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

$$dq = \left( {{q \over {\pi r}}} \right)dl$$

$$ = {q \over {\pi r}}\left( {rd\theta } \right)\,\,\,\,\,$$ (as $$dl = rd\theta $$)

$$ = \left( {{q \over \pi }} \right)d\theta $$

Electric field at $$O$$ due to $$dq$$ is

$$dE = {1 \over {4\pi { \in _0}}}.{{dq} \over {{r^2}}}$$

$$ = {1 \over {4\pi { \in _0}}}.{q \over {\pi {r^2}}}d\theta $$

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.

$$\therefore$$ $$E = \int {dE\sin \theta = \int\limits_0^\pi {{q \over {4{\pi ^2}{r^2}{ \in _0}}}} } \sin \theta d\theta $$

$$ = {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left[ { - \cos \theta } \right]_0^\pi $$

$$ = {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left( { + 1 + 1} \right)$$

$$ = {q \over {2{\pi ^2}{r^2}{ \in _0}}}$$

The directions of $$E$$ is towards negative $$y$$-axis.

$$\therefore$$ $$\overrightarrow E = - {q \over {2{\pi ^2}{r^2}{ \in _0}}}\widehat j$$