P26_054.PDF

54. (a) The potential across capacitor 1 is 10 V, so the charge on it is

q 1 = C 1 V 1 =(10 µ F)(10 V) = 100 µ C .

(b) Reducing the right portion of the circuit produces an equivalence equal to 6 . 0 µ F, with 10 V across

it. Thus, a charge of 60 µ C is on it – and consequently also on the bottom right capacitor. The

bottom right capacitor has, as a result, a potential across it equal to

C = 60 µ C

10 µ F

=6 . 0V ,

6=4 . 0 V across the group of capacitors in the upper right portion of the circuit.

Inspection of the arrangement (and capacitance values) of that group reveals that this 4 . 0Vmust

be equally divided by C 2 and the capacitor directly below it (in series with it). Therefore, with

2 . 0 V across capacitor 2, we ﬁnd

−

q 2 = C 2 V 2 =(10 µ F)(2 . 0V)=20 µ C .

V = q

which leaves 10