Question

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

• A. $\frac{3}{2}\frac{{ϵ}_{0}A{V}^2}{d}$
• B. $\frac{{ϵ}_{0}A{V}^2}{d}$
• C. $\frac{2{ϵ}_{0}A{V}^2}{d}$
• D. $\frac{{ϵ}_{0}A{V}^2}{2d}$

Answer 1

The correct option is A

$\frac{3}{2}\frac{{ϵ}_{0}A{V}^2}{d}$

$q=CV=\left(\frac{{\epsilon }_{0}A}{d}\right)V$

After charging $q$ will become constant.

$W=U_f-U_i$

$=\frac{q^2}{2C_f}-\frac{q^2}{2C_i}$

$=\frac{q^2}{2}(\frac{1}{C_f}-\frac{1}{C_i})$

$=\frac{\left({\epsilon }_{0}AV/d^2\right)}{2}(\frac{2d}{{\epsilon }_{0}A}-\frac{d}{{\epsilon }_{0}A})$

$=\frac{{\epsilon }_{0}A{V}^2}{2d}$