Question

A parallel plate capacitor has plates with area $A$ & separation $d$. A battery charges the plates to a potential difference of $V_0$. The battery is then disconnected & a di-electric slab of constant $K$ & thickness $d$ is introduced. Calculate the positive work done by the system (capacitor + slab) on the man who introduces the slab.

Answer 1

charges $Q=CV=\frac{{\epsilon }_{o}A}{d}{V}_o$

work sone= Initial energy-Final energy

$W=\frac{1}{2}{C}_{old}{V}^2-\frac{Q^2}{2C_{new}}$

$=\frac{1}{2}\left(\frac{{\epsilon }_{o}A}{d}\right)V_o^2-\frac{A{\epsilon }_o{V}_o^2}{2Kd}$

$=\frac{{\epsilon }_{o}A{V}_o^2}{2d}[1-\frac{1}{K}]$