ch25_p49 - 49. (a) Initially, the capacitance is C0 = 0A d...

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49. (a) Initially, the capacitance is C A d 0 0 12 2 885 10 012 12 10 89 == × × = ε .( . ) . C Nm 2 2 2 m m pF. di (b) Working through Sample Problem 25-7 algebraically, we find: () 2 2 12 2 C 2 0 23 8.85 10 (0.12m )(4.8) 1.2 10 pF. ( ) (4.8)(1.2 0.40)(10 m) (4.0 10 m) A C db b εκ κ −− × = × −+ + × (c) Before the insertion, q = C 0 V (89 pF)(120 V) = 11 nC. (d) Since the battery is disconnected, q will remain the same after the insertion of the slab, with q = 11 nC. (e) Eq A × × = /) ( . ) 0 91 2 11 10 10
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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