Question

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

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

Answer 1

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

REF.Image.

The diagram is :-

The electric field due to the ring an it &

axis at a distance x is given by:-

$E=\frac{kqx}{(x^2+R^2{)}^{3/2}}$

To find maximum electric field, we will use the concept of maximum and minimum :-

$\frac{dE}{dx}=kq\frac{(x^2+R^2{)}^{3/2}-3/2(x^2+R^2{)}^{1/2}.2x^2}{(x^2+R^2)}$

Now, $\frac{dE}{dx}=0\Rightarrow (x^2+R^2{)}^{3/2}=\frac{3}{2}\times 2x^2(x^2+R^2{)}^{1/2}$

$\Rightarrow x^2+R^2=3x^2$

$\Rightarrow 2x^2=R^2$

$\Rightarrow x^2=\frac{R^2}{2}$

$\Rightarrow x=\pm \frac{R}{\sqrt{2}}$

So $E_{max}=\frac{k.q.R}{\sqrt{2}{(\frac{R^2}{2}+R^2)}^{3/2}}=\frac{k.q.2R}{3\sqrt{3}{R}^3}$

$E_{max}=\frac{1}{4\pi E_0}\frac{2q}{3\sqrt{3}{R}^2}$