A parallel plate capacitor of capacitance \(12.5 \mathrm{pF}\) is charged by a battery connected between its plates to potential difference of \(12.0 \mathrm{~V}\). The battery is now disconnected and a dielectric slab \(\left(\epsilon_{\mathrm{r}}=6\right)\) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _______.\(\times 10^{-12} \mathrm{~J}\).

Change in potential energy of the capacitor

\(=\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{r}}\)

\(=\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]\)

\(=\frac{\left(\mathrm{C}_0 \mathrm{~V}\right)^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_0 \mathrm{~V}^2\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]\)

Using \(\mathrm{C}_0=12.5 \mathrm{pF}, \mathrm{V}=12 \mathrm{~V}, \epsilon_{\mathrm{r}}=6\)

\(=\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^2 \times\)

\(\frac{5}{6}\)

\(=750 \mathrm{pJ}=750 \times 10^{-12} \mathrm{~J}\)