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 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 :

[d]

\[E=\int\limits_{-\pi /2}^{\pi /2}{dE}\cos \theta =2\int\limits_{0}^{\pi /2}{\frac{k\lambda Rd\theta }{{{R}^{2}}}}\cos \theta \]

\[\overrightarrow{E}=\frac{2}{4\pi {{\varepsilon }_{0}}}\frac{qR}{\pi R{{R}^{2}}}[\sin \theta ]_{0}^{\pi /2}=\]

\[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}[sin90-\sin 0](-\hat{j})\]

\[\overrightarrow{E}=\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}(-\hat{j})\]