11.3 Solutions
527
The capacitance is given by
C =
dC =
a
0
ε 0 a dx
D + x θ
= ε 0 a
a
0
dx
D + x θ
=
ε 0 a
D
a
0
1 +
x θ
D
−1
dx
=
ε 0 a
D
a
0
1 −
xθ
D
+ · · ·
dx
ε 0 a
D
x −
x 2 θ
2D
a
0
=
ε 0 a 2
D
1 −
aθ
2D
Note that for θ = 0, capacitance reduces to that for the parallel plate capacitor.
11.90 (a) The equivalent capacitance of C 1 and C 2 in parallel is C 12 = 8 + 4 =
12 μF.
The combined capacitance of C 12 and C 3 in series is C =
C 3 C 12
C 3 + C 12
=
3 × 12
3 + 12
= 2.4 μF.
Applied charge q = C V = 2.4 × 100 = 240 μC. Therefore charge on
C 3 will be q 3 = 240 μC. PD across C 3 will be V 3 =
q 3
C 3
=
240
3
= 80 V.
The PD across C 1 and C 2 will be equal.
V 1 = V 2 = (V − V 3 ) = (100 − 80) = 20 V
(b) Now
q 1
C 1
=
q 2
C 2
(∵ V 1 = V 2 )
∴
q 1 =
C 1 q 2
C 2
=
8
4
q 2 = 2q 2
Also q 1 + q 2 = 240
∴
q 1 = 160 μC and q 2 = 80 μC
(c) U 1 =
1
2
C 1 V
2
1 =
1
2
× 8 × 10
−6
× 20
2
= 0.0016 J
U 2 =
1
2
C 2 V
2
2 =
1
2
× 4 × 10
−6
× 20
2
= 0.0008 J