Electrostatics Practice Test

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

C ∝ A^2 C ∝ 1/A C ∝ A C is independent of A

Correct Answer:

C ∝ A

Explanation:

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.

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