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 $2 a$, 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. $\frac{\lambda}{4 \pi \varepsilon_0 a}$

Answer 1

Answer: (A)

The field at $O$ due to small element $d x$ is
Hence, due to one wire,
$E_1=\int\limits_a^{\infty} \frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda d x}{x^2} $
$E_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda}{a}$ towards left.
Electric lield 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 $E=2 \times \frac{1}{4 \pi \varepsilon_0} \frac{\lambda}{a}=\frac{\lambda}{2 \pi \varepsilon_0 a}$