Two long thin charged rods with charge density λ 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 (. where .k=(1/4 π ε0))

Question

Two long thin charged rods with charge density λ 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 (. where .k=(1/4 π ε0)) (A) (k 2 λ/d) (B) (k 2 λ2/d) (C) (k 2 λ/d2) (D) (k 2 λ2/d2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ea22e958c63a712191e6c52c639d4a65-.png" /> Electric field due to rod (1) at distance ' d'=(λ/2 π ε0 d) So, force per unit length (F/I)=(q E/I)=λ[(λ/2 π ε0 d)] =(k 2 λ2/d)

Q. Two long thin charged rods with charge density $λ$ 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 $($ where $k = \frac{1}{4 π ε _{0}})$

$\frac{k 2 λ}{d}$

$\frac{k 2 λ ^{2}}{d}$

$\frac{k 2 λ}{d ^{2}}$

$\frac{k 2 λ ^{2}}{d ^{2}}$

Electric field due to rod (1) at distance ' $d^{'} = \frac{λ}{2 π ε _{0} d}$
So, force per unit length $\frac{F}{I} = \frac{qE}{I} = λ [\frac{λ}{2 π ε _{0} d}]$
$= \frac{k 2 λ ^{2}}{d}$

Q. Two long thin charged rods with charge density λ 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 ( where k=4πε0​1​)

dk2λ​

dk2λ2​

d2k2λ​

d2k2λ2​