Potential Energy of a Charged Conductor

Derivation →

Let us consider a conductor of capacitor $C$ is given a charge $+q$ then the electric potential

$V=\frac{q}{C} \qquad(1)$

The work done in charging the conductor is stored as potential energy in the electric field in the vicinity of the conductor.

Let bring a small charge $dq$ from infinity to the surface of the conductor. Then the work done will be

$dW=V\:dq \qquad(2)$

From equation $(1)$ and equation $(2)$

$dW=\frac{q}{C} \:dq$

Therefore, the work done in increasing the charge on the conductor from $0$ to $q$ is

$W=\int dW$

$W=\frac{1}{C} \int_{0}^{q} q dq$

$W=\frac{1}{C} \left[ \frac{q^{2}}{2} \right]^{q}_{0}$

$W=\frac{1}{2} \frac{q^{2}}{C}$

This work is stored in the form of electric potential energy $U$. Then

$U=\frac{1}{2} \frac{q^{2}}{C}$

$U=\frac{1}{2} C V^{2}$

$U=\frac{1}{2} q V$