Which of the following actions will increase capacitance of a parallel plate capacitor? I. Decreasing the…

Analyze the question: We need to find which actions increase the capacitance of a parallel plate capacitor.

Concept: The capacitance of a parallel plate capacitor is

C = ε₀εᵣA / d

where

• A = overlapping area of the plates
• d = distance between the plates
• εᵣ = relative permittivity (dielectric constant) of the material between the plates
• ε₀ = permittivity of free space

So C is directly proportional to A and εᵣ, and inversely proportional to d.

Check each statement:

• I. Decreasing the distance between the plates: d decreases, so C increases. ✔
• II. Increasing the distance between the plates: d increases, so C decreases. ✘
• III. Increasing the area of plate overlap: A increases, so C increases. ✔
• IV. Avoiding the use of a dielectric: The material between the plates becomes air or vacuum (εᵣ ≈ 1), which is the lowest value. So C decreases. ✘

Only I and III increase the capacitance.

Answer: A. I and III only

Exam tip: Remember "C is Big when Area is Big and distance is small." Any dielectric multiplies the capacitance, so removing it always reduces C. In questions like this, check each statement against C = ε₀εᵣA/d.