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