Geometric effect on capacitance: when the plate area of a capacitor increases (all else unchanged), what happens to its capacitance?

Introduction / Context:
Capacitance depends on geometry and dielectric properties. Designers frequently adjust plate area and spacing to tune capacitance values for filters, timing circuits, and energy storage. Recognizing the direct proportionality between plate area and capacitance is fundamental.

Given Data / Assumptions:

• Parallel-plate capacitor model applies.
• Plate separation and dielectric constant remain fixed.
• Only the effective plate area A is increased.

Concept / Approach:
The ideal relation is C = ε * A / d, where ε is the permittivity of the dielectric, A is plate area, and d is separation. With ε and d constant, capacitance scales linearly with A: larger plate area yields larger capacitance.

Step-by-Step Solution:

Verification / Alternative check:
Practical capacitors: doubling electrode area approximately doubles capacitance for the same dielectric and thickness. This is consistent with data sheets where larger packages (greater effective area) exhibit higher capacitance for identical dielectrics and thicknesses.

Why Other Options Are Wrong:

• The capacitance decreases / unaffected: Oppose the C ∝ A relation.
• The voltage it can withstand increases: Breakdown voltage depends primarily on dielectric strength and thickness (d), not area.

Common Pitfalls:

• Confusing breakdown voltage with capacitance; they are governed by different parameters.
• Ignoring fringing and construction details; however, first-order behavior remains C ∝ A.

Final Answer:
the capacitance increases