Question

The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{q}{3\sqrt{3R^2}}$
• B. $\frac{2}{4\pi {\epsilon }_0}\frac{2q}{3R^2}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{2q}{3\sqrt{3}{R}^2}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{3q}{2\sqrt{3R^2}}$

Answer 1

The correct option is C $\frac{1}{4\pi {\epsilon }_0}\frac{2q}{3\sqrt{3}{R}^2}$

Given,

Radius of ring $=R$

Total Charge on ring $=q$

Electric field on x-axis $\int dE=\int \frac{1}{4\pi {\epsilon }_o}\frac{dQ}{{\left(\sqrt{R^2+x^2}\right)}^2}\cos \theta$

$E=\frac{1}{4\pi {\epsilon }_o}\frac{q}{{\left(\sqrt{R^2+x^2}\right)}^2}\cos \theta$

$\sin \theta =\frac{R}{\sqrt{R^2+x^2}}$

Square on both side

$R^2+x^2=\frac{R^2}{{\sin }^2\theta }$

After putting value

$E=\frac{1}{4\pi {\epsilon }_o}\frac{q}{R^2}\cos \theta {\sin }^2\theta$

For point of maxima, differentiate this equation with respect to $\theta$ and equal it to zero.

$\frac{dE}{d\theta }=\frac{1}{4\pi {\epsilon }_o}\frac{q}{R^2}\frac{d\left(\cos \theta {\sin }^2\theta \right)}{d\theta }$

$\frac{dE}{d\theta }=\frac{1}{4\pi {\epsilon }_o}\frac{q}{R^2}(\cos \theta \times 2\sin \theta \times \cos \theta +{\sin }^2\theta (-\sin \theta \left)\right)$

$\frac{dE}{d\theta }=\frac{1}{4\pi {\epsilon }_o}\frac{q}{R^2}(2{\cos }^2\theta \sin \theta -{\sin }^3\theta )=0$

$2{\cos }^2\theta \sin \theta -{\sin }^3\theta =0$

$\tan \theta =\sqrt{2}$

$\sin \theta =\frac{\sqrt{2}}{\sqrt{2+1}}=\sqrt{\frac{2}{3}},$ $\cos \theta =\frac{1}{\sqrt{2+1}}=\frac{1}{\sqrt{3}}$

Double differentiation, to evaluate that it is maximum or minimum.

$\frac{d^{2}E}{d{\theta }^2}=\frac{d}{d{\theta }^2}\left[\frac{1}{4\pi {\epsilon }_o}\frac{q}{R^2}(2{\cos }^2\theta \sin \theta -{\sin }^3\theta )\right]$

$\frac{d}{d{\theta }^2}(2{\cos }^2\theta \sin \theta -{\sin }^3\theta )=2{\cos }^2\theta \cos \theta +2\times 2\cos \theta (-\sin \theta )\sin \theta -3{\sin }^2\theta \cos \theta$

$=2{\cos }^3\theta -7{\sin }^2\theta \cos \theta$

After putting values of $\sin \theta \&\cos \theta$

$\frac{d^{2}E}{d{\theta }^2}=2{\left(\frac{1}{\sqrt{3}}\right)}^3-7{\left(\sqrt{\frac{2}{3}}\right)}^2\left(\frac{1}{\sqrt{3}}\right)=-2.3$

Hence, at $\theta ={\tan }^{-1}\sqrt{2}={54.7}^o$, Electric field is maximum.

$E=\frac{1}{4\pi {\epsilon }_o}\frac{q}{R^2}\cos \theta {\sin }^2\theta$

$E=\frac{1}{4\pi {\epsilon }_o}\frac{q}{R^2}\left(\frac{1}{\sqrt{3}}\right){\left(\sqrt{\frac{2}{3}}\right)}^2$

Maximum Electric field, $E=\frac{1}{4\pi {\epsilon }_o}\frac{q}{R^2}\left(\frac{2}{3\sqrt{3}}\right)N/C$