Geometric scaling of a capacitor — effect on capacitance and selectivity Q A capacitor of capacitance C and selectivity factor Q is uniformly scaled down in all linear dimensions by a factor K (same dielectric, same frequency). What are the new capacitance and selectivity?

Introduction / Context:
Uniform geometric scaling (all lengths multiplied by K) is a common thought experiment to understand how electrical parameters change. For a parallel-plate-like capacitor with the same dielectric, we can deduce how capacitance and the quality/selectivity factor Q scale at a fixed frequency.

Given Data / Assumptions:

• All linear dimensions (length, width, separation, conductor thickness) scaled by K.
• Dielectric constant unchanged; same operating frequency.
• Q is defined by Q = X_c / ESR at that frequency (or equivalently 1/DF).

Concept / Approach:
For a parallel-plate model, C = ε A / d. With A → K^2 A and d → K d, the new capacitance is C′ = ε (K^2 A) / (K d) = K C. For ESR dominated by conductor resistance, R ∝ length / area; under uniform scaling, length → K L and cross-sectional area → K^2 A_cs, so R′ = (K L)/(K^2 A_cs) = R / K. The reactance scales as X_c′ = 1/(ω C′) = X_c / K. Thus, Q′ = X_c′ / R′ = (X_c / K) / (R / K) = X_c / R = Q.

Step-by-Step Solution:
Compute C′ via C′ = ε A′ / d′ with A′ = K^2 A, d′ = K d → C′ = K C.Assess ESR scaling: R′ = R / K for uniform geometry scaling.Evaluate reactance: X_c′ = X_c / K at fixed ω.Compute Q′ = X_c′ / R′ = Q → select “K C, Q”.

Verification / Alternative check:
Dimensionless Q depends on the product ω C R; with C′ R′ = (K C)(R / K) = C R, Q remains invariant under uniform scaling.

Why Other Options Are Wrong:
Other scalings for C conflict with C ∝ A/d; Q scaling proposed in alternatives does not follow from consistent R and X_c transformations.

Common Pitfalls:
Forgetting that both plate area and separation scale; overlooking that conductor resistance decreases with uniform downscaling.

Final Answer:
K C, Q