E = sigma / 2 eo
why:
take field along axis at distance x from uniformly charged disk radius R,
at radius r
dQ = 2 pi sigma r dr
so dEx = (1/4 pi eo) [ (2 pi sigma r dr)x ] / (x^2+r^2)^1.5
integrate from 0 to R
Ex = (sigma x/ 2eo) integral r dr/ (x^2+r^2)^1.5
= (sigma /2 eo) [ 1 - 1/ sqrt( {R^2/x^2}+1) ]
if R >>x then
Ex = sigma / 2eo
Find the electric field due to an infinite plane of positive charges with uniform surface charge density sigma
1 answer