Question

A thin semicircular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field at the centre $O$ is :-

• A. $\frac{q}{2{\pi }^2{\epsilon }_0{r}^2}\hat{J}$
• B. $\frac{q}{4{\pi }^2{\epsilon }_0{r}^2}\hat{J}$
• C. $-\frac{q}{4{\pi }^2{\epsilon }_0{r}^2}\hat{J}$
• D. $-\frac{q}{2{\pi }^2{\epsilon }_0{r}^2}\hat{J}$

Answer 1

The correct option is B $\frac{q}{4{\pi }^2{\epsilon }_0{r}^2}\hat{J}$

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

Linear charge density$=\lambda =\frac{q}{\pi R}$

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.

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

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

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

$⟹E=\frac{\lambda R}{2\pi {ϵ}_o{R}^2}$

$⟹E=\frac{q}{2{\pi }^2{ϵ}_o{R}^2}$