Surface Charge Distrib on Plane

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SUMMARY

The discussion focuses on calculating the electric potential due to a surface charge density defined as \(\sigma = \sigma_0 \sin(\alpha x) \sin(\beta y)\) on the plane \(z = 0\). The initial approach involves using Gauss's Law, but the infinite nature of the plane complicates direct application. The solution suggests employing the relationship \(\vec{E} = -\vec{\nabla} V\) to derive the potential, integrating the electric field in the z-direction to obtain the potential function.

PREREQUISITES
  • Understanding of Gauss's Law and its application to electric fields.
  • Familiarity with surface charge density concepts in electrostatics.
  • Knowledge of vector calculus, specifically gradient operations.
  • Basic principles of electric potential and its relation to electric fields.
NEXT STEPS
  • Study the application of Gauss's Law for infinite planes in electrostatics.
  • Learn about the integration of electric fields to find electric potentials.
  • Explore vector calculus techniques, particularly gradient and divergence.
  • Investigate the behavior of potential functions in the presence of varying charge densities.
USEFUL FOR

Students and professionals in physics, particularly those studying electrostatics, electrical engineers, and anyone interested in advanced topics related to electric fields and potentials.

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Homework Statement


The plane z = 0 is charged to a density [tex]\sigma = \sigma_0 sin(\alpha x) sin(\beta y)[/tex]

Find the potential.

Homework Equations





The Attempt at a Solution



Well I first thing I would normally do is use Gauss's Law to find E
[tex]E = \frac{\sigma}{2e0}[/tex] for an infinite plane, however in this case I don't appear to be able to just plug it in like that.

My next thought would be to find the total charge Q, but how does one do that when the plane is infinite?
 
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I think you might have to use [tex]\vec{E}=-\vec{\nabla}.V[/tex], where [tex]\vec{\nabla}.V=\frac{dV}{dx}\hat{x}+\frac{dV}{dy}\hat{y}+\frac{dV}{dz}\hat{z}[/tex].

So if you know the electric field acts in the z direction you can say that [tex]\frac{dV}{dz}=\frac{-\sigma}{2\epsilon_0}[/tex], so you can integrate w.r.t to z to find the potential. I think that's how you would do it anyway, hope this helps.
 

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