If the distance between the plates of a parallel plate capacitor is decreased, its capacitance

Analyze the question: We are asked what happens to the capacitance of a parallel plate capacitor when the distance between its plates is decreased.

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

C = ε₀εᵣA / d

where:

• C = capacitance
• 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) when A and the dielectric stay the same.
2. If d decreases, the fraction A/d gets larger, so C increases.
3. Example: if d is halved, C doubles. If d goes from 4 mm to 2 mm, C becomes twice its original value.

Checking the options:

• A. Becomes zero: this would only happen if d became infinitely large, so it is wrong.
• B. Decreases: this is the opposite of what happens, so it is wrong.
• C. Increases: this is correct.
• D. Is halved: this would happen if d were doubled, so it is wrong.
• E. Remains the same: C depends on d, so it is wrong.

Tip: Remember "close plates, big capacity." Capacitance is directly proportional to plate area and dielectric constant, and inversely proportional to plate separation. Bringing the plates closer allows more charge to be stored for the same voltage, because the opposite charges attract and hold each other more strongly.