Question

In the following figure, one semi infinite wire and semi circular arc is having linear charge density $+\lambda$ and the other semi infinite wire is having charge density $-\lambda$. Find the magnitude of electric field at point O:

• A. Zero
• B. $\frac{\sqrt{2}K\lambda }{R}$
• C. $\frac{2\sqrt{2}K\lambda }{R}$
• D. None of these

Answer 1

Answer: (C)

$\frac{2\sqrt{2}K\lambda }{R}$

The electric field at O due to semi infinite wire with line charge density $+\lambda$ is $\overrightarrow{E_1}=K\frac{2\lambda }{R}\hat{i}$
and electric field at O due to semi infinite wire with line charge density $-\lambda$ is $\overrightarrow{E_2}=K\frac{2(-\lambda )}{R}\hat{i}$
The electric field at O due to semi circular wire with line charge density $+\lambda$ is $\overrightarrow{E_3}={\int }_0^{\pi /2}2d{E}_1\sin \theta \hat{j}=2{\int }_0^{\pi /2}K\frac{\lambda Rd\theta }{R^2}\sin \theta \hat{j}=K\frac{2\lambda }{R}\hat{j}$
Net field at O is $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}=K\frac{2\lambda }{R}\hat{i}-(-K\frac{2\lambda }{R})\hat{i}+K\frac{2\lambda }{R}\hat{j}=K\frac{2\sqrt{2}\lambda }{R}\hat{j}$
$|\overrightarrow{E}|=K\frac{2\sqrt{2}\lambda }{R}$