Physics - Electrostatics Question with Solution | TestHub

Key Idea: It is given that the ring is conducting. As the ring is conducting, so electric field at its centre is zero, ie, ${\overrightarrow{\mathbf{E}}}_{\text{total }}=0$ $\begin{array}{rl}{\overrightarrow{\mathbf{E}}}_{AKB}+{\overrightarrow{\mathbf{E}}}_{ACDB}& =0\\ {\overrightarrow{\mathbf{E}}}_{ACDB}& =-{\overrightarrow{\mathbf{E}}}_{AKB}\\ {\overrightarrow{\mathbf{E}}}_{ACDB}& =-\overrightarrow{\mathbf{E}}\text{ (along }KO\text{ ) }\end{array}$ Therefore, the electric field at the centre due to the charge on the part $ACDB$ of the ring is $E$ along $OK$. lternative :The fields at $O$ due to $AC$ and $BD$ cancel each other. The field due to $CD$ is acting in the direction $OK$ and equal in magnitude to $E$ due to $AKB$.