A parallel plate capacitor is charged. If the plates are pulled apart  [DCE 2019]

The electric field between the plates is given by

       $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}}\text{ or }\mathrm{V}=\mathrm{Ed} \text{ or }\mathrm{V}\propto \mathrm{d}$

Hence, if the plates are pulled apart, the potential difference increases.