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 $\underline{\hspace{2cm}}$$\times10^{-12} \mathrm{~J}$.