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 diele

Question

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

TestHub · Accepted Answer

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

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