First slide

Capacitance

Question

The plate separation in a parallel plate condenser is d and plate area is A. If it is charged to V volts, then the work done in increasing the plate separation to 2 d will be

Difficult

A

$\frac{ε_{0} {AV}^{2}}{2 d}$
B

$\frac{ε_{0} A^{2} V}{2 d}$
C

$\frac{ε_{0} {AV}^{2}}{d}$
D

$\frac{ε_{0} A^{2} V}{d}$

Solution

The work done W is given by
$W = U_{2} - U_{1}$
Here, $U_{1} = \frac{1}{2} {CV}^{2}$
and $U_{2} = \frac{1}{2} (\frac{C}{2}) (2 V)^{2} = {CV}^{2}$
$\begin{aligned} ∴ W = {CV}^{2} - \frac{1}{2} {CV}^{2} = \frac{1}{2} {CV}^{2} \\ = \frac{1}{2} (\frac{ε_{0} A}{d}) V^{2} = \frac{ε_{0} {AV}^{2}}{2 d} \end{aligned}$

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