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Gauss's law, nucleus, potential proton due to other protons

  1. May 5, 2015 #1
    I'm a little out of touch with this stuff , but I'm really not getting it..

    So my book is considering: in a nucleus of ##z## protons, consider one proton in spherical charge distribution to other protons.

    So ## \rho = (z-1)e/(4/3) \pi R^{3}##, where ##R## is the radius of the nucleus, is the charge density. (this is fine)

    I have for Gauss's law:

    ##\int E.d\vec{A} = Q_{enclosed}/\Sigma_{0}##,

    So I get ##E.4\pi r^{2}=\rho (4/3) \pi r^{3} / \Sigma_{0}##
    ##=>## ##E=(z-1)e r/4\pi R^{3}##

    Whereas my book has ##q/4 \pi \Sigma_{0} r= (4/3) \pi r^{3} \rho/4 \pi \Sigma_{0}r=(z-1)e/4 \pi \Sigma_{0}(r^{2}/R^{3}##,

    I dont understand the far LHS of this? But anyway I'm a factor ##r## out, I have no idea how the book gets ##r^{2}##,

    your assistance is greatly appreciated, thank you !
     
    Last edited: May 5, 2015
  2. jcsd
  3. May 5, 2015 #2

    BvU

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    ( I hope it's OK with you if I use ##\varepsilon_o## for the permittivity of the vacuum. With your symbol, I keep wondering what gets summed :wink:)

    $${1\over 4 \pi \varepsilon_o}\; {q\over r} $$ looks more like a potential, not an electric field strength. That would explain the factor r.
     
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