Question

For the given charge distribution, net electric field at the centre of non-conducting ring is (in $\text{N/C}$)
(Assume that charge on first quadrant is neutral, second and third quadrant is positively charged, fourth quadrant is negatively charged ) (Round off to the nearest integer value)

• A. 285
• B. 285.00
• C. 285.0

Answer 1

Answer: (A), (B), (C)

Given:

${\lambda }_1=+10\text{nC/m}=+10\times {10}^{-9}\text{C/m};$

${\lambda }_2=-10\text{nC/m}=-10\times {10}^{-9}\text{C/m};$

$R=1\text{m}$


Electric field $E_1$ due to positive charged semi-circular ring $\text{CDE}$ will be along $\text{OA}$.

So, $E_1=\frac{2k{\lambda }_1}{R}=\frac{2\times 9\times {10}^9\times 10\times {10}^{-9}}{1}$

$\Rightarrow E_1=180\text{N/C}$ (along $\text{OA}$)

Electric field $E_2$ due to negative charged quartered ring $\text{ABC}$ will be along $\text{OB}$.

So,
$E_2=\frac{\sqrt{2}k\lambda }{R}=\frac{\sqrt{2}\times 9\times {10}^9\times 10\times {10}^{-9}}{1}$

$\Rightarrow E_2=90\sqrt{2}\text{N/C}$ (along $\text{OB}$)

Further, using formula of resultant vector

$E_{\text{net}}=\sqrt{E_1^2+E_2^2+2E_1{E}_2\cos \theta }$

$\Rightarrow E_{net}=\sqrt{{180}^2+(90\sqrt{2}{)}^2+2\times 180\times 90\sqrt{2}\times \cos {45}^{\circ }}$

$\Rightarrow E_{net}=284.6\approx 285\text{N/C}$

Accepted answer: $285,285.0,285.00$