Question

Find the electric field at the centre of semicircular ring shown in the figure.

• A. $\frac{4kq}{\pi R^2}\left(\hat{i}\right)$
• B. $\frac{4kq}{\pi R^2}(-\hat{i})$
• C. $\frac{2kq}{\pi R^2}(-\hat{i})$
• D. $\frac{4kq}{\pi R}(-\hat{i})$

Answer 1

The correct option is B $\frac{4kq}{\pi R^2}(-\hat{i})$


Consider an element of length $Rd\theta$ and charge on the element be $dq=\lambda Rd\theta$ , Where $\lambda$ is charge per unit length.

Here sine component of net electric field due to elements of both positive and negative charges will cancel out and only cosine components contribute in the net field. i.e,

For small element , electric field is given by :

Electric field at the center due to elemental lengths, $d{E}_0=2dE\cos \theta$

$\Rightarrow d{E}_0=\frac{2dq}{4\pi {\epsilon }_0{R}^2}\cos \theta$

Net electric field at the centre, $E=\int d{E}_0$

$\Rightarrow E=\frac{\lambda }{2\pi {\epsilon }_{0}R}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta d\theta$



$\Rightarrow E=\frac{2k\left(2q\right)}{\pi R^2}=\frac{4qk}{\pi R^2}$

As the net field is in the direction of the negative x-axis so,

$\overrightarrow{E}=\frac{4qk}{\pi R^2}(-\hat{i})$

Hence, option (b) is the correct answer.