A thin glass rod is bent into a semicircular of radius r. A charge +Q is uniformly distributed along the upper half, and a charge –Q is uniformly distributed along the lower half, as s shown in figure. The electric field E at P, the center of the semicircle, is

Take PO as the x-axis and PA as the y-axis. Consider two elements EF and $E\text{'}F\text{'}$ of width $d\theta$ at angular distance $\theta$ above and below PO, respectively. 

The magnitude of the field at P due to either element is

$dE=\frac{1}{4\pi {\epsilon }_0}\frac{rd\theta \times Q/(\pi r/2)}{r^2}=\frac{Q}{2{\pi }^2{\epsilon }_0{r}^2}d\theta$

Resolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PB. Therefore, field at P due to pair of elements is $2dE\sin \theta$.

$E=\underset{0}{\overset{\pi /2}{\int }}2dE\sin \theta =2\underset{0}{\overset{\pi /2}{\int }}\frac{Q}{2\pi {\epsilon }_0{r}^2}\sin \theta d\theta =\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$