55. (a) Initially, the capacitance is ()122220028.85 10C /N m(0.12 m )89 pF.1.2 10mACdε−−×⋅===×(b) Working through Sample Problem 25-7 algebraically, we find: 1222220238.85 10C /N m(0.12m )(4.8)1.2 10 pF.()(4.8)(1.2 0.40)(10 m)(4.0 10 m)ACdb bεκκ−−−=×−+−+ ×(c) Before the insertion, q = C0V(89 pF)(120 V) = 11 nC. (d) Since the battery is disconnected, qwill remain the same after the insertion of the slab, withq =11 nC. (e) Eq A××=⋅/)(.)091211 101001210C / (8.85mkV / m.CNm
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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.