In a parallel plate capacitor, the distance between the plates is $$d$$ and potential difference across plates is $$V.$$ Energy stored per unit volume between the plates of capacitor is

Solution :
Energy stored, in parallel plate capacitor is given by $$U = \frac{1}{2}\frac{{{q^2}}}{C}$$
$$\eqalign{ & {\text{but}}\,\sigma = \frac{q}{A}\,\,\,{\text{and}}\,\,C = \frac{{{\varepsilon _0}A}}{d} \cr & \therefore U = \frac{1}{2}\frac{{{{\left( {\sigma A} \right)}^2}}}{{\left( {\frac{{{\varepsilon _0}A}}{d}} \right)}} \cr & {\text{or}}\,\,U = \frac{{A{\sigma ^2}d}}{{2{\varepsilon _0}}} \cr & {\text{or}}\,\,U = \frac{1}{2}{\left( {\frac{\sigma }{{{\varepsilon _0}}}} \right)^2} \times {\varepsilon _0}Ad \cr & {\text{or}}\,\,U = \frac{1}{2}{E^2}{\varepsilon _0}Ad \cr} $$
Energy stored per unit volume i.e. energy density is thus given by
$$\eqalign{ & u = \frac{U}{V} = \frac{U}{{Ad}} = \frac{1}{2}{\varepsilon _0}{E^2} \cr & \Rightarrow u = \frac{1}{2}{\varepsilon _0}{\left( {\frac{V}{d}} \right)^2} = \frac{1}{2}{\varepsilon _0}\frac{{{V^2}}}{{{d^2}}} \cr} $$