Multipole expansion on a sphere

[darkpsi]

Homework Statement

A sphere of radius R, centered at the origin, carries charge density
ρ(r,θ) = (kR/r²)(R - 2r)sinθ,
where k is a constant, and r, θ are the usual spherical coordinates. Find the approximate potential for points on the z axis, far from the sphere.

Homework Equations

The multipole expansion of V and some manipulations on the charge density?

The Attempt at a Solution

So I thought I had solved the problem, but both the monopole and the dipole vanished and I believe the answer given was approximated by the dipole term. And when I decided to try the quadrupole, it got really messy. I ended up with (3πkR⁵/64ε_oz³. Is this right at all? I tried to change the charge density in terms of the vector, r', from the origin to the points on/in the sphere but I don't seem to be getting anywhere.
Here's what I did to find the dipole:

V(r,θ) = 1/4πε_o * 1/r² ∫∫∫ r'cosθ'kR(R-2r')sin²θ' dr'dθ'dφ'

Everything is fine until I integrate with respect to θ:

∫₀^πsin²θ'cosθ' dθ' = 1/4sin⁴θ' |₀^π = 0

The monopole did the same thing, but I expected that.

 

[Jhero]

The quadrupole was just a nightmare.

Thank you for your question. I understand your frustration with trying to find the potential for points on the z axis far from the sphere using the multipole expansion. However, I believe that your approach may be incorrect.

Firstly, the monopole term will always vanish since the charge density is centered at the origin, meaning that the net charge of the sphere is zero. So, it is expected that the monopole term will be zero.

For the dipole term, you have correctly set up the integral using spherical coordinates. However, when integrating with respect to θ, you need to take into account the limits of integration. Since we are looking for points on the z axis, the limits of integration for θ should be from 0 to π/2, not 0 to π. This will give a non-zero result for the dipole term.

As for the quadrupole term, it is indeed a more complicated integral and can get messy. However, if you set up the integral correctly and use the correct limits of integration, you should be able to get a solution. I recommend double checking your limits of integration and making sure you are using the correct charge density in your integral.

I hope this helps and good luck with your calculations! Remember, sometimes it takes a few tries to get the correct solution, so don't get discouraged. Keep trying and you will get there.