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Electromagnetism question (Vector integration).

  1. Nov 11, 2009 #1

    MathematicalPhysicist

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    1. The problem statement, all variables and given/known data
    Show that the force caused by external electric field on a charge distribution is given by:
    [tex]F=qE(0)+(p\cdot \nabla)E(0)+...[/tex] (it's enough to show the first two terms), where E(0) is the electric field in the origin which we choose to develop the expression around it.
    p is the electric dipole.


    2. Relevant equations



    3. The attempt at a solution
    So, [tex]F=\int d^3 r E(r)\rho(r)[/tex], from integration by parts I get:
    [tex]F= r \rho(r) E(r) -\int r[\nabla(\rho(r)) E(r) +div(E(r))\rho(r)]d^3 r[/tex]
    Now I develop E(r) around E(0) by taylor expansion: [tex]E(r)=E(0)+div(E(0))r+...[/tex], so div(E(r))=div(E(0))+...
    Now the first term in F is zero by boundary conditions, Now I know that [tex]p=\int d^3 r \rho(r) r[/tex], but if I plug div(E(r))=div(E(0))+... to the third term I get something else than the second term in the question, the first term I do get, because: [tex]\int r \nabla(rho) d^3 r=r\rho(r) - \int \rho(r) d^3r[/tex] and plugging it back and inserting E(r)=E(0)+div(E(0))r... I get that the third term is: div(E(0))p, and the second term is -2div(E(0))p+-qE(0), so in summary I get: qE(0)+p.div(E(0))+...
    But it's not the same as what I need to show here, but I don't see where did I got this wrong, can someone enlight me?
     
  2. jcsd
  3. Nov 11, 2009 #2

    gabbagabbahey

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    How exactly are you coming up with this?:confused:

    You also seem to be using the wrong formula for the Taylor expansion of a vector field...where are you getting this from?
     
  4. Nov 11, 2009 #3

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    Ok, I think I got my mistake, the taylor expansion should be:
    [tex]E(r)=E(0)+r\cdot\nabla E(0)+...[/tex]
    correct?
     
  5. Nov 11, 2009 #4

    MathematicalPhysicist

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    For the first question, I guess I take [tex]u=E(r)\rho(r)[/tex] and [tex]v=r[/tex] in integration by parts, where I use (uv)'=u'v+uv'.
     
  6. Nov 11, 2009 #5

    gabbagabbahey

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    That looks better!:smile:
     
  7. Nov 11, 2009 #6

    gabbagabbahey

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    No, integration by parts doesn't work the same way in 3 dimensions as it does in one. You are trying to compute a volume integral (over [itex]d^3\textbf{r}[/itex]), not just an integral over [itex]r[/itex].

    In vector Calculus, there are actually six product rules and none of them help you here.
     
  8. Nov 12, 2009 #7

    MathematicalPhysicist

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    OK, thanks I see how to solve it now, it's quite immediate.
     
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