If the plate separation d is doubled while Q and plate area remain the same, what is Ef/Ei for the parallel-plate capacitor in the ideal case?

• A. 1
• B. 2
• C. 0.5
• D. 4

Answer: (A)

When a parallel-plate capacitor is isolated with a fixed charge, the energy stored is U = Q^2/(2C). Doubling the plate separation lowers the capacitance by a factor of two (C → C/2). Substituting gives final energy Ef = Q^2/[2(C/2)] = Q^2/C, which is twice the initial energy Ei = Q^2/(2C). So the energy increases by a factor of two, and Ef/Ei = 2. Intuitively, the voltage doubles because V = Q/C, and with the same charge, a smaller capacitance means a larger voltage, and the energy U = (1/2) QV becomes twice as large.