Question

For the given charge distribution on the ring, the net electric field at the centre of non-conducting ring is
(Assume the part of ring in first and third quadrant is neutral, second quadrant is positively charged, fourth quadrant is negatively charged)

• A. $180\sqrt{2}\text{N/C}$
• B. $180\sqrt{3}\text{N/C}$
• C. $180\sqrt{5}\text{N/C}$
• D. $180\sqrt{7}\text{N/C}$

Answer 1

The correct option is A $180\sqrt{2}\text{N/C}$

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

Electric field due to the quarter ring at the centre,
$\alpha ={90}^{\circ }$
$\Rightarrow E=\frac{\sqrt{2}k\lambda }{R}$

Therefore, net electric field,

$E_{\text{net}}=\frac{\sqrt{2}k{\lambda }_1}{R}+\frac{\sqrt{2}k{\lambda }_2}{R}\left[\text{along OB}\right]$

$\Rightarrow E_{\text{net}}=\frac{2\sqrt{2}k\lambda }{R}(\because {\lambda }_1={\lambda }_2=\lambda )$

$\Rightarrow E_{\text{net}}=\frac{2\sqrt{2}\times 9\times {10}^9\times 10\times {10}^{-9}}{1}$

$\Rightarrow E_{\text{net}}=180\sqrt{2}\text{N/C}$

Hence, option $\left(a\right)$ is correct.