Capacitors

JEE Main level, single correct, medium

3 of 180

Question 3 of 180

A parallel plate capacitor is charged to V0V_0 and battery is disconnected. A dielectric slab of constant KK and thickness d/2d/2 is inserted. New potential difference is:

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Answer: (C) V0K+12KV_0 \frac{K+1}{2K}

Cnew=ε0Ad/2+d/(2K)=C02KK+1C_{new} = \frac{\varepsilon_0 A}{d/2 + d/(2K)} = C_0 \frac{2K}{K+1}. Since QQ is constant, Vnew=Q/Cnew=V0K+12KV_{new} = Q/C_{new} = V_0 \frac{K+1}{2K}.

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