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Induced surface charge density

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculate the surface charge density induced by a point source above an infinite conducting plane, with 0 potential.

    2. Relevant equations
    ##E=-\nabla V##
    ##V=\frac{q}{4\pi \epsilon_0 r}##

    3. The attempt at a solution
    I used the method of image charges and I calculated the potential induced by the charge and its mirror image ##V(x,y,z)##. Then I took a box around the plane and I applied the Gauss law. However, In their solution, assuming that the charge is in the ##+z## region and the surface is in the xy plane, they say that the E is 0 in the ##-z## area. From here ##\sigma = \epsilon_0 E = -\epsilon_0 \partial_z V##. I got the same result, but with a factor of 2 difference. My question is, why is the electric field 0 in the ##-z## region?
    Thank you!
     
  2. jcsd
  3. Feb 20, 2017 #2

    BvU

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    Plate is grounded. There will be a surface charge distribution topside only. The solution you found is only valid for z > 0 .
     
  4. Feb 26, 2017 #3

    rude man

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    If the potential of the plane is zero then you can think of the plane as connected to Earth by a wire. The potential of Earth is zero. So there can be no E field between Earth and the plane. No E field, no E flux lines from Earth to the plane, ergo no charges on the plane's bottom to terminate those lines.

    (The necessary charges on the top of the plane are provided by the Earth.)
     
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