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Point charge placed in the vicinity of two plates

  1. Nov 15, 2005 #1
    Hello all..

    Here's a question I need some help with...

    A point charge +Q has been placed between two infinitely large conducting plates. The charge induced on the surface of the first plate is [itex]q_{1}[/itex]. What should be the charge induced on the surface of the second? (Diagram attached; the charge is closer to the left plate)

    (A) [itex]-(q_1 + Q)[/itex]
    (B) [itex]Q[/itex]
    (C) [itex]-(q_1 - Q)[/itex]
    (D) None

    (Correct answer = A)

    If we had just one plate, then the induced charge on the side facing the charge would be -Q. But what happens if we bring another plate as well? How does the charge distribution change?

    I am not sure if the "correct" answer is indeed correct.

    Thanks for your help...

    Cheers
    Vivek
     

    Attached Files:

  2. jcsd
  3. Nov 15, 2005 #2

    Physics Monkey

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    You can solve the problem of a single point charge outside a conducting plate using the method of images, specifically with one image charge. For a point charge between two conducting plates, you can again use images but you have to use an infinite number of image charges. A solution is therefore possible, but that isn't really relevant for the problem.

    In the case of a single conducting plane with a point charge outside it, the conducting plane acquires a charge that has the same magnitude but opposite sign to the point charge, right? So if you know that one conducting plate has a charge of [tex] q_1 [/tex] and there is also a point charge of [tex] Q [/tex], how much charge would you expect to be drawn up on the second conducting plate? Hint: how would a single conducting plate respond to an arbitrary given charge distribution outside it?
     
  4. Nov 16, 2005 #3
    Hi

    Thanks!! I get it now. I don't believe how I missed that.

    My only problem with that line of reasoning was whether the induced charge is to be treated like a real charge as much as the actual point charge sitting between the plates. But I guess the induced charge as as much real as the point charge to all points in space, in particular, those on the plate.

    But shouldn't [itex]q_{1} = -Q[/itex]?

    Also, if we assume this charging process to be made up of two steps: in the first one, we bring the (left) first plate in the vicinity of the point charge. Once we've done that, images (and sheer common sense) tell us that the induced charge on the right face of this plate equals [itex]-Q[/itex]. Now we bring in the second plate and place it at the location shown in the diargam. Does the charge distribution on the left plate get affected by the presence of the plate on the right?

    [PS: My 700th post!! Yay!!]
     
  5. Nov 19, 2005 #4

    Physics Monkey

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    Yes, the induced charge is very real, it is "drawn up from infinity" if you like. However, [tex] q_1 \neq - Q [/tex], and in fact what [tex] q_1 [/tex] is depends on the distance of [tex] Q [/tex] from the conducting plate. For instance, it should be clear from symmetry that if [tex] Q [/tex] is halfway between the plates then [tex] q_1 = -Q/2 [/tex] and [tex] q_2 = -(q_1 + Q) = - Q /2 [/tex].

    This should answer you second question as well. Clearly the right plate has to effect the charge on the left plate. Let's imagine for simplicity that I am going to bring the right plate in from infinity so as to have the charge [tex] Q [/tex] halfway between the plates. The initial condition is of course [tex] q_1 = - Q [/tex] as you say. As you bring the right plate in from infinity, it is always closer to the charge [tex] Q [/tex] than to the left conducting plate therefore some charge will be induced on it. However, it is then clear that this charge (which will be negative) will also affect the charge on left plate (reducing it). The final result is [tex] -Q/2 [/tex] on each plate in the case of [tex] Q [/tex] midway between them.
     
    Last edited: Nov 19, 2005
  6. Nov 21, 2005 #5
    Thanks again. Sorry for the delayed response...I need to digest what you've said before (if need be) I post another reply to this thread.

    Cheers
    Vivek
     
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