​A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between ​

Question

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......... \(\times 10^{-12} \mathrm{~J}.\)  

Answer 1

Correct answer is : 750

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 \(\epsilon_{\mathrm{r}} \mathrm{C}_{0}.\)

Change in potential energy of the capacitor \(=\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}}\)

\(=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^{2}}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}_{0}}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]\)

\(=\frac{\left(\mathrm{C}_{0} \mathrm{~V}\right)^{2}}{2 \mathrm{C}_{0}}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_{0} \mathrm{~V}^{2}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]\)

Using \(\mathrm{C}_{0}=12.5 \mathrm{pF}, \mathrm{V}=12 \mathrm{~V}, \epsilon_{\mathrm{r}}=6\)

\(=\frac{1}{2}(12.5) \times 12^{2}\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^{2} \times \frac{5}{6}\)

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