At what distance, a, along the axis of a uniformly charged disk of radius R is the axial electric field strength equal to one half its value at the surface of the disk at the center.
electric field of disk (attached)
The Attempt at a Solution
First I derived the electric field of a disk
E=σ/2ε0 [1-(a/(x^2 +R^2)^(1/2))] (attached).
This problem wants us to look at where a<<R, Therefore the new electric field equation in that limiting case is:
If we want to know what "a" distance will half this electric field we:
(σ/4piε0) = σ/2ε0 [1-(a/(x^2 +R^2)^(1/2))] and solve for a?
I have attached my work, Attachments are labeled accordingly.
My final result for a-
a= (x^2 + R^2)^(1/2) / 2
Thank you in advanced!