The capacitance of parallel plates capacitors varies

Analyze the question: It asks how the capacitance (C) of a parallel plate capacitor depends on the distance (d) between the plates.

Concept: The capacitance of a parallel plate capacitor is given by:

C = ε₀εᵣA / d

where

• ε₀ = permittivity of free space
• εᵣ = relative permittivity of the dielectric between the plates
• A = area of overlap of the plates
• d = separation between the plates

Solution:

1. Keep A and the dielectric constant.
2. Then C ∝ 1/d, since ε₀, εᵣ and A are constants.
3. This means that if the plates are moved farther apart, the capacitance decreases. If the plates are brought closer, the capacitance increases.

Checking the options:

• A. Inversely as the square of d: wrong, since d is not squared. (That is the pattern for force in Coulomb's law.)
• B. Directly as the square of d: wrong.
• C. Inversely as d: correct, since C ∝ 1/d.
• D. Directly as d: wrong, because increasing d reduces C.

Answer: C

Tip: Remember "C is directly proportional to Area, inversely proportional to distance" (C ∝ A/d). Large plates store more charge, but a larger gap weakens the electric field coupling between the plates, so capacitance falls. Don't confuse this with the inverse-square pattern in Coulomb's law and gravitation.