- #1

- 162

- 14

## Homework Statement

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.

## Homework Equations

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:

E=σ/2ε0

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!

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!

[/B]