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ExamsJEE AdvancedPhysics

Two identical square metal plates, each of side 1 m, are held 0.01 m apart (like a parallel-plate capacitor) with one pair of edges perpendicular to and touching an oil surface. The plates are connected to a 500 V battery. They are then lowered vertically into the insulating oil at a steady speed of 0.001 m/s. Calculate the current drawn from the battery during this process. (Dielectric constant of oil K = 11, epsilon₀ = 8.85 * 10⁻¹² C² N⁻¹ m⁻¹) Express in nanoampere.

  1. 40.7 nA
  2. 45.0 nA
  3. 36.3 nA
  4. 48.7 nA

Correct answer: 40.7 nA

Solution

As the plates descend, the oil-filled portion grows at rate v = 0.001 m/s. The capacitance increases at a constant rate dC/dt = epsilon₀ * (K-1) * (1 m width) * v / d. The battery current is I = V * dC/dt.

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