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

Let us consider a differential element dl. charge on this element.

$$\begin{gathered} dq=\left(\frac{q}{\pi r}\right)dl \\ =\frac{q}{\pi r}\left(rd\theta \right) \left(\because dl=rd\theta \right) \\ =\left(\frac{q}{\pi }\right)d\theta \end{gathered}$$

Electric field at O due to dq is

$dE=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}·\frac{dq}{r^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}·\frac{q}{\mathrm{\pi }{r}^2}d\theta$

The component $dE\cos \theta$ will be counter balanced by another element on left portion. Hence resultant field at O is the resultant of the component $dE\sin \theta$ only.

$$\begin{gathered} \therefore E=\int dE\sin \theta ={\int }_0^{\mathrm{\pi }}\frac{q}{4{\mathrm{\pi }}^2{\mathrm{r}}^2{\mathrm{\epsilon }}_{\mathrm{o}}}\sin \theta d\theta \\ =\frac{q}{4{\mathrm{\pi }}^2{\mathrm{r}}^2{\mathrm{\epsilon }}_{\mathrm{o}}}{\left[-\cos \theta \right]}_0^{\mathrm{\pi }} \\ =\frac{q}{4{\mathrm{\pi }}^2{\mathrm{r}}^2{\mathrm{\epsilon }}_{\mathrm{o}}}\left[-(-1-1)\right]=\frac{q}{2{\mathrm{\pi }}^2{\mathrm{r}}^2{\mathrm{\epsilon }}_{\mathrm{o}}} \end{gathered}$$

The direction of E is towards negative y-axis.

$\therefore \overrightarrow{E}=-\frac{q}{2{\mathrm{\pi }}^2{\mathrm{r}}^2{\mathrm{\epsilon }}_{\mathrm{o}}}\hat{j}$