787_Physics ProblemsTechnical Physics

# 787_Physics ProblemsTechnical Physics - Chapter 27 *P27.75...

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Chapter 27 127 *P27.75 (a) Think of the device as two capacitors in parallel. The one on the left has κ 1 1 = , Ax 1 2 =+ F H G I K J A A . The equivalent capacitance is κκ 10 1 20 2 0 0 0 22 2 + = + F H G I K J + F H G I K J = ++− A d A dd x d x d xx A A A A A AA af . (b) The charge on the capacitor is QCV =∆ Q V d = 0 2 A a f . The current is I dQ dt dQ dx dx dt V d v Vv d == = = 00 2 0202 1 ∆∆ a f a f . The negative value indicates that the current drains charge from the capacitor. Positive current is clockwise 0 1 A Vv d . ANSWERS TO EVEN PROBLEMS P27.2 364 . h P27.32 171 P27.4 (a) see the solution; (b) 105 m A P27.34 0153 P27.6 (a) 17 0 . A ; (b) 85 0 . kAm 2 P27.36 500 A , 24 0 P27.38 448 A P27.8 (a) 99 5 2 ; (b) 800 m m P27.40 (a) 0.530; (b) 221 J; (c) 15.1°C P27.10 (a) 221 nm ; (b) no; see the solution P27.42 (a) 317 m ; (b) 340 W P27.12 30 3 . MAm 2 P27.44 (a) 0 660 . kWh ; (b) 3 96¢ . P27.14 (a) 3.75 k ; (b) 536 m P27.46 (a) 2.05 W; (b) 3.41 W; no
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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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