Question

A rod of length l with uniform charge per unit length $\lambda$ is placed at a distance d from origin along the x-axis. A similar rod is placed at the same distance along Y-axis. Determine the magnitude of net electric field intensity at the origin.

• A. $\frac{\lambda l}{4\pi {ϵ}_{0}d(d+l)}$
• B. $\frac{2\sqrt{2}\lambda l}{4\pi {ϵ}_{0}d(d+l)}$
• C. $\frac{\sqrt{2}\lambda l}{4\pi {ϵ}_{0}d(d+l)}$
• D. $\frac{3\sqrt{2}\lambda l}{4\pi {ϵ}_{0}d(d+l)}$

Answer 1

Answer: (C)

$\frac{\sqrt{2}\lambda l}{4\pi {ϵ}_{0}d(d+l)}$

Let's first find Electric field due to the rod on x-axis, say $E_1$

Consider a small length element $dx$, distance $x$ away from origin.

Field due to this $d{E}_1=\frac{k\lambda dx}{x^2}$

Integrating from $x=d$ to $x=d+l$

Gives$E_1=\frac{\lambda l}{4\pi {ϵ}_{0}d(d+l)}$ along -x direction

Now, $E_2$is along the -y direction and same in magnitude.

Thus,

$|E|=\sqrt{2}|E_1|=\frac{\sqrt{2}\lambda l}{4\pi {ϵ}_{0}d(d+l)}$