Suppose that two balls are suspended by identical threads of length .10m anchored to the same point. The pith balls have different masses and charges. The charge of one is 2.0 X 10^-7 C and the other one is 6.0 X 10^-8 C. Both threads make a 25 degree angle at equilibrium with the vertical. What is the mass of each ball?
culombs law = kq1q2/r^2
weight of ball 1 = m1g
weight of ball 2 = m2g
Tension in x direction = Tsin(25) = kq1q2/r^2
Tension in y direction of ball 1 = Tcos(25) = m1g
Tension in y direction of ball 2 = Tcos(25) = m2g
r = distance between charges = .10sin(25)
The Attempt at a Solution
I drew a free body diagram of the first ball and set the x and y components of the tension equal to the weight and the force of the like charges applied on ball one. Since both strings make a 25 degree angle with the vertical the pith balls would be the same distance away from the vertical and have equal tensions and masses. This is where I think I am making a mistake. I solve for the mass of the first ball by solving for T in one equation and subbing that into the other equation. I get that the mass of both balls are something like 3.3 x 10 ^ -3 kg. The answer in the back of the book says that the masses of both balls is 1.2 x 10 ^ -4 kg. I would really appreciate if someone could give me their input on the problem.