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ExamsJEE AdvancedPhysics

A parallel-plate capacitor has a metallic conducting plate inserted between its plates, occupying 60% of the gap (leaving 40% of the gap as dielectric/air). Without the plate the capacitance is C = 20 nF. The capacitor is connected to a DC voltage source of 100 V. The metallic plate is then slowly extracted from the gap. Find the mechanical work performed during the extraction.

  1. 150 microjoules
  2. 100 microjoules
  3. 200 microjoules
  4. 250 microjoules

Correct answer: 150 microjoules

Solution

With plate: effective gap = 0.4d, so C_with = epsilon₀*A/(0.4d) = C/0.4 = 50 nF. U_with = (1/2)*50e-9*100² = 250 microJ. U_without = (1/2)*20e-9*100² = 100 microJ. Change in charge: delta_Q = (C_without - C_with)*V = (20-50)*1e-9*100 = -3000 nC. Work by source: W_source = V*delta_Q = 100*(-3000e-9) = -300 microJ (source receives energy). Energy conservation: W_mech + W_source = delta_U => W_mech - 300 = 100 - 250 = -150. W_mech = 150 microJ.

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