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Volume charge density

  1. Sep 21, 2007 #1
    an electron is distributed around a proton according to the volume charge density [tex]\rho = A e^{-2r/a_o}[/tex] where A is a constant, a_o is the Bohr radius and r is the distance from the center of the atom.

    Find A:
    we know that [tex] Q=\int \rho dV = -e[/tex]
    i was wonder if this was the integral that i set up:
    [tex]-e=\int _0 ^{\inf} A e^{-2r/a_o} 4 \pi r^2 dV [/tex]

    Find the electric field produced by the atom at the Bohr radius?
    E4 pi a_o^2=-e/epsilon

    then solve for E, is this right?
     
  2. jcsd
  3. Sep 21, 2007 #2

    learningphysics

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    Looks right to me... except should have dR in the integral.

    Left side looks right... but on the right side you need the charge enclosed within the bohr radius... not the entire charge...
     
  4. Sep 21, 2007 #3
    the reason why its dr is because:

    dV=4pir^2dr

    right?


    For the second part, what is the charge enclosed within the bohr radius? Im not quite sure how that plays a role in the equation for flux.
     
  5. Sep 21, 2007 #4

    learningphysics

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    yes.

    integrate [tex]\rho dV[/tex] from 0 to a_o

    what does gauss law say about the flux through the spherical surface at r=a_o
     
  6. Sep 21, 2007 #5
    so the right side should really be:

    [tex]E 4 \pi a_{o}^{2} = \frac{\int_0^{a_o} A e^{-2r/a_o} 4 \pi r^2 dr}{\epsilon_o} [/tex]
     
  7. Sep 21, 2007 #6

    learningphysics

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    Right... with dR in the integral on the right side...
     
  8. Sep 21, 2007 #7
    what is the dR integral?

    isnt that what i have?
     
  9. Sep 21, 2007 #8

    learningphysics

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    oops sorry... yes, that's right... I thought I saw dV there before I clicked reply... did you change it?

    Anyway, it looks correct now.
     
    Last edited: Sep 21, 2007
  10. Sep 21, 2007 #9
    haha yes i changed it right when i saw the dV, you must have hit reply while the system was updating :P
     
  11. Sep 21, 2007 #10

    learningphysics

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    :wink:
     
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