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Electrodynamics - Finding Surface Charge

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a very thin flat plate positioned in the x-z plane. The plate is semi-infinite with an edge running along the z axis. The plate is held at potential V0. Using techniques from teh theory of complex variables a solution is determined to be:

    V(x,y) = A/(2)1/2[(x2 + y2)1/2 - x]1/2 + V0

    Where A is a constant.

    Evaluate an expression for the y component of the electric field. Then carefully take limits for x > 0 and y approaching zero to show that the surface charge on the plate is given by:

    s(x) = -A(epsilon0)/(x1/2)



    2. Relevant equations



    3. The attempt at a solution

    I've worked out a few other parts of this problem, but this part has me stumped. I take the derivative of V(x,y) with respect to y and that gives you the electric field for varying y, right? Then you would use Gauss' law, right? I can't seem to work it out though. My work is scanned and attached to this message.

    Thanks!
    Andrew
     

    Attached Files:

  2. jcsd
  3. Mar 12, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No, the electric field is given by the negative gradient of the potential;

    [tex]\textbf{E}=-\mathbf{\nabla}V=-\frac{\partial V}{\partial x}\hat{\mathbf{i}}-\frac{\partial V}{\partial y}\hat{\mathbf{j}}-\frac{\partial V}{\partial z}\hat{\mathbf{k}}[/tex]

    As for determining the surface charge, your text should tell you the boundary conditions on the electric field crossing a surface charge, use those.
     
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