Posted by Alex on .
Neutral metal sphere A, of mass 0.10kg, hangs from an insulating wire 2.0m long. An identical metal sphere B, with charge q is brought into contact with sphere A. Sphere A goes 12 degrees away from Sphere B. Calculate the initial charge on Sphere B.
Note: when one object with charge Q is brought in contact with a neutral object 1/2 the charge is transferred to the neutral object.
I don't understand how to do this.
Ans: 3.9x10^6C

Physics 
drwls,
After the spheres touch, each one acquires a charge of q/2, and the Coulomb repulsion force pushes them away from each other. If each one hangs inclined A = 12 degrees, and T is the tension in the wire,
T cos A = M g
T sin A = k (Q/2)^2/(2L sinA)^2
k is the Coulomb constant, 8.99 x 10^9 N•m^2/C^2
2L sin A is the separation of the spheres
T can be eliminated by dividind one equation by the other
tan A = k (Q/2)^2/(2L sinA)^2/(Mg)
M g tan A = k Q^2/(16 L sin A)^2
This should let you solve for Q