d
Let us consider a differential element \(d l .\) charge on this element.

\(d q=\left(\frac{q}{\pi r}\right) d l\)

\(=\frac{q}{\pi r}(r d \theta)(\because d l=r d \theta)\)

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

Electric field at \(O\) due to \(d q\) is

\(d E=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{d q}{r^{2}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q}{\pi r^{2}} d \theta\)

The component \(d E \cos \theta\) will be counter balanced by another element on left portion. Hence resultant field at \(\mathrm{O}\) is the resultant of the component \(d E \sin \theta\) only.

\(\therefore E=\int d E \sin \theta=\int_{0}^{\pi} \frac{q}{4 \pi^{2} r^{2} \epsilon_{0}} \sin \theta d \theta\)

\(=\frac{q}{4 \pi^{2} r^{2} \epsilon_{0}}[-\cos \theta]_{0}^{\pi}=\frac{q}{4 \pi^{2} r^{2} \epsilon_{0}}(+1+1)\)

\(=\frac{q}{2 \pi^{2} r^{2} \epsilon_{0}}\)

The direction of \(E\) is towards negative \(y-\) axis.

\(\therefore \vec{E}=-\frac{q}{2 \pi^{2} r^{2} \epsilon_{0}} \hat{j}\)