A parallel plate capacitor of capacitance $12.5\ \text{pF}$ is charged by a battery connected between its plates to a potential difference of $12.0\ \text{V}$. The battery is now disconnected and a dielectric slab ($\varepsilon_r = 6$) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ______ $\times10^{-12}\ \text{J}$.
Numerical value type. Enter your answer.
Answer: 750
Initial energy: $U_i = \dfrac12CV^2 = \dfrac12\times12.5\times10^{-12}\times144 = 900\times10^{-12}\ \text{J}$.
With the battery disconnected the charge stays constant, so $U = \dfrac{Q^2}{2C}$ falls by the factor $\varepsilon_r$:
$$U_f = \frac{900}{6}\times10^{-12} = 150\times10^{-12}\ \text{J}$$
Change $= 900 - 150 = 750\times10^{-12}\ \text{J}$ (a decrease).
Solution by Sreeraj P, M.Sc Physics