Re: Bad monitor discharge instructions

>As I said before, Capacitors store charge (Coulombs), not Energy (Joules)...
>Of course the charge in the tube does carry energy, so:
>
>Q = C * V
>V = E / Q
>
>so E = Q * V = C * V^2...
>
>So assuming your .001 F measurement is correct, and using V=20,000
>E = 400,000J Hmmm... that'll keep you warm at night...

Whoa! That's not quite right, you're off by a factor of 2.
...(hint, I posted this a week ago)...but, so here's the *real* derivation...

Let's look at the energy required to transfer a unit of charge (dQ) through
a potential difference in a capacitor

Q = C * V (you got this one)

so to transfer this charge through V we have work, or:

dW = dQ * V

substituting for V

dW = dQ * Q/C

now for a bit of calculus to
integrate the equation from 0 to V

E = {(Q dQ)/C} = (1/C) * (Q^2/2) = (Q^2)/(2C) = (CV^2)/2

so, energy stored in a cap at steady state is:

E = 1/2 * C * V^2 ; where V is voltage across cap at the steady state.

Enjoy!
-Anders.
Received on Mon Dec 20 10:48:46 1999