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(A) For a parallel plate capacitor,the energy density $u = \frac{1}{2} \varepsilon_0 E^2$. Since the electric field $E$ between the plates is given by

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$\frac{q^2}{2 \varepsilon_0 A^2}$

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(A) For a parallel plate capacitor,the energy density $u = \frac{1}{2} \varepsilon_0 E^2$. Since the electric field $E$ between the plates is given by $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma = \frac{q}{A}$ is the surface charge density. Substituting the value of $E$ in the energy density formula: $u = \frac{1}{2} \varepsilon_0 \left( \frac{\sigma}{\varepsilon_0} \right)^2 = \frac{\sigma^2}{2 \varepsilon_0}$. Now,substituting $\sigma = \frac{q}{A}$: $u = \frac{(q/A)^2}{2 \varepsilon_0} = \frac{q^2}{2 \varepsilon_0 A^2}$.

The mean electrical energy density between the plates of a charged air capacitor is (where $q=$ charge on capacitor,$A=$ area of capacitor plate):

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