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I don't understand Gauss' law

  1. Feb 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A hollow spherical shell carries charge density [tex]\rho=k/r^2[/tex] in the region [tex]a\leq r\leq b[/tex]. Use Gauss' Law in integral form to find the electric field in three regions: (i) r<a, (ii) a<r<b, (iii) r>b.

    2. Relevant equations

    Gauss' Law in integral form: [tex]\oint_{\text{surface}}\textbf{E}\cdot d\textbf{a}=\frac{1}{\epsilon_0}Q_{\text{enc}}[/tex]

    where [tex]Q_{\text{enc}}=\int_{\text{volume}}\rho\;d\tau[/tex] is the total charge enclosed within the surface.



    3. The attempt at a solution

    I'm looking at (ii) to start. Due to symmetry we have [tex]\textbf{E}=|\textbf{E}|\hat r[/tex] and so we can pull this out of the left term of the integral in Gauss' law. But I'm confused on how to compute Qenc. I would expect to take the radius part of the integral from a to b, this way:

    [tex]Q_{\text{enc}}=\int_{\text{volume}}\rho\;d\tau= \int_{\text{surface}}\int_a^b \rho\;dr\;d\textbf{a}[/tex]

    However the solutions manual suggests that I should leave r as a variable and take the integral from a to r, this way:

    [tex]Q_{\text{enc}}=\int_{\text{volume}}\rho\;d\tau= \int_{\text{surface}}\int_a^r \rho\;d\bar r\;d\textbf{a}[/tex]

    I don't understand this. Why do we get to ignore the field between r and b?

    Thanks.
     
  2. jcsd
  3. Feb 17, 2012 #2

    BruceW

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    Homework Helper

    well part ii) is asking for the electric field between a and b. So where will the Gaussian surface extend to?
     
  4. Feb 17, 2012 #3
    OH, I see now!

    That was silly.

    Thanks.
     
  5. Feb 17, 2012 #4

    BruceW

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    Homework Helper

    haha, no worries!
     
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