Electrodynamics - Finding Surface Charge

AI Thread Summary
The discussion focuses on evaluating the electric field from a potential function related to a semi-infinite plate held at potential V0. The electric field is determined by taking the negative gradient of the potential V(x,y), which involves partial derivatives with respect to x and y. Participants emphasize the importance of applying Gauss' law and boundary conditions for electric fields at the surface charge. The final expression for the surface charge density on the plate is derived as s(x) = -A(epsilon0)/(x1/2). The conversation highlights the mathematical techniques necessary to solve the problem effectively.
metgt4
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Homework Statement



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)



Homework Equations





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
 

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metgt4 said:
I take the derivative of V(x,y) with respect to y and that gives you the electric field for varying y, right?

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

\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}}

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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