A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are 

After isolation of capacitor, charge is constant.

Capacity,  $\mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$

$\therefore$ Capacitance decreases with increase in distance.

                         $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$

$\therefore$ Potential increases with decrease in capacitance.