740_Physics ProblemsTechnical Physics

740_Physics - 80 Capacitance and Dielectrics SOLUTIONS TO PROBLEMS Section 26.1 Definition of Capacitance e ja f e ja f Q = CV = 4.00 10 6 F 12.0 V

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80 Capacitance and Dielectrics SOLUTIONS TO PROBLEMS Section 26.1 Definition of Capacitance P26.1 (a) QCV ==× = × = −− 4 0 01 0 1 2 0 4 8 0 4 8 0 65 .. . . F V C C ej af µ (b) = × = 4 0 0 1 5 0 6 0 0 6 0 0 66 . . V C C P26.2 (a) C Q V == × = 10 0 10 100 10 100 6 6 . C 10.0 V F (b) V Q C × × = 100 10 10 100 6 6 C 1.00 F V Section 26.2 Calculating Capacitance P26.3 E kq r e = 2 : q = × ×⋅ = 490 10 0210 899 10 0240 4 2 9 . . NC m Nm C C 22 (a) σ π × = q A 0240 10 40 1 2 0 133 6 2 . . . a f Cm 2 (b) Cr =∈ = × = 4 4 885 10 0120 0 12 ππ . pF P26.4 (a) CR 4 0 R C kC e = × × = 4 8 99 10 1 00 10 8 99 0 91 2 . F mm e j (b) =∈= ×× = 4 4 8 85 10 2 00 10 0222 0 12 3 . C m Nm pF 2 2 e j (c) QC V × = × 2 2 21 0 1 0 0 2 2 0 13 11 V C P26.5 (a) Q Q R R 1 2 1 2 = QQ R R 12 1 2 13 5 0 7 0 0 += + F H G I K J C
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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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