For the same capacitor with Q fixed, if the plate side length L is doubled (so area increases by a factor of 4), what is 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

For the same capacitor with Q fixed, if the plate side length L is doubled (so area increases by a factor of 4), what is Ef/Ei?

Explanation:
With the charge fixed, increasing the plate area fourfold makes the capacitance four times larger. Since V = Q/C, the final voltage drops to one-fourth of the initial voltage. The energy stored is U = 1/2 Q V, so the final energy is U_f = 1/2 Q (V_i/4) = U_i/4. Thus the ratio Ef/Ei is 1/4, or 0.25.

With the charge fixed, increasing the plate area fourfold makes the capacitance four times larger. Since V = Q/C, the final voltage drops to one-fourth of the initial voltage. The energy stored is U = 1/2 Q V, so the final energy is U_f = 1/2 Q (V_i/4) = U_i/4. Thus the ratio Ef/Ei is 1/4, or 0.25.

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