Approximation of electric field of uniform charged disk

[yoran]

Hi,

Homework Statement

The electric field of a uniform charged disk at a point on its axis at a distance x from the disk is given by
E = 2k_e\pi\sigma(1-\frac{x}{\sqrt{x^2+R^2}})
where R the radius of the disk and \sigma the surface charge density.
In my notes it says that when x\gg R, that is when the distance x to the disk is much bigger than the radius of the disk, then
E\approx k_e\frac{Q}{x^2}
with the Q the total charge on the disk. How do they come to that result?

Homework Equations

The Attempt at a Solution

When x\gg R, then \sqrt{x^2+R^2}\approx\sqrt{x^2}=x. But then I get that E=0, which is not obviously not correct.
I guess the approximation that \sqrt{x^2+R^2}\approx\sqrt{x^2} is wrong. But how is the approximation then?

Thank you.

 

[tiny-tim]

When x\gg R, then \sqrt{x^2+R^2}\approx\sqrt{x^2}=x. But then I get that E=0, which is not obviously not correct.
I guess the approximation that \sqrt{x^2+R^2}\approx\sqrt{x^2} is wrong. But how is the approximation then?

Thank you.

Hi yoran!

Always divide so that you get a "1" at the beginning:

√(x²/(x² + R²)) = 1/√(1 + R²/x²) ≈ 1 - R²/2x².

 

[DavidWhitbeck]

Also Yoran your mistake was that basically you did a 0th order expansion!

If you are not comfy with binomial expansions, you can always MacLaurin expand the function in R/x. You are still using the same thing though-- expanding in R/x instead of x because you want a variable that is small i.e. much less than one (so you can truncate the expansion and still have it legitimately approximate the original function), which R/x satisfies, while just straight up x is very large.

 

[yoran]

Allright then thanks a lot!