Question

Two identical metal plates are given positive charges Q₁ and Q₂(< Q₁) respectively. If they are now brought close to gather to form a parallel plate capacitor with capacitance c, the potential difference between them is 

(A) [(Q₁ + Q₂)/(2c)] 

(B) [(Q₁ + Q₂)/c] 

(C) [(Q₁ – Q₂)/(2c)] 

(D) [(Q₁ – Q₂)/c]

Answer 1

The correct option (D) [(Q₁ – Q₂)/c]

Explanation:

when metal plates are brought close, charges induced on first plate is 

– Q₂ and on second plate is – Q₁.

∴ Two plates have charges:

Q₁ – Q₂ and Q₂ – Q₁

i.e. Q₁ – Q₂ and – (Q₁ – Q₂)

∴ charge on either plate is Q₁ – Q₂

∴ Potential difference = V = [(charge on either plate)/(capacitance)]

= [(Q₁ – Q₂)/c]