Question

A capacitor of capacitance 2mu
F is connected to a cell of emf 20volt.The plates are drawn apart slowly to double the distance between them. The work done by the external agent on the plates is? ​

Answer 1

Answer:

The work done by external agent is $2\times10^{-4}\ \rm J$

Explanation:

Given:

Capacitance, C=2 muF

Emf of the battery=20 volt

We know that the capacitance of a capacitor is given by

$C=\dfrac{A\epsilon_0}{d}$

When the distance is doubled then the capacitance is halved.

New capacitance= $\dfrac{C}{2}=1\ \mu F$

Work done by done by external agent is equal to the change in potential Energy stored in the capacitor. Let W be the work done

$W=\dfrac{V^2}{2}\left (C-\dfrac{C}{2} \right )\\\\W=\dfrac{CV^2}{4}\\\\W=\dfrac{2\times 10^{-6}\times 20^2}{4}\\\\W=2\times 10^{-4}\ \rm J$

Hence the work done by the external agent is calculated.