A 4.00 µF capacitor is connected to a 18.0 V battery. How much energy is stored in the capacitor?

This should be a relatively easy problem yet, I can't get it. Isn't it just E=.5*C*V^2

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  1. E = (1/2) C V^2, yes

    Remember that the "mu" symbol means "micro- or 10^-6

    E = (1/2)*4*10^-6*(18)^2
    = 6.48*10^-4 J

    Keep three significant figures.

    If your answer sheet or book does not agree with that answer, it is wrong.

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  2. awesome thanks, that's the correct answer. I was using the wrong conversion. :)

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