MHT CET202617 April 2026Evening ShiftPhysicsCapacitanceActual
Two indentical metal plates are given charges q₁ and q₂(q₂ < q₁) respectively. They are brought close together to form a parallel plate capacitor with capacitance 'C'. The potential difference 'V' between the plates is
Options
- Aq₁ - q₂ C
- Bq₁ + q₂ C
- Cq₁ - q₂ 2C
- Dq₁ + q₂ 2C
Correct answer
C. q₁ - q₂ 2C
Step-by-step solution
When two identical metal plates are given charges q₁ and q₂ , the charge on the outer surfaces of both plates is q₁ + q₂ 2 . The charge on the inner surface of the first plate is q₁ - q₁ + q₂ 2 = q₁ - q₂ 2 . The charge on the inner surface of the second plate is q₂ - q₁ + q₂ 2 = - ( q₁ - q₂ 2 ) . The charge on the capacitor is the magnitude of the charge on the inner surfaces, which is q = q₁ - q₂ 2 . The potential difference V between the plates is given by V = q C . Substituting the value of q , we get V = q₁ - q