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Derivation of bound charges

  1. May 20, 2012 #1
    Upon reading about bound charges I stumbled on something I didn't quite understand. It is not a physical thing but purely a mathematical thing.

    In the attached section my book wants to take the gradient:

    ∇'(1/r)

    with respect to the source coordinates, r'. Now, can someone by inspection of the attached file tell me what these source coordinates represent. Are they they coordinates of a point inside some charge distribution with respect to a fixed point inside the distribution? Would that then mean that in vector notation:

    r = R + r'

    where R is the distance from P to the reference point inside the distribution?

    And from all that can someone tell me how you would differentiate ∇'(1/r) with respect to
    r' to get the answer in the bottom of the attached file? :)

    thanks
     

    Attached Files:

  2. jcsd
  3. May 20, 2012 #2

    tiny-tim

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    hi aaaa202! :smile:
    no, (the diagram should say so, but doesn't :redface: …) they're the coordinates of the point marked "P" (which isn't the name of the point, it's the dipole moment density vector :rolleyes:) wrt a fixed origin (whose position doesn't matter)
    should be easy now :wink:
     
  4. May 20, 2012 #3
    I'm still a little confused on how r depends on r'. If R is the distance to the origin used for the coordinates r' isn't then, as I said:

    r = R + r'

    ? :)
     
  5. May 20, 2012 #4

    tiny-tim

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    let's see …

    in that integral, r is the outside point, and is fixed (a constant)

    r is explained as the distance from r to r',

    so r2 (the denominator) = (r - r')2

    (the notation they're using is very misleading :redface:)
     
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