Question

A semicircular arc of radius $a$ is charged uniformly and the charge per unit length is $\lambda$. The electric field at its centre is

• A. $\frac{\lambda }{2\pi {\epsilon }_0{a}^2}$
• B. $\frac{\lambda }{4\pi {\epsilon }_{0}a}$
• C. $\frac{{\lambda }^2}{2\pi {\epsilon }_{0}a}$
• D. $\frac{\lambda }{2\pi {\epsilon }_{0}a}$

Answer 1

Answer: (D)

Electric intensity at centre $O$, due to small element dl of charged ring

$dE=\frac{\lambda dl}{4\pi {\epsilon }_0{a}^2}=\frac{\lambda (ad\theta )}{4\pi {\epsilon }_0{a}^2}$
As is clear from figure, horizontal components of $dE$ will cancel out in pairs and vertical components will add.
$\therefore E=\underset{0}{\overset{\pi }{\int }}dE\sin \theta =\underset{0}{\overset{\pi }{\int }}\frac{\lambda }{4\pi {\epsilon }_{0}a}\sin \theta d\theta$
$=\frac{\lambda }{4\pi {\epsilon }_{0}a}[-\cos \theta {]}_0^{\pi }$
$E=\frac{\lambda }{4\pi {\epsilon }_{0}a}(-\cos \pi +\cos 0)=\frac{\lambda }{2\pi {\epsilon }_{0}a}$