JEE PYQ: Electrostatics Question 51
Question 51 - 2019 (09 Jan Shift 1)
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:
(1) $\frac{R}{\sqrt{5}}$
(2) $\frac{R}{\sqrt{2}}$
(3) $R$
(4) $R\sqrt{2}$
Show Answer
Answer: (2)
Solution
$E = \frac{kQh}{(R^2 + h^2)^{3/2}}$. For maximum, $\frac{dE}{dh} = 0$: $h = \frac{R}{\sqrt{2}}$.