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 :

Step-by-step solution

On bringing the charged metal plates closer, electric field E → in the intervening space is E → = E 1 → + E 2 → Where Intensity of field due plate charged by q 1 is E 1 →   = σ 1 2 ε 0   =   q 1 2 ε 0 A (directed rightwards) And Intensity of field due to plate charged by q 2 is E 2 →   = σ 2 2 ε 0   =   q 2 2 ε 0 A (directed leftwards) So, Net field is given by E → = E 1 → + E 2 →   ⇒