A parallel plate capacitor having capacitance 12   pF is charged by a battery to a potential difference of 10   V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is

Step-by-step solution

Initial energy of capacitor U i = 1 2 q 2 c = 1 2 × 120 × 120 12 = 600   pJ Since battery is disconnected so charge remain same. Final energy of capacitor U f = 1 2 q 2 c   k = 1 2 × 120 × 120 12 × 6 . 5 = 92   pJ W + U f = U i   W = 508   pJ