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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, $\frac{dE}{dh} = 0$: $h = \frac{R}{\sqrt{2}}$.


Learning Progress: Step 51 of 62 in this series