Physics - Electrostatics Question with Solution | TestHub

Electric field intensity at $O$ due to small elemental length $dl$ of charged ring, $\begin{array}{rl}& dE=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\lambda dl}{a^2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\lambda ad\theta }{a^2}\\ & dE=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\lambda }{a}d\theta \end{array}$ $\therefore$ Net electric field at centre $O$ is $\begin{array}{rl}E=\int dE\sin \theta & ={\int }_0^{\pi }\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{a}\sin \theta d\theta \\ & =\frac{\lambda }{4\pi {\epsilon }_{0}a}[-\cos \theta {]}_0^{\pi }\\ \therefore E& =\frac{\lambda }{2\pi {\epsilon }_{0}a}.\end{array}$