k|QB|/(x – 3)2 x = 4.5 m
e.
Fe = |qCEtotal| = (6 x 10-6 C)(40,000 N/C) = 0.24 N 
3.
E1 = E2 = kQ1/r12
E = (9 x 109 N•m2/C2)(5 x 10-6 C)/(5 m)2 = 1800 N/C
E1x + E2x = 0
E1y = E2y = 2(1800N/C)sin37 = 2200 N/C (in -y direction)
4.
For the adjacent charges:
E = kQ1/r2 = (9 x 109 N•m2/C2)(40 x 10-6 C)/(2 m)2
E = 9 x 104 N/C  Enet = (2(9 x 104)2)½ = 1.27 x 105 N/C
For the opposite charge:
E = kQ/r2 = (9 x 109 N•m2/C2)(40 x 10-6 C)/(8½ m)2
E = 4.5 x 104 N/C
Enet = 1.27 x 105 + 4.5 x 104 = 1.72 x 105 N/C outward
Fnet = qE = (40 x 10-6 C)(1.72 x 105 N/C ) = 6.9 N outward
5. a.
Positive. The electric field due to q1 is to the right
(toward a negative charge). The electric field due to q2
must be to the left in order to have a combined field of
zero. The field due to a positive charge is to the left.
b.
E1 + E2 = 0  kq1/r12 = -kq2/r22
q2 = -q1(r2/r1)2 = 3.0 x 10-9 C(0.40 m/0.10 m)2
q2 = 4.8 x 10-8 C
c.
Fe = kq1q2/r2
Fe = (9 x 109 N•m2/C2)(3 x 10-9 C)(4.8 x 10-8 C)/(.3 m)2
Fe = 1.4 x 10-5 N 