question_answer

Two semicircular rings lying in the same plane of uniform linear charge density \[\lambda \] have radii r and 2r. They are joined using two straight uniformly charged wires of linear charge density \[\lambda \] and length r as shown in the figure. The magnitude of electric field at common centre of semi-circular rings is

A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{3\lambda }{2r}\]                      

B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda }{2r}\]

C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2\lambda }{r}\]

D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda }{r}\]

Correct Answer: D

Solution :

[d] The electric field due to both straight wires shall cancel at common centre O. The electric field due to larger and smaller semi-circular rings at O be E and E' respectively. \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2\lambda }{2r}E'\frac{1}{4{{\pi }^{{{e}_{0}}}}}\frac{2\lambda }{r}\] \[\therefore \]Magnitude of electric field at O is \[=E-\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{2\lambda }{r}-\frac{\lambda }{r} \right)=\frac{1}{4\pi {{\varepsilon }_{0}}r}\frac{\lambda }{r}\]