# Finding electric potential with non-uniform charge density?

1. Apr 23, 2013

### solijoli

We have a spherical shell. We want to find the electric potential everywhere. Given charge density σ=k*sinθ, k is constant. Also it is wanted from direct integration. V=1/4piε ∫σ/r da. How can we do this, please help?
I thought because it is sphere da is in the direction of r so da=r^2sinθ dθ d(phi) in spherical coordinates but I couldn't solve it.

2. Apr 23, 2013

### cabrera

Have tried ∫sin² x dx = ½ x − ¼ sin 2x + C?

3. Apr 23, 2013

### solijoli

My question is what would be the integral boundaries for Vinside and Voutside. When you just integrate sin^2x dx you just get one result. How can you write for inside and for outside?

4. Apr 24, 2013

### cabrera

The electric filed is zero. Gauss theorem. The volume inside does not have a charge. You cannot have a closed surface with a charge inside, ∇ρ=0, because all the charges are distributed on the surface. That's the reason why the electric field inside a metal is zero. But the potential is cte or zero. It is the same from the physical point of view

5. Apr 24, 2013

### physwizard

There is an expansion of the Green's function in terms of Legendre polynomials. You could use that. If that doen't work, then you could use the expansion of the Green's function in terms of spherical harmonics. Refer chapter 3 of Jackson, 3rd edition, in particular, section 3.3 .
Since this problem has azimuthal symmetry, you may also want to have a look at equation 3.33 in Jackson and apply the boundary conditions that the potential is zero at infinity and should be finite at zero.
@ cabrera:
woudn't agree with you as:
1) Nothing mentioned in the problem specifies that it is a metal.
2) Charge density does not have spherical symmetry.

Last edited: Apr 24, 2013
6. Apr 24, 2013

### cabrera

physwizard, I meant inside the shell. The field should zero,shouldn't it?

7. Apr 24, 2013

### Staff: Mentor

no, not with a non-conducting shell. The charges don't move around on a non-conductor.

Suggest that we move this to the homework forum and follow the rules there?

8. Apr 25, 2013

### solijoli

If the charge density was uniform, wouldn't the potential inside be equal to the potential of the surface? V=1/4pi epsilon * Q/R . But in this case the charge density is not uniform. Also it is a conducting shell so i think the electric field inside should be zero because of gauss law, there is no q enclosed since all the q is on surface then field=0