Two identical thin metal plates has charge q _ 1 and q _ 2...

On bringing the charged metal plates closer, electric field $\overset{\to}{E}$ in the intervening space is $\overset{\to}{E} = \overset{\to}{{E}_{1}} + \overset{\to}{{E}_{2}}$

Where

Intensity of field due plate charged by ${q}_{1}$ is \left|\overset{\to}{{E}_{1}}\right| = \dfrac{{\sigma}_{1}}{2{ε}_{0}} = \dfrac{{q}_{1}}{2{ε}_{0}A}$(directed rightwards)</p><p>And Intensity of field due to plate charged by${q}_{2}$is </p><p>\left|\overset{\to}{{E}_{2}}\right| = \dfrac{{\sigma}_{2}}{2{ε}_{0}} = \dfrac{{q}_{2}}{2{ε}_{0}A}$ (directed leftwards)
So, Net field is given by

$\overset{\to}{E} = \overset{\to}{{E}_{1}} + \overset{\to}{{E}_{2}} ⇒ E = {E}_{1} - {E}_{2}$

$∴ E = \dfrac{{q}_{1}}{2{ε}_{0}A} - \dfrac{{q}_{2}}{2{ε}_{0}A} = \dfrac{\left({q}_{1} - {q}_{2}\right)}{2{ε}_{0}A} . . . \left(1\right)$

For parallel plate capacitor $C = \dfrac{{ε}_{0}A}{d} . . . \left(2\right)$

From above two equations

$E = \dfrac{\left({q}_{1} - {q}_{2}\right)}{2C}d . . . \left(3\right)$

$∵$Relation between intensity of field \left(E\right)$and potential difference \left(V\right)$ between the plates is

$E = \dfrac{V}{d} . . . \left(4\right)$

From equation \left(3\right)$and \left(4\right)$

V = \dfrac{\left({q}_{1} - {q}_{2}\right)}{2C}