ch25-p035

# ch25-p035 - U f = 1 2 C ( V ) 2 = 1 2 A d ( V ) 2 = (8 . 85...

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35 (a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by ± 0 A/d , the charge is q = CV = ± 0 AV /d . After the plates are pulled apart, their separation is d ± and the potential difference is V ± .Th en q = ± 0 AV ± /d ± and V ± = d ± ± 0 A q = d ± ± 0 A ± 0 A d V = d ± d V = 8 . 00 mm 3 . 00 mm (6 . 00 V) = 16 . 0V . (b) The initial energy stored in the capacitor is U i = 1 2 CV 2 = ± 0 AV 2 2 d = (8 . 85 × 10 12 F / m)(8 . 50 × 10 4 m 2 )(6 . 00 V) 2(3 . 00 × 10 3 mm) =4 . 51 × 10 11 J
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Unformatted text preview: U f = 1 2 C ( V ) 2 = 1 2 A d ( V ) 2 = (8 . 85 10 12 F / m)(8 . 50 10 4 m 2 )(16 . 0 V) 2(8 . 00 10 3 mm) = 1 . 20 10 10 J . (c) The work done to pull the plates apart is the difference in the energy: W = U f U i = 1 . 20 10 10 J 4 . 51 10 11 J = 7 . 49 10 11 J....
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## This test prep was uploaded on 04/15/2008 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.

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