Geometry–capacitance relation: with all else equal (dielectric and plate area), does increasing the plate separation reduce the capacitance value of a parallel-plate capacitor?

Difficulty: Easy

Correct Answer: Correct

Explanation:

Introduction / Context:Capacitance depends on physical geometry and dielectric properties. For parallel-plate capacitors, plate area, separation, and dielectric constant determine the value. This item checks the inverse relationship between separation and capacitance.

Given Data / Assumptions:

  • Parallel-plate model with uniform dielectric.
  • Plate area and dielectric constant fixed.
  • Fringing effects neglected for the basic relation.

Concept / Approach:The ideal formula is C = ε * A / d, where ε is permittivity, A is plate area, and d is plate spacing. As d increases, A and ε fixed, C decreases proportionally. This is independent of frequency in the ideal electrostatic model (though real parts exhibit parasitics at high frequency).

Step-by-Step Solution:

1) Start from C = ε * A / d. 2) Treat ε and A as constants; vary only d. 3) Increasing d increases the denominator, reducing C.

Verification / Alternative check:Doubling spacing halves capacitance in the first-order model; measurements on adjustable capacitors show inverse proportionality for modest changes.

Why Other Options Are Wrong:“Incorrect / depends only on frequency / only for electrolytics” each conflicts with the geometric dependence; the relation is technology-agnostic.

Common Pitfalls:Forgetting that dielectric constant changes also affect C; over-interpreting fringing at extreme geometries.

Final Answer:Correct

More Questions from Capacitors

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion