question_answer

  A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field\[\overrightarrow{E}\]at the centre O is ?     AIEEE  Solved  Paper-2010

A) \[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}\hat{j}\]                        

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

C)        \[-\frac{q}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}\hat{j}\]

D)        \[-\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}\hat{j}\]

Correct Answer: D

Solution :

\[E=\int\limits_{-\pi /2}^{\pi /2}{dE}\cos \theta =2\int\limits_{0}^{\pi /2}{\frac{k\lambda Rd\theta }{{{R}^{2}}}}\cos \theta \] \[\overrightarrow{E}=\frac{2}{4\pi {{\varepsilon }_{0}}}\frac{qR}{\pi R{{R}^{2}}}[\sin \theta ]_{0}^{\pi /2}=\] \[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}[sin90-\sin 0](-\hat{j})\] \[\overrightarrow{E}=\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}(-\hat{j})\]