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.
 A: 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. 

source: http://demo.webassign.net/ebooks/cj6demo/pc/c18/read/main/c18x18_7.htm
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 $qV$.
A: 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.
A: 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.
A: 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. 
