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How much energy is stored on a 23-μ F capacitor that is charged to a potential difference of 40 Volts?
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The energy in a capacitor is given by \[ U = \frac{1}{2}QV = \frac{1}{2}CV^2 \] Now you have C and V, so you have your answer.
U is the energy. C is the capacitance. V is the voltage. Just make sure you put C and V in SI units.
to get it to F from the u-F is the constant10^6?
micro Farad. Not mega Farad!
so 10^-6
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oh whoops! thanks!
I still am not getting the correct answer I am doing (23*10^-6)*40^2*.5
That's 0.0184 J. What's the supposed answer?
It is saying J isnt the right units
Nevermind I wasnt typing it in correctly i was using 1.84*10^2 instead of -2
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