Phys 203A
Homework 1 Solutions
3. Charge flows until the potential difference across the capacitor is the same as the potential
difference across the battery. The charge on the capacitor is then q C V , and this is the same
as the total charge that has passed through the battery. Thus, q 25 10 6 F 120V 3.0 10 3 C.
11. The equivalent capacitance is given by Ceq q V , where q is the total charge on all the
capacitors and V is the potential difference across each of them. For N identical capacitors in
parallel, Ceq NC, where C is the capacitance of one of them. Thus,
q
N
1.00C
110 V 1.00 10
V C
6
9090.
F
19. (a) After the switches are closed, the potential differences across the capacitors are the same
and the two capacitors are in parallel. The potential difference from a to b is given by
Vab Q Ceq , where Q is the net charge on the combination and Ceq C1 C2 4.0 F is the
equivalent capacitance. The net charge on the combination can be found by considering the
initial charges on the two capacitors. The initial charge on capacitor 1 is
q1
C1 V
1.0 F 100V
100 C
3.0 F 100V
300 C.
and the initial charge on capacitor 2 is
q2
C2
V
The two capacitors are connected in opposite orientations, so the net charge on the combination
is Q q2 q1 200 C. The potential difference is
200 C
4.0 F
Vab
50 V.
(b) The charge on capacitor 1 is now q1
C1 Vab
1.0 F 50 V
50 C.
(c) The charge on capacitor 2 is now q2
C2
3.0 F 50 V
150 C.
Vab
27. (a) In the first case U1 q2 2C , and in the second case
U2
2
q 2
2C
2
q2
4C
1
2
U1 .
So the energy after the connection is half of what it was. U 2 2.0 J.
(b) This question is addressed in Touchstone Example 28-2. We can not simply ascribe this
missing energy to heating losses P i 2 R due to the resistance of the wires. The losses are due
to electromagnetic radiation
29. (a) Let q be the magnitude of the charge on the plates. Since the capacitance of a parallelplate capacitor is given by C 0 A d , the charge is q C V
0 A V d . With the circuit
disconnected, this charge will not change when the plates are pulled apart. When the separation
between the plates becomes 2d , the potential difference will be V , where
q
V
2d
q
0A
C
2 V.
(b) The initial energy stored in the capacitor is
1
2
Ui
C
V
2
0
A
V
2
,
2d
and the final energy stored is
Uf
1 0A
( V ')2
2 2d
1 0A
4
2 2d
V
2
0
A
V
d
2
.
This is twice the initial energy.
(c) The work done to pull the plates apart is the difference in the energy:
W ext
Uf
Ui
0
A
V
2d
2
.
37. The capacitance of a cylindrical capacitor with a dielectric in the gap is given by
C
Cair
2 0L
,
ln(b / a)
where is the dielectric constant, L is the length, a is the inner radius, and b is the outer radius.
The capacitance per unit length of the cable is
C
L
2
0
ln(b / a)
2
pF m
ln 0.60 mm 0.10 mm
81 pF m.
47. During charging, the charge on the positive plate of the capacitor is given by
q C VB 1 e
t
,
where C is the capacitance, VB is the potential difference of the charging battery, and
the capacitive time constant. We require q 0.990C VB , so
t
0.990 1 e
t
e
RC is
0.010.
Taking the natural logarithm of both sides, we obtain t
ln 0.010
4.61 . .
49. (a) The potential difference across the capacitor varies as a function of time t as
V
where
VB 1 e
,
VB is the potential difference across the battery. Solving for
t
ln
(b) C
t
R
2.41 10
6
s
15.0 103
V
1.30 s
ln 1 125
VB
1.61 10
10
yields
2.41 s.
F.
51. (a) The potential difference across the capacitor varies as a function of time t as
V
V0 e
t
,
where V0 100V is the initial potential difference across the capacitor. Then, using the data
provided in the problem,
t
ln
(b) At t = 17.0 s, t
17.0 s
2.17 s
V
V
10.0 s
ln 0.0100
V0
2.17 s.
7.83, so
V0 e
t
100V e
7.83
53. (a) The energy stored in the capacitor is given by
U
q02
,
2C
3.96 10 2 V .
where C is the capacitance and q0 is the initial charge on the capacitor. Thus
q0
2 1.0 10 6 F 0.50 J
2CU
1.0 10 3 C .
(b) The magnitude of the initial current is found by evaluating Eq. 28-46 at t 0 :
q0
RC
i
1.0 10 3 C
1.0 106
1.0 10 6 F
(c) The time constant for this circuit is
difference across the capacitor is
V0
q0 C
1.0 106
RC
1.0 10
1.0 10 3 A.
3
C
1.0 10
1.0 10
6
F
6
F
1.0 s. The initial potential
1000 V.
The potential difference across the capacitor at a time t is
VC
V0 e
t
1000 V exp
t 1.0 s .
If the capacitor is being discharged through the resistor, they are connected together in a circuit.
The sum of potential changes around the circuit must be zero. Therefore, the potential difference
across the resistor at a time t is
VR
VC
1000 V exp
t 1.0 s .
(d) The resistor produces thermal energy at a rate of
P
VR
R
2
1000 V
1.0 106
2
exp
t 0.5 s
1.0 W exp
t 0.5 s .