Question

A parallel plate capacitor is charged by a battery. After some time, the battery is disconnected and a dielectric slab of dielectric constant k is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected? Justify your answer.

Answer 1

(i) The capacitance of capacitor increases to K times (since C = (Kε0A)/d ∝ K)

(ii) The potential difference between the plates becomes 1/K times. Reason: V = Q/C; Q same, C increases to K times; V' = V/K

(iii) As E  = V/d = and V is decreased; therefore, electric field decreases to 1/K times. Energy stored by the capacitor, U = Q2/2C. As Q = constant, C is increased, and so energy stored by capacitor decreases to 1/K times

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