A charged wire is bent in the form of a semi-circular arc of radius a. If charge per unit length is \[\lambda \] coulomb/metre, the electric field at the centre O is:

[C] Considering symmetric elements each of length \[dl\] at A and B, we NOTE that electric fields perpendicular to PO are cancelled and those ALONG PO are added. The electric field due to an element of length \[dl(=ad\theta )\] along PO. \[dE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{dq}{{{a}^{2}}}\COS \theta \] \[(\because dl=ad\theta )\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda dl}{{{a}^{2}}}\cos \theta \] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda (ad\theta )}{{{a}^{2}}}\cos \theta \] Net electric field at O \[E=\int_{-\pi /2}^{\pi /2}{dE=2\int_{O}^{\pi /2}{\frac{1}{4\pi {{\varepsilon }_{0}}}}}\frac{\lambda a\cos \theta \,d\theta }{{{a}^{2}}}\] \[=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\frac{\lambda }{a}[\sin \theta ]_{o}^{\pi /2}\] \[=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\cdot \frac{\lambda }{a}\cdot 1=\frac{\lambda }{2\pi {{\varepsilon }_{o}}a}\]