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Deriving the local field and Clausius Mossotti formula in a dielectric

  1. Jun 27, 2013 #1
    (My last https://www.physicsforums.com/showthread.php?p=4424810#post4424810 post did not get much attention so I try again without all these formulae. Think this will be more clear...)

    To derive the local field in a non-polar dielectric you assume a very small spherical cavity in which (since there is an applied field to it) you have made up surface charges. Integrating over those made up surface charges you get the Lorentz field from which you can derive the Clausius-Mosotti formula. My question is: when you integrate over those made up surface charges, why dont you also integrate over the real surface charges on the outside of the dielectric? Don't these two fields (made up and real) cancel each other?
     
  2. jcsd
  3. Jun 27, 2013 #2
    I think you're right, maybe the problem is that you found a simplified(or elemental) explanation that's why it doesn't make much sense. Here is how the real thing is: to calcculate the local field([itex]E_{i}[/itex]) you take,as you explained, a spherical cavity and next calculate the local field as a contribution of two parts 1)[itex]E_{out}[/itex] due to all the charges that lie outside the sphere which includes all free charges and all bound charges which can lie not only on the external surface but could also be volumetric bound charges(div P≠0) 2) [itex]E_{near}[/itex] due to charges inside the sphere and is demostrated to be zero for regular distributions so the result is:
    [itex]E_{i}=E_{out}+E{near}=E_{out}[/itex] and this is the desired result.
    You can find a good explanation in Grifiths an also in Jackson
     
  4. Jul 2, 2013 #3
    Thanks, that makes me more Confident that there is more to it. I don't Think I will read Jackson in a while (I know it is chapter 4), but when I do I will keep all this in mind.
     
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