A parallel plate capacitor has a uniform electric field '\(\vec{E}\)' in the space between the plates. If the distance between the plates is ‘d’ and the area of each plate is ‘A’, the energy stored in the capacitor is : (ε₀ = permittivity of free space)

CONCEPT:

The energy stored in the capacitor is defined as the product of the capacitance and square of the electric potential. It is written as:

\(U = \frac{1}{2}CV^2\)

Where C is the capacitance of the capacitor and V is the electric potential.

CALCULATION: 

Given that the capacitor has a uniform electric field \(\vec{E}\) and 'd' is the distance between the plates and 'A' is the area and is shown in the figure below;

The energy stored in the capacitor is defined as the product of the capacitance and square of the electric potential. Using equation (1) we have;

\(U = \frac{1}{2}CV^2\) ----(1)

and, the capacitance of a capacitor is written as;

 \(C = \frac{\varepsilon _0A}{d}\) 

and electric potential,

V = Ed

Now, on putting the value of "C" and "V" in equation (1) we get;

\(U = \frac{1}{2}(\frac{\varepsilon _0A}{d})​​(Ed)^2\)

⇒ \(U = \frac{1}{2}({\varepsilon _0A})​​(E^2d)\)

⇒ \(U = \frac{1}{2}{\varepsilon _0A}​​E^2d\)

Hence, option 4) is the correct answer.