Two infinitely long parallel wires having linear charge densities λ1, and λ2 respectively, are placed at a distance of R metres. The force per unit length on either wire will be (k=(1/4 π ε0))

Question

Two infinitely long parallel wires having linear charge densities λ1, and λ2 respectively, are placed at a distance of R metres. The force per unit length on either wire will be (k=(1/4 π ε0)) (A) k (2 λ1 λ2/R2) (B) k (2 λ1 λ2/R) (C) k (λ1 λ2/R2) (D) k (λ1 λ2/R)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fc95aa7c997b1ecdef2b5d4884074eeb-.png" /> Force on l length of the wire 2 is F2=Q E1=(λ2 l) (2 k λ1/R) ⇒ (F2/l)=(2 k λ1 λ2/R) Also, (F1/l)=(F2/l)=(F/l)=(2 k λ1 λ2/R)

Q. Two infinitely long parallel wires having linear charge densities $λ_{1}$ , and $λ_{2}$ respectively, are placed at a distance of $R$ metres. The force per unit length on either wire will be $(k = \frac{1}{4 π ε _{0}})$

$k \frac{2 λ _{1} λ _{2}}{R ^{2}}$

$k \frac{2 λ _{1} λ _{2}}{R}$

$k \frac{λ _{1} λ _{2}}{R ^{2}}$

$k \frac{λ _{1} λ _{2}}{R}$

Force on $l$ length of the wire 2 is
$F_{2} = Q E_{1} = (λ_{2} l) \frac{2 k λ _{1}}{R}$
$\Rightarrow \frac{F _{2}}{l} = \frac{2 k λ _{1} λ _{2}}{R}$
Also, $\frac{F _{1}}{l} = \frac{F _{2}}{l} = \frac{F}{l} = \frac{2 k λ _{1} λ _{2}}{R}$

Q. Two infinitely long parallel wires having linear charge densities λ1​, and λ2​ respectively, are placed at a distance of R metres. The force per unit length on either wire will be (k=4πε0​1​)

kR22λ1​λ2​​

kR2λ1​λ2​​