Energy stored in a capacitor.


by Per Oni
Tags: capacitor, energy, stored
Per Oni
Per Oni is offline
#1
Sep4-12, 03:00 PM
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?
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universal_101
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#2
Sep4-12, 03:27 PM
P: 285
Quote Quote by Per Oni View Post
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.
Per Oni
Per Oni is offline
#3
Sep5-12, 02:37 PM
P: 262
Quote Quote by universal_101 View Post
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.

Per Oni
Per Oni is offline
#4
Sep8-12, 06:02 AM
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.
Per Oni
Per Oni is offline
#5
Sep10-12, 05:45 AM
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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