# Force acting between bodies - using multipole expansion

1. Jun 16, 2012

### glumm

1. The problem statement, all variables and given/known data
Actually, this is not truly a homework, I'm just ineterested in how to solve problems, like the one below.
So, we have two conductive spheres, at a distance R from each other, the radii are r1 and r2 (r1 and r2 are comperable in size, while R is significantly larger than either of them), both speheres are insulated, and they have a net charge of Q1 and Q2, respectevly.
What is the force acting between them (up to the first order term)?

2. Relevant equations
We were to solve these kind of exercises when studying multipole expansion, so I guess, we could use the formulae of that.
http://en.wikipedia.org/wiki/Spherical_multipole_moments#General_spherical_multipole_moments
(Sorry, this is my first post, I don't really know how to write equations properly yet.)

3. The attempt at a solution
Well, I'm quite confused about it rigth now, though probably it was meant to be an easy problem, because the configuration has azimuthal symmetry (by taking the z axis to be parallel with the vector pointing from the center of one of the spheres to center of the other). Therefore, we could use the Legendre polynomials for the expansion, instead of the spherical harmonics, I presume. Also, it seems to me, that the part with importance gonna be the dipole moment, because the azimuthally symmetric effect of the other sphere.
Even if we know the formulae for the expansion and the multipole moments, I find the situation quite confusing, because both spheres are affecting the other.
One more thing that might be used during the solution is that inside the conductive sphere the electric field is 0 and the potential is constant.

Thanks!

Last edited: Jun 16, 2012
2. Jun 19, 2012

### rude man

What happened to F = Q1Q2/4πεR2?

3. Jun 21, 2012

### glumm

Yeah, that's the the simplest approximation, but the question is: how to go one step beyond (to the dipole moment, I guess). Probably the problem could be solved with applying the image charges method multiple times but I'd like to know whether there's a bit more simple or elegant way. (Actually, a more general one, so that it could be applied to similar problems, like conductive cylinder over conductive plane, or any conducting bodies near each other. Probably it has something to do with capacitances.)

4. Jun 21, 2012

### rude man

It's not an approximation, it is 100% exact. The only assumption is the charges are stationary wrt each other.

If you allow relative movement then magnetic forces come into play. Things get very hairy, involving relativistic mechanics and beyond.

cf. http://en.wikipedia.org/wiki/Coulomb's_law

5. Jun 21, 2012

### TSny

You might find the text Static and Dynamic Electricity by Smythe (1950) to be helpful. It discusses two conducting spheres using images. You can find the full text here

http://archive.org/details/StaticAndDynamicElectricity

Some relevant pages are 36-40 and 118-121.

Also, if you feel like depressing yourself, take a look at the 126 problems at the end of chapter 5 on pages 199-216.
(And we all thought the problems in Jackson's text were a nightmare )

6. Jun 21, 2012

### gabbagabbahey

The problem statement says that the spheres are conductive, so the charge will not be uniformly distributed over their surfaces, and the field outside each sphere will not be the same as that of a point charge carrying the same net charge.

At "large"distances from each sphere, the field will be approximately that of a point charge, but the effect of the field of a point charge on a surface of charge, depends on how the charge is distributed over the surface:

$$\mathbf{F}_{1 \text{on} 2} = \int_{2}\mathbf{E_1}dq_2$$

Last edited: Jun 21, 2012
7. Jun 21, 2012

### gabbagabbahey

If you want to solve Laplace's equation via Legendre polynomial expansion, you must first come up with appropriate boundary conditions. So, if you are at a point in between the two spheres, what are the boundaries of your region [there are 3 of them, and one is very far away ;0)]? What can you say about the potential on those boundaries?

Last edited: Jun 21, 2012
8. Jun 21, 2012

### rude man

Ooh, good point. I was thinking point charges.