A charge +q is distributed over a thin ring of radius r with line… - Vidyadip

MCQ

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

A
equal to $q Q / 2 \pi \varepsilon_{0} r$
✓
equal to $q Q / 4 \pi \varepsilon_{0} r$
C
equal to zero only, if the path is a straight line perpendicular to the plane of the ring
D
equal to $q Q / 8 \pi \varepsilon_{0} r$

✓

Answer

Correct option: B.

equal to $q Q / 4 \pi \varepsilon_{0} r$

b
$(b)$ 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

$=V \cdot Q=\frac{k q Q}{r}=\frac{q Q}{4 \pi \varepsilon_{0} r}$

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