Question 96: Electrostatics & Capacitance

Question

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 ________ $$	imes10^{-12} \mathrm{~J}$$.

Accepted Answer

$$\begin{aligned}
E_1 & =\frac{1}{2}\left(\frac{25}{2}ight) 	imes 10^{-12} 	imes 144 \
& =900 	imes 10^{-12} \mathrm{~J} \
E_2 & =\frac{1}{2}\left(6 	imes \frac{25}{2} 	imes 10^{-12}ight)\left(\frac{12}{6}ight)^2=150 	imes 10^{-12} \mathrm{~J} \
\Delta E & =750 	imes 10^{-12} \mathrm{~J}
\end{aligned}$$

Electrostatics & Capacitance

Integer / Numeric Answer Type