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MHT CET202525 Apr 2025Evening ShiftPhysicsCapacitanceActual

Two identical metal plates are given charges q ₁ and q ₂ ( q ₂ < q ₁ ) respectively. If they are now 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 2 C
  4. Dq₁+q₂ 2 C

Correct answer

C. q₁-q₂ 2 C

Step-by-step solution

When two identical metal plates with charges q₁ and q₂ form a parallel plate capacitor, the charges redistribute such that the outer surfaces carry equal charges Q_ outer = q₁ + q₂ 2 due to symmetry. The inner surface charges are then Q_ inner1 = q₁ - q₁ + q₂ 2 = q₁ - q₂ 2 and Q_ inner2 = q₂ - q₁ + q₂ 2 = q₂ - q₁ 2 , satisfying Q_ inner1 = -Q_ inner2 as required. The effective charge governing the electric field is Q_ eff = |Q_ inner1 | = |q₁ - q₂| 2 . Given q₂ The potential difference is then V = Q_ eff C = q₁ - q

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