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?
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ACorrect
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BIncorrect
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CTrue only if area decreases too
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DTrue only for electrolytics
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EDepends only on frequency
Answer
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