Question

A charge $+q$ is distributed over a thin ring of radius $r$ with line charge density $\lambda =q{\sin }^2\theta /(\pi r)$. Note that the ring is in the $XY-$ plane and $\theta$ is the angle made by r with the $X-$axis. The work done by the electric force in displacing a point charge $+Q$ from the centre of the ring to infinity is

• A. equal to $qQ/2\pi {\epsilon }_{o}r$
• B. equal to $qQ/4\pi {\epsilon }_{o}r$
• C. equal to zero only, if the path is a straight line perpendicular to the plane of the ring
• D. equal to $qQ/8\pi {\epsilon }_{o}r$

Answer 1

Answer: (B)

As charge distribution is over a circle, potential due to charge over ring is
$V=\frac{k}{r}\cdot q_{\text{total}}$
where, $q_{\text{total }}=\underset{0}{\overset{2\pi }{\int }}\frac{q{\sin }^2\theta }{\pi r}(rd\theta )$
$=\frac{q}{\pi }\underset{0}{\overset{2\pi }{\int }}\frac{1-\cos 2\theta }{2}\cdot d\theta$
$=\frac{q}{\pi }{\left[\left(\frac{\theta }{2}-\frac{\sin 2\theta }{4}\right)\right]}_0^{2\pi }=q$
So, potential at centre of ring $=V$
$=\frac{k{q}_{\text{total }}}{r}$
$\Rightarrow V=\frac{kq}{r}$
Also, work done in taking a charge $Q$ from centre of ring to infinity $=$ potential energy of system
$=V\cdot Q=\frac{kqQ}{r}=\frac{qQ}{4\pi {\epsilon }_{0}r}$