Question

A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected? Justify your answer in each case.

Answer 1

(i) The capacitance of capacitor increases to \(\mathrm{K}\) times \(\left(\right.\) since \(\left.C=\left(\mathrm{K} \varepsilon_{0}^{\mathrm{A}}\right) / d \propto \mathrm{K}\right)\)
(ii) The potential difference between the plates becomes \(1 / K\) times.
Reason: \(\mathrm{V}=\mathrm{Q} / \mathrm{C} ; \mathrm{Q}\) same, \(\mathrm{C}\) increases to \(\mathrm{K}\) times; \(\mathrm{V}^{\prime}=\mathrm{V} / \mathrm{K}\)
(iii) As \(E=V / d=\) and \(V\) is decreased; therefore, electric field decreases to \(1 / K\) times. Energy stored by the capacitor, \(U=Q^{2} / 2 C\). As \(Q=\) constant, \(C\) is increased, and so energy stored by capacitor decreases to \(1 /\) K times

Answer 2

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