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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

  1. Aq₁ - q₂ C
  2. Bq₁ + q₂ C
  3. Cq₁ - q₂ 2C
  4. 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

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