# Electrostatics, potential between and outside concentric spheres

1. Oct 10, 2011

### Telemachus

Hi there. I have this problem which I was trying to solve, but I couldn't fit the boundary conditions.

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

The solution are Legendre polynomials:
$$V(r,\theta)=\Sigma_{l=0}^{\infty}\left (A_lr^l+\frac{B_l}{r^{l+1}}\right )P_l(\cos\theta)$$

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: $$V(r_a)=V_0,V(r_b)=0,V \rightarrow 0 if r \rightarrow \infty$$

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.
$$V(r,\theta)=\Sigma_{l=0}^{\infty} \frac{B_l}{r^{l+1}} P_l(\cos\theta)$$
$$V(r_a,\theta)= \frac{B_l}{r^{l+1}} \frac{2}{2l+1}=V_0\int_0^{\pi}P_{l'}\cos \theta\sin\theta d\theta$$

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):

$$\frac{1}{r} \frac{\partial^2}{\partial r^2} \left ( rV \right )=0$$
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).

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Last edited: Oct 10, 2011
2. Oct 10, 2011

### praharmitra

So where you are going wrong is that there are actually two regions for you to consider. One is r > b (region I) and the other is a < r < b (region II). Since the first region contains infinity you must set A=0. But the same is not true for the second region.

Work in two of the regions separately. Apply separate boundary conditions in both regions. You should get your answer.

3. Oct 10, 2011

### Telemachus

Thanks. I just did it this way, I considered the region a<r<b. And I considered that there is no change on V not for theta neither for phi. I've expressed it in spherical this way:
$$\frac{1}{r^2} \frac{\partial }{\partial r} \left ( r^2\frac{\partial V}{\partial r} \right )+\frac{1}{r^2 \sin \theta} \frac{\partial }{\partial \theta} \left ( \sin \theta \frac{\partial V}{\partial \theta} \right )+\frac{1}{r^2 \sin^2 \theta} \left ( \frac{\partial^2 V }{\partial \Phi^2} \right )=0$$
As I considered V as isotropic for Phi and theta (I mean there is no variation of the field on the theta and Phi directions) I get this single differential equation:
$$\frac{1}{r^2} \frac{\partial }{\partial r} \left ( r^2\frac{\partial V}{\partial r} \right )=0$$
And working this just a bit:
$$\frac{\partial V}{\partial r} +\frac{r}{2} \frac{\partial^2 V}{\partial r^2}=0$$
The solution I get for this differential equation is:
$$V(r)=-\frac{C_1}{r}+C_2$$
Then under the conditions $$V(a)=V_0$$ $$V(b)=0$$ I get $$C_2=\frac{C_1}{b},C_1=\frac{V_0}{\left ( \frac{1}{b}-\frac{1}{a} \right )}$$
This is for the region a<r<b

Is this right? I should work the outer region then. But the thing that bothers me here is that I'm getting a negative potential (perhaps is just some silly mistake with the signs, but perhaps I did it all wrong).

4. Oct 10, 2011

### praharmitra

I don't see how you are getting a negative potential. I'm positive its positive!

Ya, now you should do the same for the outer region!

5. Oct 10, 2011

### Telemachus

Oh, you're right! C1 is negative, then -C1 is positive, thank you very much :)

For the region outside the inner spherical shell doesn't affect at all, right? or should I consider as it isn't at 0 potential that I have some kind of charge distributed over it?

6. Oct 10, 2011

### Telemachus

If I consider that the inner shell doesn't affect at all I get that the potential is 0 over the entire space outside the configuration, for r>b, I think I'm doing something wrong. A is zero, so it won't blow up at infinity, V(b)=0.

$$V(b,\theta)=\Sigma_{l=0}^{\infty}\frac{B_l}{b^{l+1}} P_l(\cos\theta)=0$$

Then Bl must be zero and V=0 for the outer region?

Last edited: Oct 10, 2011
7. Oct 10, 2011

### praharmitra

That's the right answer for the boundary conditions given.

8. Oct 10, 2011

### Telemachus

thanks praharmitra :)