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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 $μ$ J

b

400 $μ$ J

c

-400 $μ$ J

d

200 $μ$ J

answer is B.

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Detailed Solution

$\begin{array}{l} W_{batt} + W_{ext} = ΔU \\ ΔU = U_{f} - U_{i} = \frac{1}{2} (\frac{C}{2}) V^{2} - \frac{1}{2} {CV}^{2} \\ = - \frac{1}{2} \cdot \frac{1}{2} {CV}^{2} = - \frac{1}{4} {CV}^{2} \\ Hence, from (1) - 400 + W_{ext} \\ = - 200 \Rightarrow W_{ext} = 200 μJ \end{array}$

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