In a capacitor with fixed charge Q, increasing plate area A will have what effect on the electric field between the plates?

• A. Decreases
• B. Increases
• C. Stays the same
• D. Doubles

Answer: (A)

With fixed charge on the plates, the strength of the field between them depends on how concentrated that charge is on the plate surfaces. For a parallel-plate capacitor, the field between the plates is effectively set by the surface charge density σ = Q/A, giving E ≈ σ/ε0, or more precisely E ≈ Q/(ε0 A).

So as you increase the plate area while keeping Q the same, the charge is spread over a larger area, reducing σ, and the electric field between the plates decreases. A quick check comes from using C = ε0 A/d, V = Q/C, and E = V/d; this yields E = Q/(ε0 A), which clearly shows E decreases as A increases (doubling A halves E, etc.).