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Method of images; explanation

  1. May 9, 2008 #1
    The setup is: Suppose that we have a point charge q held a distance d from an infinite, grounded, conducting plate. Let the plate lie in the xy-plane, and suppose that the point charge is located at coordinates (0, 0, d). What is the scalar potential above the plane?

    The solution to this problem is here: http://farside.ph.utexas.edu/teaching/em/lectures/node64.html

    In the example they conclude that "the total charge induced on the plate is equal and opposite to the point charge which induces it" - so it is -q.

    My question is: The plate is a grounded conductor, so the potential at z=0, and then the electric field, must be zero at z=0. How can the electric field be zero at z=0 when the plate is negatively charged?

    I hope you can help me. I've spent most of the time in the bus home worrying about this.
     
  2. jcsd
  3. May 9, 2008 #2

    Hootenanny

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    What is the electric field inside a conductor?
     
  4. May 9, 2008 #3
    It's zero (I also wrote that - z=0 is where the conductor is).

    EDIT: In order for the electric field to be zero inside the conductor, a positve charge must reside at the bottom of the plate?
     
  5. May 9, 2008 #4

    Hootenanny

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    Correct, there's your answer. Even if a conductor has a net charge, the electric field inside it is always zero.
     
  6. May 9, 2008 #5
    How can that be? In my book (Griffith's), it says that there is as much plus charge as there is negative charge inside the conductor. But in this example we have a negative. The example doesn't mention positive charges on the plate.
     
  7. May 9, 2008 #6

    Hootenanny

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    No. Consider a simplfied analogy: What is the electric field at a point lying halfway along the line joining two electrons?

    Note that you edited your post after I had already replied.
     
  8. May 9, 2008 #7
    It's zero.
     
  9. May 9, 2008 #8
    Sorry, I'll stop editing.

    Ok, here's where I stand now. The electric field is zero inside the conductor, because the external electric field induces charges inside the conductor, so they create an electric field in the opposite direction.

    The induced charges are what we are calculating. But again, how can only negative charges be induced if the electric field is to be zero inside the conductor?
     
  10. May 9, 2008 #9

    Hootenanny

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    Firstly, there are no free charges inside a conductor, they all lie on the surface.
    When using the method of images, it is usual to assume that the plate has no thickness, therefore to talk about 'inside' the conductor is non-sensical. However, if you want something a little more of a substantial explanation consider the point I raised in post #6: The electric field and a point between two equal charges is zero. Now extend these negative charges so that you have two infinite lines of negative charges. If you placed a positive or negative charge at any point between these two lines would it move?
     
  11. May 9, 2008 #10
    No, it would not. This is because the negative charges have an electric field that points into the wires, so the net electric field at the midpoint is zero.

    Can we assume the conductor has a thickness? That would help me.
    When we calculate the induced charge, is it the charge that is inside the conductor or is it the free charges that reside on the surface of the conductor?

    And where do these free charges on the surface come from?
     
    Last edited: May 9, 2008
  12. May 9, 2008 #11
    Ok, so our charge q pushes away the protons in the conductor and attracts the electrons. This leaves a net negative charge in the area directly below q, and this is the induced charge we calculate? And this induced charge is also what creates (together with the repelled protons) the electric field that make the total eletric field inside the conductor go to zero?

    - assuming that it makes sense to say that our conductor has a thickness in this case.
     
    Last edited: May 9, 2008
  13. May 9, 2008 #12

    Hootenanny

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    Okay, I'll try and answer all your questions in one swoop. Since our plane is conducting and grounded, it must be at zero potential. The presence of the point charge above the conducting plane induces charge distribution on the conducting plane, such that the potential of the conducting plane remains zero. This induced charge distribution represents the image charge and is drawn onto the conducting plate from the ground. Note that this induced charge lies on the surface of the conducting plane and there is no net charge inside the conductor, the conduction electrons in the plate itself are not affected. If there was free charge present in the volume of the conductor, then this would mean there is also an electric field inside the conductor and ergo a non-constant potential inside the conductor, thus violating the requirement of an equipotential plane.

    Does that answer all your questions?
     
  14. May 9, 2008 #13
    Yes, it does.

    Thank you very much for solving my mystery!
     
  15. May 9, 2008 #14

    Hootenanny

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    A pleasure :smile:
     
  16. May 9, 2008 #15
    Hmm, I actually have some "post-questions".

    When the positive charges are removed (because the conductor is grounded), the electric field is still zero inside the conductor? If yes, what is the counterpart of the negative charges then?

    I hope it's ok I ask this one more question.
     
  17. May 9, 2008 #16

    Hootenanny

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    Erm, what positive charges? The positive charges inside a conductor are the nuclei, which are fixed in a lattice and do not move.
     
  18. May 9, 2008 #17
    Hmm, in turns out then that I have not understood it completely. Please correct me where I am wrong:

    We have the charge q and a conducting plate. The charge q has an electric field E_0 and it induces charges in the conductor - the positve charges in the conductor are the nuclei and the negative are the electrons from the nuclei (these are all inside the conductor). These charges produce another electric field E_1, which makes the total electric field inside the conductor zero.

    Since the plate is grounded, negative charges would come from the ground so there will be no more positive charges., so now the only induced charge on the conducting plate is a negative charge.

    If the above is correct, then how can the electric field inside the conductor be zero still?
     
  19. May 9, 2008 #18

    Hootenanny

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    Not quite, the conductor is initially uncharged: there are no net charges on or in the conductor. The external electric field induces a charge density on the surface of the conductor by drawing additional charges from the ground, it is these additional charges which form the charge distribution producing the image charge on the conducting plate. There is still not net charge inside the conductor.
     
    Last edited: May 9, 2008
  20. May 9, 2008 #19
    Ok, so my "story" will go like this (I hope it's OK I write it explicitly, but in this way I am sure that I have understood it):

    The charge q has an electric field E_0 and it induces charges in the conductor by drawing negative charges from the ground - this negative charge distribution is on the surface of the conductor.

    But also the charge q induces charges inside the conductor (the electrons and nuclei) and these charges produce another electric field E_1, which makes the total electric field inside the conductor zero.

    Am I correct now?
     
  21. May 9, 2008 #20

    Hootenanny

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    Good :approve:
    NO! As I've said many times previously, there is no net charge inside the conductor! If there were a net charge inside the conductor, then there would be an electric field, which would require a non-constant potential, which we can't have since the conductor must be equipotential!
     
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