Q. A semi-circular are of radius a is charged uniformly and the charge per unit length is $\lambda.$ the electric field at its centre is $\ $
A) $\frac{\lambda}{2\pi\varepsilon_{0}a^{2}}$
B) $\frac{\lambda^{\ }}{4\varepsilon_{0}a^{\ }}$
C) $\frac{\lambda^{2}}{4\pi\varepsilon_{0}a^{\ }}$
D) $\frac{\lambda}{2\pi\varepsilon_{0}a^{\ }}$
Explanation
(b) $E = \sum_{}^{}{{\frac{dq}{4\pi\varepsilon_{0}a^{2}} =}\ }\sum_{}^{}{{\frac{\lambda dl}{4\pi\varepsilon_{0}a^{2}} =}\ }\frac{\lambda(\pi a)}{4\pi\varepsilon_{0}a^{2}} = \frac{\lambda}{4\varepsilon_{0}a^{\ }}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.