How to Find the Potential from a Surface Charge Density?

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Homework Statement



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

Homework Equations


The Attempt at a Solution



Ok, so I tried to just integrate directly:

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

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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