Electrostatics Question 234

Question: Two infinitely long parallel wires having linear charge densities $ {\lambda _{1}} $ and $ {\lambda _{2}} $ respectively are placed at a distance of R metres. The force per unit length on either wire will be $ ( K=\frac{1}{4\pi {\varepsilon _{0}}} ) $ [MP PMT/PET 1998; DPMT 2000]

Options:

A) $ K\frac{2{\lambda _{1}}{\lambda _{2}}}{R^{2}} $

B) $ K\frac{2{\lambda _{1}}{\lambda _{2}}}{R} $

C) $ K\frac{{\lambda _{1}}{\lambda _{2}}}{R^{2}} $

D) $ K\frac{{\lambda _{1}}{\lambda _{2}}}{R} $

Show Answer

Answer:

Correct Answer: B

Solution:

Force on l length of the wire 2 is $ F _{2}=QE _{1}=({\lambda _{2}}l)\frac{2k{\lambda _{1}}}{R} $

therefore $ \frac{F _{2}}{l}=\frac{2k{\lambda _{1}}{\lambda _{2}}}{R} $

Also $ \frac{F _{1}}{l}=\frac{F _{2}}{l}=\frac{F}{l}=\frac{2k{\lambda _{1}}{\lambda _{2}}}{R} $



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