Question

For the given charge distribution, the net electric field at the centre of the non-conducting ring is

• A. $180\text{N/C}$
• B. $220\text{N/C}$
• C. $320\text{N/C}$
• D. $360\text{N/C}$

Answer 1

The correct option is D $360\text{N/C}$

Given:
${\lambda }_1=10\text{nC/m}=10\times {10}^{-9}\text{C};R=1\text{m}$
${\lambda }_2=-10\text{nC/m}=-10\times {10}^{-9}\text{C}.$



Electric field due to upper half will be directed toward $\text{OB}$ and also, electric field due to lower half will be directed toward $\text{OB}$.
Electric field due to arc of a circle,
$E=\frac{2k\lambda }{R}\sin \frac{\alpha }{2}$

Electric field due to semicircle at the centre,
$\alpha ={180}^{\circ }$
$\Rightarrow E=\frac{2k\lambda }{R}$

Therefore, net electric field at centre

$E_{\text{net}}=\frac{2k{\lambda }_1}{R}+\frac{2k{\lambda }_2}{R}$ [Along $\text{OB}$]

$\Rightarrow E_{\text{net}}=\frac{4k\lambda }{R}(\because {\lambda }_1={\lambda }_2=\lambda )$

$\Rightarrow E_{\text{net}}=\frac{4\times 9\times {10}^9\times 10\times {10}^{-9}}{1}$

$\Rightarrow E_{\text{net}}=360\text{N/C}$

Hence, $\left(d\right)$ is correct answer.