# Multipole Expansion

1. Apr 17, 2017

### grandpa2390

1. The problem statement, all variables and given/known data
Four particles are each placed a distance a from the origin
3q at (0,a)
-2q at (a,0)
-2q at (-a,0)
q at (0,-a)
find the simple approximate formula for the potential valid at points far from the origin. Express in Spherical coordinates

2. Relevant equations
P=qr
$V = \frac{1}{4*pi*ε_o}*\frac{p*r}{r^2}$

3. The attempt at a solution

the first step seems to be to say that

I don't understand this.
The way I did this was to use the law of cosines and I added the potentials up for some point r distance away From the origin due to the four point charges. This gave me a big mess. this is supposed to be the official soln.
I don't understand why the z component is just the two charges on the z axis, and y the y component is just the charges on the y axis. doesn't each charge affect each component?

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2. Apr 17, 2017

### Orodruin

Staff Emeritus
$\vec p$ is the dipole moment. Only the two charges along the $z$ axis contribute to the $z$-component of the dipole moment since the other charges are located at $z = 0$.

In more generality, the dipole moment is given by (assuming zero net charge)
$$\vec p = \sum_i q_i \vec x_i,$$
where $q_i$ is the charge and $\vec x_i$ the position of particle $i$.

3. Apr 17, 2017

### grandpa2390

position relative to the origin? That makes sense if that's true, but I am looking for it with a position far away from the origin. so if I draw a line connecting each charge to my point, each line has a y and z component.

using the traditional method, the potential at the origin should be 0 right. they are all at the same distance away, and the charges add up to 0.

Last edited: Apr 17, 2017
4. Apr 17, 2017

### Orodruin

Staff Emeritus
Yes.

You are looking for the field far away from the origin in comparison to where the particles are located. This is dominated by the lowest order non-vanishing multipole moment - in this case the dipole moment (the monopole moment is zero as the total charge is zero).

This is irrelevant. This is taken care of by the multipole expansion and (since your dipole moment is dominating at large distances) the field far away from the charge configuration will be completely determined by the dipole moment.

Yes, but I do not see how this is relevant to your problem. You need the potential far away from the origin, not at the origin.

5. Apr 17, 2017

### grandpa2390

Can you tell me how they got from the second line to the third line here:

http:// https://www.physicsforums.com/attachments/230510-6-png.195282/?temp_hash=43fa2e3a321f0257bb4467f1b059eff3 [Broken]
Why is the Pcos(theta) = 1/2 5cos^3 - 3cos^2
Is it a half angle formula type deal or what?

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Last edited by a moderator: May 8, 2017
6. Apr 17, 2017

### Orodruin

Staff Emeritus
Did you move to another problem? The octopole is not dominating your setup at large distances. The $P_\ell$ is a Legendre polynomial.

7. Apr 17, 2017

### grandpa2390

I'm sorry. it is a different problem. I just didn't want to spam the forum with multiple threads. This is a different problem.

Do you know how they expanded the P*cos or do you need the whole problem? I figured it might have been just an identity or something, and I didn't want to create a whole new thread for something that I thought was just a simple rule. I can create one though if need be.

and it is actually P_n

Last edited: Apr 17, 2017
8. Apr 17, 2017

### Orodruin

Staff Emeritus
$P_n$ or $P_\ell$, it is completely irrelevant what you call the index. It is a Legendre polynomial all the same. The Legendre polynomials (or more generally, associated Legendre functions) evaluated at $\cos\theta$ contain the typical angular dependence of the multipole expansions (they form part of the spherical harmonics). $\ell = 0$ is the monopole with $P_0(x) = 1$, $\ell = 1$ is the dipole with $P_1(x) = x$, $\ell = 2$ is the quadrupole with $P_2(x) = (3x^2 -1)/2$ and so on and so forth.

By the way, you should make a new thread for a new problem - filling in the homework template and showing your effort.

9. Apr 17, 2017

Thanks.