Question

The electric field intensity due to a thin infinitely long wire of uniform linear charge density $\lambda$ at $\text{O}$ as shown in the figure is

• A. $\frac{\lambda }{2\pi {ϵ}_{0}R}$
• B. $\frac{\sqrt{2}\lambda }{2\pi {ϵ}_{0}R}$
• C. $\frac{\lambda \sqrt{5}}{2\pi {ϵ}_{0}R}$
• D. $0$

Answer 1

The correct option is A $\frac{\lambda }{2\pi {ϵ}_{0}R}$


As the charge lies along the same line joining the infinite line charge distribution, Electric field due to infinite line charge distribution will be zero.

Electric field at the point $\text{O}$ will be due to only the arc of radius $R$.

Electric field intensity $\left(E\right)$ due to the arc at $O$ is,

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

Substituting $K=\frac{1}{4\pi {ϵ}_0}$, we get

$E=2\left(\frac{1}{4\pi {ϵ}_0}\right)\frac{\lambda }{R}$

$\therefore E=\frac{\lambda }{2\pi {ϵ}_{0}R}$

Hence, option (a) is correct answer.