The capacitance of a parallel-plate capacitor

Analyze the question: We are asked which statement correctly describes the capacitance of a parallel-plate capacitor.

Concept: The capacitance of a parallel-plate capacitor is

C = ε₀εᵣA / d

where

• A = area of each plate
• d = separation between the plates
• εᵣ = relative permittivity (dielectric constant) of the material between the plates (εᵣ = 1 for air or vacuum, and greater than 1 for any dielectric)
• ε₀ = permittivity of free space

Check each option:

• A. Decreases when the separation decreases. Wrong. C is inversely proportional to d, so decreasing d increases C.
• B. Increases when the potential difference is increased. Wrong. Capacitance depends only on the geometry (A, d) and the medium (ε). When V increases, the stored charge Q increases proportionally, so C = Q/V stays the same.
• C. Greater without a dielectric than with one. Wrong. It is the other way round, since εᵣ > 1 for a dielectric.
• D. Greater with a dielectric than without. Correct. Inserting a dielectric multiplies the capacitance by εᵣ, so C_with = εᵣ × C_without, which is larger.

Why the dielectric helps: The dielectric's molecules become polarised and reduce the effective electric field between the plates. For the same charge, the potential difference drops, and since C = Q/V, the capacitance rises.

Exam tip: Remember "C is for Close, Constant-free, and Clever material." In other words, capacitance goes up with larger area, smaller separation, and a dielectric, and it does not depend on the voltage or charge.