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