Capacitor plate area and stored charge — choose the correct completion: The ______ the area of the plates, the greater the number of electrons and ______ that can be stored (all else equal).

Introduction / Context:
Capacitance measures a component’s ability to store electric charge per unit voltage. Two key geometric factors influence capacitance in the parallel-plate model: the plate area and the separation distance. This item asks you to complete a sentence about how plate area affects stored charge.

Given Data / Assumptions:

• Parallel-plate intuition: C ≈ epsilon * A / d.
• Voltage across the capacitor is fixed for comparison.
• Material permittivity (epsilon) and plate separation (d) unchanged.

Concept / Approach:
From C = epsilon * A / d, increasing the plate area A increases capacitance C linearly if other variables are constant. For a given applied voltage V, stored charge is Q = C * V, so larger C means proportionally larger Q. Therefore, a larger plate area allows more charge to be stored.

Step-by-Step Solution:

Verification / Alternative check:
Practical capacitors that target higher capacitance within the same volume often increase effective surface area (e.g., etched/roughened foils, multi-layer structures) to realize greater charge storage per volt.

Why Other Options Are Wrong:

• Smaller, charge: Reducing area reduces capacitance and stored charge.
• “Polarity” options: Polarity refers to terminals’ sign, not quantity of stored charge; plate area does not change “polarity.”

Common Pitfalls:
Confusing energy with charge; forgetting that charge scales with capacitance at fixed voltage; ignoring that dielectric properties and distance also affect C.

Final Answer:
larger, charge.