Q. For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is :

Solution:

Electric field on axis of ring
$E = \frac{kQh}{\left(h^{2}+R^{2}\right)^{3/2}} $
for maximum electric field
$\frac{dE}{dh} = 0 $
$ \Rightarrow h = \frac{R}{\sqrt{2}} $

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions