JEE Main · 2010 · 11 Apr 2010 · Offline · Electrostatics

Spread the charge uniformly over the semicircle, so the charge per unit length is $\lambda =q/(\pi r)$. A small element subtending angle $d\theta$ at the centre gives a field of size $\frac{1}{4\pi {\epsilon }_0}\frac{\lambda d\theta }{r}$ at O. By symmetry the components along the diameter cancel in pairs and only the perpendicular components survive, all pointing the same way. Adding them, $E=\frac{\lambda }{4\pi {\epsilon }_{0}r}{\int }_0^{\pi }\sin \theta d\theta =\frac{\lambda }{4\pi {\epsilon }_{0}r}\times 2$. Putting $\lambda =q/(\pi r)$ gives $E=\frac{q}{2{\pi }^2{\epsilon }_0{r}^2}$, directed along $-\hat{j}$ for the positive charge, so $\vec{E}=-\frac{q}{2{\pi }^2{\epsilon }_0{r}^2}\hat{j}$.