c
Before inserting dielectric capacitance is given $\mathrm{C}_0=12.5 \mathrm{pF}$ and charge on the capacitor $\mathrm{Q}=\mathrm{C}_0 \mathrm{~V}$ After inserting dielectric capacitance will become $\in_{\mathrm{s}} \mathrm{C}_0$.
Change in potential energy of the capacitor
$=\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{r}}$
$=\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\in_{\mathrm{r}}}\right]$
$=\frac{\left(\mathrm{C}_0 \mathrm{~V}\right)^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\in_{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_0 \mathrm{~V}^2\left[1-\frac{1}{\in_{\mathrm{r}}}\right]$
Using $\mathrm{C}_0=12.5 \mathrm{pF}, \mathrm{V}=12 \mathrm{~V}, \in_{\mathrm{r}}=6$
$=\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^2 \times$
$\frac{5}{6}$
$=750 \mathrm{pJ}=750 \times 10^{-12} \mathrm{~J}$
