Why does surface charge density change for an inner surface?

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SUMMARY

The discussion centers on the change in surface charge density for an inner surface, specifically regarding the negative value associated with the surface charge density of the inside of an outer shell. The key takeaway is that the change in sign of the surface charge density arises from the reversal of the surface normal direction, which is conventionally represented by a change of sign in the area. This understanding clarifies that the charge itself does not change; rather, it is the coordinate system and the orientation of the surface area that influence the sign of the charge density.

PREREQUISITES
  • Understanding of surface charge density concepts
  • Familiarity with electric fields and charge distributions
  • Knowledge of vector calculus, particularly surface normals
  • Basic grasp of electrostatics and Gauss's law
NEXT STEPS
  • Study the implications of surface normals in electrostatics
  • Learn about Gauss's law and its application to surface charge densities
  • Explore the mathematical representation of charge distributions
  • Investigate coordinate systems and their effects on physical quantities in electromagnetism
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This discussion is beneficial for physics students, electrical engineers, and anyone studying electromagnetism, particularly those interested in understanding surface charge density and its mathematical implications.

Ascendant78
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Rather than try to explain what I am talking about, I am going to link to an image of it below. My question is in regards to the negative value I marked in the answer (the one for the surface charge density for the inside of the outer shell). While I get the concept that it would be negative in this type of scenario, I don't know exactly why? I am assuming it has something to do with the positive charge facing outwards and the surface area being analyzed facing inwards, but I am curious about how this pops up in the math so that I will know on more complicated problems that won't be so obvious?

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The change in sign arises because the direction of the surface normal is reversed, which is conventionally represented by a change of sign in the area.

Claude.
 
Claude Bile said:
The change in sign arises because the direction of the surface normal is reversed, which is conventionally represented by a change of sign in the area.

Claude.

Oh, thank you. What was confusing me was that I thought it meant that the charge was actually different at that point, which was the part that was throwing me off. I wasn't understanding at what point the charge would shift from negative to positive, but now I see that I was looking at it wrong.

So if I understand you correctly, the only reason the sign changes at that point for the charge isn't because the charge itself is changing, but simply because the coordinate system causes it to based off the comparison of the surface area and the charge, right?
 

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