Uniformly charged disk and the E field some distance Z from the center

[anban]

Homework Statement

Hi,

I have a problem that describes a uniformly charged disk and the electric field a distance z from the center.

I have found an equation that describes the E field at any point z already. Now I have to find out how the E field decreases as z increases-- as 1/r^2, I assume, but I am not sure.

Homework Equations

E = 2\pikσz(1-\frac{z}{\sqrt{z^{2}+R^{2}}}) where R is the radius of the disk and z is the distance away from the disk.

The Attempt at a Solution

I know I have to do a binomial expansion of (R/z). I think I am having mathematical issues rather than physical issues.

Before the expansion, I need to get some term (R/z). Does taking out a z^{2) from the denominator leave a z^{4} out front? Or a z^{3} out front? If I can figure this out then I know the rest.

 

[TSny]

$\sqrt{z^2+R^2} = \sqrt{z^2(1+(R/z)^2)}$ and $\sqrt{(a b)} = \sqrt{a}\sqrt{b}$

 

[TSny]

E = 2\pikσz(1-\frac{z}{\sqrt{z^{2}+R^{2}}})

I believe you must have a typo in this expression. The dimensions on the right do not match the dimensions of electric field.

 

[anban]

Good catch: The expression should be E = 2\pikσ(1-\frac{z}{\sqrt{z^{2}+R^{2}}}).

So, the denominator of the last term can be rewritten as z\sqrt{1-(R^{2}/z^{2})}. After a binomial expansion, I got that the E field decreases as 1/z.

 

[TSny]

I got that the E field decreases as 1/z.

I don't think that's correct. Check your binomial expansion simplification. Your original idea that the disk should behave as a point charge for large distances is right.