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Question

Three equal charges, 2.0 × 106 C each, are held at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the other two.

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Solution

Since all the charges are of equal magnitude, the force on the charge at A due to the charges at B and C will be of equal magnitude. (As shown in the figure)

That is, FBA=FCA=Fsay


The horizontal components of force cancel each other and the net force on the charge at A,
F'=FBAsinθ+FCAsinθF'=2Fsinθ
Given: r = 5 cm =0.05 m
By Coulomb's Law, force,
F=14πε0q1q2r2
F'=2×9×109×2×10-62×sin60°0.052F'=2×9×109×2×10-62×320.052
F' = 24.9 N

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