78. (a) Initially, the capacitance isC0=ε0Ad=³8.85×10−12C2N·m2´(0.12 m2)1.2×10−2m=89pF.(b) Working through Sample Problem 26-6 algebraically, we fnd:C=ε0Aκκ(d−b)+b=³8.85×10−12C2N·m2´(0.12 m2)(4.8)(4.8)(1.2−0.40)(10−2m) + (4.0×10−3m)= 120 pF.(c) Be±ore the insertion,q=C0V(89 pF)(120 V) = 11 nC.Since the battery is disconnected,qwillremain the same a±ter the insertion o± the slab.(d)E=q/ε0A=11×10−9C/³8.85×10−12C2N·m2´(0.12 m2)=10kV/m.(e)E0=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.