Q.
A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field E at the centre $O$ is
A) $\frac{q}{4\pi^{2}\varepsilon_{0}r^{2}}\ \widehat{\mathbf{i}}$
B) $- \frac{q}{4\pi^{2}\varepsilon_{0}r^{2}}\ \widehat{\mathbf{j}}$
C) $- \frac{q}{2\pi^{2}\varepsilon_{0}r^{2}}\ \widehat{\mathbf{i}}$
D) $\frac{q}{2\pi^{2}\varepsilon_{0}r^{2}}\ \widehat{\mathbf{j}}$
Explanation
(c) Linear charge density
[image:1_image389.emf]
$\ \ \ \lambda = \left( \frac{q}{\pi r} \right)$
$E = \int_{}^{}{dE\sin{\theta\left( - \widehat{\mathbf{J}} \right)}}$
$\ \ \ \ = \int_{}^{}{\frac{K.dq}{r^{2}}\sin{\theta\left( - \widehat{\mathbf{J}} \right)}}$
$E = \frac{K}{r^{2}}\int_{}^{}{\frac{qr}{\pi r}d\ \theta\sin{\theta\left( - \widehat{J} \right)}}$
$\ \ \ \ = \frac{K}{r^{2}}\frac{q}{\pi}\int_{0}^{\pi}{\sin{\theta\left( - \widehat{\mathbf{J}} \right)}}$
$= \frac{q}{2\pi^{2}\varepsilon_{0}r^{2}}\left( - \widehat{\mathbf{J}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.