Parallel-plate formula check: Capacitance C is directly proportional to the dielectric constant (relative permittivity) and the plate area A, and inversely proportional to the plate separation d (C = ε * A / d). Is the claim “directly proportional to dielectric constant and distance, inversely to area” correct?

Difficulty: Easy

Correct Answer: Incorrect

Explanation:


Introduction / Context:
The geometry of a capacitor determines its capacitance. For the common parallel-plate model, a simple formula relates material properties and dimensions. This question tests whether the stated proportionalities match the well-known relationship used in design and analysis.


Given Data / Assumptions:

  • Parallel-plate approximation with negligible fringing.
  • Dielectric fully fills the region between plates.
  • Relative permittivity ε_r multiplies ε_0 to yield ε.


Concept / Approach:
The canonical formula is C = ε * A / d, where ε = ε_r * ε_0. Capacitance increases with plate area A and dielectric constant ε, and decreases as the separation d increases. Therefore, any statement claiming direct proportionality to distance and inverse to area is the exact opposite and is incorrect for this standard geometry.


Step-by-Step Solution:

State the formula: C = ε * A / d. Identify proportionalities: C ∝ ε and C ∝ A; C ∝ 1/d. Compare with the claim: it reverses the roles of A and d. Conclude the claim is incorrect for parallel plates.


Verification / Alternative check:
Dimensional and physical intuition: larger area gives more field-coupled region (higher C); greater distance weakens the field coupling (lower C). Measurements of variable capacitors confirm these trends.


Why Other Options Are Wrong:
Correct only for cylindrical/fringing cases: the sign of dependence does not reverse for standard configurations; while exact formulas differ, C does not increase with distance in such ideal models.


Common Pitfalls:
Mixing up proportionalities; overlooking that dielectric constant boosts capacitance by concentrating the electric field (higher ε reduces effective field for the same charge, enabling more Q per volt).


Final Answer:
Incorrect

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