A parallel plate capacitor of capacitance 12.5 p F is charged by a battery connected between its plates to potential difference of 12.0 V. The…

A parallel plate capacitor of capacitance $\displaystyle 12.5 p F$ is charged by a battery connected between its plates to potential difference of $\displaystyle 12.0$ V . The battery is now disconnected and a dielectric slab $\displaystyle \left(\epsilon_{\mathrm{r}}=6\right)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is $\displaystyle \_\_\_\_$× $\displaystyle 10^{-12} \mathrm{~J}$.