Question

If the half portion $\text{ACB}$ of the uniformly charged ring $\text{ADBCA}$ is cut-out then the net electric field due to remaining half portion will be

• A. $90\text{N/C}$ along $\text{OC}$
• B. $90\text{N/C}$ along $\text{OD}$
• C. $180\text{N/C}$ along $\text{OC}$
• D. $180\text{N/C}$ along $\text{OD}$

Answer 1

The correct option is B $90\text{N/C}$ along $\text{OD}$

Given:
$\lambda =-10\text{nC/m}=-10\times {10}^{-9}\text{C/m};R=2\text{m}$

If half portion $\text{(ACB)}$ of the uniformly charged ring $\text{(ADBCA)}$ is cut-out then remaining portion $\text{(ADB)}$ will be a semi- circular ring whose net electric field is given by

$E=\frac{2k\lambda }{R}$

$\Rightarrow E=\frac{2\times 9\times {10}^9\times 10\times {10}^{-9}}{2}$

$\Rightarrow E=90\text{N/C}$

Its direction will be along $\text{OD}$ as the charge is of negative nature.