Electrostatic Potential and Capacitance Part 2

A parallel plate capacitor of capacitance 12.5‌pF is charged by a battery connected between its plates to potential difference of 12.0V. The battery is now disconnected and a dielectric slab (Er=6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ ×10−12J.