Potential due to a dipole layer

In summary, The conversation is asking for an explanation on how to arrive at equation (1.24) and the attempt at a solution involves trying to make the second integral zero by performing a gradient and using a given Taylor approximation. The expert advises against assuming the second integral is zero and suggests using the Taylor approximation with some substitutions to get the desired result.
  • #1
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Homework Statement



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.


Homework Equations



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

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
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:

1. What is a dipole layer?

A dipole layer is a distribution of electric charges with equal and opposite charges separated by a small distance. It can be formed by two point charges or by a continuous distribution of charges.

2. How is potential due to a dipole layer calculated?

The potential due to a dipole layer can be calculated by using the formula V = (k * p * cosθ) / r^2, where k is the Coulomb's constant, p is the magnitude of the dipole moment, θ is the angle between the dipole moment and the point at which potential is being calculated, and r is the distance from the dipole layer to the point.

3. What is the relation between potential due to a dipole layer and distance from the dipole layer?

The potential due to a dipole layer is inversely proportional to the square of the distance from the dipole layer. As the distance increases, the potential decreases.

4. Can a dipole layer have a net charge?

No, a dipole layer cannot have a net charge. It always consists of equal and opposite charges, resulting in a net charge of zero.

5. What are some real-life examples of dipole layers?

Some examples of dipole layers in everyday life include water molecules, which have a dipole moment due to the uneven distribution of charges, and magnets, which have a north and south pole that act as a dipole layer.

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