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Method of images to find potential near a grounded plane

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Have my E&M final tomorrow. Trying to work old test questions. Please check my answer for this question.
    Uniformly charged sphere of charge Q and radius R, center is a distance d above a grounded plane.


    2. Relevant equations
    V(r) in the interior of a sphere is given in the problem to be Q/8πε0[3/R-r2/R3]


    3. The attempt at a solution
    Using method of images, let there be a second sphere of charge -Q a distance d below the plane. Let d be along the z-axis
    for sphere above: r2=x2+y2+(z-d)2
    sphere below: r2=x2+y2+(z+d)2

    my final answer
    Q/8ε0(4zd/R3)
     
    Last edited by a moderator: Dec 6, 2012
  2. jcsd
  3. Dec 6, 2012 #2

    haruspex

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    Is this relevant? The title says 'potential near grounded plane', which suggests it's at points external to the sphere.
    Seems right.
    How do you get that? Please post your working.
     
  4. Dec 6, 2012 #3
    Yeah sorry my post title was too brief. The question asks for the potential at the center of the sphere.
    The V(r) that I gave was for this sphere in empty space. So the problem was to find how this changed with the inclusion of the grounded plane.

    V1=Q/8πε0(3/R - r12/R3) (first sphere)
    V2=-Q/8πε0(3/R - r22/R3) (second sphere)

    the total V will be the sum of potentials from each sphere so

    V(r)total = Q/8πε0(3/R - r12/R3- 3/R + r22/R3)

    where r1 and r2 are the r's I gave in the initial post
    Then I just simplified this expression to my final answer.
    Now that I think about it if I'm trying to find V at the center of the sphere then z = d and my answer will be Q/8πε0(4d2/R3)
     
  5. Dec 6, 2012 #4
    This is all wrong. I cant add these to potentials together because they are the potentials in each sphere, and the point is only inside one of them. Should be the potential inside the first sphere plus the potential outside the second sphere.
    Something like

    V(r) = Q/8πε0[3/R-r12/R3]-Q/4πε0[1/r2]
     
  6. Dec 6, 2012 #5

    haruspex

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    Yes, that looks right. But you said you wanted the potential in the centre, so you can substitute for r1 and r2, right?
     
  7. Dec 6, 2012 #6
    Potential at the center of the first sphere was what I was trying to find. Im getting confused about my position vectors.
    r1 is the distance from the center of the first sphere. Since we want the potential at that point r1=0.
    r2 is the distance from the center of the second sphere. So r2 = 2d

    then V(r) reduces to Q/8πε0[3/R - 1/d]
     
  8. Dec 6, 2012 #7

    haruspex

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    Yes, that looks good.
     
  9. Dec 6, 2012 #8
    Thanks for your help!
     
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