Potential of charged disk for x R

[qamptr]

I understand the answer to this part of the problem intuitively but there's definitely something in the math that I'm missing--

Homework Statement

Along the axis through the center of a charged disk, for a point P at a distance x from the center, the potential will be

$$\frac{Q}{2\pi\epsilon_{0}R^{2}}\left[\sqrt{R^{2}+x^{2}}-x\right]$$

Homework Equations

For x>>R, I'm supposed to understand that this reduces to

$$V\approx\frac{Q}{2\pi\epsilon_{0}R^{2}}\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right]=\frac{Q}{4\pi\epsilon_{0}x}$$

The Attempt at a Solution

As I said, I understand that this must work out, but I'm misunderstanding something about the math. It seems as though the term under the radical above ought to become just $$x^{2}$$ for x>>R, and thus inside the brackets x-x=0 and then V=0, which is untrue. Could someone please explain how $$\left[\sqrt{R^{2}+x^{2}}-x\right]$$ becomes $$\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right]$$ ?

 

[buffordboy23]

Could someone please explain how $$\left[\sqrt{R^{2}+x^{2}}-x\right]$$ becomes $$\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right]$$ ?

Bring the x^2 term outside the radical, then apply the general binomial expansion formula for (1+y)^(1/2) = 1 + (1/2)y - ...

 

[Mindscrape]

Binomial approximation will get you there.

$$\sqrt{R^2+x^2}=\sqrt{\frac{R^2 x^2}{x^2}+x^2}$$

Now use the approx
$$x \sqrt{\frac{R^2}{x^2}+1}=x(1+\frac{1}{2} \frac{R^2}{x^2})$$

 

[qamptr]

Binomial approximation will get you there.

$$\sqrt{R^2+x^2}=\sqrt{\frac{R^2 x^2}{x^2}+x^2}$$

Now use the approx
$$x \sqrt{\frac{R^2}{x^2}+1}=x(1+\frac{1}{2} \frac{R^2}{x^2})$$

*thumps forehead* thanks.