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A Capacitor C is charged by a battery of EMF V volts and battery is disconnected. If now capacitor plates are pulled apart to increase the separation to 3d/2 from d and a dielectric constant $k = 3$ is filled between the plates. Find final energy stored in this capacitor. Take $C = 10 μ F$ and $V = 20$ Volt.

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$1 μ J$

$1 n J$

$1 m J$

$1 p J$

answer is C.

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Detailed Solution

$C = \frac{ε_{0} A}{d}, V_{f} = \frac{q^{2}}{2 C}$

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