The distance between the plates of a charged parallel-plate...

For an isolated charged capacitor, the charge $Q$ on the plates remains constant.

The capacitance of a parallel-plate capacitor is given by $C = \dfrac{\epsilon_0 A}{d}$. As the distance $d$ between the plates increases, the capacitance $C$ decreases.

The potential difference across the capacitor is $V = \dfrac{Q}{C}$. Since $Q$ is constant and $C$ decreases, the potential difference $V$ increases.

The energy stored in the capacitor is $U = \dfrac{Q^2}{2C}$. Since $C$ decreases, the energy $U$ increases.

The electric field between the plates is $E = \dfrac{V}{d} = \dfrac{Q}{A \epsilon_0}$. Since $Q$ and $A$ are constant, the electric field $E$ remains constant.

The energy density between the plates is $u = \dfrac{1}{2} \epsilon_0 E^2$. Since $E$ is constant, the energy density remains constant.

Therefore, the energy of the capacitor and the potential difference across it will change.