QJEEM24Ph1879: JEE Main 2024 Physics - Class 12 JEEM24Ph1879 Solution: A parallel plate capacitor of capacitance 12.5   pF is…

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 $$({ϵ}_{\mathrm{r}}=6)$$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ $$\times {10}^{-12}\mathrm{J}$$.