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Charging of Capacitors

If the distan...

Question

If the distance between plates of a capacitor having capacity $C$ $&$ charge $Q$ is increased to double then work done will be :-

A

Q^{2} / 4 C

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B

Q^{2} / 2 C

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C

Q^{2} / C

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D

{2 Q}^{2} / C

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Solution

The correct option is C $Q^{2} / C$
$\begin{matrix} V_{1} = \frac{Q^{2}}{2 C} V_{f} = \frac{Q^{2}}{2 C / 2} = \frac{Q^{2}}{C} W = V_{f} - V_{1} = \frac{Q^{2}}{C} = \frac{Q^{2}}{2 C} = \frac{Q^{2}}{2 C} A n s . (B) \end{matrix}$

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