Charge placed between two metal plates

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Homework Help Overview

The problem involves two infinite conducting plates separated by a distance L, with a point charge q located between them at a distance x from one of the plates. The task is to find the charges induced on each plate, but the problem lacks information about the potential of the plates, leading to confusion regarding the setup and solution.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the applicability of the method of images, questioning whether image charges would be necessary for both plates given the lack of information about their potentials.
  • Some participants express uncertainty about how to set up image charges without knowing if the plates are grounded or have a specific potential difference.
  • Concerns are raised about the implications of the induced charges being negative, with discussions on the neutrality of the plates and the expected behavior of charges in response to the external point charge.
  • Alternative methods, such as Green's reciprocity theorem, are suggested as potentially more effective than the method of images.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Participants are questioning assumptions and considering different approaches without reaching a consensus. Some guidance has been offered regarding the use of potentials and the implications of charge distribution.

Contextual Notes

Participants note the ambiguity in the problem regarding the potentials of the plates, which affects the setup of the equations for image charges. There is also a discussion about the implications of the induced charges being negative and the neutrality of the plates.

issacnewton
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Hi

Here is the problem.

Two infinite conducting plates 1 and 2 are separated by distance L. A point charge q is located between the plates , at a distance x from plate 1. Find the charges induced on each plate.

This is a problem from "Problems in general physics" by Igor Irodov. The answers given are

q_1=-\frac{q(L-x)}{L} ,\;\;q_2=-\frac{q x}{L}

Now I am a bit confused by this problem. Problem doesn't say anything about the plates 1 and 2. We don't know what potential they are maintained at. Whether they are grounded or not. So that makes it more confusing. But the answers given are very simple. Can we use method of images here ? any hints ?

thanks
 
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I'd use the method of images at first sight ... but this is a bit like being between two mirrors: would the image charge in one plate also have an image in the other plate? Or am I just being mean?
 
Simon, we don't know the potentials of the two plates. How can we set up the equations for image charges ?...plates are not necessarily grounded...
 
Use metadata...
Does it make sense in the context of the question for the plates to be grounded and/or have an arbitrary potential difference maintained between them? Surely you've done this sort of problem before?
 
Well I have done standard image problems...like ones in Griffiths. But here how do we set up the image charges ?... Usually in image problems we have some information on the potentials...
 
Strange - that's seldom been the case for me.
Like I said - use the meta data: what makes sense from the context of the question?

This sort of decision making is real common IRL - you very seldom have all the information explicitly laid on for you. But you always have more implicit information than you think.

eg. one bit of metadata is that Igor Irodov is unlikly to be lying about the answer and his problem implies that you are able to solve it with the information provided and your understanding of physics.

Clearly you are not used to this ... so practise: get used to it. Just propose some potentials out of the air and see what you get. Hint: don't work too hard - try making the plates have the same potential as each other first.
 
one more problem which I see with the solution is that both q_1,\;q_2 are negative. Now if the metal plates are neutral initially, and if we bring this charge q from some where else and place it there, the total induced charge is negative. Which does not seem right. Plates being electrically neutral, there must be positive charge on the plates
on the other side. Would this be correct ?
 
IssacNewton said:
one more problem which I see with the solution is that both q_1,\;q_2 are negative. Now if the metal plates are neutral initially, and if we bring this charge q from some where else and place it there, the total induced charge is negative. Which does not seem right. Plates being electrically neutral, there must be positive charge on the plates
on the other side. Would this be correct ?

This should not be so troublesome for you. With an infinite perfect conductor, an external field (like that of a point charge) with force the charge on the conductor to move around so that it no longer experiences any net force. This means that some of the positive charge may be moved off to the edges of the plate (at infinity), essentially removing it from the picture. If you add the two charges given in the answer together, you will get -q, which is exactly what you should expect by a quick application of using the uniqueness theorem to solve the potential on the other side of either plate.


As for solving the problem at hand, you could use the method of images, but I know from experience that you will end up with 2 infinite sums for the potential, and so obtaining the charge distribution will be mathematically non-trivial.

A better method is to use Green's reciprocity theorem; for the first distribution, use the one given in the problem, and for the second distribution, remove the point charge and have one of the plates at a higher potential V_0.
 
As for solving the problem at hand, you could use the method of images, but I know from experience that you will end up with 2 infinite sums for the potential, and so obtaining the charge distribution will be mathematically non-trivial.
Hence the hint in my first post ... I had hoped to get there sooner than this :)
 

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