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Gauss Law

  1. Sep 14, 2006 #1
    A hollow spherical shell carries a charge density
    [tex] \rho = \frac{k}{r^2} [/tex]

    in the region a<= r <= b. As in the figure
    Find the elctric field in these three regions
    i) r <a
    ii) a<r<b
    iii) r>b

    SOlution:
    for r<a it simple.. no exclosed charge for any gaussian sphere within that region so E = 0

    for a<r<b
    the thing which stumps is the charge density... isnt charge density given in coulombs per cubic metre usually??

    what im concerned about in the enclosed charge in teh gaussian sphere of radius a<r<b... would i integrate rho from r' = a to r' = r??

    rho dot dr would give the total charge enclosed, no??

    doing that gives
    [tex] q_{enc} = k\left(\frac{1}{a} - \frac{1}{r}\right) [/tex]

    [tex] E (4 \pi r^2) = \frac{k}{epsilon_{0}} \left(\frac{1}{a} - \frac{1}{r}\right)

    for the third part that is r>b would i do something similar but integrate from r' =a to r'=b??

    Please help!

    Help is always greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Sep 14, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Absolutely. Which means that k has what units?

    Yes.

    No! An element of charge would be rho dV, not rho dr. rho is the charge density per unit volume.
     
  4. Sep 14, 2006 #3
    so K has units of C/m

    so i should be integrating rho dV but still r'=a to r'=r

    thanks for the help
     
  5. Sep 14, 2006 #4

    Doc Al

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    Staff: Mentor

    Yes and yes.
     
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