How would the capacitance of a parallel plate capacitor be affected if the distance of separation of its plates is decreased? It will

Analyze the question: We want to know what happens to the capacitance (C) of a parallel plate capacitor when the plate separation (d) is decreased.

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 material between the plates
• A = area of overlap of the plates
• d = distance between the plates

Solution:

1. From the formula, C is inversely proportional to d (C ∝ 1/d), provided A and the dielectric stay the same.
2. If d decreases, the denominator becomes smaller, so the value of C becomes larger.
3. Example: if d is halved, C doubles.

Physical reason: when the plates are closer, the electric field from one plate is more effective at attracting opposite charge on the other plate. The capacitor can therefore store more charge for the same potential difference, and since C = Q/V, the capacitance rises.

Checking the options:

• A. Increase in value ✔ (matches C ∝ 1/d)
• B. Decrease slightly ✘ (opposite of the true effect)
• C. Remain unchanged ✘ (d is in the formula, so C changes)
• D. Drop to zero ✘ (C would only approach zero if d became very large)

Exam tip: Remember "Area Aids, Distance Dims", meaning capacitance goes up with larger Area and down with larger Distance. So decreasing the distance increases capacitance. Also, inserting a dielectric with a higher εᵣ increases C too.