ch25_p46 - 46. (a) The electric field E1 in the free space...

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46. (a) The electric field E 1 in the free space between the two plates is E 1 = q / ε 0 A while that inside the slab is E 2 = E 1 / κ = q / κε 0 A . Thus, VE d bE b q A db b 01 2 =− + = F H G I K J −+ F H G I K J b g εκ 0 , and the capacitance is () ( ) ( ) 2 2 12 4 2 C Nm 0 0 8.85 10 115 10 m 2.61 13.4pF. 2.61 0.0124m 0.00780m 0.00780m A q C Vd b b −− ×× == = = + (b) q = CV = (13.4 × 10 –12 F)(85.5 V) = 1.15 nC. (c) The magnitude of the electric field in the gap is
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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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