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Phys 181b Problem Set 4 Solution

# Phys 181b Problem Set 4 Solution - 6(a We use Eq 25-17 C =...

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6. (a) We use Eq. 25-17: C ab b a = = × = 4 40 0 38 0 8 99 10 40 0 38 0 84 5 0 9 π ε . . . . . . mm mm mm mm pF. N m C 2 2 b gb g d ib g (b) Let the area required be A . Then C = ε 0 A /( b – a ), or ( ) ( )( ) ( ) 2 2 2 12 C 0 N m 84.5pF 40.0mm 38.0mm 191cm . 8.85 10 C b a A ε = = = ×

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8. The equivalent capacitance is C C C C C C eq F F F F F F. = + + = + + = 3 1 2 1 2 4 00 10 0 500 10 0 500 7 33 . . . . . . µ µ µ µ µ µ b gb g
10. The charge that passes through meter A is q C V CV = = = = eq F V C. 3 3 250 4200 0 315 . . µ b gb g

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12. (a) The potential difference across C 1 is V 1 = 10.0 V. Thus, q 1 = C 1 V 1 = (10.0 µ F)(10.0 V) = 1.00 × 10 –4 C. (b) Let C = 10.0 µ F. We first consider the three-capacitor combination consisting of C 2 and its two closest neighbors, each of capacitance C . The equivalent capacitance of this combination is 2 eq 2 1 50 C C C C . C. C C = + = + Also, the voltage drop across this combination is 1 1 1 0 40 1 50 eq CV CV V . V . C C C . C = = = + + Since this voltage difference is divided equally between C 2 and the one connected in series with it, the voltage difference across C 2 satisfies V 2 = V /2 = V 1 /5. Thus ( ) 5 2 2 2 10 0V 10 0 F 2 00 10 C. 5 . q C V . . µ § · = = = × ¨ ¸ © ¹
31. (a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by 0 i A d ε , the charge is 0 i i q CV AV d ε = = . After the plates are pulled apart, their separation is f d and the potential difference is V f . Then 0 2 f f q AV d ε = and 0 0 0 . f f f f i i i i d d d A V q V V A A d d ε ε ε = = = With 3 3.00 10 m i d = × , 6.00 V i V = and 3 8.00 10 m f d = × , we have 16.0 V f V = . (b) The initial energy stored in the capacitor is (in SI units) 2 12 4 2 2 11 0 3 1 (8.85 10 )(8.50 10 )(6.00) 4.51 10 J. 2 2 2(3.00 10 ) i i i i AV U CV d ε × × = = = = × × (c) The final energy stored is 2 2 2 0 0 0 1 1 . 2 2 f f f i f f i i f f i i i i d d d A A AV U V V U d d d d d d ε ε ε § · § · = = = = ¨ ¸ ¨ ¸ © ¹ © ¹ With / 8.00/3.00 f i d d = , we have 10 1.20 10 J. f U = × (d) The work done to pull the plates apart is the difference in the energy: W = U f U i = 11 7.52 10 J. ×

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42. The capacitor can be viewed as two capacitors C 1 and C 2 in parallel, each with surface area A
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Phys 181b Problem Set 4 Solution - 6(a We use Eq 25-17 C =...

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