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We assume that the electron enters the plate capacitor with a constant velocity. If we also assume that the electric field is homogeneous and edge effects can be neglected, a force along the y-axis accelerates the electron and increases its kinetic energy. However, as the plates are stationary and cannot move towards the electron, where does this energy come from? Of course electrical potential energy is converted into kinetic energy, but as the electron exits its source infinitely far away, its potential energy is zero. and as it moves perpendicularly to the homogeneous field lines, no work is done. Are these assumptions invalid?

I know that similar questions were already asked, but there were no comments on the effects at the edges of the capacitor, where the field is inhomogeneous, if it is not idealised. Is this an important fact to be considered? image of the setup

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    $\begingroup$ The electron moves in the y direction, and the electric field lines are in the y direction. So the work would just be W = EeΔy. $\endgroup$
    – John Rennie
    Commented Nov 27, 2022 at 11:40

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the electron comes out of the source with a constant velocity from infinitely far away - that is true, but once it enters the homogeneous electric field, there will be a constant force acting on the charge as F=qE. This force does work on the electron, accelerating it and providing a velocity in the y direction, we can also calculate the difference in the equipotential lines the electron passes through to find the loss of potential energy and the subsequent increase in kinetic energy.

Yes, from infinitely far away as you have shown, the electron is at the 0 V line, and it will start of with no EPE - an increasing then constant force will cause the electron to gain an increasingly negative EPE, thus having a gain in kinetic energy. An offset start will allow the electron to begin with a negative EPE in the first place as equipotential lines stretch out infinitely.The effects at the corner for decently strong charged plates is negligible, as the change in curl of field lines will not be too much, so it is not important to practically consider it

You can see the equipotential lines and the field lines

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I now found the exact same question from user Chronial, answered by user Mark Eichenlaub in 2013. The answer is great and uses the edge effects to explain why energy conservation is not violated. The field is compared to a dipole field. As the question is the same, I propose to close this thread (I might not have a high enough reputation score.). I thank everyone for their contribution!

It can be found here: Energy Gain with capacitor?

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