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Question

A parallel-plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart until their separation is 2d. What is the work required to separate the plates?

A
20AV2d
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B
0AV2d
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C
30AV22d
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D
0AV22d
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Solution

The correct option is D 0AV22d
Let C1 and C2 be the capacitance of capacitor of separation d and 2d respectively.
C1=Aϵ0d and C2=Aϵ02d
As there is no transfer of charge in this case thus, charge will remain conserved.
q1=q2=q
We know that q = CV so,
q=Aϵ0d×V---------(i)
We know that work done is given by change in potential energy
W.D=ΔU=UfUi
W.D=q22C2q22C1
using equ (i) and putting the values of C1 and C2 we get
W.D = ϵ0AV22d

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