How to Find the Potential from a Surface Charge Density?

[quasar_4]

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?

 

[tiny-tim]

Hi quasar_4!

have you tried simplifying the maths by changing variables, to X = (ax+by/√(a²+b²), Y = (bx-ay/√(a²+b²)) ?

 

[Antiphon]

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.

 

[quasar_4]

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

 

[paranormal]

First, it's always a good idea to check if there are any symmetries in the problem that can simplify the integration. In this case, since the surface charge density is a cosine function, we can use the fact that cosine is an even function to simplify the integral. This means that we can rewrite the integral as:

\Phi = \frac{2}{4 \pi \epsilon0} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sigma(x,y)}{x^2 + y^2 + z^2} dx dy

Next, we can use the substitution u = x^2 + y^2 to simplify the integral further. This will result in a double integral in terms of u and z.

\Phi = \frac{2}{4 \pi \epsilon0} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sigma(u)}{u + z^2} \frac{du}{\sqrt{u}} dz

Now, we can use separation of variables by writing \sigma(u) as \sigma(u) = \sigma0 \cos{(a \sqrt{u})} and using the fact that \cos{(a \sqrt{u})} can be expanded as a series of Bessel functions. This will result in a series of integrals that can be evaluated using standard techniques.

Alternatively, if you are allowed to use a computer or calculator, you can use numerical integration methods to solve the integral. This will give you a more accurate solution without the need for hand calculations.