Question

A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net electric field $\overrightarrow{E}$ at the centre $O$ is :

• A. $\frac{q}{4{\pi }^2{ϵ}_0{r}^2}\overrightarrow{j}$
• B. $-\frac{q}{4{\pi }^2{ϵ}_0{r}^2}\overrightarrow{j}$
• C. $-\frac{q}{2{\pi }^2{ϵ}_0{r}^2}\overrightarrow{j}$
• D. $\frac{q}{2{\pi }^2{ϵ}_0{r}^2}\overrightarrow{j}$

Answer 1

The correct option is C $-\frac{q}{2{\pi }^2{ϵ}_0{r}^2}\overrightarrow{j}$

Lets consider Linear charge density, $\lambda =\frac{q}{\pi r}$. . . . .(1)

Consider a small element AB of length $dl$ subtending an angle

$d\theta$ at the center O as shown in the figure.

Charge on the element, $dq=\lambda dl$

$dq=\lambda rd\theta$ $(d\theta =\frac{dl}{r})$

The electric field at the center O due to the charge element is

$dE=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{r^2}=\frac{\lambda rd\theta }{4\pi {\epsilon }_0{r}^2}$

Resolve $dE$ into two rectangular components,

By symmetry, $\int dEcos\theta =0$

The net electric field at O is,

$\overrightarrow{E}={\int }_0^{\pi }dEsin\theta (-\hat{j})={\int }_0^{\pi }\frac{\lambda rd\theta }{4\pi {\epsilon }_0{r}^2}sin\theta (-\hat{j})$

$\overrightarrow{E}=-{\int }_0^{\pi }\frac{qrsin\theta d\theta }{4{\pi }^2{\epsilon }_0{r}^3}\hat{j}$ ( from equation 1)

$\overrightarrow{E}=-{\int }_0^{\pi }\frac{qsin\theta d\theta }{4{\pi }^2{\epsilon }_0{r}^2}\hat{j}=-\frac{q}{4{\pi }^2{\epsilon }_0{r}^2}[-cos\theta {]}_0^{\pi }\hat{j}$

$\overrightarrow{E}=-\frac{q}{2{\pi }^2{\epsilon }_0{r}^2}\hat{j}$

The correct option is C.