Question

A semi-circular rod of radius $R$ is charged uniformly with a total charge $QC$. The electric field intensity at the centre of the curvature is
(where $K=\frac{1}{4\pi {ϵ}_0}$)

• A. $\frac{2KQ}{\pi R^2}$
• B. $\frac{3KQ}{\pi R^2}$
• C. $\frac{KQ}{\pi R^2}$
• D. $\frac{4KQ}{\pi R^2}$

Answer 1

The correct option is A $\frac{2KQ}{\pi R^2}$

Let us take an element at angle $\theta$ from X-axis of angular width $d\theta$.

Linear charge density$=\lambda =\frac{q}{\pi R}$

Length of ring=$Rd\theta$

$dq=\lambda Rd\theta$

Field at origin due to this element of charge-

$dE=\frac{Kdq}{R^2}=\frac{K\lambda Rd\theta }{R^2}$ along shown direction.

Now, due to symmetry, horizontal components of field from left part of ring will cancel field from right part of ring and vertical components will add up.

$E={\int }_0^{\pi }dE\sin \theta$

$E={\int }_0^{\pi }\frac{K\lambda R}{R^2}\sin \theta d\theta$

$⟹E=\frac{K\lambda }{R}[-\cos \theta {]}_0^{\pi }$

$⟹E=\frac{2K\lambda }{R}$

$⟹E=\frac{2KQ}{\pi R^2}$

Answer-(A)