Question

An infinite long thread carries a charge, uniformly spread over its length. It is bent in the form as shown. If the charge per unit length is $\lambda$, the electric field at the point $O$ is:

• A. $\frac{\lambda }{2\pi {\epsilon }_{0}r}$
• B. $\frac{\lambda }{\sqrt{2}\pi {\epsilon }_{0}r}$
• C. $\frac{\lambda }{2\sqrt{2}\pi {\epsilon }_{0}r}$
• D. Zero

Answer 1

The correct option is A $\frac{\lambda }{2\pi {\epsilon }_{0}r}$

R.E.F image

The above figure is divided into 3 parts

(1) & (3) are semi-infinite line charges

(2) is ar arc with central angle ${90}^o$

electric field is calculated separately $\to$

For (1)

$Ex=\frac{k\lambda }{R}$ $Ey=\frac{k\lambda }{R}$

For (2)

by substituting $\theta =\frac{\pi }{2}$ in the formula

$E=\frac{2k\theta \sin \frac{\theta }{2}}{\theta R^2}$; where $\theta =$ total charge

$\theta =$ angle subtended at the center

we get $E=\frac{2\lambda k}{R}$

For (3),

$Ex=-\frac{k\lambda }{R}$ $Ey=\frac{k\lambda }{R}$

By adding electric fields of all 3 cases

$E=\frac{2k\lambda }{R}=\frac{\lambda }{2\pi {\in }_{0}R}$