ch25-p055 - 55. (a) Initially, the capacitance is C0 = 0 A...

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55. (a) Initially, the capacitance is () 12 2 2 2 0 0 2 8.85 10 C /N m (0.12 m ) 89 pF. 1.2 10 m A C d ε ×⋅ == = × (b) Working through Sample Problem 25-7 algebraically, we find: 12 2 2 2 2 0 23 8.85 10 C /N m (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 10 012 10 C / (8.85 m kV / m. C Nm
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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