Consider a parallel plate capacitor of area  $$A$$  (of each plate)  and separation  $$d$$ between the plates. If $$E$$ is the electric field and $$\varepsilon_0$$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is: 

To express the potential energy stored in a parallel plate capacitor in terms of the electric field $$E$$, plate area $$A$$, separation $$d$$, and permittivity $$\varepsilon_0$$, recall that the general formula for the energy is$$U = \frac{1}{2}CV^2$$where $$C$$ is the capacitance and $$V$$ the potential difference across the plates. For a parallel plate capacitor, the capacitance is given by$$C = \frac{\varepsilon_0 A}{d}$$and the uniform electric field relates to the potential difference via$$V = Ed\,. $$Substituting these expressions into the energy formula leads to$$U = \frac{1}{2}\,\frac{\varepsilon_0 A}{d}\,(Ed)^2 = \frac{1}{2}\varepsilon_0 E^2 A d\,. $$

An alternative derivation begins with the energy density of an electric field,$$u = \frac{1}{2}\varepsilon_0 E^2\,, $$and notes that the volume between the plates is $$A\,d$$. Multiplying gives$$U = u\times(A d) = \frac{1}{2}\varepsilon_0 E^2 A d\,, $$which agrees with the previous result.

Thus, the energy stored in the parallel plate capacitor can be written as$$\frac{1}{2}\varepsilon_0 E^2 A d\,. $$