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…

On bringing the charged metal plates closer, electric field $\vec{E}$E in the intervening space is $\vec{E}=\vec{{E}_{1}}+\vec{{E}_{2}}$E=E1​​+E2​​

Where

Intensity of field due plate charged by ${q}_{1}$q1​ is $|\vec{{E}_{1}}|=\frac{{\sigma }_{1}}{2{\epsilon }_{0}}=\frac{{q}_{1}}{2{\epsilon }_{0}A}$∣E1​​∣=2ϵ0​σ1​​=2ϵ0​Aq1​​(directed rightwards)

And Intensity of field due to plate charged by ${q}_{2}$q2​ is

$|\vec{{E}_{2}}|=\frac{{\sigma }_{2}}{2{\epsilon }_{0}}=\frac{{q}_{2}}{2{\epsilon }_{0}A}$∣E2​​∣=2ϵ0​σ2​​=2ϵ0​Aq2​​ (directed leftwards)

So, Net field is given by

$$\vec{E}=\vec{{E}_{1}}+\vec{{E}_{2}}\Rightarrow E={E}_{1}-{E}_{2}$$E=E1​​+E2​​⇒E=E1​−E2​

$$\therefore E=\frac{{q}_{1}}{2{\epsilon }_{0}A}-\frac{{q}_{2}}{2{\epsilon }_{0}A}=\frac{({q}_{1}-{q}_{2})}{2{\epsilon }_{0}A}...(1)$$∴E=2ϵ0​Aq1​​−2ϵ0​Aq2​​=2ϵ0​A(q1​−q2​)​...(1)

For parallel plate capacitor $C=\frac{{\epsilon }_{0}A}{d}...(2)$C=dϵ0​A​...(2)

From above two equations

$$E=\frac{({q}_{1}-{q}_{2})}{2C}d...(3)$$E=2C(q1​−q2​)​d...(3)

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

$$E=\frac{V}{d}...(4)$$E=dV​...(4)

From equation $(3)$(3) and $(4)$(4)

$$V=\frac{({q}_{1}-{q}_{2})}{2C}$$V=2C(q1​−q2​)​