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A parallel plate capacitor has area of each plate A, the separation between the plates is d . It is charged to a potential V and then disconnected from the battery. The amount of work done in the filling the capacitor Completely with a dielectric constant k is

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$\frac{1}{2} \frac{V^{2} ε_{0} A}{kd}$

$\frac{1}{2} \frac{ε_{0} {AV}^{2}}{d} [1 - \frac{1}{k^{2}}]$

$\frac{1}{2} \frac{V^{2} ε_{0} A}{k^{2} d}$

$\frac{1}{2} \frac{ε_{0} {AV}^{2}}{d} [1 - \frac{1}{K}]$

answer is D.

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Detailed Solution

Work done = decrease in energy

$ie w = E_{1} - E_{2} = \frac{1}{2} \frac{ε_{0} A}{d} v^{2} - \frac{ε_{0} {Av}^{2}}{2 d} [1 - \frac{1}{k}]$

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