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Electrostatic Self-Energy of a uniform charged sphere

  1. Nov 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello, I have to calculate the self-energy of an uniform charged electron with radius R. The distributed charge is e.

    2. Relevant equations
    The SE is given as:
    [itex] E=\frac{1}{2}\int dV \int dV' \frac {\rho(\vec r)\rho(\vec r')}{ |\vec r - \vec r'|}[/itex]
    according to the problem sheet.

    3. The attempt at a solution
    Since it's uniformly charged I guess [itex] \rho(\vec r)=\rho(\vec r')=\frac{e}{(4/3)\pi R^3}[/itex]

    And from here I'm stuck, I tried to evaluate the (ugly) integral: [itex] \int_{0}^{2\pi} \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{R} \int_{0}^{R} \frac{e^2r'^2r^2sin(\theta')sin(\theta)}{2[(4/3)\pi R^3]^2\sqrt{(r-r')^2+(\theta-\theta')^2+(\phi-\phi')^2}} \, drdr'd\theta d\theta' d\phi d\phi' [/itex] (In spherical coordinates.)

    But maple just crashed when I put it in.
    What am I doing wrong? Did I misunderstand the meaning of the prime? (Can I set [itex]\vec r' = \vec 0 [/itex]?)

    Any hints are very appreciated.
    Kind regards Alex
     
  2. jcsd
  3. Nov 14, 2014 #2

    ZetaOfThree

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    It looks like you are approaching this problem in a very difficult, brute force way. Suppose you have a uniformly charge sphere of radius ##r##. What is the energy required to add a thin, uniformly charge spherical shell of thickness ##dr## to it? See if you can use the answer to this question to help solve your problem.
     
  4. Nov 14, 2014 #3
    I think the work to add the shell is given by:
    [itex]W=\frac{1}{2} \int \rho(\vec r)\phi(\vec r)dr [/itex]
    The potential [itex]\phi(\vec r)[/itex] is caused by the sphere with radius [itex]r[/itex]. I think it can be viewed as a point charge potential, when seen from the outside. [itex](>r)[/itex]
    Then I would add up these shells ([itex]dr [/itex]) until I reach R.
    Am I correct so far or am I confusing myself?
     
  5. Nov 14, 2014 #4

    ZetaOfThree

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    I think that you have the right idea, but this formula is wrong. You should be able to see that the units are wrong. Recall that ##dq=\rho dV## where ##dV## is a differential volume element. So you need to find ##dV## for a thin spherical shell.
     
  6. Nov 14, 2014 #5
    That means r is constant:
    [itex]dV=r^2sin(\theta)d\theta d\phi[/itex]
    If it's right I put it in W and integrate w from 0 to R over r?
     
  7. Nov 14, 2014 #6

    ZetaOfThree

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    That's right. Remember to integrate over ##\theta## and ##\phi## as well.
     
  8. Nov 14, 2014 #7
    Thanks a lot for the help. Will evaluate the integral tomorrow and post the result, it's getting late :) NN
     
  9. Nov 15, 2014 #8
    Good morning, I'm stuck again :)
    The work to add the spherical shell with radius [itex] dr [/itex] to the sphere with radius [itex] r [/itex]:
    [itex]W=\frac{1}{2}\int\rho(\vec r)\phi(\vec r)r^2sin(\theta)d\theta d\phi[/itex]

    Since it's uniformly charged: [itex]\rho(\vec r)=\frac{e}{(4/3)\pi r^3}[/itex]
    And the potential is: [itex]\phi(\vec r)=\frac{e}{r}[/itex]
    (I'm not sure about the charge, because we would end up with more than e charge on the sphere when we add all the shells to the sphere with radius [itex] r [/itex]?)

    The work is then: [itex]W=\frac{1}{2}\int_{0}^{2\pi} \int_{0}^{\pi} \frac{e}{(4/3)\pi r^3}\frac{e}{r}r^2sin(\theta)d\theta d\phi=\frac{3e^2}{2r^2}[/itex]

    Then I wanted to integrate through all the thin shells from [itex]0[/itex] to [itex]R[/itex] but [itex]\int_{0}^{R} \frac{3e^2}{2r^2} dr [/itex] is division by zero. Are the limits of the integral wrong?
     
  10. Nov 15, 2014 #9

    ZetaOfThree

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    Both of these are incorrect. If the sphere is uniformly charged, then ##\rho## is constant, i.e. independent of ##r## right? The potential is ##
    \phi(\vec r)=\frac{q(r)}{r}## where ##q(r)## is the charge of a uniformly charged sphere of radius ##r##. ##q(r) \neq e##!
     
  11. Nov 15, 2014 #10
    Correct: ##\rho(\vec r)=\frac{e}{(4/3)\pi R^3}## , R as radius. And ##\phi(\vec r)=\frac{Q}{R} (\frac{3}{2}-\frac{r^2}{2R^2})##
     
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