Q 12-01-142JEE MainJEE Main 2019 (9 Jan, Shift 1)Easy
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
Answer: (A) $\dfrac{R}{\sqrt2}$
On the axis, $E = \dfrac{kQh}{(R^2+h^2)^{3/2}}$. Setting $\dfrac{dE}{dh} = 0$:
$$(R^2+h^2)^{3/2} - h\cdot\frac32(R^2+h^2)^{1/2}\cdot 2h = 0 \Rightarrow R^2 + h^2 = 3h^2$$
$$h = \frac{R}{\sqrt2}$$
Solution by Sreeraj P, M.Sc Physics