Question 121 of 311

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:

A$\dfrac{R}{\sqrt{2}}$B$\dfrac{R}{\sqrt{5}}$C$R\sqrt{2}$D$R$

Show answer and solution▾

Answer: (A) $\dfrac{R}{\sqrt{2}}$

Electric field at a distance $h$ from centre on axis

$E=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{Qh}{{\left({R}^{2}+{h}^{2}\right)}^{\dfrac{3}{2}}}$

To get maximum $E, \dfrac{dE}{dh}=0$

$0=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{Q{\left({R}^{2}+{h}^{2}\right)}^{\dfrac{3}{2}}-Qh\dfrac{3}{2}{\left({R}^{2}+{h}^{2}\right)}^{\dfrac{1}{2}}2h}{{\left({R}^{2}+{h}^{2}\right)}^{3}}$

$\Rightarrow h=\pm \dfrac{R}{\sqrt{2}}$

JEE Main Physics practice question on Electrostatics (Electric Field and Electric Field Lines), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.