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 $2a$, as shown in figure. The electric field intensity at point $O$ is

• A. $\frac{\lambda }{2\pi {\epsilon }_{0}a}$
• B. $0$
• C. $\frac{\lambda }{\pi {\epsilon }_{0}a}$
• D. $\frac{\lambda }{4\pi {\epsilon }_{0}a}$

Answer 1

Answer: (A)

The field at $O$ due to small element $dx$ is
Hence, due to one wire,
$E_1=\underset{a}{\overset{\infty }{\int }}\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\lambda dx}{x^2}$
$E_1=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\lambda }{a}$ towards left.
Electric lield at $O$ due to other wire,
$E_2=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\lambda }{a}$ towards left
$\therefore$ Net field at $O$ is $E=2\times \frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{a}=\frac{\lambda }{2\pi {\epsilon }_{0}a}$