The plates of a parallel plate capacitor have an area of $90 \,cm^{2}$ each and are separated by $2.5 \,mm$. The capacitor is charged by connecting it to a $400 \,V$ supply.
$(a)$ How much electrostatic energy is stored by the capacitor?
$(b)$ View this energy as stored in the electrostatic field between the plates,and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.

(N/A) Given:
Area $A = 90 \,cm^{2} = 90 \times 10^{-4} \,m^{2}$
Distance $d = 2.5 \,mm = 2.5 \times 10^{-3} \,m$
Potential $V = 400 \,V$
Permittivity $\epsilon_{0} = 8.85 \times 10^{-12} \,C^{2} \,N^{-1} \,m^{-2}$
$(a)$ Capacitance $C = \frac{\epsilon_{0} A}{d} = \frac{8.85 \times 10^{-12} \times 90 \times 10^{-4}}{2.5 \times 10^{-3}} = 3.186 \times 10^{-11} \,F$
Energy $U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 3.186 \times 10^{-11} \times (400)^{2} = 2.55 \times 10^{-6} \,J$
$(b)$ Volume $V' = A \times d = 90 \times 10^{-4} \times 2.5 \times 10^{-3} = 2.25 \times 10^{-5} \,m^{3}$
Energy density $u = \frac{U}{V'} = \frac{2.55 \times 10^{-6}}{2.25 \times 10^{-5}} = 0.113 \,J \,m^{-3}$
Relation: $u = \frac{U}{Ad} = \frac{\frac{1}{2} (\frac{\epsilon_{0} A}{d}) V^{2}}{Ad} = \frac{1}{2} \epsilon_{0} (\frac{V}{d})^{2} = \frac{1}{2} \epsilon_{0} E^{2}$