If plate area is doubled in a vacuum parallel-plate capacitor, what happens to the capacitance?

If plate area is doubled in a vacuum parallel-plate capacitor, what happens to the capacitance?

• A. It becomes four times as large
• B. It stays the same
• C. It becomes half as large
• D. It becomes twice as large

Answer: (D)

In a vacuum, the capacitance of a parallel-plate capacitor is C = ε0 A / d. If you double the plate area while keeping the separation d the same, the ratio A/d doubles, so C becomes twice as large. This also means that for the same voltage you can store twice as much charge (Q = C V). The separation and the vacuum permittivity stay constant, so area is the factor that changes capacitance here. If a dielectric were present, the relation would include the relative permittivity, but the dependence on area would remain the same.

In a vacuum, the capacitance of a parallel-plate capacitor is C = ε0 A / d. If you double the plate area while keeping the separation d the same, the ratio A/d doubles, so C becomes twice as large. This also means that for the same voltage you can store twice as much charge (Q = C V). The separation and the vacuum permittivity stay constant, so area is the factor that changes capacitance here. If a dielectric were present, the relation would include the relative permittivity, but the dependence on area would remain the same.