# Potential inside a hollow sphere

1. Sep 30, 2006

### Logarythmic

I have a problem finding the potential inside a hollow sphere. The problem is formulated as:
"Determine the electric potential inside a hollow sphere of radius a given that the potential on its surface is f(theta)"
I have found a function V(r,theta) through separation of variables and all I have left is to apply the boundary conditions. One is given, but I guess I need another one when my function has two unknown constants. How do I find that second boundary condition?

2. Sep 30, 2006

### siddharth

Your radial solution will be of the form

$$f(r) = Ar^n + B/r^{n+1}$$

So, what happens when r goes to 0? What does that tell you about B?

You'll need only one boundary condition for $V(r,\theta)$ as a result. (Although you'll have infinite constants. Use the orthogonality property of Legendre polynomials to calculate them)

Last edited: Sep 30, 2006
3. Sep 30, 2006

### Logarythmic

So this forces B to equal zero, right? Then my potential function looks like V(r,theta) = A_n r^n P-n(cos(theta)) and I should somehow use that the integral over P_m* P_n equals zero if m is different from n, ok? I just don't know how..?
(Damn, it would be easier if I knew how to use LaTeX) ;)

4. Sep 30, 2006

### siddharth

Yeah

You've almost got it. Your potential function will be
$$V(r,\theta)= \sum_{n=0}^\infty A_n r^n P_n(\cos \theta)$$

and so at r=R

$$f(\theta)= \sum_{n=0}^\infty A_n R^n P_n(\cos \theta)$$

(Check out this LaTeX tutorial, or click the image to see the code)

Since the legendre polynomials are orthogonal, multiply both sides of the above equation by $$P_m (\cos \theta) \sin \theta$$ (this is important. Do you see why?) and integrate (can you tell me the limits?). So, can you take it from here?

Last edited: Sep 30, 2006
5. Sep 30, 2006

### Logarythmic

I understand why I should multiply both sides with P_m(cos theta) but not why to include sin theta. The limits should be from 0 to pi, right?

6. Sep 30, 2006

### Logarythmic

I got it now. sin(theta) is for the substitution cos(theta) = x to work out. Then I get an expression for A_m with an integral over P_m(cos theta)f(theta)sin(theta)d(theta). But then what?

7. Sep 30, 2006

### siddharth

Yes. Now, what about the other side of the equation? Remember that your aim is to get the value of $A_n$.

If you found that as an integral, then you're done! You know the solution to the potential at any point in the sphere.

Last edited: Oct 1, 2006
8. Oct 1, 2006

### Logarythmic

But the correct answer should be
$$V(r,\theta) = -\frac{u_0}{3} + \frac{4u_0}{3}(\frac{r}{R})^2P_2(\cos \theta)$$.

The given potential at R is
$$V(R,\theta) = u_o \cos(2\theta)$$.

(I'm using LaTeX!)

9. Oct 1, 2006

### siddharth

Since you know $f(\theta)$, evaluate the integral using the orthogonal property!

Where exactly are you stuck at? Can you post that here?

10. Oct 1, 2006

### Logarythmic

I substituted $$\cos \theta = x$$ and used the orthogonal property. Then I re-substituted to get the $$\theta$$-dependence again and now my expression looks like

$$A_m = -\frac{1+2m}{2} a^\textit{-m} \int_{0}^{\pi} P_m(\cos \theta) u_0 cos(2\theta) sin \theta d\theta$$

I dunno where to go from here.

Last edited: Oct 1, 2006
11. Oct 1, 2006

### siddharth

To simplify the integral on the RHS, as I've said before, you're going to need the orthogonal property . Have you tried writing $\cos(2\theta)$ in terms of $P_2(\cos \theta)$? (Also, $1=P_0(\cos \theta)$)

12. Oct 1, 2006

### Logarythmic

Hey, I solved it! Thanks alot! This has really helped me in my understanding of Legendre polynomials.