Question

Prove that the energy stored in a parallel plate capacitor is $\frac{q^2}{2C}$

Answer 1

The capacitor is a charge storage device work has to be done to store the charges in a capacitor. This work done is stored as electrostatic potential energy in the capacitor.
Let $q$ be the charge and $V$ be the potential difference between the plates of the capacitor. If $dq$ is the additional charge given to the plate, then work done is, $dw=Vdq$.
$dw=\frac{q}{C}dq$ $(\therefore V=\frac{q}{C})$
Total work done to charge a capacitor is
$w=\int dw={\int }_0^q\frac{q}{C}dq=\frac{1}{2}\frac{q^2}{C}$
This work done is stored as electrostatic potential energy $\left(U\right)$ in the capacitor.
$U=\frac{1}{2}\frac{q^2}{C}=\frac{1}{2}C{V}^2$ $(\therefore q=CV)$
$U=\frac{q^2}{2C}$. Hence proved.