Is the Charge on the Inner Wall of a Conducting Sphere Zero?

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SUMMARY

The charge on the inner wall of a conducting sphere is definitively zero when the sphere is isolated and has an excess charge. According to Gauss's law, the electric field inside the conductor is zero, which implies that there is no charge present on the inner surface. All excess charge resides on the outer surface of the conductor, confirming that the inner wall does not hold any charge, regardless of the sign of the excess charge.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with the properties of conductors in electrostatic equilibrium
  • Basic knowledge of electric fields and charge distribution
  • Concept of isolated conductors and their behavior under excess charge
NEXT STEPS
  • Study the implications of Gauss's Law in different geometries, such as cylindrical and planar conductors
  • Explore the concept of electric field lines and their relation to charge distribution
  • Investigate the behavior of conductors in external electric fields
  • Learn about the applications of electrostatics in real-world scenarios, such as capacitors
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Students studying physics, particularly those focusing on electrostatics, as well as educators seeking to clarify concepts related to charge distribution in conductors.

Toddy
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Hello guys,

Homework Statement


I am currently learning about Gauss's law in my physics class. I am having trouble understanding the concept of a charge placed on an isolated conductor. For example, one of my homework problems asked for the charge on the inner wall of a conducting sphere giving that the sphere has a certain excess charge. According to my book, the excess charge will move entirely to the surface of the conductor. So does that mean that the charge on the inner wall of the sphere would be the opposite sign of the excess charge? Or would it be zero, since all of the charge on a conductor is located on its outer surface? Thank you advance.
 
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It may be easy to explain in equation. Use Gauss law inside the sphere. Now look at the equation. As e-field inside the sphere is zero, it means the LHS is zero. Therefore, the RHS is zero. What does it mean?
 

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