16.
(a) The potential difference across C
1
is V
1
= 10 V. Thus, q
1
= C
1
V
1
= (10 µF)(10 V) = 1.0
× 10
−4
C.
(b) Let C = 10 µF. We first consider the three-capacitor combination consisting of C
2
and its two
closest neighbors, each of capacitance C. The equivalent capacitance of this combination is
C
eq
= C +
C
2
C
C + C
2
= 1.5C .
Also, the voltage drop across this combination is
V =
CV
1
C + C
eq
=
CV
1
C + 1.5C
=
2
5
V
1
.
Since this voltage difference is divided equally between C
2
and the one connected in series with it,
the voltage difference across C
2
satisfies V
2
= V /2 = V
1
/5. Thus
q
2
= C
2
V
2
= (10 µF)
10 V
5
= 2.0
× 10
−5
V .