Question 79 of 180
A parallel plate capacitor of capacitance is charged by a battery connected between its plates to potential difference of . The battery is now disconnected and a dielectric slab is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _____ .
Show answer and solution
Answer: 750
Before inserting dielectric capacitance is given and charge on the capacitor After inserting dielectric capacitance will become .
Change in potential energy of the capacitor
$\begin{aligned}
& =\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}} \\
& =\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right] \\
& =\frac{\left(\mathrm{C}_0 \mathrm{~V}\right)^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_0 \mathrm{~V}^2\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]
\end{aligned}$
Using
$\begin{aligned}
& =\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^2 \times \frac{5}{6} \\
& =750 \mathrm{pJ}=750 \times 10^{-12} \mathrm{~J}
\end{aligned}$
Change in potential energy of the capacitor
$\begin{aligned}
& =\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}} \\
& =\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right] \\
& =\frac{\left(\mathrm{C}_0 \mathrm{~V}\right)^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_0 \mathrm{~V}^2\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]
\end{aligned}$
Using
$\begin{aligned}
& =\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^2 \times \frac{5}{6} \\
& =750 \mathrm{pJ}=750 \times 10^{-12} \mathrm{~J}
\end{aligned}$
