[SOLVED] Electric Force II 1. The problem statement, all variables and given/known data You have a lightweight spring whose unstretched length is 4.0 cm. You're curious to see if you can use this spring to measure charge. First, you attach one end of the spring to the ceiling and hang a 1.0 g mass from it. This stretches the spring to a length of 5.0 cm. You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.5 cm. What is the magnitude of the charge (in nC) on each bead? 2. Relevant equations F = -k1 * x F = k2 * q1q2 / r^2 g = a = -9.81 k1 = Spring constant. Unknown. k2 = Electrostatic constant = 9*10^9 3. The attempt at a solution In the first experiment F = ma = -k1*x 0.001*-9.81 = -k1*(0.05 - 0.04) Therefore the spring constant k1 = 0.981 Now the second experiment: F = -k1*x F = -(0.981)(0.045 - 0.04) F = -0.004905 This is the force applied on a spring of k1 = 0.981, to stretch it 0.5cm Similarily, this should also be the force between the two beads: F = k2*q1q2 / r^2, however q1 = q2 according to the question. F = k2*(2q) / r^2, solving for q. Fr^2/(2*k2) = q (-0.004905)(0.045)^2/(2*9*10^9) = q q = -0.0000000000000005518125 C q = -0.0000005518125 nC This answer is incorrect. Where is my mistake?