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+3 votes
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in Physics by (150k points)

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.

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(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, increases to K times

∴ V' = V/K

As E =  V/d = and V is decreased; therefore, electric field decreases to 1/K times.

(iii) 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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