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Understanding Gaus's law

  1. Oct 7, 2014 #1
    Hello guys,
    1. The problem statement, all variables and given/known data
    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.
     
  2. jcsd
  3. Oct 7, 2014 #2

    td21

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    Gold Member

    Welcome to physicsforums.
    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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