Jackson 6.4 (Multipole Expansion)

1. Jan 18, 2009

shehry1

1. The problem statement, all variables and given/known data
Jackson 6.4b

2. Relevant equations
Multipole expansion especially Eq 4.9 in Jackson which is for a Quadrupole

3. The attempt at a solution
I found the result in 6.4a. The rho over there tells us that there is a charge density inside the sphere. Since the charge density is uniform about the x,y and z axis hence the dipole part of the expansion will go to zero.

Now for the quadrupole Q33: I convert the formula 4.9 to spherical coordinates and integrate the resulting expression:
(3z^2 - r^2 )*r^2 dr * d(Cos theta) * d(phi).

I get 0 because of the Cos thing. I think I need something like 5/3 from the Cosine to make the answer correct.

2. Jan 19, 2009

gabbagabbahey

The charge density calculated in part (a) is accurate everywhere inside the sphere. But don't you also need to account for the surface charge density when calculating the quadrapole moment?

3. Jan 19, 2009

shehry1

I was (implicitly) under the impression that the rho calculated would be valid for the surface as well. Could you kindly give me a physical reason for its not being valid at the surface.

Regards

4. Jan 19, 2009

gabbagabbahey

Well, the problem states that the sphere is neutral...is that possible if you have a non-zero constant charge density throughout the sphere and on the surface?

5. Jan 19, 2009

shehry1

Ahh..so:
(induced surface charge) + (induced volume charge) = 0.

Thanks a lot.

6. Jan 19, 2009

gabbagabbahey

To be clear; since part (c) asks you to calulate the surface charge density-- and that part comes after this part of the question--- I'd assume you are expected to use an entirely different method to determine the quadrapole moment. You are probably expected to first determine E outside the sphere (using your knowledge of what the electric field of an oscillating magnetic dipole looks like) and then use that to determine the potential and then the quadrapole moment tensor.

Last edited: Jan 19, 2009