The energy conservation issue with parallel charged plates with a hole.

[Harperchisari]

A while back I thought of an issue with parallel charged plates. Imagine this: a set of opposite charged resistive plates with holes in the center. In theory, there is a finite amount of energy required to push a positive charged particle through the hole in the positive plate (in theory it should be comparatively small). However, due to the relatively uniform electric field in between the plates, the charged particle could garner up to an infinite amount of kinetic energy if the distance is infinite. This creates the issue that it takes a set, finite energy to push the particle into the plate, but, depending on the distance, one could get more energy out than in. While work has to be put into separate the plates, the potential energy of the plates doesn't change if a charged particle enters and exists the system. Even if heat is created by the force on the plates, that doesn't change their charge, and thereby the potential.

I have attached a document to this post which gives a more specific example as well as an image to clarify the problem. I have since run out of people to ask as no seems to be able to figure out what specifically is conceptually the source of the energy.

 

[Harperchisari]

I was considering writing up a triple integral which describes the electric field but from what I remember the equation for a finite plane isn't simplify-able , and wouldn't result in a very useful answer. I have seen quite a few posts regarding conservation of energy within a capacitor, similar to this however those do not seem to pose the "hole" scenario.

 

[Vagn]

I think you are neglecting the electric field outside the capacitor plates. So although released at an initial kinetic energy, the kinetic energy as it passes through the first plate would be less as it feels the field due to the first plate. It is then accelerated whilst in the capacitor, but then loses kinetic energy after passing through the second plate due to the field on this side of the capacitor.

 

[Harperchisari]

I think you are neglecting the electric field outside the capacitor plates. So although released at an initial kinetic energy, the kinetic energy as it passes through the first plate would be less as it feels the field due to the first plate. It is then accelerated whilst in the capacitor, but then loses kinetic energy after passing through the second plate due to the field on this side of the capacitor.

In theory there should be minimal charge on the outside. For an explanation see this Quora article. If the plates didn't affect the field on the outside of the other one, yes all energy would be lost as the particle travels to a point about half the width of the capacitor away from the exit plate. However they do effect one another and the field at the two points w/2 away from each plate on the outside is less than the one point in the center on the inside.

In the above above screenshots from the Phet Charges and Fields simulation ( which can be downloaded here for free) you can see that at the same distance, on the outside you have about 12 V/m, while in the center you have 28.7 V/m. This is 24 on the outside with 28 on the inside, resulting in a net imbalance in energy in and out.

 

[Vagn]

View attachment 241806
In theory there should be minimal charge on the outside. For an explanation see this Quora article.

In reality you have the fringe fields, because the plates are not infinite, which in this case shouldn't be neglected. Instead you end up with equipotentials that look somewhat like the following:

 

[Harperchisari]

In reality you have the fringe fields, because the plates are not infinite, which in this case shouldn't be neglected. Instead you end up with equipotentials that look somewhat like the following:
View attachment 241811

I modified my last statement, look at the excerpts from the Phet simulation. Basically, you're "helped" by the other plate by 2 V/m, with 2 V/m pulling before it reaches the first plate, and 2 V/m pushing after it leaves the second one.

 

[Harperchisari]

In reality you have the fringe fields, because the plates are not infinite, which in this case shouldn't be neglected. Instead you end up with equipotentials that look somewhat like the following:
View attachment 241811

Specifically, note how the equipotential lines are much less dense outside of the capacitor. (Sorry about the rapid edits and replies, Ⅰ just keep thinking of things)

 

[Vagn]

Specifically, note how the equipotential lines are much less dense outside of the capacitor. (Sorry about the rapid edits and replies, Ⅰ just keep thinking of things)

You still have the angular part to consider though (i.e. the equipotential lines are more dense off to the side of the capacitor than directly behind it. Other things to consider would be if the fringe fields would result in a net field between the oppositely charged plates of capacitor 1 and 2 and how that would affect the kinetic energy of the charge.

 

[Ibix]

The mere existence of potential lines is enough to show you what's wrong. The "return" leg in your pdf involves crossing all the same equipotential lines you crossed in the other direction on the "forward" leg.

 

[Harperchisari]

The mere existence of potential lines is enough to show you what's wrong. The "return" leg in your pdf involves crossing all the same equipotential lines you crossed in the other direction on the "forward" leg.

See above comments, and the field in the return portion is significantly mitigated.

 

[Ibix]

See above comments, and the field in the return portion is significantly mitigated.

You cross the same equipotential lines in both directions. You cannot get back to the start otherwise. So it follows trivially that the decrease in energy in one direction is the same as the increase in the other. The rate at which you cross the lines (edit: and the path taken to cross them) is immaterial.

I don't know what mental model you are using, but it's wrong. You are describing a circumstance similar to sledging down a hill then climbing back up the hill to sledge back down. You could theoretically do that as a closed loop if you could eliminate friction entirely, but you cannot extract energy from it.

 

[Harperchisari]

You cross the same equipotential lines in both directions. You cannot get back to the start otherwise. So it follows trivially that the decrease in energy in one direction is the same as the increase in the other. The rate at which you cross the lines (edit: and the path taken to cross them) is immaterial.

I don't know what mental model you are using, but it's wrong. You are describing a circumstance similar to sledging down a hill then climbing back up the hill to sledge back down. You could theoretically do that as a closed loop if you could eliminate friction entirely, but you cannot extract energy from it.

alright, makes sense. Thank you.

 

[Dale]

While work has to be put into separate the plates, the potential energy of the plates doesn't change if a charged particle enters and exists the system

On the contrary, the potential energy does change. With Poynting’s theorem you can easily prove that the amount that it changes is greater than or equal to the work done on the charge.