Potential between 2 concentric spheres

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Homework Help Overview

The problem involves finding the electric potential between two concentric spheres with specified boundary conditions involving Legendre polynomials. The spheres have radii a and b, and their potentials are defined by constants multiplied by the 3rd and 5th Legendre functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of orthogonality of Legendre functions to simplify the problem and eliminate terms in the series. There are attempts to express constants in terms of each other based on boundary conditions. Questions arise about how to handle the constants without the typical limits of r approaching zero or infinity.

Discussion Status

Some participants have offered guidance on how to approach the boundary conditions and the superposition principle. There is recognition that the problem may have multiple interpretations, and while progress is being made, there is no explicit consensus on the final approach.

Contextual Notes

Participants note the challenge of determining constants due to the lack of traditional limits in this specific problem setup. There is also mention of the need to satisfy multiple boundary conditions simultaneously.

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The problem: Find the potential between 2 concentric spheres, of radii a & b, where the potential of the spheres are held at Ca*P3(cos(t)) & Cb*P5(cos(t)), where P3 & P5 are the 3rd & 5th Legendre functions, t is theta & Ca & Cb are constants.

The general solution to Laplace with azimuthal symetry:

V = sum over l of (Al*(r^l)+Bl*r^(-(l+1)))*Pl(cos(t))

I can use the orthoganality of the Legendre functions to elimimate all but one of the series terms when I plug in the boundary conditions, giving:

V(a,t) = (A*a^3 + B*a^(-4))P3(cos(t)) = Ca*P3(cos(t))

V(b,t) = (C*b^5 + D*b^(-6))P5(cos(t)) = Ca*P5(cos(t))

The solution should be a combination of these, but I'm a bit stumped about how to find the constants when I can't use r=zero or r=infinity to zero one of them (as is done in single shell problems). I have a feeling this may become obvious after a good sleep, but I'd be grateful for a nudge in the right direction.
 
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When using the orthogonality above, you see that you can just get rid of the polynomials and you are left with 4 constants; but if you go back and use orthogonality again, this time multiplying by say, P0(cost), you can get rid of the RIGHT sides, which reduce to zero. With that, you can solve A in terms of B and C in terms of D: when you plug the values into your first equations, this leaves you with 2 constants to determine...
 
Sorry, disregard the above! i went a little too fast.. will get back to you
 
So:
we know that potentials obey the law of superposition and that there should be only one solution satisfying the boundary conditions. Thus, if we have a potential V1 at radius a and a potential V2 at radius b, we can think that V1 vanishes at b and V2 vanishes at a, so:
V1(a) = Ca*P3(cos(t)), V1(b) = 0
V2(b) = Cb*P5(cos(t)), V2(a) = 0
right? I think this gives you enough boundary conditions to solve for the constants; your final solution is then in terms of: V(r) = V1(r) + V2(r), for a<r<b
 
Thanks Goddar, as soon as I left it alone for a while the same answer came to me. Interestingly, your first answer sounds similar to what I was initially thinking as well, great minds & all that ;)
 

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