# Electric dipole potential (Taylor expansion)

In the x-y plane, I have a charge of $-e$ at $\mathbf{r} = x \mathbf{i} + y \mathbf{j}$, and another of $+e$ at some point a distance of $\mathbf{s} = s\mathbf{i}$ from $\bf{r}$, such that the resulting potential is:

$$4\pi \epsilon_0 F= -\dfrac{e}{\bf{|r|}}+\dfrac{e}{|\bf{r}-\bf{a}|} = -\dfrac{e}{\sqrt{x^2+y^2}}+\dfrac{e}{\sqrt{(x-s)^2 + y^2}}$$

If we assume that $s\to 0$, my lecturer has said that "we may expand the second term in a Taylor series for small $s$", such that:

$$4\pi \epsilon_0 F\approx -\dfrac{e}{r} + \dfrac{e}{\sqrt{x^2 + y^2}} \dfrac{1}{\sqrt{1-\frac{2xs}{r}+\frac{s^2}{r}}}$$

Edit: The lecturer then goes on to write that the above:

$$=\dfrac{e}{r}\left[-1 + \left(1+ \dfrac{xs}{r^2}\right) \right]$$

However, I do not understand how the transition to this line has come about. Could anyone kindly explain?

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There's no need to use an approximation sign there yet; if you fill in the dots, you get an equality. With $r=\sqrt{x^2+y^2}$, we have
\begin{align} 4\pi\epsilon_0F &= -\frac er+\frac e{\sqrt{(x-s)^2+y^2}} \\ &= -\frac er+\frac e{\sqrt{x^2+y^2-2xs+s^2}} \\ &= -\frac er+\frac e{\sqrt{x^2+y^2}}\frac1{\sqrt{1+(-2xs+s^2)/(x^2+y^2)}} \\ &= -\frac er+\frac er\frac1{\sqrt{1-\frac{2xs}{r^2}+\frac{s^2}{r^2}}}\;. \end{align}
@dplanet: So your question now is how to Taylor-expand the inverse square root? Which part of that are you having trouble with? Do you know how to obtain the Taylor expansion of $1/\sqrt{1+x}$? –  joriki Nov 25 '12 at 22:49
I know how to obtain the Taylor expansion of that. From your answer the lecturer neglects the $s^2$ term since $s\to 0$, then takes out a common factor or $\dfrac{e}{r}$ to get $-\dfrac{e}{r} (-1 + \dfrac{1}{\sqrt{1- \dfrac{2xs}{r}}})$. I understand how to get here, but then the lecturer rearranges this to obtain $\dfrac{e}{r}\left[ -1 + (1+\dfrac{xs}{r^2})\right]$. This bemuses me. –  dplanet Nov 25 '12 at 22:57
Sorry, there was an error in the last line of my answer. Is it clearer now? (I wouldn't say "neglects the $s^2$ term since $s\to0$" -- since the goal is to Taylor-expand up to first order in $s$, any terms quadratic in $s$ can be dropped, since they can't contribute to the linear term.) –  joriki Nov 25 '12 at 23:01