Question

Two infinitely long parallel wires having linear charge densities

Answer 1

Answer:

answer : \frac{\lambda_1\lambda_2}{2\pi\epsilon_0R}

Electric field due to infinitely long wire of linear charge density \lambda_1 at R distance is given by, E=\frac{\lambda_1}{2\pi\epsilon_0R}

it is given that , another infinity long wire is placed at R distance parallel to first wire. so, find charge on the wire.

as we know, Q=\lambda l, where l is length of wire.

so, charge on 2nd wire, Q=\lambda_2l

from formula, F = QE

here, Q=\lambda_2l and E=\frac{\lambda_1}{2\pi\epsilon_0R}

so, force experienced by 2nd wire , F = \frac{\lambda_1\lambda_2l}{2\pi\epsilon_0R}

hence, force per unit length, \frac{F}{l}=\frac{\lambda_1\lambda_2}{2\pi\epsilon_0R}