To solve this problem, we need to find the energy stored in a parallel plate capacitor.$\\$Given:$\\$• Electric field between the plates $E. \\$• Distance between the plates $d. \\$• Area of each plate $A. \\$• Permittivity of free space $\varepsilon_0. \\$The energy stored in a capacitor is given by the formula:$\\ U = \frac{1}{2} C V^2 \\$Where:$\\$• $C$ is the capacitance of the capacitor.$\\$• $V$ is the potential difference between the plates.$\\$For a parallel plate capacitor, the capacitance $C$ is given by:$\\ C = \frac{\varepsilon_0 A}{d} \\$The potential difference $V$ between the plates is related to the electric field $E$ by:$\\ V = E \cdot d \\$Substituting these into the energy formula:$\\ U = \frac{1}{2} \left(\frac{\varepsilon_0 A}{d}\right) (E \cdot d)^2 \\$Simplify the expression:$\\ U = \frac{1}{2} \cdot \frac{\varepsilon_0 A}{d} \cdot E^2 \cdot d^2 \\ U = \frac{1}{2} \varepsilon_0 A E^2 d \\$Thus, the energy stored in the capacitor is:$\\ U = \frac{1}{2} \varepsilon_0 E^2 A d \\$This corresponds to Option 3.