Question

What is the electric field at the centre $(x=0,y=0)$ of a charged semi-circle of radius $R$ and linear charge density $\lambda$?

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

Answer 1

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

Let us take an element at angle $\theta$ from X-axis of angular width $d\theta$.

Length of ring=$Rd\theta$

$dq=\lambda Rd\theta$

Field at origin due to this element of charge-

$dE=\frac{dq}{4\pi {ϵ}_o{R}^2}=\frac{\lambda Rd\theta }{4\pi {ϵ}_o{R}^2}$ along shown direction.

Now, due to symmetry, horizontal components of field from left part of ring will cancel field from right part of ring and vertical components will add up.

$E={\int }_0^{\pi }dE\sin \theta$

$E={\int }_0^{\pi }\frac{\lambda R}{4\pi {ϵ}_o{R}^2}\sin \theta d\theta$

$⟹E=\frac{\lambda R}{4\pi {ϵ}_o{R}^2}[-\cos \theta {]}_0^{\pi }$

$⟹E=\frac{\lambda }{2\pi {ϵ}_{o}R}$