P26_016

# P26_016 - 16(a The potential diﬀerence across C1 is V1 =...

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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 three-capacitor 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.

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