Solved Griffiths Problem 3.28 - What to Conclude?

  • Thread starter Thread starter ehrenfest
  • Start date Start date
  • Tags Tags
    Griffiths
ehrenfest
Messages
2,001
Reaction score
1
[SOLVED] Griffiths Problem 3.28

Homework Statement


Please stop reading unless you have Griffiths E and M book.

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?


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
It is a pure dipole.
The cos\theta tells you it is pure P_1(cos\theta).
 
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.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top