Question

Find the electric field at the centre of a uniformly charged semicircular ring of radius R and linear charge density $\lambda$ :

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

Answer 1

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

Consider an element having charge $dq$ at angle $\theta$ of angular width $d\theta$.

Length of element$=Rd\theta$

$dq=\lambda Rd\theta$

Field due to this element$=dE=\frac{dq}{4\pi {ϵ}_o{R}^2}$

$⟹dE=\frac{\lambda Rd\theta }{4\pi {ϵ}_o{R}^2}=\frac{\lambda d\theta }{4\pi {ϵ}_{o}R}$

Now, from symmetry we can see that the horizontal component of field of right side get cancelled by that of left side.

And, the vertical component of force add up.

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

$⟹E={\int }_0^{\pi }\frac{\lambda }{4\pi {ϵ}_{o}R}\sin \theta d\theta$

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

$⟹E=\frac{2\lambda }{4\pi {ϵ}_{o}R}$

$⟹E=\frac{\lambda }{2\pi {ϵ}_{o}R}$

Answer-(A)