Homework #9 solutions.
c)
2
1
1
2
CV2 =
0.14 × 10−6 F (22.9 V)
2
2
−5
= 3.66 × 10 J
U=
V
V
= 3 × 106 m
.
d) Dielectric breakdown of air occurs at about 3000 mm
Vmax = Emax d
6 V
5 × 10−5 m 0.14 × 10−6 F
Qmax = Vmax C = Emax d C = 3 × 10
m
−4
= 2.10 × 10 C
Prob. 9-4. Tipler 25-52.
Solution
The first capacitor has a charge of
Q1 = C1 V1 = (2.0 µF) (12 V) = 24.0 µC
When the two capacitors are connected in parallel, the total charge on the equivalent capacitor is just 24.0 µC. Then
Qeq
24.0 µC
Ceq =
=
Veq
4.0 V
= 6.0 µF
Ceq = C1 + C2
C2 = Ceq − C1 = 6.0 µF − 2.0 µF = 4.0 µF
Prob. 9-5. Tipler 25-54.
Solution
a) The capacitors are originally in series. The equivalent capacitance of the circuit is Ceq =
3 µF. The charge on each capacitor is the charge on the equivalent capacitor.
Qeq = Ceq V = (3 µF) (12 V)
= 36 µC = Q1 = Q2
The capacitors are disconnected and reassembled in parallel. The equivalent capacitance is
now C1 + C2 = 16 µF; the total charge on the equivalent capacitance is Q1 + Q2 = 72 µC.
In parallel, both capacitors have the same potential.
Qeq
72 µC
= 4.5 V
=
V1 = V2 =
Ceq
16 µF
b) In the original series circuit,
1
1
2
Ui = Ceq V 2 = (3 µF) (12 V) = 2.16 × 10−4 J
2
2
After reassembly into the parallel circuit, the stored energy is
1
1
2
Uf = Ceq V 2 = (16 µF) (4.5 V) = 1.62 × 10−4 J
2
2
Note that during reassembly, 2.16 × 10−4 J − 1.62 × 10−4 J = 5.4 × 10−5 J has been lost. In
a real circuit, most of this energy would be converted to heat in the wires.