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Energy in dielectric systems

  1. Sep 29, 2014 #1
    Hello all.

    I have a doubt about the derivation of energy in dielectrics formula (Griffiths pages 191 - 192).

    In a certain step of the formula derivation, we encounter the following operation:

    (view formula below).

    I do not undertand that operation.

    Can someone help me? Dibujo.JPG
     
    Last edited by a moderator: Sep 30, 2014
  2. jcsd
  3. Sep 29, 2014 #2
    What means the [itex]\bigtriangleup[/itex]? It looks like it only works on the first term of D·E.
     
  4. Sep 29, 2014 #3
    Laplace operator
     
  5. Sep 29, 2014 #4
    Thanks zoki85, I used to see for the Laplacian [itex]\bigtriangledown^2[/itex].

    Almarpa. The link you have set it's a bit out of context. It only shows that you can conmute the lapace operator and the dot product since D and E differ only by a constant [itex]\epsilon[/itex].
     
  6. Sep 29, 2014 #5
    2(E.E)=▽2E.E+E.▽2E

    you can think the ▽2 as a scalar, but it also is a differential operator like d/dx
     
  7. Sep 30, 2014 #6
    It is not the laplacian operator. It represents an incremental variation of the quantity this symbol goes with.
     
    Last edited: Sep 30, 2014
  8. Sep 30, 2014 #7
    It is not the laplace operator. It is just an increment.
     
  9. Sep 30, 2014 #8

    Meir Achuz

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    ##\nabla^2({\bf E\cdot E})## is not that simple.
     
  10. Sep 30, 2014 #9

    Meir Achuz

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    ##\Delta## is just an infinitesimal variation. That step is only valid if ##\epsilon##
    does not vary with position anywhere in space. Then the step just says
    ##\Delta({\bf E\cdot D)=E\cdot(\Delta D)+(\Delta E)\cdot D}##.
     
  11. Sep 30, 2014 #10
    If so, he should wrote it d , not Δ
     
  12. Sep 30, 2014 #11

    Meir Achuz

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    ##\Delta## is commonly used, with the limit ##d=lim\Delta\rightarrow 0##.
     
  13. Oct 1, 2014 #12
    Sorry, but I still do not get it.

    What happens with the 1/2 term? It vanishes, but I can not see why.
     
  14. Oct 1, 2014 #13

    Meir Achuz

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    ##\Delta({\bf E\cdot D)=E\cdot(\Delta D)+(\Delta E)\cdot D}=2\epsilon{\bf E\cdot E}##
    if ##\epsilon## is constant.
     
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