Two positively charged particles of masses m₁ = 2m and m₂ = m are suspended from a common point by two light strings of unequal lengths. In equilibrium both particles are at the same horizontal level and the angle between the strings is 45°. If F is the electrostatic force exerted by m₁ on m₂, then F/(mg) is:
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Question

Two positively charged particles of masses m₁ = 2m and m₂ = m are suspended from a common point by two light strings of unequal lengths. In equilibrium both particles are at the same horizontal level and the angle between the strings is 45°. If F is the electrostatic force exerted by m₁ on m₂, then F/(mg) is:

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G1 Physics · Accepted Answer

Let the strings make angles α and β with the vertical for masses 2m and m respectively. tanα = F/(2mg) tanβ = F/(mg) Also, α + β = 45°. Put x = F/(mg). Then tan(α+β) = (x/2 + x)/(1 - x²/2) = 1. Therefore, x² + 3x - 2 = 0. The positive root is x = (√17 - 3)/2. Hence, F = [(√17 - 3)/2]mg.