The direction (q) of  $\overrightarrow{\mathrm{E}}$ at point P due to uniformly charged finite rod will be

$\begin{array}{l}\mathrm{dE}=\frac{\mathrm{dq}}{4{\mathrm{\pi \epsilon }}_0\left({\mathrm{x}}^2+{\mathrm{r}}^2\right)}\\ \mathrm{d}\overline{\mathrm{E}}=\mathrm{dEcos}\mathrm{\theta }\widehat{\mathrm{i}}+\mathrm{dEsin}\mathrm{\theta }\widehat{\mathrm{j}}\end{array}$

$\begin{array}{l}{\mathrm{E}}_{\mathrm{x}}=\int {\mathrm{dE}}_{\mathrm{x}}={\int }_0^{\mathrm{\pi }/3} \mathrm{dEcos}\mathrm{\theta }=\mathrm{K}\int \frac{\mathrm{dq}}{\left({\mathrm{r}}^2+{\mathrm{x}}^2\right)}\times \frac{\mathrm{r}}{\sqrt{{\mathrm{r}}^2+{\mathrm{x}}^2}}\\ {\mathrm{E}}_{\mathrm{x}}=\mathrm{K}\int \frac{\mathrm{r\lambda dx}}{{\left({\mathrm{r}}^2+{\mathrm{x}}^2\right)}^{3/2}}\\ \text{ put }\Rightarrow \mathrm{x}=\mathrm{rtan}\mathrm{\theta }\mathrm{dx}={\mathrm{rsec}}^2\mathrm{\theta d\theta }\\ {\mathrm{E}}_{\mathrm{x}}=\mathrm{K}{\int }_0^{\mathrm{\pi }/3} \frac{\mathrm{r\lambda }{\mathrm{rec}}^2\mathrm{\theta d\theta }}{{\mathrm{r}}^3{\sec }^2\mathrm{\theta }}=\frac{\mathrm{K\lambda }}{\mathrm{r}}{\int }_0^{\mathrm{\pi }/3} \cos \mathrm{\theta d\theta }\\ \Rightarrow {\mathrm{E}}_{\mathrm{x}}=\frac{\mathrm{k\lambda }}{\mathrm{r}}\frac{\sqrt{3}}{2}\widehat{\mathrm{i}}\mathrm{similarly}{\mathrm{E}}_{\mathrm{y}}=\frac{\mathrm{K\lambda }}{\mathrm{r}}\left(+\frac{1}{2}\widehat{\mathrm{j}}\right)\\ \tan \mathrm{\theta }=\left(\frac{{\mathrm{E}}_{\mathrm{y}}}{{\mathrm{E}}_{\mathrm{x}}}\right)\Rightarrow \tan \mathrm{\theta }=(\sqrt{3}{)}^{-1}\Rightarrow \mathrm{\theta }={30}^0\end{array}$

at angle $\mathrm{\theta }=90º$from x-axis