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
4.
When the rod is removed, the charges return to normal
(equal on both spheres).
a.
conduction
induction
b.
Place a grounding wire on the right side.
– 1 C of excess charge on the piece of silk.
6.
positively charged
conduction
negatively charged
induction
Like charges repel, which for B is away from A.
Fe = kQAQB/r2; since QB is doubled, while all else is the
same, then Fe is 2 x 1 N = 2 N.
The force is the same for both spheres (Newton's third
law).
kQAQB/r2;
Fe =
since QA and QB are both doubled, while
all else is the same, then Fe is 2 x 2 x 1 N = 4 N.
2
Fe = kQAQB/r ; since r is doubled, while all else is the
same, then Fe is 1/22 x 1 N = ¼ N.
Fe = kQAQB/r2; since r is halved, while all else is the
same, then Fe is 1/½2 x 1 N = 4 N.
The electric force is the same for both the electron and
proton (Newton's third law).
Opposite charges attract.
kQAQB/r2;
Fe =
increase.
since r is decreasing the force will
F = ma; since the electron is lighter it will have greater
acceleration.
Since the electron has greater acceleration, it will
travel a greater distance before collision.
18.
B
Sphere A is attracted to the wall and opposite charges
attract  the wall must be negative.
19.
B
F3 = Fg = mg = (0.4 kg)(10 m/s2) = 4 N
20.
C
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.
B
The electric field equals zero closest to |smaller|
charge and between when charges have same sign.
A
The electric field equals zero closest to |smaller|
charge and outside when charges have different signs.
D
The electric field is away from the positive charges.
B
The electric field is away from the positive charge and
toward the negative charge.
C
The electric field is toward the negative charges.
A
The electric field is away from the positive charges and
toward the negative charges.
D
The electric fields cancel each other out.
D
The electric fields cancel each other out.
38. a.
17.
A
D
37.
16.
A
The electric field is toward -Q () and away from +Q
( )   .
36.
15.
A
A
35.
14.
B
27.
34.
13.
B
The electric field is away from +Q () and toward –Q
( )   .
33.
12.
D
B
The two electric fields cancel each other out.
32.
11.
A
C
31.
10.
D
25.
30.
9.
C
e = kQ/r2; since r is doubled, while all else is the same,
then E is 1/22 x E= ¼E..
29.
8.
C
A
28.
7.
A
E = kQ/r2; since Q is doubled, while all else is the
same, then E is 2 x E = 2E.
24.
26.
5.
The direction of the electric field is away from the
positive charge.
23.
C
3.
F3 = F1sin53 = 3 N
22.
A
2.
B
21.
F1 = F3/cos53 = 5 N (This is a 3-4-5 triangle)
Fe = kQ1Q2/r2
Fe = (9.0 x 109 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
39. a.
Fe = kQ1Q2/r2 = (9 x 109)(10 x 10-6)(8 x 10-6)/(0.4)2
Fe = 4.5 N (attraction)
b.
(-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)
40. a.
Fg
52. 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  x = 20 m
53.
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
54. a.
Fe = qE = (1 x 10-6 C)(1000 N/C) = 1 x 10-3 N
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)2tan30/(9 x 109)]½
Q = 2.2 x 10-6 C
41.
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
55.
E = kQ/r2
E = (9 x 109 N•m2/C2)(10 x 10-6C)/(5 m)2 = 3600 N/C
The strength of E inside a good conductor is zero.
42.
56.
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
The net charge is located on the surface.
43.
The electric field is perpendicular to the surface.
44. a.      
b. + + + + + + + +
 +
  
–
  

––––––––
45. a.
The 50 C generates the stronger field.
b.
The charges exert equal force.
46.
E = kQ/r2
E = (9.0 x 109 N•m2/ C2)(10 x 10-6 C)/(3 m)2 = 1 x 104 N/C
47.
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 
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)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)
49.
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)
50.
Fe = qE
Fe = (3 x 10-7 C)(500 N/C) = 1.5 x 10-4 N east
51.
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 
57.
Fe = qE = (5 x 10-7 C)(200 N/C) = 1 x 10-4 N west
58. 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
59. 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
60. 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
61.
K = W = Fd; Since the electron and proton are subject
C
to the same force for the same distance, K is the same.
62.
D
Ue = kQ1Q2/r; when charges have the same sign, Ue is
positive.
63.
B
W = Ue; since Ue is positive when the charges have
the same sign and initial Ue = 0, B is positive.
64.
B
V = kQ/r and is scalar  only B has V1 + V2 + V3 + V4 >
0
65.
D
V = kQ/r and is scalar  A and C have V1 + V2 + V3 + V4
= 0 since two V's are positive and two V's are negative.
66.
C
Since E is vector, the position of +Q and –Q has to
balance for E = 0  only C has V = 0 and E = 0.
67.
B
W = F||d = qEd: For a positive charge positive work is
done when d is opposite E.
68.
B
W = F||d = qEd: For a negative charge negative work is
done when d is opposite the opposite of E.
69.
A
Ue = qV = (-1.6 x 10-19 C)(750 J/C) = -1.2 x 10-16 J
W = F||d = 0 along path C because it is  to F.
70.
A
77. a.
V-5 = kQ-5/r = (9 x 109)(-5 x 10-6)/1 = -4.5 x 104 V
V+10 = kQ+10/r = (9 x 109)(10 x 10-6)/3 = 3 x 104 V
Vtot = V-5 + V+10 = -4.5 x 104 + 3 x 104 = -1.5 x 104 V
b. (1)
W = qV
W = (-1.6 x 10-19 C)(-1.5 x 104 V) = 2.4 x 10-15 J
(2)
W = qV = (-1e)V
W = (-1e)(-1.5 x 104 V) = 1.5 x 104 eV
c.
V-5 = kQ-5/r = (9 x 109)(-5 x 10-6)/1 = -4.5 x 104 V
V+10 = kQ+10/r = (9 x 109)(10 x 10-6)/3 = 3 x 104 V
Vtot = V-5 + V+10 = -4.5 x 104 + 3 x 104 = -1.5 x 104 V
78. a.
V40 = kQ2/r2 = (9 x 109)(40 x 10-6)/5 = 7.2 x 104 V
V-10 = kQ1/r1 = (9 x 109)(-10 x 10-6)/5 = -1.8 x 104 V
Vtot = 7.2 x 104 V + (-1.8 x 104 V) = 5.4 x 104 V
b.
V40 + V-10 = 0  kQ40/r = -kQ-10/(10 – r)
40/x = 10/(10 – x) x = 8 m
79. a.
A is on the same equipotential line as X. B and C are on
corresponding lines, but around –Q.
71.
Ue = kqQ/r
Ue = (9.0 x 109)(-1 x 10-6)(10)/25 = -3600 J
72. a.
Ue = kqQ/r
Ue = 0 J
b.
Ue = kqQ/r
Ue = (9.0 x 109)(+2 x 10-6)(+20 x 10-6)/2 = 0.18 J
c.
W = U2 m – U
W = 0.18 J – 0 J = 0.18 J
73. a.
Ue = kqQ/r
Ue = (9.0 x 109)(10 x 10-6)(-1)/10 = -9000 J
b.
Ue = kqQ/r
Ue = (9.0 x 109)(10 x 10-6)(-1)/2 = -45000 J
c.
-Ue = K = ½mv2
-(-45000 J – -9000 J) = ½(0.2 kg)v2  v = 600 m/s
74.
V = W/Q
V = 10 J/5 C = 2 V
75.
V = kQ/r
V = (9.0 x 109 N•m2/C2)(10 x 10-6 C)/0.25 m = 3.6 x 105 V
76. a.
r (m)
E
V
1 (9x109)(1x10-6)/12 = 9000 (9x109)(1x10-6)/1 = 9000
2
(9x109)(1x10-6)/22 = 2250
(9x109)(1x10-6)/2 = 4500
3
(9x109)(1x10-6)/32 = 1000
(9x109)(1x10-6)/3 = 3000
(9x109)(1x10-6)/42
= 563 (9x109)(1x10-6)/4 = 2250
4
c.
The slope of the V vs. r graph at r = 3 m is equal to the
magnitude of E at 3 m.
b.
-Ue = K = ½mv2
1.2 x 10-16 J = ½(9.11 x 10-31 kg)v2v = 1.6 x 107 m/s
80. a. The electron (gains or loses) potential energy.
b.
Ue = qeV = (-1)(10 V) = -10 eV
81.
the electric field equal to zero
Arrangement 1 only
the voltage equal to zero
Arrangement 1 and 2
82.
3
Has positive potential energy
1, 2
Has negative potential energy
1
Requires the most work to separate
83. a.
V = kQ/r = (9 x 109 N•m2/C2)(10 x 10-6 C)/0.7 m
Vnet = 1.3 x 105 V x 3 = 4 x 105 V
b.
V3 = kQ/r = (9 x 109)(-25 x 10-6)/0.7 = -3.2 x 105 V
Vnet = 2(1.3 x 105 V) + (-3.2 x 105 V) = -6 x 104 V
84. a.
V1 = kQ/r = (9 x 109)(50 x 10-6)/0.30 = 1.5 x 106 V
V2 = kQ/r = (9 x 109)(-50 x 10-6)/0.60 = -7.5 x 105 V
Vtot = 1.5 x 106 V – 7.5 x 105 V = 7.5 x 105 V
b.
V1 = kQ/r = (9 x 109)(50 x 10-6)/0.40 = 1.13 x 106 V
V2 = kQ/r = (9 x 109)(-50 x 10-6)/0.40 = -1.13 x 105 V
Vtot = 1.13 x 106 V – 1.13 x 106 V = 0 V
85. a.
V1 = -V2  kq1/r1 = -kq2/r2
3.0/d = 1.5/(0.30 – d)  d = .20  x = -0.10 + 0.20 = 0.1 m
b.
W = qV = 0 J because V = 0.
86.
Q = CV; since V is doubled, while C remains the same,
C
then Q is doubled (2 x Q).
87.
Uc = ½CV2; since V is doubled, while C remains the
D
same, then Uc is quadupled (4 x Uc).
88.
C
d.
v = dx/t
4.9 x 107 m/s = 0.028 m /t  t = 5.7 x 10-10 s
e.
v = at
v = (5.1 x 1016 m/s2)(5.7 x 10-10 s) = 2.8 x 107 m/s
f.
v = dx/t
t = dx/v = 0.22 m/4.9 x 107 m/s = 4.5 x 10-9 s
g.
v = dy/t
dy = vt = (2.8 x 107 m/s)(4.5 x 10-9 s) = 0.13 m
C = єoA/d; since A is doubled, while d remains the
same, then C is doubled (2 x C).
89.
B
C = єoA/d; since d is doubled, while A remains the
same, then C is halved (½ x C).
90.
B
The voltage is constant because the battery remains
connected.
91.
C
C = єoA/d and Q = CV; since d is halved, then C is
doubled, but V is constant  Q is doubled (2 x Q).
92.
B
Once the battery is disconnected the charge can't
change.
Practice Multiple Choice
1.
93.
C = єoA/d and Q = CV; since A is doubled, then C is
doubled, but Q is constant  V is halved (½ x V).
94. a. Plate (A or B) has the higher electric potential.
b. The dashed lines are (equipotentials or electric field).
c. The solid arrows are (equipotentials or electric field).
d. Each dashed line equals 5 V. The potential difference
between the two plates is (0, 5, 10 or 20) V.
e. The plates are 0.01 m apart. The electric field is
(1000, 2000, 4000 or 8000) V/m.
95. a.
A
Q = CV = (330 x
10-12
F)(12 V) = 3.96 x
10-9
C
b.
Uc = ½QV = ½(3.96 x 10-9 C)(12 V) = 2.38 x 10-8 J
96. a.
Q = CV
C = Q/V = 10 x 10-6 C/12 V = 8.3 x 10-7 F
b.
Uc = ½QV = ½(10 x
10-6
C)(12 V) = 6 x
10-5
1
3.7
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
3.
D
Fe = k|Q1Q2|/r2
If r is ½  (½)2 = ¼ in the denominator, makes Fe 4 x.
4.
D
Fg = GMm/r2 and Fe = kQq/r2 have r2 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
J
97.
K
A
C
C = KєoA/d
C = (1)(8.85 x 10-12)(0.075)/0.005 = 1.3 x 10-10 F
C = (3.7)(8.85 x 10-12)(0.075)/0.005 = 4.9 x 10-10 F
98. a.
Fg = mg
Fg = (2.8 x 10-15 kg)(10 m/s2) = 2.8 x 10-14 N
b.
E = V/d
E = 340 V/0.01 m = 3.4 x 104 V/m
c.
Fe = qE  q = Fe/E = Fg/E
q = 2.8 x 10-14 N/3.4 x 104 V/m = 8.24 x 10-19 C
d.
N = q/e
N = 8.24 x 10-19 C/1.6 x 10-19 C = 5
99. a.
K = Ue = qV
K = (1.6 x 10-19 C)(7.0 x 103 J/C) = 1.1 x 10-15 J
b.
K = ½mv2
1.1 x 10-15 J = ½(9.11 x 10-31 kg)v2  v = 4.9 x 107 m/s
c.
Fe = qE = ma
(1.6 x 10-19 C)(2.9 x 105 N/C) = (9.11 x 10-31 kg)a
a = 5.1 x 1016 m/s2
E is away from positive charges (EA  and EB ). The
horizontal components cancel  .
8.
B
Electric field, E, is toward negative charges  .
9.
C
All action and reaction forces are equal and in the
opposite direction (Newton's third law).
10.
B
Fe on –q is opposite E. Since the force is up the field
must be down; toward plate B.
11.
A
Fe = |q|E
Fe = (1.6 x 10-19 C)(3 x 103 N/C) = 4.8 x 10-16 N
12.
A
Electric forces exist only between objects that have a
net charge  no charge, no force.
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
16.
D
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.
Fe = |q|E: qe = qp  Fe is the same, but opposite
directions. v and d are different (ap  ae).
17.
A
The electric field inside a conductor, whether
charged, uncharged, hollow or solid, is always zero.
18.
A
40.
E = k|Q|/r2
E = (9 x 109 N•m2/C2)(4 x 10-6 C)/(2m)2 = 9 x 103 N/C
19.
D
26.
B
28.
B
E = kQ/d2  Etot = [(kQ/d2)2 + (kQ/d2)]½ = 2(kQ/d2)
Fe = qE = q(2kQ/d2) = 2kqQ/d2
D
C
B
D
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
D
F1 is  (like charges repel)
F2 is  (opposite charges attract)  
Fe = |q|E
Fe = (4 x 10-6 C)(1.3 x 104 N/C) = 5.2 x 10-2 N
2.
W = Ue = qV
W = (4 x 10-6 C)(36 V) = 144 x 10-6 J = 1.44 x 10-4 J
33.
C
V = kQ/r = (9 x 109 N•m2/C2)(-5 x 10-6 C)/(0.5 m)
V = (9 x 109)(-5 x 10-6)/(5 x 10-1)  V = -9 x 104 V
34.
D
Ue = qV = (1 x 10-6 C)(-9 x 104 J/C) = -9 x 10-2 J = -0.09 J
35.
D
E is constant  V/d is constant. If point P is ¼ of the
distance between 2 V and 10 V, then it must be at 4 V.
36.
D
E = V/d = 8 V/0.04 m = 200 V/m
37.
B
Disconnected: Q' = Q. Doubling d: C' = ½C.
Q' = C'V'  V' = 2V
38.
C
Connected: V' = V. Doubling A: C' = 2C
Q' = C'V'  Q' = 2Q
39.
A
Ue = kQAQB/r
Ue = (9 x 109)(1 x 10-6)(1 x 10-6)/0.3 = 0.03 J
0 = mAvA + mBvB  vA = vB
Ue = KA + KB = mv2  v = (UE/m)½ = (0.03//⅓)½ = 0.3 m/s
K = qV: Q' = 2Q and V' = V  K' = 2K
K = ½mv2: K' = 2K and m' = 2m  v' = v
Connected: V' = V. Doubling A and d: C' = C
Q' = C'V'  Q' = Q
Vtot = kQ/d + kQ/d = 2kQ/d
Ue = qV = q(2kQ/d) = 2kqQ/d
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 
c.
V = Ed = (10,000 N/C)(0.2 m) = 2,000 V
W = Ue = qV 60 J = (5 C)V  V = 60 J/5 C = 12 V
32.
B
VA = VB = kQA/rA = (9 x 109)(1 x 10-6 C)/0.15 m = 6 x 104 V
Vtot = VA + VB = 2VA = 2(6 x 104 V) = 12 x 104 V
Practice Free Response
1.
31.
C
With like charges, V is zero only at an infinite
distance away from the two charges.
48.
30.
A
E = V/d, if V' is 2V, then E' is 2E and if d' = 1/5d then E'
is 5E. 5E x 2E = 10E = 10(2,000 N/C) = 20,000 N/C
49.
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
29.
C
D
47.
F1 = k|Q1Q2|/r2 = k(+2Q)(1Q)/d2 = 2kQ2/d2
touch: Q1 = Q2 = ½Q  F2 = k|½Q|2/d2 = ¼kQ2/d2 (1/8)
27.
A
D
46.
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
C =єoA/d  Increasing A increase C. Increasing d
decreases C capacitance. Increasing V, Q: no effect.
45.
E = kQA/rA2 = kQB/rB2
4/x2 = 1/(x – 6)2 x = 12
24.
B
C
44.
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.
D
E = kQA/rA = kQB/rB
4/x = 1/(6 – x) x = 4.8 m
43.
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
B
42.
The electric field is away from the positive charge 
and toward the negative charge   .
21.
A
Disconnected: Q' = Q. Doubling A, halving d: C' = 4C
Q' = C'V'  V' = ¼V.
41.
With opposite charges, E is zero outside the two
charges and closest to the smaller charge.
20.
D
D
a.
Fe = k|QAQB|/r2
Fe = (9 x 109)(9 x 10-6)(1 x 10-6)/(3)2 = 0.009 N
b.
Ue = kQAQB/r = (9 x 109)( 9 x 10-6)(-1 x 10-6)/3 = -.027 J
c.
Fe = ma
0.009 N = (0.01 kg)a  a = 0.9 m/s2 
d.
Ue = kQAQB/r = (9 x 109)(+9 x 10-6)(-1 x 10-6)/1 = -.081 J
e.
-Ue = K = -(-0.081 J – -0.027 J) = 0.054 J
f.
K = ½mvB2
0.054 J = ½(0.01 kg)v2  v = 3.3 m/s
g.
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 
h.
VA = kQA/rA = (9 x 109)(9 x 10-6)/1.5 = 54,000 V
VB = kQB/rB = (9 x 109)(-1 x 10-6)/1.5 = -6,000 V
Vtot = 54,000 V + (-6,000 V) = 48,000 V
i.
EA = EB to the right of B
k|QA|/x2 = k|QB|/(x – 3)2 x = 4.5 m
j.
VA + VB = 0 between the two spheres
kQA/x = -kQB/(3 – x) x = 2.7 m
k.
Fe = |qCEtotal| = (6 x 10-6 C)(40,000 N/C) = 0.24 N 
l.
Ue = qCVtotal = (6 x 10-6 C)( 48,000 J/C) = 0.288 J
m.
When C is infinitely far away, Ue is converted to K.
-Ue = K = ½mv2
-(0 – 0.288 J) = ½(0.004)v2  v = 12 m/s
3.
4.
5.
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)
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 
a.
C = єoA/d = (8.85 x 10-12 C2/N•m2)(0.2 m2)/0.01 m
C = 1.77 x 10-10 F
b.
Q = CV = (1.77 x 10-10 F)(50 V) = 8.85 x 10-9 C
c.
Uc = ½QV = ½(8.85 x 10-9)(50 V) = 2.2 x 10-7 J
6.
d.
d' = 2d  C' = ½C = ½(1.77 x 10-10 F) = 8.85 x 10-11 F
Q' = C'V  Q' = ½Q = ½(8.85 x 10-9 C) = 4.43 x 10-9 C
e.
A' = 2A  C' = 2C = 2(1.77 x 10-10 F) = 3.54 x 10-10 F
Q = C'V'  V' = ½V = ½(50 V) = 25 V
a.
½mv2 = -qV
½(9.11 x 10-31 kg)(6.0 x 107 m/s)2 = (1.6 x 10-19 C)V
V = 10,000 V
b.
E = V/d = 600 V/0.012 m = 5 x 104 V/m (N/C)
c.
|qE| = ma
(1.6 x 10-19 C)(5 x 104 N/C) = (9.11 x 10-31 kg)a
a = 8.8 x 1015 m/s2
d.
v = d/t
t = d/v = (0.04 m)/(6.0 x 107 m/s) = 6.7 x 10-10 s
e.
vt = vo + at
vt = 0 + (8.8 x 1015 m/s2)(6.7 x 10-10 s) = 5.9 x 106 m/s
f.
vy/vx = dy/dx
(5.9 x 106 m/s)/(6.0 x 107 m/s) = dy/0.5 m  dy = 0.05 m