Capacitance of a parallel-plate capacitor with dielectric constant κ, plate area A, and separation d is:

Capacitance of a parallel-plate capacitor with dielectric constant κ, plate area A, and separation d is:

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

Answer: (A)

Capacitance is the amount of charge stored per unit voltage, Q/V. For a parallel-plate capacitor with a dielectric between the plates, the electric displacement is D = ε E, where ε = κ ε0. The field between plates is E = V/d, and the surface charge density is σ = Q/A = D. Combining these gives Q = σ A = ε E A = κ ε0 (V/d) A. Therefore the capacitance is C = Q/V = κ ε0 A / d. This shows how inserting a dielectric increases the ability to store charge by a factor of κ compared to vacuum, where C would be ε0 A / d. The remaining expressions would not have the correct κ dependence or would mix the factors in a way that changes the dimensions or the physical meaning.

Capacitance is the amount of charge stored per unit voltage, Q/V. For a parallel-plate capacitor with a dielectric between the plates, the electric displacement is D = ε E, where ε = κ ε0. The field between plates is E = V/d, and the surface charge density is σ = Q/A = D. Combining these gives Q = σ A = ε E A = κ ε0 (V/d) A. Therefore the capacitance is C = Q/V = κ ε0 A / d.

This shows how inserting a dielectric increases the ability to store charge by a factor of κ compared to vacuum, where C would be ε0 A / d. The remaining expressions would not have the correct κ dependence or would mix the factors in a way that changes the dimensions or the physical meaning.