Question

Electric field at center $O$ of semicircle of radius $a$ having linear charge density $\lambda$ is given as :

• A. $\frac{2\lambda }{{ϵ}_{o}a}$
• B. $\frac{\lambda \pi }{{ϵ}_{o}a}$
• C. $\frac{\lambda }{2\pi {ϵ}_{o}a}$
• D. $\frac{\lambda }{\pi {ϵ}_{o}a}$

Answer 1

The correct option is C $\frac{\lambda }{2\pi {ϵ}_{o}a}$

Consider an element having charge $dq$ at angle $\theta$ of angular width $d\theta$
Length of element $=ad\theta$
$dq=\lambda ad\theta$
Field due to this element $=dE=\frac{dq}{4\pi {ϵ}_0{a}^2}$
$\to dE=\frac{\lambda ad\theta }{4\pi {ϵ}_0{a}^2}=\frac{\lambda d\theta }{4\pi {ϵ}_{0}a}$

Now, from symmetry, we can see that the horizontal component of field of right side gets cancelled by that of left side.
And, the vertical compound of force add up,

$E=\int d{E}_y=\int dE\sin \theta$

$\Rightarrow E={\int }_0^x\frac{\lambda }{4\pi {ϵ}_{0}a}\sin \theta d\theta$

$\Rightarrow E=\frac{\lambda }{4\pi {ϵ}_{0}a}|-\cos \theta {|}_0^x$

$\Rightarrow E=\frac{2\lambda }{4\pi {ϵ}_{0}a}$

$\Rightarrow E=\frac{\lambda }{2\pi {ϵ}_{0}a}$