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

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, 560 pJ, 600 pJ, 508 pJ, 692 pJ

MARKS App · 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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