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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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1 Answer 1

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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

$$ \begin{align} 4\pi\epsilon_0F &= -\frac er+\frac e{\sqrt{(x-s)^2+y^2}} \\ &= -\frac er+\frac e{\sqrt{x^2+y^2-2xs+s^2}} \\ &= -\frac er+\frac e{\sqrt{x^2+y^2}}\frac1{\sqrt{1+(-2xs+s^2)/(x^2+y^2)}} \\ &= -\frac er+\frac er\frac1{\sqrt{1-\frac{2xs}{r^2}+\frac{s^2}{r^2}}}\;. \end{align} $$

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Ah, I see, so that's not the Taylor expansion at all. The Taylor expansion must come in the line she writes after this (sorry, was confused about this, don't see why she had to write ellipsis so presumed that was the expansion!). Please see my edit. –  dplanet Nov 25 '12 at 22:22
    
@dplanet: Hehe -- I almost wrote "ellipsis", then wondered whether it might be confusing, then wrote "dots" instead :-) –  joriki Nov 25 '12 at 22:47
    
@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

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