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Gauss's Law in differential form

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Gauss's Law is often given as:

    [tex]\nabla \cdot \vec{E} = \rho/ \epsilon_0 [/tex]

    However E is, in general a function of position, so the equation is really
    [tex]\nabla \cdot \vec{E}(\vec{r}) = \rho(\vec{r}) /\epsilon_0[/tex]
    correct?
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 26, 2008 #2

    pam

    User Avatar

    Yes. The (r) is often left out, but understood.
    Just apply the divrgence theorem to get Gauss's integral law.
     
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