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Two identical small balls, each of mass 0.1 g, are suspended from a…
Question
Two identical small balls, each of mass 0.1 g, are suspended from a common point by silk threads of length 0.50 m. Each ball carries an equal charge q, and at equilibrium the centres of the balls are 0.06 m apart. Taking g = 10 m/s^2 and k = 9 x 10^9 N·m^2/C^2, find the charge q on each ball, in nanocoulomb, to three decimal places.
✓ Verified answer: 4.903checked by our engine — not a guess
Step-by-step solution
For each ball: tension along thread, weight mg down, and Coulomb repulsion Fc horizontal.
Equilibrium gives tan(theta) = Fc/(mg), where theta is the half-angle with sin(theta) = (x/2)/L.
Here x/2 = 0.03, so tan(theta) = 0.03/sqrt(0.5^2 - 0.03^2) = 0.03/0.49910 = 0.060108.
Then Fc = mg tan(theta) = (1e-4)(10)(0.060108) = 6.0108e-5 N.
Also Fc = k q^2 / x^2, so q = sqrt(Fc x^2 / k) = sqrt(6.0108e-5 x 0.0036 / 9e9) = 4.903e-9 C ≈ 4.903 nC.
Final answer4.903
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