MCQ
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
    $\frac{k 2 \lambda}{d}$
  • $\frac{k 2 \lambda^2}{d}$
  • C
    $\frac{k 2 \lambda}{d^2}$
  • D
    $\frac{k 2 \lambda^2}{d^2}$

Answer

Correct option: B.
$\frac{k 2 \lambda^2}{d}$
b
(b)

Electric field due to rod $(1)$ at distance $d=\frac{\lambda}{2 \pi \varepsilon_0 d}$

So, force per unit length $\frac{F}{l}=\frac{q E}{l}=\lambda\left[\frac{\lambda}{2 \pi \varepsilon_0 d}\right]$

$=\frac{k 2 \lambda^2}{d}$

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