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58.7k views
in Electrostatics by (75.1k points)
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A parallel plate capacitor has plate of area A and separation d. It is charged to a potential difference V0. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

(A) [(A∈0V02) / d]

(B) [(A∈0V02) / 2d]

(C) [(A∈0V02) / 3d] 

(D) [(A∈0V02) / 4d]

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1 Answer

+1 vote
by (70.5k points)

The correct option (A) [(A∈0V02)/d]

Explanation:

Work done = change in energy

W = (Q2/2C1) – (Q2/2C)

Now C = (A∈0/d) & C1 = (A∈0/d)

i.e. C1 = (C/3)

∴ W = [Q2/(2C/3)] – (Q2/2C) = (2Q2/2C) = (Q2/C)

∴ W = (Q2/C) = [(C2V02)/C] = CV02 --------- Q = CV

∴ W = [(V02A∈0) / d] --------- C = (A∈0/d)

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