JEE Main 2024 — 4 April, Shift 2 — Question 52

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 $\in_{\mathrm{r}} \mathrm{C}_{0}$.

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 $\mathrm{C}_{0}=12.5 \mathrm{pF}, \mathrm{V}=12 \mathrm{~V}, \epsilon_{\mathrm{r}}=6$ $=\frac{1}{2}(12.5) \times 12^{2}\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^{2} \times \frac{5}{6}$

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