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2 Capacitors and energy density
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[QUOTE="Shreya, post: 6825105, member: 689347"] 1) Considering the extreme case, I can see that ##E## of ##C_1## must be small, since the same charge will be spread over a larger area ##E = \frac \sigma \epsilon##. And since ##E## is squared (unlike Volume(##=Ad##) in Equation 1, low ##E## would mean lower ##U## :oldsurprised:. $$C= \epsilon \frac {A}{d}$$ 2) I know this formula for Capacitance of a Parallel plate capacitor. So greater area would mean greater ##C## and therefore lower ##U##. 3)Lower ##E## would mean lower potential drop for the same plate seperation. And since ##V## is squared, its effect overrides that of increase in C, implying lower Energy. [USER=681522]@Steve4Physics[/USER], I have modified my arguments using your suggestion. Could you please verify them. [/QUOTE]
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Introductory Physics Homework Help
2 Capacitors and energy density
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