Finding potential for a sphere, when to use laplace equation

1. Nov 26, 2014

cshanny

1. The problem statement, all variables and given/known data
This is example 3.9 in Griffiths Electrodynamics.
"A specified charge density σ(θ) (inclination angle) is glued over the surface of a spherical shell of radius R. Find the resulting potential inside and outside the sphere." The problem suggests that although it is possible to compute this by using the integral for potential V=(1/4πε)int[σ(θ)dA], it is much easier to do so by separation of variables.

2. Relevant equations
The solution to this example uses separation of variables, using the general solution to Laplace's equation in spherical coordinates.

3. The attempt at a solution
At the suggestion of the book, I correctly found the potential inside and outside of the sphere using the Laplace equation. My question is what in the problem signifies that this is most easily solved with Laplace equation and not, say, Gauss' law. I could certainly say that because the charge is on the surface of the sphere that the electric field inside is 0 and therefore the potential is constant. I could also find the electric field outside by integrating the charge density to find the total enclosed charge outside of the sphere (which yields zero...). I'm obviously missing something here.

2. Nov 26, 2014

Orodruin

Staff Emeritus
The enclosed charge for a gaussian surface outside the sphere is not zero, it is the total charge of the sphere. This way you would have to argue by symmetry that the electric field must be radial and integrate to find the potential.

3. Nov 26, 2014

cshanny

Sorry I left something out in my post. They actually give the charge distribution as kcosθ...so I calculated Qenclosed by integrating the charge density over the surface of the sphere. The fact that theres a sinθcosθ in the integral made it go to zero.

Also, I don't quite understand why Gauss' law is not applicable.

4. Nov 26, 2014

Orodruin

Staff Emeritus
Gauss law is applicable. Your assumption that the field is purely radial and constant is not.

5. Nov 26, 2014

cshanny

Got it. How else can I approach this problem?

6. Nov 26, 2014

Orodruin

Staff Emeritus
What was wrong with solving the Laplace equation?

7. Nov 26, 2014

Orodruin

Staff Emeritus
Posting again to make sure you see it rather than miss my edit.

Of course, there are different methods available to solve the Laplace equation. One of them would be to integrate to sum up the contributions from each part of the surface. (Aka, Green's function methods)

8. Nov 26, 2014

cshanny

Nothing it worked just fine and I got the correct answer. I was just curious if there was any other obvious way of doing this.