Which statement correctly describes how capacitance depends on plate area and separation for a parallel-plate capacitor?

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Multiple Choice

Which statement correctly describes how capacitance depends on plate area and separation for a parallel-plate capacitor?

Explanation:
Capacitance measures how much charge a capacitor can hold per unit voltage, and for a parallel-plate capacitor the geometry sets this relationship directly. If the plates have area A and are separated by distance d in a dielectric with permittivity ε, the charge on a plate creates a nearly uniform field between the plates. The field is E = Q/(ε A), and the potential difference is V = E d = Q d /(ε A). Solving for capacitance C = Q/V gives C = ε A / d. This means you increase the plate area to store more charge at the same voltage, and you decrease the separation to boost the storage, hence C is proportional to A and inversely proportional to d. In vacuum, ε is ε0 (or ε = ε0 εr in a dielectric). The other statements would imply no dependence on distance, no dependence on area, or dependence on distance in the wrong direction, which contradicts this relationship.

Capacitance measures how much charge a capacitor can hold per unit voltage, and for a parallel-plate capacitor the geometry sets this relationship directly. If the plates have area A and are separated by distance d in a dielectric with permittivity ε, the charge on a plate creates a nearly uniform field between the plates. The field is E = Q/(ε A), and the potential difference is V = E d = Q d /(ε A). Solving for capacitance C = Q/V gives C = ε A / d. This means you increase the plate area to store more charge at the same voltage, and you decrease the separation to boost the storage, hence C is proportional to A and inversely proportional to d. In vacuum, ε is ε0 (or ε = ε0 εr in a dielectric). The other statements would imply no dependence on distance, no dependence on area, or dependence on distance in the wrong direction, which contradicts this relationship.

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