Potential inside sphere with surface charge density

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Homework Help Overview

The problem involves a spherical shell with a specified surface charge density, sigma(theta) = kcos(theta), and requires finding the resulting electric potential both inside and outside the sphere. The context is rooted in electrostatics and the application of Gauss's Law.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Gauss's Law to determine the electric field inside the shell, questioning why the expected result does not align with their findings regarding the potential. Participants discuss the implications of the non-uniform charge distribution on the electric field and potential.

Discussion Status

Participants are exploring the nuances of applying Gauss's Law in this scenario, with some guidance provided on the need for additional reasoning beyond the law itself. There is recognition of the importance of considering symmetry and the relationship between electric field lines and Gaussian surfaces.

Contextual Notes

There is an emphasis on the non-uniformity of the charge distribution and its impact on the electric field, which is a key aspect of the discussion. The original poster's reference to a specific example in Griffiths suggests a reliance on established texts for further understanding.

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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?
 
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It is not that Gauss law does not give the right answer; it is because you misunderstood something :smile:
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 :smile:
 
Last edited:
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?
 
Yup :approve:
 

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