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. $\frac{q}{4\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$
• B. $-\frac{q}{4\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$
• C. $-\frac{q}{2\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$
• D. $\frac{q}{2\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$

Answer 1

Answer: (C)

Linear charge density $\lambda = \left(\frac{q}{\pi r}\right)$
$E = \int dE\,sin\,\theta \left(-\hat{j}\right) =\int \frac{K.dq}{r^{2}}sin\,\theta \left(-\hat{j}\right)$
$E = \frac{K}{r^{2}}\int\frac{qr}{\pi r} d\theta\,sin\,\theta \left(-\hat{j}\right)$
$= \frac{K}{r^{2}} \frac{q}{\pi } \int\limits^{\pi}_{0} \,sin\,\theta \left(-\hat{j}\right)$
$= \frac{q}{2\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \left(-\hat{j}\right)$