Question

A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.

Answer 1

Initial Energy = $\frac{1}{2}C{V}^2$

and $C=\frac{{\in }_{0}A}{d}$

when the battery removed and dielectric induced

$q=C_1{V}_1$

but $C_1=\frac{{\in }_{0}AK}{d}$

Then $V_1=\frac{V}{K}$

So, Final Energy

$=\frac{1}{2}\frac{K{\in }_{0}A}{d}.{\left(\frac{V}{K}\right)}^2$

$=\frac{1}{2}\frac{{\in }_{0}A{V}^2}{d}$

Work done = Change in energy

$=\frac{1}{2}\frac{{\in }_{0}A{V}^2}{Kd}-\frac{1}{2}\frac{{\in }_{0}A{V}^2}{d}$

$=\frac{1}{2}\frac{{\in }_{0}A{V}^2}{d}(\frac{1}{K}-1)$