Basic capacitor design principle (electronics): Which action increases the capacitance of a parallel-plate capacitor, all else being equal (same plate area and dielectric unless noted)?

Introduction / Context:Capacitance measures a capacitor’s ability to store charge per unit voltage. For a parallel-plate capacitor, it depends on plate area, separation, and the dielectric constant. Understanding how geometry affects capacitance is fundamental in analog design, filters, and energy storage applications.

Given Data / Assumptions:

• Parallel-plate capacitor model.
• Plate area A and dielectric constant epsilon are fixed unless explicitly changed.
• We consider geometric separation d between plates.

Concept / Approach:The ideal formula is C = epsilon * A / d, where epsilon = epsilon0 * epsilon_r. Thus, for fixed A and epsilon, the only geometric variable is distance d. Reducing d increases C proportionally; increasing d reduces C. Applied voltage does not change capacitance in the linear region (it changes charge Q = C * V, not C itself).

Step-by-Step Solution:Start with C = epsilon * A / d.Identify how to increase C: decrease d or increase A or epsilon_r.Among the given choices, only “move plates closer” decreases d, increasing C.

Verification / Alternative check:Consider a numeric example: let epsilon * A = k. If d = 2 mm, C = k/2. Halve the distance to 1 mm → C doubles to k/1. This matches the inverse relationship.

Why Other Options Are Wrong:(a) Increases d → reduces C. (c) and (d) alter V, which affects Q but not C for linear dielectrics. (e) is invalid because a correct action is listed.

Common Pitfalls:Confusing stored charge with capacitance; thinking that a higher voltage increases capacitance. Capacitance is a geometry and material property, not a function of applied V in ideal capacitors.

Final Answer:Moving the plates closer together.