A parallel plate capacitor has insulated plates of area \( A \) and separation \( d_{1} \). It is filled with a dielectric of dielectric constant \( K \). If the plates have a charge density \( \sigma \), then, the work done in changing the distance between the plates from \( d_{1} \) to \( d_{2}\ (\gt d_{1}) \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{\sigma^{2}A}{2\varepsilon_{0}K}(d_{2} - d_{1}) \)
The plates are insulated, so the charge stays constant, and the work done in separating the plates equals the increase in stored energy. The dielectric is taken to fill the gap between the plates in both positions.
\( Q = \sigma A \)
\( \displaystyle C = \frac{K\varepsilon_{0}A}{d} \)
\( \displaystyle U = \frac{Q^{2}}{2C} = \frac{\sigma^{2}A^{2}d}{2K\varepsilon_{0}A} = \frac{\sigma^{2}A\,d}{2\varepsilon_{0}K} \)
\( \displaystyle W = U_{2} - U_{1} = \frac{\sigma^{2}A}{2\varepsilon_{0}K}(d_{2} - d_{1}) \)
Shortcut: The attractive force between the plates is constant at constant charge, \( F = Q^{2}/(2K\varepsilon_{0}A) = \sigma^{2}A/(2\varepsilon_{0}K) \); work = force × displacement \( (d_{2} - d_{1}) \).