The expression for the capacitance of a parallel-plate capacitor with plate area A, separation d, and a dielectric with permittivity ε is:

The expression for the capacitance of a parallel-plate capacitor with plate area A, separation d, and a dielectric with permittivity ε is:

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

Answer: (A)

Capacitance in a parallel-plate arrangement depends on how much charge you can store per volt, and it grows with the plate area while shrinking as the plates get farther apart. The dielectric between the plates changes the field through its permittivity ε. Between the plates, the electric field is E = σ/ε, where σ is the surface charge density Q/A. The potential difference is V = E d = (Q/(A ε)) d. The capacitance is C = Q/V, which gives C = ε A / d. If the dielectric is vacuum, ε becomes ε0, yielding C = ε0 A / d. The other forms misplace the factors of ε, A, or d, so they don’t give the correct proportionalities for C.

Capacitance in a parallel-plate arrangement depends on how much charge you can store per volt, and it grows with the plate area while shrinking as the plates get farther apart. The dielectric between the plates changes the field through its permittivity ε.

Between the plates, the electric field is E = σ/ε, where σ is the surface charge density Q/A. The potential difference is V = E d = (Q/(A ε)) d. The capacitance is C = Q/V, which gives C = ε A / d.

If the dielectric is vacuum, ε becomes ε0, yielding C = ε0 A / d.

The other forms misplace the factors of ε, A, or d, so they don’t give the correct proportionalities for C.