How to Find the Potential from a Surface Charge Density?

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Homework Help Overview

The problem involves finding the electric potential from a surface charge density defined on the z=0 plane as \(\sigma(x,y) = \sigma0 \cos{(ax+by)}\). Participants are exploring methods to compute the potential in space based on this charge distribution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to integrate the surface charge density directly to find the potential but finds this approach challenging. They consider using separation of variables but express uncertainty about how to proceed with the given surface charge density.
  • One participant suggests simplifying the mathematics through a change of variables, while another points out a potential issue with the integral setup regarding the coordinates of the observation point versus the source location.
  • The original poster reflects on the physical implications of the surface charge density and its effect on the electric field, questioning how to visualize the situation given the non-uniform nature of the charge distribution.
  • There is a discussion about the integration process and the need to clarify variable usage in the integral.

Discussion Status

Contextual Notes

Participants are working within the constraints of a sample exam problem, where only analytical methods are permitted, and they are grappling with the complexities introduced by the non-uniform surface charge density.

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Homework Statement



The z=0 plane has a surface charge density [tex]\sigma(x,y) = \sigma0 \cos{(ax+by)}[/tex]. Find the potential everywhere in space.

Homework Equations


The Attempt at a Solution



Ok, so I tried to just integrate directly:

[tex]\Phi = \frac{1}{4 \pi \epsilon0} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{\sigma(x,y)}{x^2 + y^2 + z^2} dx dy[/tex]

but this proves to be formidable by hand (it's a sample exam problem, so nothing but brain power can be used to integrate). My guess is this isn't the best approach to take. Can I try just using separation of variables? How do I do this knowing sigma, and not V, on the plane?
 
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Hi quasar_4! :smile:

have you tried simplifying the maths by changing variables, to X = (ax+by/√(a2+b2), Y = (bx-ay/√(a2+b2)) ?
 
You integral is not quite right. Your observation point has coordinates which are not the same variables as the source location used as the dummy differential variables.
 
Hm, ok, let me take a step back.

So if I understand physically, the surface charge density should induce an electric field. If the surface charge density were uniform, the field would be all in the z-direction by symmetry. I'm having a hard time picturing in my mind what's happening with the surface charge density not being uniform... seems that I can't easily just do symmetry argument here.

Here's what I *was* thinking: If I want to find the potential at some point P above the z=0 plane, I can use the vector v pointing from the origin to P (I figured if the plane's infinite, I can put the origin wherever I want) and with magnitude sqrt(x^2+y^2+z^2). Then I should be able to integrate the thing above, though I guess I should have written all the x and ys as primed variables and integrated over primed variables only. How is the integral wrong? I don't quite see it...

tiny-tim (hi!), you're right, I shouldn't have given up without trying a substitution - I just assumed automatically that it looked more complicated than it should :P
 

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