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

In summary, the potential energy of the capacitor doesn't change if a charged particle enters and exists the system.
  • #1
Harperchisari
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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.
 

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  • The “Redstone Torch” - The Parallel Plate Problem_.pdf
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  • #2
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.
 
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  • #3
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.
 
  • #4
Vagn said:
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.
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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.
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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.
 
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  • #5
Harperchisari said:
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:
capacitor-equipotential-lines-jpg.jpg
 
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  • #6
Vagn said:
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.
 
  • #7
Vagn said:
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) 😂
 
  • #8
Harperchisari said:
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.
 
  • #9
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.
 
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  • #10
Ibix said:
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.
 
  • #11
Harperchisari said:
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.
 
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  • #12
Ibix said:
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.
 
  • #13
Harperchisari said:
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.
 

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

1. What is the energy conservation issue with parallel charged plates with a hole?

The energy conservation issue with parallel charged plates with a hole is that there is a loss of energy due to the presence of the hole. This is because the electric field lines are disrupted and not all of the electric potential energy is converted into kinetic energy of the charges.

2. How does the presence of a hole affect the energy conservation in parallel charged plates?

The presence of a hole in parallel charged plates disrupts the electric field lines, causing a loss of energy. This is because the electric potential energy is not fully converted into kinetic energy of the charges, leading to a decrease in the overall energy conservation.

3. Can the energy conservation issue be solved by simply filling the hole in the parallel charged plates?

No, filling the hole in the parallel charged plates will not completely solve the energy conservation issue. While it may reduce the loss of energy, it will not eliminate it entirely. This is because the electric field lines will still be disrupted to some extent, resulting in a loss of energy.

4. How does the distance between the plates affect the energy conservation issue?

The distance between the plates does not directly affect the energy conservation issue. However, it can indirectly impact the issue by affecting the strength of the electric field and the amount of disruption caused by the hole. A larger distance between the plates may result in a weaker electric field and a smaller loss of energy.

5. Are there any practical applications of the energy conservation issue with parallel charged plates with a hole?

Yes, the energy conservation issue with parallel charged plates with a hole has practical applications in various fields such as electronics, particle accelerators, and capacitors. Understanding and controlling the loss of energy in these systems is crucial for their efficient functioning.

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