Question

Two long thin charged rods with charge density $\lambda$ each are placed parallel to each other at a distance $d$ apart. The force per unit length exerted on one rod by the other will be $\left(\right.$ where $\left.k=\frac{1}{4 \pi \varepsilon_{0}}\right)$

• A. $\frac{k 2 \lambda}{d}$
• B. $\frac{k 2 \lambda^{2}}{d}$
• C. $\frac{k 2 \lambda}{d^{2}}$
• D. $\frac{k 2 \lambda^{2}}{d^{2}}$

Answer 1

Answer: (B)

Electric field due to rod (1) at distance ' $d'=\frac{\lambda}{2 \pi \varepsilon_{0} d}$
So, force per unit length $\frac{F}{I}=\frac{q E}{I}=\lambda\left[\frac{\lambda}{2 \pi \varepsilon_{0} d}\right]$
$=\frac{k 2 \lambda^{2}}{d}$