A thin semi-circular ring ofradius r has a po | vedclass

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Question

Let us consider a differential element $d l .$ charge on this element.

$d q=\left(\frac{q}{\pi r}ight) d l$

$=\frac{q}{\pi r}(r d 	heta)(\bec

Vedclass · Accepted Answer

$-$$\;\frac{q}{{2{\pi ^2}{\varepsilon _0}{r^2}}}\hat j$

Vedclass · Answer

Let us consider a differential element $d l .$ charge on this element.

$d q=\left(\frac{q}{\pi r}ight) d l$

$=\frac{q}{\pi r}(r d 	heta)(\because d l=r d 	heta)$

$=\left(\frac{q}{\pi}ight) d 	heta$

Electric field at $O$ due to $d q$ is

$d E=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{d q}{r^{2}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q}{\pi r^{2}} d 	heta$

The component $d E \cos 	heta$ will be counter balanced by another element on left portion. Hence resultant field at $\mathrm{O}$ is the resultant of the component $d E \sin 	heta$ only.

$	herefore E=\int d E \sin 	heta=\int_{0}^{\pi} \frac{q}{4 \pi^{2} r^{2} \epsilon_{0}} \sin 	heta d 	heta$

$=\frac{q}{4 \pi^{2} r^{2} \epsilon_{0}}[-\cos 	heta]_{0}^{\pi}=\frac{q}{4 \pi^{2} r^{2} \epsilon_{0}}(+1+1)$

$=\frac{q}{2 \pi^{2} r^{2} \epsilon_{0}}$

The direction of $E$ is towards negative $y-$ axis.

$	herefore \vec{E}=-\frac{q}{2 \pi^{2} r^{2} \epsilon_{0}} \hat{j}$

A thin semi-circular ring ofradius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec E$ at the centre $O$ is

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