Question

A thin glass rod is bent in a semicircle of radius R.A charge is non-uniformly distributed along the rod with a linear charge density $\lambda=\lambda_0sin\theta$ ($\lambda_0$ is a positive constant). The electric field at the center P of the semicircle is

• A. $-\frac{\lambda_0}{8\pi\epsilon_0R}\hat{j}$
• B. $\frac{\lambda_0}{8\pi\epsilon_0R}\hat{j}$
• C. $\frac{\lambda_0}{8\pi\epsilon_0R}\hat{i}$
• D. $-\frac{\lambda_0}{8\pi\epsilon_0R}\hat{i}$

Answer 1

Answer: (C)

$\frac{\lambda_0}{8\pi\epsilon_0R}\hat{i}$

Answer 2

The correct answer is option (C): $\frac{\lambda_0}{8\pi\epsilon_0R}\hat{i}$
The magnitude of the field = $dE=\frac{1}{4\pi\epsilon_0}\frac{rd\theta\times Q/(\pi r/2)}{r^2}=\frac{Q}{2\pi^2\epsilon_0r^2}d\theta$
$E=\int_{0}^{\frac{\pi}{2}}2dEsin\theta =2\int_{0}^{\frac{\pi}{2}}\frac{Q}{2\pi\epsilon_0r^2}sin\theta d \theta =\frac{Q}{\pi^2\epsilon_0r^2}$=