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Homework Help: Griffiths Example 3.6

  1. Feb 17, 2008 #1
    [SOLVED] Griffiths Example 3.6

    1. The problem statement, all variables and given/known data
    Please stop reading unless you have Griffiths E and M book.

    In Example 3.6, we must assume that there is no charge inside the sphere even though it does not explicitly state that? Or is that what "hollow" means?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 17, 2008 #2
    No hollow simply means in this case the potential is only in tin shell on the surface rather then being throughout which would make [tex] A_l [/tex] obsolete. As long as it not specifically stated that there is net charge or not we assume there isn't. All the charges are positioned homogeneously throughout
  4. Feb 17, 2008 #3
    So you are saying that we do assume there is no charge inside the sphere, correct?

    He explicitly says that the charge is on the surface? What does "hollow" add to the problem?
  5. Feb 17, 2008 #4
    No, what happens is that there is no NET charge. There is still charge around, but the overall contribution is such that the charges chancel each other out. look at example 3.8 and see how the electric field causes the charges to accumulated and "charge" even though it is an "uncharged" metal sphere.

    Hollow is simply saying that there is a potential within the thing, else there would not be from what i understand.
  6. Feb 17, 2008 #5
    But Laplace's equation does not apply unless there not only no net charge but exactly 0 charge density everywhere inside the sphere. And if Laplace's equation does not apply, neither do the solutions (eqn 3.65) that he uses.
  7. Feb 17, 2008 #6
    Read Section 3.1.1 and look at the example 3.3, 3.4, 3.5. It states that we don't need a charge to have a potential. There is no net charge. LaPlace is equal to 0. The same is true for this section if you look at Eqn 3.53 from which Eqn 3.65 are derived
  8. Feb 17, 2008 #7
    I don't think you understand my question.

    In Example 3.7, Griffiths says "assuming there is no charge there". Why does he not say that in Example 3.6? I contend that he made a mistake and that he should have said it in Example 3.6. If that assumption is not made, then everything in Example 3.6 makes no sense.

    Please confirm this.
  9. Feb 17, 2008 #8
    He means no free charge OUTSIDE the sphere, else it would influence the result, as the boundary conditions would change. With only the sphere as a source of potential the normal boundary conditions can be used
  10. Feb 17, 2008 #9
    I still think you don't understand my question.

    I am asking about the assumptions we are making in Example 3.6. In Example 3.7, Griffiths says explicitly that there is no charge anywhere outside of the sphere. In Example 3.6, he does not say that there is no charge inside the sphere. But that is an assumption that is necessary in order to use the solutions to Laplace's equations.
    Someone please confirm this.
  11. Feb 17, 2008 #10
    Yes there is no charge inside, cause else he would state it. The problem of the charge inside comes later in a different version, dipole in dielectric
  12. Feb 17, 2008 #11
    OK. That's all. Thanks.
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