For a parallel-plate capacitor, the capacitance is C = ε0 A / d. If the plate separation d is halved while area stays the same, what happens to C?

• A. Doubles
• B. Halves
• C. Quarters
• D. Remains the same

Answer: (A)

This question tests how a parallel-plate capacitor’s capacitance changes with plate separation. Since C = ε0 A / d, the capacitance is inversely proportional to the distance between the plates. Keeping the area the same and halving the separation makes the denominator half as large, so the capacitance becomes ε0 A / (d/2) = 2 ε0 A / d = 2C. In other words, the capacitor can store twice as much charge for the same voltage. The result would not be the same, nor would it decrease; it doubles.