# Potential due to a dipole layer

1. Jul 4, 2009

### Azorspace

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.

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2. Jul 5, 2009

### gabbagabbahey

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 $\mathbf{x}\to \mathbf{x}-\mathbf{x}'$ and $\mathbf{a}\to \mathbf{n}d$ :

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

To first order in $d$, 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 $D(\mathbf{x}')$.

Last edited: Jul 5, 2009