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Q.

In the figure a capacitor of capacitance 2 μF is connected to a cell of emf 20 volt. The plates of the capacitor are drawn apart slowly to double the distance between them. The work done by the external agent on the plates is :

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a

-200 μJ

b

400 μJ

c

-400 μJ

d

200 μJ

answer is B.

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Detailed Solution

Wbatt +Wext =ΔUΔU=Uf−Ui=12C2V2−12CV2=−12⋅12CV2=−14CV2 Hence, from (1)−400+Wext =−200⇒Wext =200μJ

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