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Potential inside sphere with surface charge density

 
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Nov20-10, 05:48 PM   #1
 

Potential inside sphere with surface charge density


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


2. Relevant equations



3. 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?
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Nov20-10, 10:38 PM   #2
 
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: [tex]\oint _S \vec{E}d\vec{S}=0[/tex] 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
Nov21-10, 12:46 AM   #3
 
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?
Nov21-10, 02:39 AM   #4
 

Potential inside sphere with surface charge density


Yup
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