P26_078 - 78. (a) Initially, the capacitance is 2 8.85 1012...

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78. (a) Initially, the capacitance is C 0 = ε 0 A d = ³ 8 . 85 × 10 12 C 2 N · m 2 ´ (0 . 12 m 2 ) 1 . 2 × 10 2 m =89pF . (b) Working through Sample Problem 26-6 algebraically, we fnd: C = ε 0 κ ( d b )+ b = ³ 8 . 85 × 10 12 C 2 N · m 2 ´ (0 . 12 m 2 )(4 . 8) (4 . 8)(1 . 2 0 . 40)(10 2 m) + (4 . 0 × 10 3 m) = 120 pF . (c) Be±ore the insertion, q = C 0 V (89 pF)(120 V) = 11 nC . Since the battery is disconnected, q will remain the same a±ter the insertion o± the slab. (d) E = q/ε 0 A =11 × 10 9 C / ³ 8 . 85 × 10 12 C 2 N · m 2 ´ ( 0 . 12 m 2 ) =10kV / m . (e) E 0 = E/κ
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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