Question

The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?
(a) Charge on the capacitor
(b) Potential difference across the capacitor
(c) Energy of the capacitor
(d) Energy density between the plates

Answer 1

(b) Potential difference across the capacitor
(c) Energy of the capacitor.

Because the charge always remains conserved in an isolated system, it will remain the same.
Now,
V=$\frac{Qd}{{\in }_{0}A}$
Here, Q, A and d are the charge, area and distance between the plates, respectively.
Thus, as d increases, V increases.
Energy is given by
E = $\frac{qV}{2}$
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.