Force acting between bodies - using multipole expansion

In summary: Laplace's equation.TheAttempt at a SolutionWell, 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 going to be the dipole moment, because the azimuthally symmetric effect of the other sphere.Even if we know the formulae for the expansion and
  • #1
glumm
2
0

Homework Statement


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)?

Homework 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.)

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 going to 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!
 
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  • #3
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
glumm said:
Yeah, that's the the simplest approximation, .)

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
glumm said:

Homework Statement


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)?

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 :frown:)
 
  • #6
rude man said:
It's not an approximation, it is 100% exact.

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:

[tex]\mathbf{F}_{1 \text{on} 2} = \int_{2}\mathbf{E_1}dq_2[/tex]
 
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  • #7
glumm said:
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 going to 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.

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?
 
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  • #8
gabbagabbahey said:
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:

[tex]\mathbf{F}_{1 \text{on} 2} = \int_{2}\mathbf{E_1}dq_2[/tex]

Ooh, good point. I was thinking point charges.
 

Related to Force acting between bodies - using multipole expansion

1. What is a multipole expansion?

A multipole expansion is a mathematical technique used to describe the force acting between two bodies. It involves representing the force as a series of terms, with each term corresponding to a different type of interaction between the bodies.

2. How does a multipole expansion differ from other methods of calculating forces?

Multipole expansion is unique in that it takes into account the spatial distribution of charges or masses within a body, rather than just considering the overall charge or mass. This allows for a more accurate and detailed description of the force between bodies.

3. Can a multipole expansion be used to calculate forces between any type of bodies?

Yes, a multipole expansion can be used to calculate forces between any type of bodies, as long as they interact through electromagnetic, gravitational, or other fundamental forces.

4. What are the main applications of multipole expansion in science and engineering?

Multipole expansion is commonly used in fields such as electromagnetism, astrophysics, and fluid dynamics to calculate forces between bodies. It is also used in engineering to model interactions between structures and to design efficient systems for energy transfer.

5. Are there any limitations to using multipole expansion for calculating forces?

While multipole expansion is a powerful tool for describing forces between bodies, it becomes less accurate at larger distances and with more complex systems. In these cases, other methods such as numerical simulations may be more appropriate.

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