Question

If the electric field at the centre of the semicircular ring shown in figure. is $\frac{-xkq\hat{i}}{\pi R^2}$, then find $x$.

Answer 1

Consider an element of length $Rd\theta$ and charge on the element $dq=\lambda Rd\theta$ where $\lambda =$ charge per unit length.
Here sin component of net electric filed due to element of both positive and negative charge will cancel each other and only cosine component contribute the filed. i.e,
$d{E}_O=2dE\cos \theta =2\frac{kdq}{R^2}\cos \theta =2\frac{k\lambda Rd\theta }{R^2}\cos \theta$
$E=\frac{2k\lambda R}{R^2}{\int }_0^{\frac{\pi }{2}}\cos \theta d\theta =\frac{2k\lambda R}{R^2}$
Since $q=\lambda \left(\frac{\pi }{2}\right)R,$ thus $E=\frac{2k\left(2q\right)}{\pi R^2}=\frac{4qk}{\pi R^2}$
As net field in the direction of negative x axis so $\overrightarrow{E}=\frac{4qk}{\pi R^2}(-\hat{i})$
thus, $x=4$