Question

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

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

Answer 1

Answer: (C)

Linear charge density $\lambda =\left(\frac{q}{\pi r}\right)$
$E=\int dEsin\theta \left(-\hat{j}\right)=\int \frac{K.dq}{r^2}sin\theta \left(-\hat{j}\right)$
$E=\frac{K}{r^2}\int \frac{qr}{\pi r}d\theta sin\theta \left(-\hat{j}\right)$
$=\frac{K}{r^2}\frac{q}{\pi }\underset{0}{\overset{\pi }{\int }}sin\theta \left(-\hat{j}\right)$
$=\frac{q}{2{\pi }^2{\epsilon }_0{r}^2}\left(-\hat{j}\right)$