Taylor Expansion for a configuration of $2$ point charges on a line

Question

Was getting back into physics and reading a chapter on electrostatics which sets up the following situation. We have a configuration of point charges - one $-q$ at the point ($-d,0,0$) and one $+q$ at the point ($d,0,0$). The potential of this configuration is then just $$V(x) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{\lvert x-d \rvert} - \frac{q}{\lvert x + d\rvert}\right)$$ Now the book then says "Taylor expanding this expression for $x\gt\gt d$ tells you approximately how the field behaves far away". It's my understanding that performing a Taylor expansion at a point gives one the ability to approximate how a function behaves at that point when simply calculating the value might be too cumbersome or difficult. However, it's been a while since I've actually done a Taylor Expansion and was looking for some help in doing this for myself.

So, I know the general formula for a Taylor Expansion is the following $$f(x)\vert_a = \sum \frac{f^{(n)}(a)}{n!}(x-a)^n$$ Here in this case I'm a bit confused as to how to approach this given the information that we are expanding about a point $x$ such that $x\gt\gt d$ I get that it means we are investigating how the function behaves far away from the point $(d,0,0$) but not sure how this condition meaningfully changes out function. It's my first instinct to say that since $x \gt\gt d$ that adding or subtracting $d$ from $x$ doesn't really make a difference and thus $x + d = x - d =x$ but then that leaves us with $$V(x) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{x} - \frac{q}{x}\right) = 0$$ Which I know isn't what the example is prompting.

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TL;DR

I want to know what the condition $x \gt\gt d$ meaningfully changes about how we proceed with the Taylor expansion of the potential function and then would also like some guidance in carrying out said expansion.

Answer 1

(Finite) Taylor expansion needs a small parameter that is sometimes obtained as the reciprocal of a large parameter. Here that small parameter is $d/x$, and you get that to appear in the formulas by dividing top and bottom by $x$. So you deal with $(q/x)/(1 \mp d/x)$ which is $q/x \pm qd/x^2$ to first order in $d/x$. (Here I am assuming you are doing all this on the line containing the charges.) As you noticed, first order expansion is the simplest approximation that actually says anything interesting about $V$.

In a more general situation in which $x$ and $d$ are just arbitrary vectors, you can write

$$|x \pm d|^{-1}=((x \pm d) \cdot (x \pm d))^{-1/2}=(|x|^2 \pm 2 (x \cdot d) + |d|^2)^{-1/2} \\=|x|^{-1} (1 \pm 2 (x/|x| \cdot d/|d|) (|d|/|x|) + |d|^2/|x|^2)^{-1/2}$$

and expand that.

Answer 2

I'm not sure exactly what they meant, but here's what I would do. Of course $x>\!\!>d$ implies $x\pm d>0$, so the absolute values aren't needed.

$$\frac1{x-d}-\frac1{x+d}$$ $$=\frac{(x+d)-(x-d)}{(x-d)(x+d)}$$ $$=\frac{2d}{x^2-d^2}$$ $$=\frac{2d/x^2}{1-d^2/x^2}$$

We can use the Taylor expansion of $1/(1-u)$ around $u=0$:

$$=(2d/x^2)\left(1+\left(\frac{d^2}{x^2}\right)+\left(\frac{d^2}{x^2}\right)^2+\left(\frac{d^2}{x^2}\right)^3+\left(\frac{d^2}{x^2}\right)^4+\cdots\right)$$ $$\approx\frac{2d}{x^2}\left(1+\frac{d^2}{x^2}\right)$$ $$\approx\frac{2d}{x^2}$$