Hi there. I have this problem which I was trying to solve, but I couldn't fit the boundary conditions.(adsbygoogle = window.adsbygoogle || []).push({});

The exercise says: Two conducting spherical shells with radii a and b are concentrically disposed and charged at potentials φa and φb respectively. If rb>ra, find the potential between the shells and for points r>rb.

Well, at first I tried to attack this as Griffith does in chapter 3. At first considered the Laplacian in spherical coordinates with azimuthal symmetry

[tex]\frac{1}{r} \frac{\partial^2}{\partial r^2} \left ( rV \right ) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left ( \sin \theta \frac{\partial V}{\partial \theta} \right )=0[/tex]

The solution are Legendre polynomials:

[tex]V(r,\theta)=\Sigma_{l=0}^{\infty}\left (A_lr^l+\frac{B_l}{r^{l+1}}\right )P_l(\cos\theta)[/tex]

Then I tried to fit the boundary conditions, at first I tried to use all of them. I've defined the potentials as V0 at ra, and V=0 at rb.

Boundary conditions: [tex]V(r_a)=V_0,V(r_b)=0,V \rightarrow 0 if r \rightarrow \infty[/tex]

So, as I thought V should accomplish all of this, because of the last condition I assumed A=0 for all l's.

And then tried to use the first condition.

[tex]V(r,\theta)=\Sigma_{l=0}^{\infty} \frac{B_l}{r^{l+1}} P_l(\cos\theta)[/tex]

[tex]V(r_a,\theta)= \frac{B_l}{r^{l+1}} \frac{2}{2l+1}=V_0\int_0^{\pi}P_{l'}\cos \theta\sin\theta d\theta[/tex]

Now I think this is wrong, perhaps I can make this fit one of the conditions (the one that gives potential zero for rb), but I think I can't make it fit the other. I think I should considered the problem in two parts and find the potential between the shells using two boundary conditions, and outside the bigger shell using another two boundary conditions. I need some help with this. Should I attack the problem as I think I should, finding the potential for the inner configuration and the outer configuration as different problems or it can be found an expression for V on the entire space on a straight way using the three boundary conditions simultaneously?

I actually thought that I could get a simpler differential equation by considering that the configuration gives an isotropic conditions for phi, but also for theta, I think this is wrong, but I'm not sure.

I would get then (dv/dtheta=0):

[tex]\frac{1}{r} \frac{\partial^2}{\partial r^2} \left ( rV \right )=0[/tex]

and thats easier to solve, but I think its wrong because then I wouldn't distinguish from the symmetry on this problem and the one involving two infinite concentric cylinders (perhaps the only difference is the third condition for the potential at infinity).

Bye there, thanks in advance.

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# Electrostatics, potential between and outside concentric spheres

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