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Homework Help: Electric potential above plane

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Calculate potential.
     
    Last edited: Sep 28, 2010
  2. jcsd
  3. Sep 28, 2010 #2

    kuruman

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    Are you given the dimensions of the plane or is it an infinite plane? If it is infinite, why not use Gauss's Law?
     
  4. Sep 28, 2010 #3
    I would normally, but the question explicitly states to solve for V(z) using the first integral. In addition, I'm not given any dimensions.
     
  5. Sep 28, 2010 #4

    kuruman

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    If you are not given any dimensions, then the best you can do is assume that it is an infinite plane. Define an origin anywhere on the plane and consider an element of charge of area dx'dy' at (x', y') from the origin. Then the contribution to the potential at point (x,y,z) from this charge would be

    [tex]dV=\frac{1}{4 \pi \epsilon_0}\frac{n \: dx'dy'}{((x-x')^2+(y-y')^2+z^2))^{1/2}}[/tex]

    Integrate over x' and y' to get the total potential due to all the charges on the plane. Since the plane is infinite, you may choose x=0 and y=0 without loss of generality. You may also convert to polar coordinates if that's easier for you. When you get your final answer, compare with the result from Gauss's Law to check your work.
     
  6. Sep 28, 2010 #5
    Cheers, great help! :)
     
  7. Sep 29, 2010 #6
    Anyone?
     
  8. Sep 29, 2010 #7

    kuruman

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    Why does the integral have to be evaluated over these limits? Do you have a ring of charge with inner diameter r1 and outer diameter r2? For an infinite plane, r goes to infinity which means that your potential diverges. This is to be expected for an infinite charged plate.
     
  9. Sep 29, 2010 #8
    Oh, I don't mean r1 and r2 as being two separate radii, I just mean they are the limits of integration, and no two limits of integration could give me the correct answer for the potential if that last step is correct. Is it only by evaluating over the surface that the potential diverges?
     
    Last edited: Sep 29, 2010
  10. Sep 29, 2010 #9

    kuruman

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    Limits of integration have physical significance. To integrate over the entire plane, you need to set the lower limit to zero and the upper limit to infinity.
    And where is the above potential zero, i.e. what is your reference point? It seems that it is zero on the plane at z=0. The summation over point charges assumes that the potential is zero at infinity, not at the origin. So where do you want your potential to be zero, at the plate or at infinity? No matter which you choose, you will end up with a very large (infinite) term to consider. This problem is artificial, in that there is no such thing as an infinite charge plate. One way to get around this is to integrate over r from zero to some radius R to get the potential V(z). Then consider what happens in the limit z/R << 1, i.e. do a Taylor expansion and see what you get.
     
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