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 $(\epsilon_{\mathrm{r}}=6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ $\times10^{-12} \mathrm{~J}$.
Answer (integer)
750
Solution
<p>$$\begin{aligned}
E_1 & =\frac{1}{2}\left(\frac{25}{2}\right) \times 10^{-12} \times 144 \\
& =900 \times 10^{-12} \mathrm{~J} \\
E_2 & =\frac{1}{2}\left(6 \times \frac{25}{2} \times 10^{-12}\right)\left(\frac{12}{6}\right)^2=150 \times 10^{-12} \mathrm{~J} \\
\Delta E & =750 \times 10^{-12} \mathrm{~J}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Dielectrics
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