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 :

[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

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

\[=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\frac{\lambda }{a}[\sin \theta ]_{o}^{\pi /2}\]

\[=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\cdot \frac{\lambda }{a}\cdot 1=\frac{\lambda }{2\pi {{\varepsilon }_{o}}a}\]