Unformatted text preview: 16. (a) The potential diﬀerence across C1 is V1 = 10 V. Thus, q1 = C1 V1 = (10 µF)(10 V) = 1.0 × 10−4 C.
(b) Let C = 10 µF. We ﬁrst consider the threecapacitor combination consisting of C2 and its two
closest neighbors, each of capacitance C . The equivalent capacitance of this combination is
Ceq = C + C2 C
= 1. 5 C .
C + C2 Also, the voltage drop across this combination is
V= CV1
CV1
2
=
= V1 .
C + Ceq
C + 1. 5 C
5 Since this voltage diﬀerence is divided equally between C2 and the one connected in series with it,
the voltage diﬀerence across C2 satisﬁes V2 = V /2 = V1 /5. Thus
q2 = C2 V2 = (10 µF) 10 V
5 = 2.0 × 10−5 V . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Capacitance

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