A parallel-plate capacitor has square plates of side length L with area A = L^2 and plate charge ±Q. The electric field between plates is E = Q/(ε0 A). If Q is doubled (A fixed), what is the ratio Ef/Ei?

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Study for the UCF PHY2054 General Physics Exam. Use flashcards and multiple-choice questions complete with hints and explanations. Boost your understanding and get exam-ready!

Multiple Choice

A parallel-plate capacitor has square plates of side length L with area A = L^2 and plate charge ±Q. The electric field between plates is E = Q/(ε0 A). If Q is doubled (A fixed), what is the ratio Ef/Ei?

Explanation:
The key idea is that the electric field between the plates depends on the charge per area. For a parallel-plate capacitor with area A and plate charge ±Q, the field is E = Q/(ε0 A). If you double the charge while keeping the area fixed, the field becomes E' = (2Q)/(ε0 A) = 2E. So the ratio Ef/Ei is 2. Since V = Ed for fixed separation, the voltage would also double.

The key idea is that the electric field between the plates depends on the charge per area. For a parallel-plate capacitor with area A and plate charge ±Q, the field is E = Q/(ε0 A). If you double the charge while keeping the area fixed, the field becomes E' = (2Q)/(ε0 A) = 2E. So the ratio Ef/Ei is 2. Since V = Ed for fixed separation, the voltage would also double.

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