# Potential problem

1. May 5, 2007

### neelakash

1. The problem statement, all variables and given/known data

Find the potential in the interior of a sphere of unit radius when the potential on the surface f(θ)=cos^2(θ).

2. Relevant equations

3. The attempt at a solution

I think the correct procedure is to apply uniqueness theorem.We know when the potential at every point of the surface is given,and the potential in that region obeys Laplace's (Here, Poisson'sequation),the potential function is unique.

So,I think it would be the same inside the sphere.

Please check if it is correct.

2. May 6, 2007

### siddharth

That's not right.
What the uniqueness theorem says is that, if you find a function V so that
(i) It satisfies Laplace's (or Poisson's) equation inside the region
(ii) It satisfies the given boundary conditions
then, V is the unique solution inside the region.

In your case, you took V as $$\cos^2 \theta$$. Does it satisfy Laplace's equation inside the region?

You can find the potential insde the sphere, by solving laplace's equation in spherical coordinates by the separation of variables technique, and then fitting the answer to your boundary conditions.

Last edited: May 6, 2007
3. May 6, 2007

### neelakash

I see.The uniqueness theorem is for the "solution" of Laplace's equation.That should be determined first.

4. May 6, 2007

### tnho

Yes. After you have find a solution satisfying the Laplace equation and the boundary condition(s), then you can use the Uniqueness theorem to claim that solution is the only solution allowed.

5. May 6, 2007

### neelakash

Should we use Laplace's equation or Poisson's equation.How is Poisson's equation solved?

6. May 7, 2007

### tnho

In your case, there is no net charge in the sphere, right?
so, Laplace is enough.

For Poisson's equation, you have to do the homogeneous part (Laplace part, solved by separating variables) and inhomogeneous part (the additonal term on the right, use Green function usually) separatly.

=)

Last edited: May 7, 2007
7. May 7, 2007

### neelakash

in question there is no specification whether or not there is charge inside or not.So,we better neglect it.

Thank you for clarification.