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fluidistic
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Homework Statement
Consider 2 conductor hollow spheres that share the same center, of radii a and b (a>b). The hollow sphere of radius b is at zero potential while the hollow sphere of radius a has a potential of the form [itex]V(\theta, \phi ) =V_0 \sin \theta \cos \phi[/itex]. Where [itex]V_0[/itex] is a constant and theta and phi are the spherical coordinates. Find the potential in all the space.
Homework Equations
Tons.
The Attempt at a Solution
For the region [itex]0 \leq r \leq b, \Phi (r, \theta , \phi )=0[/itex] because the E field there is 0 (due to the conductivity of the inner sphere) and thus the potential is constant and worth the same as the surface of the inner sphere, namely 0.
Now I attack the region between the 2 spheres.
I notice that there isn't any azimuthal symmetry, hence the book of Griffith is of no big help, unless I missed out a chapter. I therefore use Jackson's book only to get some help.
I propose a solution of the form [itex]\Phi (r, \theta , \phi ) = \sum _{l=0}^ \infty \sum _{m=-l}^l [A_{lm}r^l+B_{lm }r^{-(l+1 )}] Y_{lm} (\theta , \phi )[/itex] where [itex]Y _{lm} (\theta, \phi )[/itex] are the spherical harmonics.
The first boundary condition is [itex]\Phi (a,\theta , \phi ) =0 \Rightarrow \sum _{l=0}^ \infty \sum _{m=-l}^l [A_{lm}b^l+B_{lm }b^{-(l+1 )}] Y_{lm}=0[/itex].
Here I'm not 100% of what I've done. Namely, I say that since the spherical harmonics are linearly independent, the only way to get the infinite series as vanishing, is the condition that [itex]A_{lm}b^l+B_{lm }b^{-(l+1 )}=0[/itex]. Thus [itex]A_{lm}=-B_{lm}b^{-2l-1}[/itex].
Now my goal is to determine [itex]B_{lm}[/itex], I'd be done with the problem if I succeed in doing so.
I use the second boundary condition: [itex]V_0\sin \theta \cos \phi =\sum _{l=0}^ \infty \sum _{m=-l}^l B_{lm} [a^{-(l+1)}-b^{-2l-1 }a^l] Y_{lm } (\theta , \phi ) [/itex].
I call [itex]C_{lm}=B_{lm}[a^{-(l+1)}-b^{-2l-1}a^l][/itex] for convenience.
Seeking help in Jackson's book, [itex]P_l (\cos \gamma )=\sum _{m=-l}^l C_ m (\theta ', \phi ')Y_{lm} (\theta , \phi )=V_0 \sin \theta \cos \phi[/itex] and [itex]C_{ml}=C_m[/itex] doesn't depend on l. Gamma is the angle between 2 arbitrary vectors [itex]\vec x[/itex] and [itex]\vec x '[/itex] inside the region, namely [itex]\cos \gamma = \cos \theta \cos \theta ' + \sin \theta \sin \theta ' \cos (\phi - \phi ') [/itex].
If I understood well Jackson's book, [itex]C_m (\theta ', \phi ' ) = \int Y^* _{lm} P_l (\cos \gamma ) d \Omega[/itex] (so that it's a double integral-surface integral- and the * denotes the complex conjugate).
Therefore [itex]B_ {lm} = \frac{C_m (\theta ' , \theta ) }{a^{-(l+1 )}-b^{-2l-1}a^l}[/itex]. Replacing this into the original ansatz yields [tex]\Phi (r, \theta , \phi ) = \sum _{l=0}^ \infty \sum _{m=-l}^l \frac{\int Y ^ * _{lm}(\theta , \phi ) P_l ( \cos \gamma ) d \Omega }{a^{-(l+1 )}-b^{-2l-1}a^l} [r^{-(l+1)}-b^{-2l-1}r^l]Y_{lm} (\theta , \phi )[/tex] which would be my final answer.
Since [itex]P_l (\cos \gamma ) =V_0 \sin \theta \cos \phi[/itex], at first glance when r=b, Phi is worth 0 as it should while when r=a, I think that [itex]\Phi[/itex] is worth [itex]V_0 \sin \theta \cos \phi[/itex] as it should too. I wonder if my answer is correct. That was a pretty tough problem to me and it's the first in the long list (~30) I should do.
Thanks for any comment.