A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference Vo. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plate is

Here C=ε0Ad and Q=CV=ε0AV0d∴Energy stored, E1=Q22C=ε02A2V022d2CWhen battery is disconnected, Q remains the same. As d increases to 3 times, C decreases to 3 times.∴ E2=3Q22C=3ε02A2V022d2CNow, Work done = E2 - E1∴ W=3ε02A2V022d2C−ε02A2V022d2C=ε02A2V02d2C            =ε02A2V02d2×ε0A/d=ε0AV02d