752_Physics ProblemsTechnical Physics

# 752_Physics ProblemsTechnical Physics - 92 Capacitance and...

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92 Capacitance and Dielectrics P26.48 (a) CC A d == = ×× × = −− κ 0 0 12 4 3 173 8 85 10 1 00 10 0100 10 153 a f ej e j .. . . m m nF 2 (b) The battery delivers the free charge QCV × = af 153 10 120 184 9 . F V n C . (c) The surface density of free charge is σ × × Q A 18 4 10 10 184 10 9 4 4 . . C 1.00 m Cm 2 2 . The surface density of polarization charge is σσ p =− F H G I K J F H G I K J 1 1 1 1 173 183 10 4 . C m 2 . (d) We have E E = 0 and E V d 0 = ; hence, E V d × = 12 0 100 10 694 4 . . V 173 m Vm . P26.49 The given combination of capacitors is equivalent to the circuit diagram shown to the right. Put charge Q on point A . Then, QV V V AB BC CD === 40 0 10 0 40 0 ... F µµµ bgbgbg ∆∆∆ . FIG. P26.49 So, VVV BC AB CD 4 4 , and the center capacitor will break down first, at V BC = 15 0 . V . When this occurs, ∆∆ VV V AB CD BC = 1 4 375 bg V and V V AD AB BC CD =++= + + = 150 225 . . V V V . Section 26.6 Electric Dipole in an Electric Field P26.50 (a) The displacement from negative to positive charge is
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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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