question_answer

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, as shown in figure. The electric field intensity at point O is

A) \[\frac{\lambda }{2\pi {{\varepsilon }_{0}}a}\]

B) 0

C) \[\frac{\lambda }{\pi {{\varepsilon }_{0}}a}\]

D) \[\frac{\lambda }{4\pi {{\varepsilon }_{0}}a}\]

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 \[E=2\times \frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{\lambda }{a}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}a}\]