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Potential due to a dipole layer

  1. Jul 4, 2009 #1
    1. The problem statement, all variables and given/known data

    I wonder if anyone could explain me how to arrive at the equation (1.24). I have attached the part of the book where this appears.


    2. Relevant equations

    I have attached the part of the book where this appears.

    3. The attempt at a solution

    I have tried to make the second integral zero by performing the gradient and then the remaining algebra but i can't get to this, i suppose that the second integral is zero because in the expression (1.24) it doesn't appear. Using the term that has been expanded i tried to arrive at the integrand in (1.24) but i think that i have to make the second integral zero before this.
     

    Attached Files:

  2. jcsd
  3. Jul 5, 2009 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    No, you do not assume that the second integral is zero. If you are going to make assumptions like this, you need to come up with a good physical or mathematical argument for your assumption.

    Instead, notice that the given Taylor approximation can be used on the second integral with [itex]\mathbf{x}\to \mathbf{x}-\mathbf{x}'[/itex] and [itex]\mathbf{a}\to \mathbf{n}d[/itex] :

    [tex]\frac{1}{|\mathbf{x}-\mathbf{x}'+\mathbf{n}d|}=\frac{1}{|\mathbf{x}-\mathbf{x}'|}+\mathbf{n}d\cdot \mathbf{\nabla}\left(\frac{1}{|\mathbf{x}-\mathbf{x}'|}\right)+\ldots[/tex]

    To first order in [itex]d[/itex], the second integral will give you two terms; one of which will cancel the first integral, and the other produces the desired result as you take the limit in the definition of [itex]D(\mathbf{x}')[/itex].
     
    Last edited: Jul 5, 2009
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