Electrostatics Question 715
Question: Two very long line charges of uniform charge density $ +\lambda $ and $ -\lambda $ are placed along same line with the separation between the nearest ends being 2a, . The electric field intensity at point O is
Options:
A) $ \frac{\lambda }{2\pi {\varepsilon _{0}}a} $
B) 0
C) $ \frac{\lambda }{\pi {\varepsilon _{0}}a} $
D) $ \frac{\lambda }{4\pi {\varepsilon _{0}}a} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The field at O due to small element
$ dx $ is $ dE=\frac{1}{4\pi {\varepsilon _{0}}}\cdot \frac{\lambda dx}{x^{2}} $
Hence, due to one wire, $ E _{1}=\int\limits _{a}^{\infty }{\frac{1}{4\pi {\varepsilon _{0}}}\cdot \frac{\lambda dx}{x^{2}}E _{1}=\frac{1}{4\pi {\varepsilon _{0}}}\cdot \frac{\lambda }{a}} $ towards left.
Electric field at O due to other wire, $ E _{2}=\frac{1}{4\pi {\varepsilon _{0}}}\cdot \frac{\lambda }{a} $ towards left $ \therefore $ Net field at O is $ E=2\times \frac{1}{4\pi {\varepsilon _{0}}}\cdot \frac{\lambda }{a}=\frac{\lambda }{2\pi {\varepsilon _{0}}a} $
