518
11 Electrostatics
11.69 C 1 + C 2 = 9
(parallel)
(1)
C 1 C 2
C 1 + C 2
= 2
(series)
(2)
(C 1 − C 2 )
2
= (C 1 + C 2 )
2
− 4C 1 C 2 = (C 1 + C 2 )
2
− 8(C 1 + C 2 ) = 9
where we have used (1)
∴
C 1 − C 2 = 3
( 3 )
Solving (1) and (3), C 1 = 6 μF and C 2 = 3 μF.
11.70 (a) U 1 =
1
2
C 1 V
2
=
1
2
× 2 × 10
−6
× (100)
2
= 0.01 J
U 2 =
1
2
C 2 V
2
=
1
2
× 4 × 10
−6
× (100)
2
= 0.02 J
(b) C =
C 1 C 2
C 1 + C 2
=
2 × 10 −6 × 4 × 10 −6
(2 + 4) × 10 −6
=
4
3
× 10
−6 F
U =
1
2
C V
2
=
1
2
×
4
3
× 10
−6
× (100)
2
= 0.0067 F
(c) C = C 1 + C 2 = (2 + 4) × 10
−6
= 6 × 10
−6 F
U =
1
2
C V
2
=
1
2
× 6 × 10
−6
× (100)
2
= 0.03 J
11.71 C 0 =
ε 0 A
d
=
8.85 × 10 −12 × 1
0.01
= 8.85 × 10
−10 F
U 0 =
1
2
Q 2
C 0
=
1
2
×
(10 −6 ) 2
8.85 × 10 −10 = 5.65 × 10
−4 J
U =
1
2
Q 2
C
=
1
2
Q 2
C 0 K
=
U 0
K
=
5.65 × 10 −4
2
= 2.83 × 10
−4 J
The energy is decreased by U = U 0 − U = (5.65 − 2.83) × 10 −4 =
2.82 × 10 −4 J, that is, by a factor of 2.
11.72 C =
ε 0 K A
d
∴ K =
Cd
ε 0 A
=
0.1 × 10 −6 × 0.001
8.85 × 10 −12 × 1.0
= 11.3