Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electric Charges and Fields

Question:

Options:

a

b

c

d

Correct Answer:

c

Explanation:

$\text{Linear charge density }\lambda = \frac{q}{l} = \frac{q}{\pi r}$

$\text{Charge on small portion of length }rd\theta \text{ is } dq = \lambda rd\theta$

$ E = \int_\frac{-\pi}{2}^\frac{\pi}{2}{dEcos\theta} = \int_\frac{-\pi}{2}^\frac{\pi}{2}{\frac{k\lambda rd\theta}{r^2}cos\theta} $

$E = \frac{q}{2\pi^2 \epsilon_0 r^2} \text{Along -Y direction}$