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} $
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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} $
BETA
