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Homework Help: Charged particle between two conducting plates

  1. Sep 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Two large conducting plates are separated by a distance 'L', and are connected together by a wire. A point charge 'q' is placed a distance 'x' from one of the plates. Show that the proportion of the charge induced on each plate is 'x/L' and '(L-x)/L'.

    (Hint: pretend the point charge is instead a sheet charge of magnitude 'q' and distance 'x' from one of the plates)

    2. Relevant equations

    None provided. I suppose the field of a sheet charge is required, which is

    'E=q/2[tex]\epsilon[/tex]A'


    3. The attempt at a solution

    I'm not sure where to start. I'm guessing the hint to use a sheet charge is because the field lines from a point charge to a plate are always tangents to the plate.

    Since the plates are connected together by a wire, the potential of both plates is the same and so the potential difference is zero.

    How do I calculate the potential of each plate (I keep getting stuck at this part)? Do the induced charges also have a potential (I think so).

    I believe the sum of the charge on each plate 'qL' + 'qR' = 0 in order for the plates to be at equal potentials.

    Please help! I've been stuck on this for 9 months or so.
     
  2. jcsd
  3. Sep 4, 2010 #2
    The plates start out as electrically neutral, and the two are connected. That means that the total charge on both of them together must stay 0 (Conservation of charge).

    The induced charge on the plate does not spread itself out evenly. You will not get the field of a sheet of charge. The formula you posted is irrelevant.
     
    Last edited: Sep 4, 2010
  4. Sep 4, 2010 #3
    The sheet of charge is mentioned in the question. Maybe the plates do not have even charge distributions, but the set-up can be simulated using the sheet charge as an image of the point charge?
     
  5. Sep 4, 2010 #4
    That's not how you use the method of image charges. You use a point image charge to replace the conducting sheet's charge distribution. In this case, I'm a bit confused since the one solution I could come up with, and was the one I found on google, involved an infinite series of point charges.
     
  6. Sep 5, 2010 #5
    It really is an infinite series of image charges.
    Just the same way that light creates infinite number of images when reflecting between 2 parallel mirrors, the real charge q first creates 2 images of the same charge -q on 2 plates. Then each of the -q charge creates another image on the opposite plate. The cycle goes on infinitely. The total induced charge on each plate = sum of image charges of that plate
    I haven't come up with an easy way to calculate the induced charge. A suggestion: First, from the system of charges, calculate electric field E at one point on the plate. We have the surface charge density at that point = epsilon x E. Then integrate... But I'm not sure if it's possible or not :biggrin:
     
  7. Sep 5, 2010 #6
    OK people, I've got it.

    The two plates are connected by a wire, so they are at equal potential. Note that as the plates are rectangular shaped, so are the equipotentials at each plate. We can call this potential zero.

    The charges induced are qL and qR.

    By the uniqueness theorem, the potential is specified everywhere on the surface(s), and the total charge in the volume is known (q).

    Now we look for the appropriate image problem.

    Imagine then that you have the two plates as before, but now each are grounded. And now you have a sheet charge at a distance 'b' from one of the plates.

    This sheet charge induces a charge qL on the plate at distance b and qR on the other. These induced charges must sum to zero.

    Now we calculate the potential at the left plate due to the sheet charge q and the induced charge qR. We know that this potential is zero from before.

    You'll find this is V= -1/2epsilonA ( q*-b qR*D ) = 0. You can work this out by integrating the electric field from the sheet charge and from the induced charge on the other plate.

    Similarly for the right plate: V= -1/2epsilonA ( q* (D-b) - qL*D) = 0

    hence qR / q = b/D and qL / q = (D-b) / D.

    (I might have some minus signs mixed up on the potential, work through it yourselves!)
     
  8. Sep 5, 2010 #7
    It seems that either you or the question statement are totally wrong. Please check the question statement you posted.
     
  9. Sep 6, 2010 #8
    That's all she wrote boy! :-) Perhaps I should clarify that the plates are large, which I guess can mean that they can be considered tending to infinite area.

    Why is it "totally wrong"? Produce your own solution if you see the error.
     
  10. Sep 7, 2010 #9

    ehild

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    I thought a lot about this problem and I might be wrong, but I feel that the problem can not be solved on the way you presented.

    The plates were connected at the beginning, so the potential is the same on both of them. But the whole system is neutral. There is some induced charge distribution on the inside of both plates, with whole charge -QL on the left plate and -Qr on the right one, and QL+Qr=q on the outer surface. This outer surface charge is distributed evenly producing the surface charge density q/(2A) on the outer surfaces. The magnitude of the electric field is q/(2Aε 0) at both sides, pointing away from the plates. This would be the solution for the outside electric field if a sheet of charge is placed anywhere between the plates. So you can not say anything about the induced charges separately.

    If you ground the plates the neutrality of the plates is broken. There will be no charge at the outer surface, but the induced charges -QL and -Qr stay the same. But you do not have any information, except that QL+Qr=q.

    There is a theorem in electrostatics that the inside of an equipotential surface can be filled with a metal without any change in the field outside the surface. The inside of such block is at the same potential and the electric field is zero, there is no induced charge.

    Your method works if the plates are not connected at the beginning but you happen to know that they are at equal potential.


    ehild
     
  11. Sep 7, 2010 #10
    From the question statement, the situation I understand is like the attached picture. And the solution guide, I have already posted it earlier. But reading your solution, I thought you were referring to some other problem! So there is either something wrong with the problem statement or something wrong with your understanding.

    @ehild: I don't see why the outer charge is evenly distributed.
     

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  12. Sep 7, 2010 #11

    ehild

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    Assume that you have a point charge q at one side of a very large conducting plate.It attracts the opposite charges which accumulate nearby and repels the like ones which appear at the other surface of the plate. What is the electric field in that outer half of space? The electric field is zero inside the conductor. The outer charges do not feel any force from the charges which are at the other side. They repel each other and spread apart as much as possible.

    Now you have two plates at opposite sides of the point charge, and they are electrically connected so the charges can move from one to the other. What do you think about the outer surface charge densities of the plates and the field in outer space? I ask it seriously as that question made the most problem to me.


    ehild
     
  13. Sep 7, 2010 #12
    If you take the hint that that q can be considered as a sheet of charge then the arrangement can be considered as two capacitors in parallel,the sheet of charge being a common plate for both capacitors.
    The capacitance of each capacitor is given by:
    C1=k/x
    C2=k/(L-x)
    (The question stated that the plates are large so I am assuming the overlap area is equal for both capacitors and k is a constant=epsilon*A)
    V is the same for both capacitors and the charge on each is given by:
    Q1=K/x
    Q2=K/(L-x)
    (above from Q=CV K= constant)
    Q1+Q2=q from which we show that K is given by:
    K=qx(L-x)/L and

    Q1=q(L-x)/L
    Q2=qx/L
     
  14. Sep 7, 2010 #13

    ehild

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    That is the problem: why can a point charge considered as a sheet of uniformly distributed charges. If I accept it, everything is easy. But I do not see why should it be.

    ehild
     
  15. Sep 7, 2010 #14
    I took the easy way because I think the question is flexible enough to allow this but,I agree with you and I think it becomes a far more challenging problem if you consider it to be a point charge.
     
  16. Sep 7, 2010 #15

    ehild

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    Well, an infinite series or virtual charges on both sides...


    ehild
     
  17. Sep 7, 2010 #16
    Why do the outer charges feel no force? They are repelled by q (and also attracted by the inner charges). The electric field inside the conductor, which is zero, is the sum of electric fields due to q, the inner charges and also outer charges.
     
  18. Sep 7, 2010 #17

    ehild

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    How do you define an ideal conductor? So what are the field inside a conductor due to any charge outside? And what is the relation between force and electric field intensity?

    ehild
     
  19. Sep 7, 2010 #18
    The inner charges we were talking about are also "outer charges" if we understand those as charges on the surface of the conductor. The terms "inner" and "outer" we referred to are only to discriminate the charges near q and the charges far from q.

    When the system reaches electrostatic equilibrium, the charges on the conductor feel forces whose sum is zero (or maybe this is what you meant by "feel no force"). But the forces are not essentially electric forces. There is also the bond in the crystal structure.

    Consider an element charge on the surface dQ>0. Let E denote the electric field due to dQ. Because the total electric field inside the conductor = 0, the electric field due to others at the position of dQ is also E, but in the opposite direction. Therefore dQ does feel electric force = EdQ. The force to balance it out is due to the bond.

    The total electric field inside the conductor is zero, i.e. E(by q) + E(by inner charges) + E(outer charges) = 0. But we cannot conclude anything from that.
     
    Last edited: Sep 7, 2010
  20. Sep 7, 2010 #19

    ehild

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    One charge on the outer surface of the plate just opposite to q can be much closer than an "inner" charge far away from q. So the terms "near" and "far" do not properly characterize the position of the charges.

    You know that the conduction electrons are free to move inside the metal. That "bond because of the crystal structure" just does not let them to get out. That is why the excess charge on a metal accumulates at the surface.

    Do you mean the electric field due to a point charge dQ inside the metal? How do you get it?


    ehild
     
  21. Sep 7, 2010 #20
    Well, from what I understand, here is the diagram:
    |.....................|
    |{outer}...........|.............q
    |...........{inner}|
    |.....................|
    That's what I meant by "near" and "far". They are still the surface charges anyway.


    I don't really understand your question :P
    The electric field E is formed only by dQ. It's not the total electric field. And dQ is not inside the conductor, it's on the surface.
     
  22. Sep 7, 2010 #21

    ehild

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    And what is the electric field formed only by dQ inside the metal? Show an equation.

    ehild
     
  23. Sep 7, 2010 #22
    @ehild: I'm not sure if you got my point. Maybe my writing is a bit confusing?
    See the attached picture. Consider a charge element dQ on the surface of the conductor. By Gauss's law, we can prove that the electric field due to dQ only is E(dQ) = dQ/(2eodA). The red vectors correspond to the electric field of dQ. We know that inside the conductor in equilibrium state, the total electric field = 0. So the electric field due to the others charges (all excluding dQ), which is represented by the blue vector, is E(others) = E(dQ). This should also answer your question.

    My point when speaking about this is that dQ experiences the electric force, because there is E(others). That means, back to the problem, there is electric field due to q and the inner charge layer at the position of the outer layer. The outer layer's charges cannot be evenly distributed because they are not in their own electric field.
     

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  24. Sep 7, 2010 #23
    I've checked in on this after a day and I'm amazed and grateful at the responses I've gotten.

    First of all, I'd like to just say that if anyone has any issues with the question itself, I'm not the one who came up with it, nor the supposed value for the induced charge. I was merely stuck on that problem, and couldn't see a way out.


    @ehild I think understand your misgivings about my solution. Grounding the plates makes them no longer electrically neutral. I think also that you imagine the plates to have some 'thickness' so that charge can be on the 'in-facing' surface and 'out-facing' surface, which I too considered but I could not work out how this would affect the electric field outside the system. Also, I think you refer to the charge in a hollow cavity situation, where if a charge is in a hollow cavity in a conductor, outside the conductor we can tell what the magnitude of that charge is, but not it's distribution. If you were suggesting this, I'm not sure that it applies here, because the plates may perhaps not be infinite but finitely large (unlike what I previously suggested) and so the point charge q isn't completely surrounded by conductor material.

    And I am quite sure that I wouldn't be able to solve it if I considered the charge q to be point-like.

    @hikaru1221 Yes that image is what I too had in mind. There was no diagram given with the question, so hopefully both you and I are correct in thinking that is the situation intended.


    @dadface I think my method is probably the same as yours, although far more convoluted as I didn't consider them capacitors! I do believe that your method is probably the intended one as the question came up in a sheet referring to capacitance.



    That's all I have to say at this point, and I think that the question must be inherently ill-designed. Feel free to use some technique to solve the point charge problem, but I think any more work on this will produce more heat that light.
     
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