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I was reading Classical Electrodynamics by J.D Jackson and stuck at a point. He considers the potential due to a dipole with charge density $\sigma$ and distance between then d such that: $$\lim_{n\to\infty} \sigma(\bf x')\vec d= \vec D$$

Now consider diagram below,showing coordinates. enter image description here

One surface is positive and another is negative. He then considers evaluating potential function:

$$\phi =\int{ \frac{\sigma(\bf x')}{|\bf x-\bf x'|}da' }-\int{ \frac{\sigma(\bf x')}{|\bf x-x'+\vec n d|}}$$

Where $\bf x$ is the point where we are calculating potential. Now he considers $d<<|\bf x-\bf x'|$,the usual approximation. Now using binomial expansion: $$\frac {1}{\bf |x-x'+\vec n d|}= \frac {1}{\bf |x-x'|} -\frac { \bf \vec nd.(\vec \bf (x-x')}{\bf|\vec (x-x')|}$$

My problem comes at next step. He substitutes this expansion in potential function and arrives at(Eq 1.25 in picture):

$$\phi = \int{\vec D. \nabla' \frac{1}{\bf |x-x'| }da'}$$

Can you justify this last formula? Please be detail.

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First of all the first order term in the expansion should be: $-\frac{d\bf{n}.(\bf{x}-\bf{x'})}{|\bf{x}-\bf{x'}|^3}$ (notice the 'to the power three' in the denominator)

As pointed out in the book this can be written as: $-d\bf{n}.\nabla'(\frac{1}{|\bf{x}-\bf{x'}|})$ just try it out (you should get two minus signs in taking this derivative).

Inserting this in the expression for the potential gives you eq 1.25 after using your very first expression.

edit: I assume you understand that the zeroth order Taylor term disappears in the expression for the potential?

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  • $\begingroup$ Is the derivative taken w.r.t x or x' coordinates? This was my real source of confusion. Also can you explain how first term disappeared(I have explanation but I want to make sure I am correct on it). Thanks $\endgroup$ – Abhi7731756 Sep 11 at 12:18

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