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Charge placed between two metal plates

  1. Jun 28, 2012 #1
    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

    [tex]q_1=-\frac{q(L-x)}{L} ,\;\;q_2=-\frac{q x}{L}[/tex]

    Now I am a bit confused by this problem. Problem doesn't say anything about the plates 1 and 2. We dont 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
     
  2. jcsd
  3. Jun 28, 2012 #2

    Simon Bridge

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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?
     
  4. Jun 28, 2012 #3
    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.....
     
  5. Jun 28, 2012 #4

    Simon Bridge

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    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?
     
  6. Jun 28, 2012 #5
    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.......
     
  7. Jun 28, 2012 #6

    Simon Bridge

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    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.
     
  8. Jun 28, 2012 #7
    one more problem which I see with the solution is that both [itex]q_1,\;q_2[/itex] 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 ?
     
  9. Jun 28, 2012 #8

    gabbagabbahey

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    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 [itex]-q[/itex], 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 [itex]V_0[/itex].
     
  10. Jun 28, 2012 #9

    Simon Bridge

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    Hence the hint in my first post ... I had hoped to get there sooner than this :)
     
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