Question

A thin semi-circular ring of radius $R$ has a positive charge $q$ distributed uniformly over it. The net field $\overrightarrow{E}$ 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)

Given,

$k=\frac{1}{4\pi {\epsilon }_o}and\lambda =\frac{q}{\pi R}(where,\lambda =\text{linear}\text{charge}\text{density})$

$\text{Small}\text{charge}\text{on}\text{small}\text{length(Rd}\theta )is,dq=\lambda Rd\theta$

$\overrightarrow{E}=\underset{-\pi /2}{\overset{\pi /2}{\int }}dE\cos \theta =2\underset{0}{\overset{\pi /2}{\int }}\frac{k\left(\lambda Rd\theta \right)}{R^2}\cos \theta (-\hat{j})$

$\Rightarrow \overrightarrow{E}=2\times \left(\frac{1}{4\pi {\epsilon }_0}\right)\left(\frac{q}{\pi R}\right)\frac{R}{R^2}[\sin \theta {]}_0^{\pi /2}(-\hat{j})$

$\Rightarrow \frac{q}{2{\pi }^2{\epsilon }_0{R}^2}[sin90-\sin 0](-\hat{j})$

$\Rightarrow \overrightarrow{E}=\frac{q}{2{\pi }^2{\epsilon }_0{R}^2}(-\hat{j})$

Hence, Electric field at point O is $\frac{q}{2{\pi }^2{\epsilon }_0{R}^2}(-\hat{j})$