Question

A parallel plate capacitor $C$ has a charge $q$ and potential $V$ between the plates. Work required to double the distance between the plate is

• A. $\frac{1}{2}C{V}^2$
• B. $\frac{1}{4}C{V}^2$
• C. $\frac{1}{2}C\left(\frac{V}{2}\right)$
• D. $C{V}^2$

Answer 1

The correct option is B $\frac{1}{4}C{V}^2$

Initial capacitance $=\frac{{ϵ}_{o}A}{d}=C$
Final capacitance $=\frac{{ϵ}_{o}A}{2d}=C/2$
Initial Energy$=\frac{1}{2}C{V}^2$
Fianl Energy$=\frac{1}{2}\left(\frac{C}{2}\right)V^2$
Change in energy= Work done
$\frac{1}{2}\left(\frac{C}{2}\right)V^2=\frac{1}{4}C{V}^2$