14.
(a) and (b) The original potential difference V
1
across C
1
is
V
1
=
C
eq
V
C
1
+ C
2
=
(3.16 µF)(100 V)
10.0 µF + 5.00 µF
= 21.1 V .
Thus ∆V
1
= 100 V
− 21.1 V = 79 V and ∆q
1
= C
1
∆V
1
= (10.0 µF)(79 V) = 7.9
× 10
−4
C.
14.
(a) and (b) The original potential difference V
1
across C
1
is
V
1
=
C
eq
V
C
1
+ C
2
=
(3.16 µF)(100 V)
10.0 µF + 5.00 µF
= 21.1 V .
Thus ∆V
1
= 100 V
− 21.1 V = 79 V and ∆q
1
= C
1
∆V
1
= (10.0 µF)(79 V) = 7.9
× 10
−4
C.