How does increasing plate area A affect the capacitance of a parallel-plate capacitor at fixed separation and dielectric?

How does increasing plate area A affect the capacitance of a parallel-plate capacitor at fixed separation and dielectric?

• A. C ∝ A^2
• B. C ∝ 1/A
• C. C ∝ A
• D. C is independent of A

Answer: (C)

In a parallel-plate capacitor, the ability to store charge per volt scales with the overlapping plate area. With fixed separation d and dielectric ε, the capacitance is C = εε0 A / d. This means doubling the plate area doubles the capacitance, because each unit area behaves like its own tiny capacitor and these add in parallel. So increasing A increases C linearly, and the constant of proportionality εε0/d sets how large C becomes for a given area. The other options would require nonlinear, inverse, or no dependence on area, which contradicts the formula and the parallel arrangement.

In a parallel-plate capacitor, the ability to store charge per volt scales with the overlapping plate area. With fixed separation d and dielectric ε, the capacitance is C = εε0 A / d. This means doubling the plate area doubles the capacitance, because each unit area behaves like its own tiny capacitor and these add in parallel. So increasing A increases C linearly, and the constant of proportionality εε0/d sets how large C becomes for a given area. The other options would require nonlinear, inverse, or no dependence on area, which contradicts the formula and the parallel arrangement.