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_______.$\times 10^{-12} \mathrm{~J}$.
JEE MAIN 2024, Diffcult
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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 $\epsilon_{\mathrm{s}} \mathrm{C}_0$.
Change in potential energy of the capacitor
$=\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{r}}$
$=\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]$
$=\frac{\left(\mathrm{C}_0 \mathrm{~V}\right)^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_0 \mathrm{~V}^2\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]$
Using $\mathrm{C}_0=12.5 \mathrm{pF}, \mathrm{V}=12 \mathrm{~V}, \epsilon_{\mathrm{r}}=6$
$=\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^2 \times$
$\frac{5}{6}$
$=750 \mathrm{pJ}=750 \times 10^{-12} \mathrm{~J}$
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