A thin glass rod is bent into a semicircle 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 shown in Fig. Calculate electric field E at P, the centre of semicircle

Take PO as the x-axis and PA as the y-axis. Consider two elements EF and E'F' of width dθ at angular distance θ above and below PO, respectively.

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

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

Resolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PA.

Therefore, field at P due to pair of elements is 2 E sin θ

$\mathrm{E}={\int }_0^{\pi /2} 2\mathrm{Esin}\theta$

$=2{\int }_0^{\pi /2} \frac{\mathrm{Q}}{2{\pi }^2{\epsilon }_0{\mathrm{r}}^2}\sin \theta \mathrm{d}\theta =\frac{\mathrm{Q}}{{\pi }^2{\epsilon }_0{\mathrm{r}}^2}$