Phys 181b Problem Set 4 Solution

Phys 181b Problem Set 4 Solution - 6. (a) We use Eq. 25-17:...

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6. (a) We use Eq. 25-17: C ab ba = = ×− = 4 40 0 38 0 899 10 400 380 84 5 0 9 π ε .. . . mm pF. Nm C 2 2 b gb g di b g (b) Let the area required be A . Then C = 0 A /( b – a ), or () ( ) 2 2 2 12 C 0 84.5pF 40.0mm 38.0mm 191cm . 8.85 10 Cb a A −− == = ×
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8. The equivalent capacitance is CC eq F FF F. =+ + + = 3 12 400 10 0 500 10 0 733 . .. . µ µµ b gb g
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10. The charge that passes through meter A is qCV C V === = eq FV 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 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 2 150 CC . C . =+ = + Also, the voltage drop across this combination is 11 1 040 eq CV CV V. V . CC 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 22 2 10 0V 10 0 F 2 00 10 C. 5 . qC V . . §· = × ¨¸ ©¹
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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 qC V A Vd == . After the plates are pulled apart, their separation is f d and the potential difference is V f . Then 0 2 f f qA V d = and 0 00 . ff f f ii dd d A Vq V V AA d d εε = 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 AV UC V d −− ×× = = × × (c) The final energy stored is 2 2 2 0 11 . 22 f i f fi i i i i i d A V UV V U d d d d §· § · = = ¨¸ ¨ ¸ ©¹ © ¹ With / 8.00/3.00 = , 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 /2 and plate separation d , filled with dielectric materials with dielectric constants κ 1 and 2 , respectively. Thus, (in SI units),
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This note was uploaded on 07/18/2008 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.

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Phys 181b Problem Set 4 Solution - 6. (a) We use Eq. 25-17:...

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