748_Physics ProblemsTechnical Physics

748_Physics ProblemsTechnical Physics - 88 Capacitance and...

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88 Capacitance and Dielectrics P26.33 UC V = 1 2 2 af The circuit diagram is shown at the right. (a) CCC p =+= + = 12 25 0 5 00 30 0 .. . F F µ U = 1 2 30 0 10 100 0 150 6 2 ej a f J (b) C CC s =+ F H G I K J F H G I K J = 11 1 25 0 1 500 417 1 1 . F F µµ V V U C = == × = 1 2 2 20150 417 10 268 2 6 . . V FIG. P26.33 P26.34 Use U Q C = 1 2 2 and C A d = 0 . If dd 21 2 = , 1 2 = . Therefore, the stored energy doubles . *P26.35 (a) QCV × × −− 150 10 10 10 1 50 10 12 3 6 V C e j . (b) V = 1 2 2 a f V U C × × 2 2250 10 150 10 183 10 6 12 3 J F V . P26.36 u U V E ==∈ 1 2 0 2 100 10 1 2 885 10
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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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