question_answer

                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:

A)                 \[\frac{\lambda }{2\pi {{a}^{2}}{{\varepsilon }_{0}}}\]    

B)                 \[\frac{\lambda }{4{{\pi }^{2}}{{\varepsilon }_{0}}a}\]    

C)                 \[\frac{\lambda }{2{{\pi }^{2}}{{\varepsilon }_{0}}a}\]

D)                 zero

Correct Answer: C

Solution :

                                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                 \[E=\int_{-\pi /2}^{\pi /2}{dE=2\int_{O}^{\pi /2}{\frac{1}{4\pi {{\varepsilon }_{0}}}}}\frac{\lambda a\cos \theta \,d\theta }{{{a}^{2}}}\]                 \[=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\frac{\lambda }{a}[\sin \theta ]_{o}^{\pi /2}\]                 \[=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\cdot \frac{\lambda }{a}\cdot 1=\frac{\lambda }{2\pi {{\varepsilon }_{o}}a}\]