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Potential Energy of two protons

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the electric field of two protons b meters apart.
    The potential energy of the system is equal to:

    [tex] U = \frac{\epsilon_0}{2}\int {\bf E}^2dv = \int({\bf E}_1+ {\bf E}_2)^2dv [/tex]
    [tex] = \frac{\epsilon_0}{2}\int {\bf E}_1^2dv + \frac{\epsilon_0}{2}\int {\bf E}_2^2dv + \epsilon_0 \int{\bf E}_1\cdot {\bf E}_2dv [/tex]

    The third integral is not hard to evaluate if you set it up in spherical coordinates with
    one proton at the origin and the other along the polar axis (z axis) and perform the integration
    over r first. Show that it integrates to
    [tex] e^2/4\pi\epsilon_0b [/tex]
    2 The attempt at a solution

    I have been trying to solve this problem for hours, but cannot find an expression for E1 dot E2 in spherical coordinates in a way that would make this integral easy. Any guidance would be appreciated.
     
  2. jcsd
  3. Oct 2, 2013 #2

    tiny-tim

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    Hi ChowPuppy! :smile:

    If the protons are at O and P, and a typical point is at X,

    then you have cosOPX/PX2,

    which you can write as PX*cosOPX/PX3

    then find a simple expression for PX*cosOPX, write it all in terms of r (= OX), and you should get a perfect integral :wink:

    show us what you get :smile:
     
  4. Oct 2, 2013 #3
    Hey tiny-tim,

    Thanks so much for the help!
    I was finally able to get it, I think you meant cosPXO instead of cosOPX possibly, but
    basically I was able to write PX*cosPXO = [itex] r - bcos\phi [/itex] where [itex] \phi [/itex] is
    the usual spherical coordinate phi, and then using the pythagorean theorem I rewrote PX and came up with
    this integral:
    [tex] \int \frac{r - bcos\phi}{((bsin\phi)^2 + (r-bcos\phi)^2)^{\frac{3}{2}}}dr d\phi d\theta [/tex]

    And from there I was only a few substitutions away from my answer!:smile:
     
    Last edited: Oct 2, 2013
  5. Oct 2, 2013 #4

    tiny-tim

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    :biggrin: Woohoo! :biggrin:

    good puppy! :smile:
     
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