Griffiths Problem 3.28

1. Feb 24, 2008

ehrenfest

[SOLVED] Griffiths Problem 3.28

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

In this problem, I found that the approximation agrees with the exact potential. I am not sure what to conclude about higher multipoles. Are they all identically zero or do they all cancel? Is there something about this charge distribution that makes that happen? Could I have predicted that without calculating it?

2. Relevant equations

3. The attempt at a solution

2. Feb 24, 2008

pam

It is a pure dipole.
The cos\theta tells you it is pure P_1(cos\theta).

3. Feb 24, 2008

ehrenfest

4. Feb 25, 2008

pam

That post gives the definition of a dipole used in elementary texts.
It describes one simple model of a dipole.
What I meant is that the field outside the sphere has only the dipole term as in the expansion. If you calculate the multipole moments of the sphere, you will find only a diple moment because of the P_1 dilstribution of charge.
Use Griffith's or some other text for definitions, not the web.