Energy Gain with capacitor?

I have a question about energy gain in capacitors. Assume the following system:

As the electron gets accelerated inside the capacitor, it will have more kinetic energy coming out than going in. But the capacitor has not lost any energy. Where is the energy coming from?

Update: Sry, too hasty – forgot important infos: All this is (of course ^^) happening in vacuum and the plates are connected by a non-conducting material. So they can only move together. But there is no way for them to exchange (or in any other way gain/loose) electrons.

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I know very little about this, but wouldn't you expect that the capacitor would have started moving towards the right (very, very slowly)? – Keep these mind Apr 24 '13 at 11:27
@Gugg While I can't speak for all capacitors, an electron tends to have a lot less mass. I mean, to a degree that we can ignore the capacitors movement. – Alan Rominger Apr 24 '13 at 11:33
Yes it will and also Capacitance will go down , since it will be pulling the +ve plate less than pushing the -ve plate and that will further decrease the potential energy stored in the capacitor configuration – user23503 Apr 24 '13 at 11:36
@nonagon Newton's Third Law - it doesn't matter what the relative charges are. The forces are the same. You could say that the electron accelerates so fast that the capacitor also accelerates appreciably, but not really. The center of mass stays in the same place and it's impossible for the capacitor to move an appreciable amount because it has nuclei which universally dwarf the mass of an electron. – Alan Rominger Apr 24 '13 at 11:49
The plates can’t get closer because they are attached to something that is keeping them apart. And as I said – the plate construct and the electron are accelerated in opposite directions. – Chronial Apr 24 '13 at 12:03

Edge effects. After the electron leaves the capacitor, the electric field winds up slowing it back down.

Let's assume the capacitor is infinitely-massive and that the acceleration of the electron is small enough that we can ignore radiation.

Then if you were to idealize the electric field of the capacitor, treating it as a uniform field between the plates and zero elsewhere, then the electron that comes in from the side would pick up some energy and we'd have a violation of energy conservation.

However, the idealized E-field does not obey Maxwell's equations. The true E-field can be written as a gradient of some potential, and that potential in free space is smooth because it's a solution to Laplace's equation. The E-field that abruptly goes from zero outside the capacitor to a constant inside the capacitor clearly does not derive from the gradient of a smooth potential.

Since the E-field is the gradient of a potential, obviously energy is conserved. By the time the electron gets far away from the capacitor, it is back to the same kinetic energy it had to begin.

That's the answer to the question - energy is conserved for the electron because it's conserved in general for charged particles moving in a potential - but to see it in some detail, think of the field far from the capacitor as a dipole.

If you superimpose your drawn electron trajectory with this picture, you'll see that when the electron leaves the capacitor, it moves roughly the same direction as the field lines point. That is, the dot product of the field lines and the velocity is positive. Since an electron has negative charge, this means the electron is losing energy.

So the electron will pick up energy as it comes from far away and enters the capacitor, but lose that energy as it leaves again. This is because the capacitor field is an electrostatic field and can be described by a potential $V$, and basic EM tells us that energy is conserved in such a situation if we give the electron potential energy $qV$.

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I also assume that the correct explanation for this question has something to do with edge effects. But I’m not quite convinced by your answer: If it’s all about potential, wouldn’t I create a massive amount of potential by charging a capacitor in a universe full of electrons? And when I superimpose my image on yours, the velocity and field lines look very orthogonal to me ^^. – Chronial Apr 24 '13 at 22:52
Your comment about a universe full of electrons does not make sense to me. Are you saying that charging a capacitor gives potential energy to electrons that are far away? No, because the potential goes to zero far away. The velocity of the electron is not orthogonal to the field lines. I don't know how you can believe that. It obviously is not orthogonal. – Mark Eichenlaub Apr 24 '13 at 23:05
We are both talking about β here, right? sic-instincts.de/img/fb81bcfd.png – If charging the capacitor does not create potential for everything in the universe and the whole thing is all about potential, I could still bring an infinite amount of electrons to my capacitor without much of an energy investment, couldn’t I?. I don’t really understand where, when and how potential energy comes into play here to solve the problem. – Chronial Apr 25 '13 at 8:14
If you try really hard, and pick one particular special point, then yes you can find one point where the velocity and field are perpendicular. Trace the path a little further and they won't be. Yes, you could repeatedly send electrons through the capacitor. Each one would speed up then slow down again. – Mark Eichenlaub Apr 25 '13 at 14:06
Looking at the picture I would say the are quite perpendicular right after they leave the capacitor and will then slowly get tangential towards infinity. But won’t the field be really weak before that happens? – Chronial Apr 25 '13 at 15:05

The energy is of course coming from the electric field of the capacitor. The energy of any capacitor is always stored in it's electric field. If an electron is initially positioned very far away and then moves close to the capacitor, it's being pulled by the field and that means energy is being transferred. The electric field get's a little weaker - loosing energy - while the electron gains kinetic energy. After it passed by and moves away it slows down, it's KE is transferred back into the EF. Eventually it's KE will be the same value as it had initially.

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Why would it slow down once it’s outside the Field of the capacitor? – Chronial Apr 24 '13 at 14:52
Then as capacitor energy got restored who gave the energy to the particle. How is this different from my post ? , I simply said adding to the fact, the energy is restored when the particle escapes to infinity ,making its energy again 0 and then capacitor's electric field got restored. – user23503 Apr 24 '13 at 15:52
@nonagon And again the last line is incorrect. While spider's ans can be considered using a concept that you have said you don't consider. Well spidy has also not give that in his answer,. – ABC Apr 24 '13 at 16:13

When we're calculating the energy stored in a capacitor we normally assume it is isolated i.e. there are no other charges nearby to affect it. This makes the calculation nice and simple: the energy is proportional to $Q^2$ and the energy is stored in the electric field around the capacitor.

However in your question you are introducing another charge, your electron, and this charge will generate it's own electric field. So the field will be the sum of the field from the capacitor and the field generated by the electron. When the electron is accelerated by the field between the capacitor plates the kinetic energy it gains comes from the energy of the total electric field.

To take a simpler example than the one you give, suppose we put the electron near the negative plate and move it towards the positive plate. To make life easy we'll suppose we extract the kinetic energy so the electron ends up near the positive plate. Using your diagram this looks like:

In it's starting position the field from the electron reinforces the field from the capacitor, but in it's final position the field from the electron opposes the field from the capacitor. That means the total field strength, and therefore the energy stored in it, has been reduce by moving the electron. The reduction in the field energy is equal to the energy we took out.

The situation you described in your question is more complicated than this because although the electron starts near the negative plate, as in my simplified example, now the electron starts in motion and finishes in motion in a different direction and with greater energy. To be honest I don't know how to calculate the field from a moving electron, but I'm confident that that if you add up the field from the capacitor and the field from the electron you'll find that the total field energy has decreased by an amount equal to the extra kinetic energy acquired by the electron.

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The point is that the electron doesn't gain any kinetic energy, at least not when it's far away from the capacitor again. In fact it would lose some kinetic energy due to radiation. – Mark Eichenlaub Apr 24 '13 at 20:38
The field energy can’t decrease as there are still the same electrons on the plates and the moving electron is still the same. Once the electron is far enough from the plates, the fields will be exactly the same as before. @Mark Eichenlaub: What kind of radiation would that be? But still if there is radiation we have radiation energy – still more energy than at the beginning? – Chronial Apr 24 '13 at 22:34
@Chronial The radiation is electromagnetic. If we account for the radiation, the electron will lose kinetic energy that goes into the radiation. Energy is conserved. There won't be any more than at the beginning. The electron will actually end up with less kinetic energy than it started with. – Mark Eichenlaub Apr 24 '13 at 23:03
@MarkEichenlaub: a little OT, but how does the electron loose speed to em radiaton? I see that it has to give off em, but how does that slow down the electron? Does that effect maybe have a nice name that I can wiki? :) (btw: I know that energy is conserved ^^. My question is not if, but how that works in the given situation) – Chronial Apr 25 '13 at 8:15

Capacitor is losing energy, potential has changed as field is created even by this charge which is moving under the influence of force between capacitor plates . Take the point charge's potential , and then assume distance between capapcitor plate is d, now as -ve charge approaches +ve plate, it decreases the potential of the +ve capacitor plate more than it compensates for the -ve plate (applying $-kq/r$). As $V$ of capacitor goes down, $CV^2/2$ goes down as well.

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Comments are not intended for extended discussion, and all the comments under this post have been removed. Please (1) be nice and (2) take it to chat if you can't settle it in a few comments. – dmckee Apr 24 '13 at 17:45
The charge on each plate does not change. Ignore the field for a moment. Assuming the capacitor isn't connected to anything else, the plates are completely isolated. There is nowhere for the charge to go. You can change the field with the electron and push / pull the charge on the plates to move the charge around a bit but the total charge on each plate isn't going anywhere until the plates are connected and charge can flow between them. I think this system can be thought of as the electric equivalent of a magnet where no net work is done by the field itself. – Brandon Enright Apr 24 '13 at 23:24
Kindly read it is potential,not charge . potential due to moving charge and principle of superposition. and I conveyed the exact same reasoning , somehow people didn't agree to that but did when somebody else posted the same . and a lot of comments were deleted which carried more reasoning regarding the answer . – user23503 Apr 25 '13 at 20:59