Question

A thin semi-circular ring of radius ‘r’ has a positive charge ‘q’ distributed uniformly over it. The net electric field at the centre O is

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

Answer 1

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

The electric field can be written as,
$E=\int dEsin\theta$ $q\to \pi r$
$dq\to dr$
$=\int \frac{Kdq}{r^2}sin\theta$ $\frac{q}{\pi r}=\frac{dq}{dr}\Rightarrow dq=\frac{q}{\pi r}dr$
$=\frac{K}{r^2}\int dqsin\theta$ $rd\theta =dr$ $=\frac{q}{\pi }d\theta$
$=\frac{K.q}{\pi r^2}{\int }_0^{\pi }sin\theta d\theta$
$=\frac{2kq}{\pi r^2}=\frac{1}{2\pi {ϵ}_0}\frac{q}{\pi r^2}(-\hat{J})$