Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If E is the electric field and ${\epsilon }_0$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-

Step 1: Find Capacitance

The capacitance of a parallel plate capacitor is given by:

$$C = \frac{\varepsilon_0 A}{d}$$

Step 2: Relation Between Electric Field and Voltage

The potential difference $V$ across the plates is:

$$V = E d$$

Step 3: Energy Stored in the Capacitor

The energy stored in a capacitor is given by:

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

Substituting $C = \frac{\varepsilon_0 A}{d}$ and $V = E d$:

$$U = \frac{1}{2} \times \frac{\varepsilon_0 A}{d} \times (E d)^2$$

 $$U = \frac{1}{2} \varepsilon_0 A d E^2$$

Final Answer:

$$U = \frac{1}{2} \varepsilon_0 A d E^2$$

This represents the total electrostatic energy stored in the capacitor.