- #1

cshanny

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## Homework Statement

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.**[/B]

## Homework Equations

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

## 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.