Question

A parallel plate condenser with plate separation ‘d’ and plate area A is connected to a battery and charged to a potential V. Then the battery is disconnected and with insultated handles the plate separation is increased to 2d. Work done in the process will be?

• A. $\frac{{ϵ}_{0}A}{4d}{V}^2$
• B. $\frac{{ϵ}_{0}A}{d}{V}^2$
• C. $\frac{{ϵ}_{0}A}{2d}{V}^2$
• D. $\frac{2{ϵ}_{0}A}{d}{V}^2$

Answer 1

The correct option is C $\frac{{ϵ}_{0}A}{2d}{V}^2$

$U_1=\frac{1}{2}\frac{Q^2}{c}-------\left(1\right)$
$U_2=\frac{1}{2}\frac{Q^2}{c_1}=\frac{1}{2}\frac{Q^2}{\frac{c}{2}}=\frac{Q^2}{c}-----\left(2\right)$
$W=U_2-U_1,=\frac{1}{2}\frac{Q^2}{c}[2-1]$
$=\frac{1}{2}\frac{C^2{V}^2}{c}=\frac{1}{2}C{V}^2=\frac{1}{2}\frac{{ϵ}_{0}A}{d}{V}^2$