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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 $X Y$ - 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
equal to $q Q / 2 \pi \varepsilon_{0} r$
equal to $q Q / 4 \pi \varepsilon_{0} r$
equal to zero only, if the path is a straight line perpendicular to the plane of the ring
equal to $q Q / 8 \pi \varepsilon_{0} r$
Solution
$(b)$ As charge distribution is over a circle, potential due to charge over ring is
$V=\frac{k}{r} \cdot q_{ total }$
$\text { where, }$
$q_{\text {total }} =\int \limits_{0}^{2 \pi} \frac{q \sin ^{2} \theta}{\pi r}(r d \theta)$
$=\frac{q}{\pi} \int \limits_{0}^{2 \pi} \frac{1-\cos 2 \theta}{2} \cdot d \theta$
$=\frac{q}{\pi}\left[\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)\right]_{0}^{2 \pi}=q$
So, potential at centre of ring $=V$
$=\frac{k q _{\text {total }}}{r}$
$\Rightarrow \quad V =\frac{k q}{r}$
Also, work done in taking a charge $Q$ from centre of ring to infinity $=$ potential energy of system
$=V \cdot Q=\frac{k q Q}{r}=\frac{q Q}{4 \pi \varepsilon_{0} r}$
