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Q. The energy stored in a capacitor of capacitance $C$ having a charge $Q$ under a potential $V$ is

KEAMKEAM 2014

Solution:

By energy stored in a charged conductor Suppose, a conductor of capacity $C$ is charged to a potential $V$ and $Q$ the charge on conductor at a instant.
The potential of the conductor
$V=\frac{Q}{C}$
Now, work done in bringing a small charge $d Q$ at this potential is,
$d W=V d Q=\frac{Q}{C} d Q$
$\therefore $ Total work done in charging it from $0$ to $Q$ is,
$W=\int\limits_{0}^{Q} d W =\int\limits_{0}^{Q} \frac{Q}{C} d Q $
$=\frac{1}{2} \frac{Q^{2}}{C}$
This work is stored as the potential energy,
$U=\frac{1}{2} \frac{Q^{2}}{C} \,(\because Q=V C) $
$U=\frac{1}{2} C V^{2}=\frac{1}{2} Q V$