(c) Energy of the capacitor.

Because the charge always remains conserved in an isolated system, it will remain the same.

Now,

$V=\frac{Q d}{\epsilon_{0} A}$

Here, $\mathrm{Q}, \mathrm{A}$ and $\mathrm{d}$ are the charge, area and distance between the plates, respectively.

Thus, as d increases, V increases. Energy is given by $E=\frac{q V}{2}$

So , it will also increase.

Energy density u, that is, energy stored per unit volume in the electric field is given by $u=\frac{1}{2} \in_{0} E^{2}$

So, u will remain constant with increase in distance between the plates.