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

Answer 1

Answer: (C)

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

Answer 2

The electric field at a point on the axis of a uniformly charged ring $E=\frac{1}{4\pi\varepsilon_{0} }\frac{Q x}{\left(x^{2}+R^{2}\right)^{3/ 2}}$ where R is the radius of the ring and x is the distance of the point on the axis from the centre of the ring. $At \, x=R, E=\frac{1}{4\pi\varepsilon_{0} }\frac{QR}{\left(R^{2}+R^{2}\right)^{3/ 2}}=\frac{1}{4\pi \varepsilon_{0}}\frac{Q R}{\left(2R^{2}\right)^{3 /2}}$ $At \, x=\frac{R}{\sqrt{2}} 'E'=\frac{1}{4\pi \varepsilon_{0}}\frac{Q\left(\frac{R}{\sqrt{2}}\right)}{\left(\left(\frac{R}{\sqrt{2}}\right)^{2}+R^{2}\right)^{3 /2}}=\frac{1}{4\pi \varepsilon_{0}}\frac{2 QR}{\left(3R^{2}\right)^{3 /2}}$ $\therefore\, \frac{E}{E'}=\frac{\left(3\right)^{3/2}}{2\left(2\right)^{3/ 2}}=\frac{3\sqrt{3}}{4\sqrt{2}}$