# Energy stored in a capacitor.

by Per Oni
Tags: capacitor, energy, stored
 P: 262 In the thread about permanent magnets it is stated that power per unit volume is E.J As you perhaps saw it is quite a job to prove that fact in the case of magnets. I thought it should be a lot easier to prove that in the electrical equivalent case of 2 opposite charged plates. My back of envelope calculation went as follows: Let one plate approach the other with a constant (low) velocity and collide. Now, according to Gauss’ law E between the plates is q/Aε. E remains constant until the gap is nearly closed, I will ignore the last micro meter of distance where E vanishes. Next: J=q/At. Put together: (E is parallel to J) P=E.J x vol=E x q/At x Vol so that energy W=E x q x d , where Vol=A x d. But here E x d = U, then W=qU. So at first sight not a bad result except that the result should be 1/2qU. Where’s the rub? I think I know but what do you think?
P: 303
 Quote by Per Oni In the thread about permanent magnets it is stated that power per unit volume is E.J As you perhaps saw it is quite a job to prove that fact in the case of magnets. I thought it should be a lot easier to prove that in the electrical equivalent case of 2 opposite charged plates. My back of envelope calculation went as follows: Let one plate approach the other with a constant (low) velocity and collide. Now, according to Gauss’ law E between the plates is q/Aε. E remains constant until the gap is nearly closed, I will ignore the last micro meter of distance where E vanishes. Next: J=q/At. Put together: (E is parallel to J) P=E.J x vol=E x q/At x Vol so that energy W=E x q x d , where Vol=A x d. But here E x d = U, then W=qU. So at first sight not a bad result except that the result should be 1/2qU. Where’s the rub? I think I know but what do you think?
The problem is, you are putting energy in the system unknowingly, when you are making the plate to move at constant velocity !

Otherwise you would have to integrate to get the result, which would have put 1/2, that your calculation is missing.
P: 262
 Quote by universal_101 The problem is, you are putting energy in the system unknowingly, when you are making the plate to move at constant velocity ! Otherwise you would have to integrate to get the result, which would have put 1/2, that your calculation is missing.
That was one of my thoughts as well. But consider the fall of raindrops. They are falling after a short while with a constant velocity without putting energy in the system!

However, I should of course have done this calculation properly and used integration. Perhaps I will do in the weekend. In the meantime I’m convinced that the answer lays elsewhere.

 P: 262 Energy stored in a capacitor. I had a go doing the calculation without keeping v constant, but my maths is not up to it. With v constant we have: (say the +ve plate is travelling, E parallel with J, ignore edge fields) P/Vol = E.J dP = E I dl = E dq/dt dl = E dq v dF = E dq Since q remains constant, so does F therefore: F = E q W = int F dl = q int E dl = qU I hope somebody will do this calculation with v as a function of time or distance, perhaps it will give the correct answer after all.
 P: 262 http://www.youtube.com/watch?v=7NUbsQt-G9U For the explanation. The lecturer states that the charge is emerged in an E field which goes from E max to zero and therefore the average value is ½ E, which is one way of looking at it. I prefer the view that the +ve charge can only be attracted by the –ve plate and the field of the –ve plate is only ½ E max.

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