The capacitance of a parallel-plate capacitor with vacuum, area A, and plate separation d is?

The capacitance of a parallel-plate capacitor with vacuum, area A, and plate separation d is?

• A. C = ε0 A / d
• B. C = d /(ε0 A)
• C. C = ε0 /(A d)
• D. C = ε0 A d

Answer: (A)

Capacitance measures how much charge a device can store per volt, and for a parallel-plate capacitor the ability to store charge depends on the plate area and the distance between plates, with a vacuum between them. The electric field between the plates is uniform: E = σ/ε0 = Q/(ε0 A). The potential difference is V = E d = Q d /(ε0 A). The capacitance is C = Q/V, which gives C = ε0 A / d. This matches the intuition: larger area provides more surface to hold charge, so larger C; larger separation weakens the field coupling, so smaller C. The units work out to farads, since ε0 has units F/m, multiply by A (m^2) and divide by d (m) yields F. The other proposed forms would invert the distance dependence or give incorrect units, so they don’t describe the capacitor in vacuum.

Capacitance measures how much charge a device can store per volt, and for a parallel-plate capacitor the ability to store charge depends on the plate area and the distance between plates, with a vacuum between them. The electric field between the plates is uniform: E = σ/ε0 = Q/(ε0 A). The potential difference is V = E d = Q d /(ε0 A). The capacitance is C = Q/V, which gives C = ε0 A / d. This matches the intuition: larger area provides more surface to hold charge, so larger C; larger separation weakens the field coupling, so smaller C. The units work out to farads, since ε0 has units F/m, multiply by A (m^2) and divide by d (m) yields F. The other proposed forms would invert the distance dependence or give incorrect units, so they don’t describe the capacitor in vacuum.