Question

A ring of radius R consists two equal half parts of different charge densities. One of the half parts has charge q and other has charge 2q. The electric potential at a point on its axis at a distance of $2\sqrt{2}$ R from its centre is:

(Assume the charge density on each half part be uniform)

$k=\frac{1}{4\pi {\epsilon }_0}$

• A. $\frac{3kq}{R}$
• B. $\frac{kq}{3R}$
• C. $\frac{kq}{R}$
• D. $\frac{kq}{\sqrt{3}R}$

Answer 1

Answer: (C)

$\frac{kq}{R}$

For half part of ring with charge q: consider an element of charge $dq=\lambda Rd\theta$ where $\lambda =q/\pi R=$ line charge density.

Potential at P due to this element is $d{V}_1=\frac{kdq}{r}=\frac{k\lambda Rd\theta }{\sqrt{x^2+R^2}}$

thus, $V_1=\frac{k\lambda R}{\sqrt{x^2+R^2}}{\int }_{-\pi /2}^{\pi /2}d\theta =\frac{k\lambda R}{\sqrt{x^2+R^2}}\pi =\frac{kq}{\sqrt{x^2+R^2}}$

similarly, the potential at P due to other part of ring with charge 2q is

$V_2=\frac{k\left(2q\right)}{\sqrt{x^2+R^2}}$

Total potential at P is $V=V_1+V_2=\frac{3kq}{\sqrt{x^2+R^2}}=\frac{3kq}{\sqrt{(2\sqrt{2}R{)}^2+R^2}}=\frac{3kq}{3R}=\frac{kq}{R}$