Question

Original Question:
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

Posted by 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


Question - I don't understand what Mg is supposed to be. Also how did you get 16? and on the right side? Is that supposed to be the tension?

I found the distance between A and B to be 0.4158. Can you please explain the rest? I'm not getting the right answer.