Two long thin charged rods with charge density λ each are placed parallel to each other at a distan

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1. Electric Charges and Fields

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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

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

Solution

(b)

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

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

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

Standard 12

Physics

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