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An ideal, isolated, air-filled parallel-plate capacitor is not connected to a battery but has equal and opposite charges of 3.9 nC on its plates. The separation between the plates initially is 1.2mm, and for this separation the capacitance is 3.1 x 10^-11 F. How much work mud be done to pull the plates apart until their seperation becomes 7.7mm? (e0= 8.85 x 10^-12 C^2/N x m^2)
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Added Mon, 03 Aug '15
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C = e*A/d

A = 4.2*10^-3

work done = change in potential difference

initial V1

3.9 *10^-9 = V1 * 3.1 x 10^-11

V1 = 125.8

after changing distance, new capacitance will be:

C' = e*A/d1 = 4.82*10^-12

Q = C' * V2

V2 = 807.9

work done = 682.109
Edit
Added Mon, 03 Aug '15
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