How to Prove Electric Potential Formula for a Charged Disc?

[sheepcountme]

Homework Statement

I need to prove that the formula for electric potential (V) of a charged disc (with radius r) and with a uniform charge density, that
V=2kpisigma\(sqrt{x^2+r^2}-x)

when x is very large compared to r, and Q is the total charge of the disk, that this formula reduces to V=kQ/x

Homework Equations

Shown above

The Attempt at a Solution

I tried multiplying by a fraction of (sqrt{x^2+r^2}+x) as both numer and denom, and worked through it to get V=2kpisigma(4xsqrt{x^2+r^2})/(sqrt{x^2+r^2}+x)

Here I assumed that all under the square root was approaching x, so simplified as

V=kpisigma4x^2/x

But now I don't know if I've been doing this right..

 

[Mandeep Deka]

Take a differential ring in the charged plate, with a radius say 'r' and thickness say 'dr', then what charge does it posses?
Assuming the surface charge density to be sigma, you can find the charge covered by the differential ring. Now you must know the formula for the potential at a point due to a ring (on the axis). If you know that, then you can find the potential at the required point (at a distance 'x' as per your question) due to the differential ring. It will be only in terms of a single differential, 'dr'. Just integrate it from r=0 to r=R (where i am assuming R to be the radius of the plate), you will have the required expression!