Question

A ring of radius $R$ carries a charge $Q$, uniformly distributed along its circumference. What is the ratio of the electric field strength at a distance $R$ to that at a distance $\frac{R}{\sqrt{2}}$ along the axis?

• A. $\frac{\sqrt{3}}{8}$
• B. $\frac{3\sqrt{2}}{8}$
• C. $\frac{3\sqrt{3}}{4\sqrt{2}}$
• D. $\frac{2\sqrt{2}}{3\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{3\sqrt{3}}{4\sqrt{2}}$

Electric field strength along the x-axis due to charged ring is $E=kQ\frac{x}{\sqrt{(x^2+R^2{)}^3}}$ where $k=\frac{1}{4\pi {ϵ}_0}$
For $x=R$, $E_1=kQ\frac{R}{\sqrt{(R^2+R^2{)}^3}}=kQ\frac{1}{2\sqrt{2}{R}^2}$
for $x=\frac{R}{\sqrt{2}},E_2=kQ\frac{\frac{R}{\sqrt{2}}}{\sqrt{(\frac{R^2}{2}+R^2{)}^3}}=kQ\frac{2}{3\sqrt{3}{R}^2}$
$\therefore \frac{E_1}{E_2}=\frac{1}{2\sqrt{2}{R}^2}\times \frac{3\sqrt{3}{R}^2}{2}=\frac{3\sqrt{3}}{4\sqrt{2}}$