Potential inside sphere with surface charge density

[bcjochim07]

Homework Statement

A specified charge density sigma(theta)=kcos(theta) is glued on the surface of a spherical shell of radius R. find the resulting potential inside and outside of the sphere.

Homework Equations

The Attempt at a Solution

This is a worked example in Griffiths. The Vinside happens to be k/(3epsilon) * rcos(theta). I was thinking about Gauss's Law. If I draw a Gaussian sphere inside the shell, it encloses no charge, which seems to say that the E-field inside is 0, but this cannot be since the gradient of Vinside is not zero. Why does Gauss's Law not give the right answer here? Is it perhaps due to the fact that the surface charge distribution is not uniform?

 

[hikaru1221]

It is not that Gauss law does not give the right answer; it is because you misunderstood something
If there is no charge inside the Gaussian surface, Gauss law gives this result: $$\oint _S \vec{E}d\vec{S}=0$$ and that's all. So we need another argument to derive E=0 inside a uniformly charged conducting sphere. What is that argument? When you get this, you will see that the argument cannot apply to this case.
Observe the E-field inside the sphere of such charge configuration in page 169, example 4.2 of the book (I use the 3rd edition; maybe the other editions are not of much difference) and you may get the hint

 

[bcjochim07]

Ah. So it is that all the electric field lines that enter the Gaussian surface also exit it, and thus the net flux is zero. Even though the net flux is zero, the electric field inside the sphere is constant and nonzero. This example, I believe, shows why one ought think carefully about symmetry and the angle between the area vector and E-field vectors before immediately using Gauss's Law. Did I understand correctly?

 

[hikaru1221]

Yup

 

[paranormal]

As you correctly pointed out, Gauss's Law cannot be applied in this situation because the surface charge density is not uniform. Gauss's Law only applies to situations where the charge distribution is symmetric and uniform, allowing for a simplified calculation of the electric field. In this case, the charge distribution is not uniform and therefore Gauss's Law cannot be used.

To find the potential inside the sphere, you can use the method of images or the method described in Griffiths, which involves using the linearity of Laplace's equation to solve for the potential at any point inside the sphere. This potential will be a combination of the potential due to the surface charge and the potential due to the induced charges inside the sphere.

Therefore, the resulting potential inside the sphere will be k/(3ε) * rcos(θ) + constant, where the constant can be determined by setting the potential at the surface of the sphere equal to the potential due to the surface charge distribution.

Outside the sphere, the potential will be the same as the potential due to a point charge located at the center of the sphere, which can be found using Coulomb's Law.