JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 30)
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
750
Explanation
$$\begin{aligned}
E_1 & =\frac{1}{2}\left(\frac{25}{2}\right) \times 10^{-12} \times 144 \\
& =900 \times 10^{-12} \mathrm{~J} \\
E_2 & =\frac{1}{2}\left(6 \times \frac{25}{2} \times 10^{-12}\right)\left(\frac{12}{6}\right)^2=150 \times 10^{-12} \mathrm{~J} \\
\Delta E & =750 \times 10^{-12} \mathrm{~J}
\end{aligned}$$
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