JEE Main 2024 — Capacitance Question with Solution

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 ${ϵ}_{\mathrm{r}}{\mathrm{C}}_0$. Change in potential energy of the capacitor $\begin{aligned} & =\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}} \\ & =\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right] \\ & =\frac{\left(\mathrm{C}_0 \mathrm{~V}\right)^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_0 \mathrm{~V}^2\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right] \end{aligned}$<br/>Using$\mathrm{C}_0=12.5 \mathrm{pF}, \mathrm{V}=12 \mathrm{~V}, \epsilon_{\mathrm{r}}=6$ $\begin{aligned} & =\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^2 \times \frac{5}{6} \\ & =750 \mathrm{pJ}=750 \times 10^{-12} \mathrm{~J} \end{aligned}$