Two identical balls, each possessing a charge of 2.00 10⁻⁷ C and a mass of 100 g , are suspended by two 50 cm long insulating strings from a common point. The balls are initially held 5.0 cm apart and subsequently released. Determine the tension in the string at the instant just after release.

Step-by-step solution

Let be the angle made by the string with the vertical. The distance between the balls is x = 5.0 cm = 0.05 m , and the length of the string is L = 50 cm = 0.5 m . From the geometry of the suspension: = x/2 L = 0.025 0.5 = 0.05 = 1 - ^2 = 1 - (0.05)^2 = 0.9975 0.99875 The electrostatic force of repulsion between the balls is: F_e = 1 4 ₀ q^2 x^2 = 9 10^9 (2.00 10⁻⁷)^2 (0.05)^2 F_e = 9 10^9 4 10⁻¹⁴ 0.0025 = 0.144 N The weight of each ball is: mg = 0.1 kg 9.8 m/s ^2 = 0.98 N Just after release, the velocity of the bal