question_answer

Charge \[q\] is uniformly distributed over a thin half ring of radius\[R\]. The electric field at the centre of the ring is

A) \[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]                    

B) \[\frac{q}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]

C) \[\frac{q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\]                   

D) \[\frac{q}{2\pi {{\varepsilon }_{0}}{{R}^{2}}}\]

Correct Answer: A

Solution :

From figure\[dl=Rd\theta \] Charge on\[dl=\lambda Rd\theta \left\{ \lambda =\frac{q}{\pi R} \right\}\] Electric field at centre due to\[dl\]is                 \[dE=k\cdot \frac{\lambda Rd\theta }{{{R}^{2}}}\] We need to consider only the component\[dE\,\,\cos \,\theta \], as the component \[dE\,\sin \,\theta \] will cancel out because of the field at \[c\] due to the symmetrical element\[dl\], Total field at centre \[=2\int_{0}^{\pi /2}{dE\,\,\cos \theta }\]                 \[=\frac{2k\lambda }{R}\int_{0}^{\pi /2}{\cos \theta \,\,d\theta }\]                 \[=\frac{2k\lambda }{R}=\frac{q}{2\pi {{r}^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]