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

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Infinity Learn · Accepted Answer

Here $$C=$$ε$$0Ad$$ and $$Q=CV=$$ε$$0AV0d$$∴Energy stored, $$E1=Q22C=$$ε$$02A2V022d2CWhen$$ battery is disconnected, Q remains the same. As d increases to $$3$$ times, C decreases to $$3$$ times.∴ $$E2=3Q22C=3$$ε$$02A2V022d2CNow$$, Work done $$=$$ $$E2$$ $$-$$ $$E1$$∴ $$W=3$$ε$$02A2V022d2C$$−ε$$02A2V022d2C=$$ε$$02A2V02d2C$$            $$=$$ε$$02A2V02d2$$×ε$$0A/d=$$ε$$0AV02d$$

A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V_o. 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

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