On bringing the charged metal plates closer, electric field E in the intervening space is E=E1+E2
Where
Intensity of field due plate charged by q1 is ∣E1∣=2ϵ0σ1=2ϵ0Aq1(directed rightwards)
And Intensity of field due to plate charged by q2 is
∣E2∣=2ϵ0σ2=2ϵ0Aq2 (directed leftwards)
So, Net field is given by
E=E1+E2⇒E=E1−E2
∴E=2ϵ0Aq1−2ϵ0Aq2=2ϵ0A(q1−q2)...(1)
For parallel plate capacitor C=dϵ0A...(2)
From above two equations
E=2C(q1−q2)d...(3)
∵Relation between intensity of field (E) and potential difference (V) between the plates is
E=dV...(4)
From equation (3) and (4)
V=2C(q1−q2)