- #1

- 16

- 0

sigma(theta)=sigma(0)cos(theta)

If surface charge wasn't present and it was just a point charge I would be OK but I need a few pointers on how to do it with the above surface charge density.

Thanks in advance guys........

- Thread starter babtridge
- Start date

- #1

- 16

- 0

sigma(theta)=sigma(0)cos(theta)

If surface charge wasn't present and it was just a point charge I would be OK but I need a few pointers on how to do it with the above surface charge density.

Thanks in advance guys........

- #2

- 574

- 1

What have you tried so far? For arbitary charge densities the moments are

[tex]\int {\rho(\vec{r'}) dV'}[/tex]

and

[tex]\int {\vec{r'}\rho(\vec{r'}) dV'}[/tex]

The monopole moment is just the total charge on the surface. So integrate your surface charge density over the surface of the sphere. For the dipole moment I'm not that sure but I think you have to do the same for [tex]R\sigma(\theta)[/tex] where R is the radius of the sphere. Don't quote me on this though.

edit: change the second intergation over all components of the r vector over the sphere's surface. that would make much more sense than what I previously wrote.

[tex]\int {\rho(\vec{r'}) dV'}[/tex]

and

[tex]\int {\vec{r'}\rho(\vec{r'}) dV'}[/tex]

The monopole moment is just the total charge on the surface. So integrate your surface charge density over the surface of the sphere. For the dipole moment I'm not that sure but I think you have to do the same for [tex]R\sigma(\theta)[/tex] where R is the radius of the sphere. Don't quote me on this though.

edit: change the second intergation over all components of the r vector over the sphere's surface. that would make much more sense than what I previously wrote.

Last edited:

- #3

- 16

- 0

I was using the multipole expansion formula of phi(r) in spherical polars.

My working matches what you have said so thanks for confirming that!

:rofl:

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