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## 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

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

## Homework Equations

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

[tex]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}[/tex]

## 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 [tex]x^{2}[/tex] for x>>R, and thus inside the brackets x-x=0 and then V=0, which is untrue. Could someone please explain how [tex]\left[\sqrt{R^{2}+x^{2}}-x\right][/tex] becomes [tex]\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right][/tex] ?