Calculating Separated Distance after Coulomb's Law Interaction

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Suy
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Homework Statement


I just have a quiz, but i don't get this question
this is all i can remember
A=-4.0*10^-6C
B=+2.5*10^-6C
mass of sphere: 51g
initial separated distance:10.0cm
If A and B touch and then repel, what is the new separated distance
After they touch,
Tension force on the string is 0.578N
this is the picture after they touch
http://img10.imageshack.us/img10/5527/26792246.jpg

Homework Equations



Fe=(kq1q2)/r2
F=mg?

The Attempt at a Solution


I don't get this question at all, but i know i need to find the force first...
After they touch, the electron divide equally, so +2.5*10^-6 neutralize -2.5*10^-6, 1.5*10^-6 left, and divide by 2 is 0.75*10^-6
am i right?
 
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Discussion

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Yes, I think you have correctly averaged the charges.
The next step is to make a careful diagram of the forces involved. On each charge, there is an electric force, gravitational force and the pull of the string. Sketch these vectors for one of the charges and then get expressions for their horizontal and vertical components.

After that, you use the fact that the total force on the charge must be zero in both the vertical and horizontal directions because the charge is not accelerating. This should give you two equations to solve and find the angle.
 
How do you sketch electric force, can you show me?
http://img43.imageshack.us/img43/962/72025656.jpg
 

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0=tan(30)mg+sin(30)*0.578N+(-Fe/cos(30))
solve for radius,
the answer is 0.1m, 10 cm
am i right?
 
I don't follow your solution. The force diagram for the charge on the right should show the electric force going to the right. The gravity force down. The tension force along the string. The equation for the horizontal forces would have only two forces in it.
The equation for the vertical forces would have only two forces.
Your equation has three forces.
 
Why only have two force acting on it,? but isn't there are 3 force(tension,gravity,electric force)
can you show me the equation?
ty!
 
You must write that the sum of the forces is zero in the horizontal direction. And in the vertical direction. For the horizontal direction, there is no force of gravity so only two forces. I don't think I'm supposed to spoil your experience by giving away the equations before you attempt them.
 
ok,but do i really need vertical direction ?