Q25. A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 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 $\_\_\_\_$ $10 ^ { - 12 } \mathrm {~J}$.