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Say you have a parallel plate setup, each plate is charged with ±Q of charge, and is then disconnected from the power source and is insulated from the environment. There is no way the plates can be discharged and there is an electric field between the plates.

Now lets say you fire an electron inside the plates, since the electron is charged it will be affected by the electric field and will accelerate towards the positive plates.

Since there is work done on the electron by the electric field, energy of the parallel plate/electric field must be lost, but how can that be since the plates are not moving or losing charge?

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    $\begingroup$ What does you fire an electron inside the plates mean? Do you shoot an electron on the plate? Or is it just suddenly moving around inside the plate? $\endgroup$ – Steeven Feb 7 '16 at 22:45
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    $\begingroup$ If you understand why you don't ask yourself the same question in other situations, you will understand how to answer it here: Consider some object falling to earth. There is work done on it by the gravitational field, yet the Earth does not move or lose mass. $\endgroup$ – ACuriousMind Feb 7 '16 at 22:45
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    $\begingroup$ @ACuriousMind Ohhh I think I understand now. (potential energy from the location of the electron). Thanks for the hint. Should I delete the question now that I know or do I leave it here? $\endgroup$ – Spongebob Squareroot-pants Feb 7 '16 at 22:49
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    $\begingroup$ If you think this question will not be useful to someone else, delete it - if you think someone could benefit from reading the explicit answer, write the answer yourself :) $\endgroup$ – ACuriousMind Feb 7 '16 at 22:50
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    $\begingroup$ "there is work done on the electron by the electric field" Says who? If the electron enters from far away / infinity (zero potential energy) and makes it through the capacitor without hitting anything, it will leave the region headed toward another region of zero potential energy. I.e., it'll have the same energy entering and leaving the region, no net work done. $\endgroup$ – aquirdturtle Feb 7 '16 at 23:45
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Comments from @aquirdturtle have led me to rewrite my answer and to realise that it was a question worth asking.

@ACuriousMind has likened the situation to a mass falling on the Earth. In that case the mass and Earth system loses gravitational potential energy and they both gain kinetic energy although almost all of it resides with the mass.

Carrying on with the gravitational analogy then perhaps it is an example of gravity assist or gravitational slingshot? That idea can be ruled out because the slingshot effect works because not only is the satellite moving but so is the planet.

This sort of set up described was used by Thomson to measure the specific charge $\frac e m$ of the electron where electrons were first accelerated and then deflected by an E-field (and a magnetic field).

So assume that an electron is travelling at constant velocity along a straight line which is parallel to the plates and that a long way away from the plates the electron has zero electric potential energy and some kinetic energy.

The electron is under the influence of the E-field due to the parallel plate arrangement and so experiences a force which accelerates it; thus increasing the kinetic energy of the electron whilst decreasing the potential energy of the electron.

Since the E-field due to a charged parallel plate arrangement is small outside the region of the parallel plates, this change in kinetic energy and the resulting change in the direction of the electron’s velocity are only significant when the electron is between the parallel plates.

As the electron leaves the parallel plate arrangement it will gain potential energy and lose kinetic energy. Eventually the potential energy of the electron will become zero.

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  • $\begingroup$ This is not true if the electron comes in from infinity / far away, i.e. zero potential. The electron's motion would change in the vicinity of the plates, but it would enter and leave the region with the same kinetic energy. $\endgroup$ – aquirdturtle Feb 7 '16 at 23:42
  • $\begingroup$ You are correct that when the electron is a long way away it will not feel the E-field due to the parallel plates but once under the influence of the E-field it will be accelerated and gain kinetic energy. Think of the motion of the electron as a component velocity at right angles to the E-field which does not change because there is no force in that direction and an increasing velocity parallel to the E-field just like projectile motion in a gravitational field with the projectile being thrown horizontally from the top of a cliff. $\endgroup$ – Farcher Feb 8 '16 at 0:12
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    $\begingroup$ I agree that it temporarily gains energy as it enters the region. The point is that as it leaves the region it will lose that kinetic energy as it's potential energy increases, so no net work is done. In the short term the way the capacitor plate loses energy is because the charged particles inside the plate will move around slightly in response to the nearby electron. $\endgroup$ – aquirdturtle Feb 8 '16 at 0:27
  • $\begingroup$ I now understand. I was thinking more of the charge whilst it was between the plates. That is why I used the reverse path situation by starting from between the plates and moving away from them. $\endgroup$ – Farcher Feb 8 '16 at 1:34
  • $\begingroup$ I think you should mention this in your answer. It sounds to me like the OP might think that the capacitor plates do net work on the particle even after the particle leaves the region. Maybe it's just me though. $\endgroup$ – aquirdturtle Feb 8 '16 at 1:39

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