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

Question

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

Aicharya · Accepted Answer

Let us consider a differential element $d l$. charge on this element.

$
\begin{array}{l}
d q=\left(\frac{q}{\pi r}\right) d l \\
=\frac{q}{\pi r}(r d \theta) \quad(\because d l=r d \theta) \\
=\left(\frac{q}{\pi}\right) d \theta
\end{array}
$
Electric field at $\mathrm{O}$ due to $d q$ is
$
d E=\frac{1}{4 \pi \in_{0}} \cdot \frac{d q}{r^{2}}=\frac{1}{4 \pi \in_{0}} \cdot \frac{q}{\pi r^{2}} d \theta
$
The component $d E \cos \theta$ will be counter balanced by another element on left portion. Hence resultant field at $\mathrm{O}$ is the resultant of the component $d E \sin \theta$ only.
$
\begin{aligned}
\therefore E &=\int d E \sin \theta=\int_{0}^{\pi} \frac{q}{4 \pi^{2} r^{2} \in_{0}} \sin \theta d \theta \\
&=\frac{q}{4 \pi^{2} r^{2} \in_{0}}[-\cos \theta]_{0}^{\pi} \\
&=\frac{q}{4 \pi^{2} r^{2} \in_{0}}(+1+1)=\frac{q}{2 \pi^{2} r^{2} \in_{0}}
\end{aligned}
$
The direction of $E$ is towards negative y-axis. $\therefore \vec{E}=-\frac{q}{2 \pi^{2} r^{2} \in_{0}} \hat{j}$

A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net…

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