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Electric Potential for neutral conducting sphere

  1. May 13, 2010 #1
    1. The problem statement, all variables and given/known data
    An electrically neutral conducting sphere of inner radius Ra and outer radius Rb is centered on the origin. Two equal, positive point charges of magnitude Q are located on the x-axis on each side a distance d from the origin. Determine the electrical potential in the regions:
    a) 0≤r≤R_a
    b) R_a≤r≤R_b
    c) R_b≤r≤∞

    2. Relevant equations
    (1)∮▒〖E∙da〗= Q_enc/ϵ_o
    (2)V= -∫▒〖E ∙dl〗

    3. The attempt at a solution

    I used Gauss' Law in integral form (1) to find the electric field in each of the regions then used equation (2) to find the electric potential. My question lies on the third region in regards to the two point charges and how they come into play.

    Thanks
     
  2. jcsd
  3. May 13, 2010 #2

    gabbagabbahey

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    Oh? How exactly did you do that? Does this problem really possess any of the symmetries that allow you to extract |E| from the flux integral?

    Instead, you'll want to either use the method of images (easiest method), or solve Laplace's equation using separation of variables.
     
  4. May 13, 2010 #3
    Well its a sphere, so the only thing not symmetrical would be the two point charges, which is what is really bugging me about this whole thing. About the method of images, wouldn't I be placing a negative point charge opposite the positive one, which in turn is where the other positive point charge lies?

    Thanks for your response.
     
  5. May 13, 2010 #4

    gabbagabbahey

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    Keep in mind that the two point charges will induce some charge density (the net charge will still be zero, but negative and positive charges will separate) on the sphere, and that charge density need not be spherically symmetric (in fact, it isn't).

    It depends on which region you are trying to calculate the potential in... is [itex]d[/itex] larger or smaller than [itex]R_a[/itex] and [itex]R_b[/itex]?
     
  6. May 13, 2010 #5
    d is smaller than both Ra and Rb.
    Thanks for explaining the symmetry.
     
  7. May 13, 2010 #6

    gabbagabbahey

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    Okay, so when calculating the potential in region (a) you will place two image charges in the region r>Ra....where, and of what magnitude?
     
  8. May 13, 2010 #7
    They would be -Q in magnitude, where is what I'm not understanding. The book would say on the opposing side of the axis, which is where the other positive point charge lies. You said putting them in r > Ra which would be in the conductor. I guess a picture will help this I tried attaching one
     

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  9. May 13, 2010 #8

    gabbagabbahey

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    No, their magnitude won't be -Q....surely you have an example with a grounded spherical shell and a point charge in your text? Use that.
     
  10. May 13, 2010 #9
    Ok the text has an example but their sphere is solid, and the point charge is outside the sphere. I could utilize Q' which would be equal to the negative ratio of two distances times Q. Q' would have to be in a different region from the one I am calculating V for. I'll give it a go.
    Thanks
     
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