An experimental supercapacitor has a capacitance density defined by $C = \frac{\varepsilon A}{d}$, where $\varepsilon$ is permittivity, $A$ is area, and $d$ is thickness. If $\varepsilon = 8.85 \times 10^{-12} \text{ F/m}$, $A = 0.5 \text{ m}^2$, and $d = 100 \text{ nm}$, what is the energy $E$ stored in the capacitor at a potential difference of $5 \text{ Volts}$ ($E = \frac{1}{2}CV^2$)?