A parallel plate capacitor has a uniform electric field ‘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)

The correct answer is: (1/2)ε₀E² Ad

Step-by-step derivation:
Let us calculate the total electrostatic energy stored in a parallel plate capacitor.

The energy density (energy per unit volume) $u$ in a region with a uniform electric field $E$ is given by the formula:
$u=\frac{1}{2}{\epsilon }_0{E}^2$
where ${\epsilon }_0$ is the permittivity of free space.

The total volume $V$ of the space between the parallel plates of the capacitor, where the electric field exists, is the product of the area of each plate $A$ and the distance between the plates $d$:
$V=A·d$

The total energy $U$ stored in the capacitor is the energy density multiplied by the total volume of the region containing the electric field:
$U=u·V$

Substituting the expressions for $u$ and $V$ into the equation:
$U=\left(\frac{1}{2}{\epsilon }_0{E}^2\right)·\left(Ad\right)$
$U=\frac{1}{2}{\epsilon }_0{E}^{2}Ad$

Therefore, the total energy stored in the capacitor is $\frac{1}{2}{\epsilon }_0{E}^{2}Ad$.