The capacitance of a parallel plate capacitor

Step 1: Analyze the question We need to find the true statement about the capacitance of a parallel plate capacitor.

Step 2: The concept For a parallel plate capacitor:

C = ε₀εᵣA / d

• A = area of the plates
• d = separation between the plates
• εᵣ = relative permittivity (dielectric constant) of the material between the plates
• For a vacuum or air, εᵣ ≈ 1. For any dielectric, εᵣ > 1.

Capacitance depends only on the geometry (A and d) and the material between the plates (εᵣ). It does not depend on the charge or the potential difference.

Step 3: Check each option

• A. Decreases when temperature decreases. Temperature is not a factor in the formula, so this is not the standard answer. ✗
• B. Increases when p.d. increases. Capacitance is a constant for a given capacitor. Increasing V increases the charge Q proportionally (C = Q/V), but C stays the same. ✗
• C. Greater without a dielectric. This is the opposite of the truth. Removing the dielectric sets εᵣ = 1, which lowers C. ✗
• D. Greater with a dielectric than without. Since εᵣ > 1 for a dielectric, C = ε₀εᵣA/d is larger than the value without it. ✓

Step 4: Answer The correct option is D.

Tip: Remember "dielectric = bigger capacitance." The dielectric reduces the effective electric field between the plates, so the plates can hold more charge for the same voltage. Also remember that C never depends on Q or V. It depends only on A, d and the dielectric.