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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.
- 150 microjoules
- 100 microjoules
- 200 microjoules
- 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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