A parallel-plate capacitor with plate area A and separation d in vacuum carries charges ±Q. Which set of expressions correctly gives its capacitance C and the potential difference V?

A parallel-plate capacitor with plate area A and separation d in vacuum carries charges ±Q. Which set of expressions correctly gives its capacitance C and the potential difference V?

• A. C = ε0 A / d; V = Q / C
• B. C = A / d; V = Q / ε0 A
• C. C = ε0 d / A; V = Q / C
• D. C = ε0 A d; V = Q / C

Answer: (A)

Capacitance is how much charge per unit voltage a capacitor can store, determined by the plate area, the separation, and the medium between the plates. For a parallel-plate capacitor in vacuum, the field between the plates is uniform: E = σ/ε0 with σ = Q/A. The potential difference is V = E d, giving V = (Q/(ε0 A)) d. Since C = Q/V, this leads to C = ε0 A / d. Consequently, V = Q / C = Q d /(ε0 A). So the correct expressions are C = ε0 A / d and V = Q / C. The other forms either omit ε0, invert A and d, or mix them incorrectly, which would not match the established relation V = Q/C.

Capacitance is how much charge per unit voltage a capacitor can store, determined by the plate area, the separation, and the medium between the plates. For a parallel-plate capacitor in vacuum, the field between the plates is uniform: E = σ/ε0 with σ = Q/A. The potential difference is V = E d, giving V = (Q/(ε0 A)) d. Since C = Q/V, this leads to C = ε0 A / d. Consequently, V = Q / C = Q d /(ε0 A). So the correct expressions are C = ε0 A / d and V = Q / C. The other forms either omit ε0, invert A and d, or mix them incorrectly, which would not match the established relation V = Q/C.