A semi-circular arc of radius ‘a’ is charged uniformly and the charge per unit length is λ. The electric field at the center of this arc is

$\lambda =$linear charge density;
Charge on elementary portion $dx=\lambda dx$

Electric field at O, $dE=\frac{\lambda dx}{4\mathrm{\pi }{\epsilon }_o{a}^2}$

Due to symmetry horizontal electric field, i.e., perpendicular to AO, will be cancelled.
Hence, net electric field = addition of all electrical fields in direction of  AO

$=\underset{ }{\sum dE\cos \theta }$

      $\Rightarrow E=\int \frac{\lambda dx}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{a}}^2}\cos \theta$

Also, $d\theta =\frac{dx}{a}$ or $dx=ad\theta$

         $E={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}\frac{\lambda \cos \theta d\theta }{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{a}}=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{a}}{\left[\sin \theta \right]}_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}$

$\frac{\lambda }{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{a}}\left[1-\left(-1\right)\right]=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{a}}$