Gauss' law, method of images, and force induced on conductor plates

  • #1
Mainframes
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Hi, I would very much appreciate some guidance on the below.

Consider a one-dimensional world as depicted in the attached figure.
  • We have two (lets say positively charged particles enclosed by two conductor plates.
  • One plate is at ##x=0##, the other at ##x=L##. The particles are at ##x_1## and ##x_2##, with respective charges ##q_1## and ##q_2##

From my understanding:
  • The postively charged particles will induce a charge on the conductor plates
  • The positively charged particles will not feel a net force from the charge induced on the conductor plates due to Gauss' law (the conductor plates enclose the particles in this one-dimensional model)
  • The conductor plates will experience a force from the charged particles which can be decuded using the method of images. This is calculated by assuming a ficticious oppostively charged particle is at the opposite side of the plate. The conductor plate feels a force from each of the particles by effectively assuming the particle was doulbe the distance away from the conductor plate (and the conudctor plate has opposite charge)
  • ##k## is Coulombs constant in the attached image

The part I am struggling with is:
  • Does the conductor plate also feel a force from the other opposite conductor plate? If so, how do I calculate it? Would it somehow be equal to the force the ficticuous mirror charges would have on each other?

Your assistance is greatly appreciated.
 

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  • #2
With two parallel conductor plates, you will need to keep mirroring. You will effectively end up with an infinite set of mirror charges.
 
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  • #3
Orodruin said:
With two parallel conductor plates, you will need to keep mirroring. You will effectively end up with an infinite set of mirror charges
Im assuming you’re not suggesting that is the answer for the force is infinite (unless I’ve misunderstood)? Do you know if there is a force and how it is calculated or if it is zero by some argument? Appreciate the guidance
 
  • #4
I hope he's suggesting that Method of Images is not going to be so helpful ion this situation.

Method of Images is a trick to save you from a bunch of nasty sums and integrals when establishing the boundary conditions. If it too has a bunch of nasty sums and integrals, you might think about other techniques. (And I am saying this despite being a big fan of Method of Images)
 
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  • #5
Vanadium 50 said:
I hope he's suggesting that Method of Images is not going to be so helpful ion this situation.

Method of Images is a trick to save you from a bunch of nasty sums and integrals when establishing the boundary conditions. If it too has a bunch of nasty sums and integrals, you might think about other techniques. (And I am saying this despite being a big fan of Method of Images)
Thank you for the reply and makes sense (I will rule out method of images). So am I correct that you agree the conductor plates will indeed feel a force from each other (given the induced charges from the particles)? And that in order to calculate this I will need to do some more detailed analysis. Feedback is appreciated.
 
  • #6
Mainframes said:
Im assuming you’re not suggesting that is the answer for the force is infinite (unless I’ve misunderstood)? Do you know if there is a force and how it is calculated or if it is zero by some argument? Appreciate the guidance
An infinite sum can have a finite result. In a reflection with two conductors you need to also appropriately mirror the other conductor for appropriate boundary conditions so you end up with a new case of two conductors with twice the distance in between - so you keep mirrorring and end up with an infinite set of mirror charges, but the force sum is finite.

Vanadium 50 said:
I hope he's suggesting that Method of Images is not going to be so helpful ion this situation.
Well, that depends, it ”works” in the sense that it gives you an expression for the solution and you might be able to get some insight from it. The problem of course being that it is an infinite sum. However, other methods such as Fourier analysis also result in an (different) infinite sum. By uniqueness of the solution, those infinite sums must be equal, which can also lead to insight and it is not immediately obvious which is easier. All infinite sums are not born equal.
 
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  • #7
Sure, it "works" in that it gives you the correct answer, but trading one infinite series for another might or might not be a good thing.

I would, in this case, go back to boundary conditions - that's what this trick is supposed to help you with. Inside the plates you have a field ~1/r2 and outside it is zero. I would work out the energy density and this the energy in the field and then the force on a plate is dE/dz.

I would deal with one plate at a time, and try and use some symmetries.

I'd still have an integral to do, and it is not as nice as it might be as the field has spherical symmetry and the plates do not. So be it.
 
  • #8
Orodruin said:
An infinite sum can have a finite result. In a reflection with two conductors you need to also appropriately mirror the other conductor for appropriate boundary conditions so you end up with a new case of two conductors with twice the distance in between - so you keep mirrorring and end up with an infinite set of mirror charges, but the force sum is finite.


Well, that depends, it ”works” in the sense that it gives you an expression for the solution and you might be able to get some insight from it. The problem of course being that it is an infinite sum. However, other methods such as Fourier analysis also result in an (different) infinite sum. By uniqueness of the solution, those infinite sums must be equal, which can also lead to insight and it is not immediately obvious which is easier. All infinite sums are not born equal.
Thank you for this reply and I believe I have understood your point. Am I correct you are saying that the force on each plate is corrected by adding an additional term of the form (see attached image) $$\sum_{i=1}^{\infty}\frac{kQ_i}{[(i+1)L]^2}$$
Here ##Q_i## is the induced charge from the ##i-1##-th mirror charge. ##Q_1## is the induced charge from the two original particles.

Your argument here is that the denominator of the above sum is increasing and therefore the sum converges (where I'd mistakenly believed it to be infinite leading me to question my understanding of the method of images). In the below image I have added superscript ##LHS## and ##RHS## to denote the conductor plate.

Assuming I have understood your argument (and if I have then thank you very much), then is it fair to say that ##Q_i## also decreases as ##1/i##, potentially making the sum converge in ##i^3##
 

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  • #9
As I said, I am less than sure Method of Images is best here, but you can see what happens best with numbers. I have plates at z= 1 and the positive charge at z = 0. I then have a negative image at z = ±2. That gives me a positive charge at ±5. And so on.

So your force dur to the images will be 1/4 - 1/25 ...

It will be finite even though the series is infinite. Sometimes things are clearer when you put numbers in.
 
  • #10
Vanadium 50 said:
As I said, I am less than sure Method of Images is best here, but you can see what happens best with numbers. I have plates at z= 1 and the positive charge at z = 0. I then have a negative image at z = ±2. That gives me a positive charge at ±5. And so on.

So your force dur to the images will be 1/4 - 1/25 ...

It will be finite even though the series is infinite. Sometimes things are clearer when you put numbers in.
Sincere apologies but I am still unclear of what is happening (though the infinite summ converging is clear). It appears my previous post was incorrect where I was assuming the induced charge on the plate would induce extra charge on the opposing plate and so forth.

I'm trying to understand your method more closely. I believe you are starting from the charge and then finding an infinite series of mirror charges from there. I understand that the z=0 positive charge has negative image at z=+/-2. I am not sure how you then get a positive image at z=+/-5. Did you use the conductor plate at z=-1 to generate that (and would that not give you a positive image at z=+/-4? In that case, the force due to the images would be 1/4+1/16 though. I'm not sure how you get 1/4-1/25 (the minus sign and the denominator)

Am conscious of your time spent, specially as my questions might be trivial. Happy to be referred to do some reading if you think that is wiser. Thank you for the help so far
 
  • #11
You are insisting on using a particular technique which is bad in this case.

Yes, with multiple conductors you have images of images (or images).

If I have misunderstood your scenario and not put the charges and plates in the right places, I am sorry. But this reinforces my position that this is not the best way to solve this.
 
  • #12
Thanks, I am sure any misunderstanding will be on my part. I will look into your method (any reading or tips would be appreciated).

One reason why the Method of Images solution (well images of images) was appealing is that I am looking at a 1-dimensional toy-model. If I can give a formulaic solution to the problem (without a full calculation) and argue that the image of images term is small enough and does not qualitatively impact my result, then that would be a step forward. That is why the infinite sum approach appealed to me.

To give a bit more colour, I want to compare the force on the plates for two different configurations for the particles. Given that each configuration would involve images of images, then I might have a stronger argument why this level of detailed calculation can be deferred for now.
 
  • #13
Mainframes said:
1-dimensional toy-model
I don't know what that even means. In 1-D di you even have Coulomb's Law/.
 
  • #14
Vanadium 50 said:
I don't know what that even means. In 1-D di you even have Coulomb's Law/.
This is a fair comment. Perhaps it would have been more accurate to say a 3-D model where I am interested in resolving the force in 1-D only.
 
  • #15
Here is a simpler question that will still help me. Imagine I have a single positively charged particle between two conductor plates. Each conductor plate is square and the particle is constrained to move between the two plates (if the particle was touching one of the plates it would be at the center of the square).

Where does the particle need to be so that the sum of the forces on the conductor plates due to each other is a maximum (please do not take into account the force on the conductor plate due to the charged particle itself as I understand that part)? Is it when the particle is closest to one of the plates or is it when the the particle is half-way between the plates?
 
  • #16
If the particle is half way between the plates, in what direction is the force? What does that tell you about the magnitude of the force?
 
  • #17
Vanadium 50 said:
If the particle is half way between the plates, in what direction is the force? What does that tell you about the magnitude of the force?
He is talking about the force on the plates, not on the particle. That force is not necessarily zero by symmetry.

That said, unless the plates are approximately infinite, mirrorring will definitely not be a good idea. So I cannot se the point of specifying the plates to be squares.
 
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  • #18
Orodruin said:
He is talking about the force on the plates, not on the particle. That force is not necessarily zero by symmetry.

That said, unless the plates are approximately infinite, mirrorring will definitely not be a good idea. So I cannot se the point of specifying the plates to be squares.
I am happy to also investigate the case where the plates are approximately infinite (at least to check if the form of the infinite sum I posted earlier is correct). I have been doing some reading based on all the insightful comments here which has helped - thank you.
 

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