In triangle ABC
Sin5 = BC/AC
BC = (AC) Sin5
BC = (50) Sin5 = 4.36 cm
r = distance between the two balls = 2 BC = 2 x 4.36 = 8.72 cm = 0.0872 m
q = charge on each ball
m = mass of each ball = 5 g = 0.005 kg
electric force between the two balls is given as
F = [tex] \frac{kq^{2}}{r^{2}} [/tex]
using equilibrium of force in vertical direction
T Cos5 = mg eq-2
Using equilibrium of force in horizontal direction
T Sin5 = F eq-3
dividing eq-3 by eq-2
T Sin5 /(T Cos5) = F/mg
F = mg tan5
using eq-1
[tex] \frac{kq^{2}}{r^{2}} [/tex] = mg tan5
inserting the values
[tex] \frac{(9 \times 10^{9})q^{2}}{(0.0872)^{2}} [/tex] = 0.005 x 9.8
q = 2.03 x 10⁻⁷ C
sign of charge is same on both the balls. either it is negative or positive
