A charge +q is distributed over a thin ring o | vedclass

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Question

$(b)$ As charge distribution is over a circle, potential due to charge over ring is

$V=\frac{k}{r} \cdot q_{ total }$

$	ext { where, }$

$q_{

Vedclass · Accepted Answer

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

Vedclass · Answer

$(b)$ As charge distribution is over a circle, potential due to charge over ring is

$V=\frac{k}{r} \cdot q_{ total }$

$	ext { where, }$

$q_{	ext {total }} =\int \limits_{0}^{2 \pi} \frac{q \sin ^{2} 	heta}{\pi r}(r d 	heta)$

$=\frac{q}{\pi} \int \limits_{0}^{2 \pi} \frac{1-\cos 2 	heta}{2} \cdot d 	heta$

$=\frac{q}{\pi}\left[\left(\frac{	heta}{2}-\frac{\sin 2 	heta}{4}ight)ight]_{0}^{2 \pi}=q$

So, potential at centre of ring $=V$

$=\frac{k q _{	ext {total }}}{r}$

$\Rightarrow \quad V =\frac{k q}{r}$

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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