Two identical thin metal plates has charge $q_1$ and $q_2$ respectively such that $q_1>q_2$. The plates were brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is :

On bringing the charged metal plates closer, electric field $\overrightarrow{E}$ in the intervening space is $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$

Where

Intensity of field due plate charged by $q_1$ is $\left|\overrightarrow{E_1}\right| =\frac{{\sigma }_1}{2{\epsilon }_0} = \frac{q_1}{2{\epsilon }_{0}A}$(directed rightwards)

And Intensity of field due to plate charged by $q_2$ is 

$\left|\overrightarrow{E_2}\right| =\frac{{\sigma }_2}{2{\epsilon }_0} = \frac{q_2}{2{\epsilon }_{0}A}$ (directed leftwards)
So, Net field is given by 

$\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2} \Rightarrow E = E_1- E_2$

$\therefore E =\frac{q_1}{2{\epsilon }_{0}A}-\frac{q_2}{2{\epsilon }_{0}A}=\frac{\left(q_1-q_2\right)}{2{\epsilon }_{0}A} ...\left(1\right)$

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

From above two equations 

$E=\frac{\left(q_1-q_2\right)}{2C}d ...\left(3\right)$

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

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

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

$V = \frac{\left(q_1-q_2\right)}{2C}$