The plates of a parallel plate capacitor have an area of 90 cm² 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.

Given,  
Area of the parallel plate capacitor, A=90 cm²= $90 \times {10}^{-4} \mathrm{m}$
Distance between the parallel plates, d=2.5 mm = $2.5 \times {10}^{-3 }\mathrm{m}$
Potential applied across the capacitor, V= 400 V

(a) Electrostatic energy stored in the capacitor, U
       
              $$\begin{gathered} \boxed{\mathrm{U} = \frac{1}{2}{\mathrm{CV}}^2} \\ = \frac{1}{2}\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}{\mathrm{V}}^2 \\ = \frac{1}{2}\frac{\left(8.85 \times {10}^{-12}\right) 90 \times {10}^{-4} (400{)}^2}{2.5 \times {10}^{-3}} \\ = \frac{8.85 \times 90 \times 16}{2 \times 2.5}\times {10}^{-9} \\ = 2.55 \times {10}^{-6}\mathrm{J} \end{gathered}$$ 

(b) Volume of the medium in between the plates
                            $$\begin{gathered} = \mathrm{A} \times \mathrm{d} \\ = 90 \times {10}^{-4} \times 2.5 \times {10}^{-3} \\ = 225 \times {10}^{-7}{\mathrm{m}}^3 \end{gathered}$$

Energy stored per unit volume, u =$\frac{2.55 \times {10}^{-6}}{225 \times {10}^{-7}}$
                                            $\mathrm{u} = 0.113 {\mathrm{Jm}}^{-3}$

So, relation between magnitude of electric field and u
          
         $u = \frac{\mathrm{U}}{\mathrm{A}. \mathrm{d}} = \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{CV}}^2}}{\mathrm{A}. \mathrm{d}} = \frac{1}{2}\frac{{\mathrm{V}}^2}{\mathrm{A} .\mathrm{d}}\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}} \left[\because C={\epsilon }_{\circ }A/d\right]$             

                       $u = \frac{1}{2}{\mathrm{\epsilon }}_0 \frac{{\mathrm{V}}^2}{{\mathrm{d}}^2}$          

                      $\boxed{\mathrm{U} = \frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2}$             [$\because$$\mathrm{E} = \frac{\mathrm{V}}{\mathrm{d}}$]