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?

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2ε0AV2/d

ε0AV2/d

3ε0AV2/2d

ε0AV2/2d

answer is D.

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Detailed Solution

W=U2−U1=q221C2−1C1C1=ε0Ad,C2=C12=ε0A2dq=C1V=ε0AVd⇒W=12ε0AV2dAlternatively,W=Fd=Q22Aε0d=C12V22ε0Ad=12ε0AV2d

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