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Physics · Electrostatic Potential
हिन्दीA parallel plate capacitor is charged to a potential \( V \) by a battery. The battery is then disconnected and a dielectric slab of dielectric constant \( K \) is completely inserted between the plates. If the initial electrostatic energy stored was \( U_0 \), what is the work done by the external agent in inserting the slab slowly?
Correct answer · Option 3
\( -U_0 (1 - \frac{1}{K}) \)
Explanation
Since the capacitor is isolated, charge remains constant: \( U_f = \frac{Q^2}{2KC_0} = \frac{U_0}{K} \). Work done by external agent is \( W_{\text{ext}} = \Delta U = U_f - U_0 = U_0(\frac{1}{K} - 1) = -U_0(1 - \frac{1}{K}) \).
