757_Physics ProblemsTechnical Physics

# 757_Physics ProblemsTechnical Physics - Chapter 26 P26.61(a...

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Chapter 26 97 P26.61 (a) C A d 1 10 2 = κ ; C A d 2 20 2 2 = ; C A d 3 30 2 2 = 11 2 23 1 0 1 1 012 3 CC A d A d + F H G I K J = + = + F H G I K J =+ + F H G I K J = + + F H G I K J κκ FIG. P26.61 (b) Using the given values we find: C total F p F = 176 10 176 12 .. . *P26.62 The initial charge on the larger capacitor is QCV == = 10 150 F15 V C µµ af . An additional charge q is pushed through the 50-V battery, giving the smaller capacitor charge q and the larger charge 150 C µ + q . Then 50 5 150 10 V F C F + qq 500 2 150 117 C C C + = q So across the 5- F capacitor V q C = 117 23 3 C 5 F V Across the 10- F capacitor V = + = 150 117 26 7 C 10 F V P26.63 (a) Put charge Q on the sphere of radius a and – Q on the other sphere. Relative to V = 0 at infinity, the potential at the surface of
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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