Askiitians_Expert Yagyadutt
Last Activity: 14 Years ago
Hello raju....
This is very simple...Just it needs a few steps of derivation...I am solving it with derivation...from the begining Let me know if u get stick with any line ....
It is :
Electric field due to:
Charged ring of charge Q and radius R.
1) The Center=0.
Derivation:
Linear charge density (l) on ring =
Consider the field at center due to any element = 
But the field due to point diametrically opposite =
in opposite direction.
Net field at center = 0 (By symmetry)
2) On the axis = 
On axis of ring at distance x.
Derivation:
Fig (5)
As obvious from the diagram the field component along the line gets added due to opposite element.


Now ...We have to find at which x the field is maximum...so calculate dE/dx ...
dE/dx = Z [ (R^2 + x^2)^3/2 - 3/2(R^2+x^2)^1/2 . 2(x).(x) ] / [R^2+x^2]^3 ; where Z = Q/4pi*epsilon
For maximum let dE/dx = 0
then ..(R^2x +x^2)^3/2 - 3x^2(R^2+x^2)^1/2 = 0
(R^2 +x^2)^1/2 [ (R^2 +x^2) - 3x^2 ] = 0
hence ...R^2 = 2x^2
or .... x = + R/root(2) or - R/root(2)
Means If ring is kept symmetrically in x-y plane with center at origin then..field will be maximum at ...a distance ..of +R/root(2) above the +z-axis...and below the negative z axis...
Is it clear
With regards
Yagya
askiitians_expert