Electrodynamics - Finding Surface Charge

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SUMMARY

The discussion centers on evaluating the surface charge on a semi-infinite flat plate held at potential V0, positioned in the x-z plane. The potential is defined as V(x,y) = A/(2)1/2[(x2 + y2)1/2 - x]1/2 + V0, where A is a constant. The y component of the electric field is derived from the negative gradient of the potential, and the surface charge density is expressed as s(x) = -A(epsilon0)/(x1/2) for x > 0 and y approaching zero. Key techniques include the use of Gauss' law and boundary conditions for electric fields across surfaces.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with complex variable theory
  • Knowledge of Gauss' law
  • Basic calculus for differentiation
NEXT STEPS
  • Study the application of Gauss' law in electrostatics
  • Explore the theory of complex variables in electrostatic problems
  • Learn about boundary conditions for electric fields at surfaces
  • Investigate the derivation of electric fields from potentials
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Students and professionals in physics, particularly those focusing on electromagnetism, electrical engineers, and anyone involved in electrostatics and potential theory.

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 the 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;

[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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