Multipole expansion of point charge placed inside a cube

[amccaffrey]

Homework Statement

A cube of side a is fi lled with a uniform charge density distribution of total charge Q. A point
charge +Q is placed at the center of the cube.

Show all odd electrostatic multipole moments vanish. (i.e., 2^l poles with odd l). Show that
among the even moments, those with l = 0; 2, vanish.

Homework Equations

Classical Electrodynamics, Jackson 4.3

http://tinypic.com/view.php?pic=wh4z5&s=6

The Attempt at a Solution

I can't seem to figure out how to go about this. There's two main things that I'm stuck with:

How to deal with the point charge within the cube

How to solve the multipole moments in cartesian (Jackson)

I believe I would have to split the integrals into two parts where one goes from 0 to the cubes surface and the other goes from the cubes surface to infinity but it doesn't seem to be working out.

I've been trying to find an example of solving the multipole moments in Cartesian and finding the moments of a volume with a charge in the middle but I've had no luck.

Any help with this is greatly appreciated.

 

[mark726]

Dear forum post author,

Thank you for your question. Let me try to provide some guidance on how to approach this problem.

Firstly, when dealing with a point charge within a cube, we can treat it as a separate point charge outside of the cube, but located at the center of the cube. This will simplify our calculations and allow us to use the standard formula for the multipole moments.

Secondly, to solve for the multipole moments in Cartesian coordinates, we can use the formula given in Jackson (4.3) for a charge distribution with spherical symmetry. In this case, we can consider the cube to be a sphere of radius a, with the charge distribution being uniform within the sphere.

Now, let's start with the odd moments. As per the formula in Jackson (4.3), the odd moments (l=1,3,5,...) will vanish if the charge distribution is symmetric with respect to inversion about the origin. In this case, since the charge distribution is uniform within the cube, it is symmetric with respect to inversion about the center of the cube. Therefore, all odd moments will vanish.

Moving on to the even moments, we can use the formula in Jackson (4.3) to calculate the moments with l=0 and l=2. For l=0, the formula reduces to the total charge Q. For l=2, we will have to integrate over the volume of the cube using the charge density distribution. Here, you can split the integral into two parts as you mentioned, one from 0 to the surface of the cube and the other from the surface to infinity.

I hope this helps you in solving the problem. If you have any further questions, please feel free to ask. Good luck with your calculations!