How Do Electric Balls Approach Each Other as They Lose Charge?

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The discussion revolves around calculating the initial relative speed of two electric balls losing charge at a rate of 1.20 nC/s. The participant is uncertain about how to approach the problem, considering using Coulomb's law to find the force and subsequently the acceleration. They note that the charge loss complicates the electrostatic force but believe it can still be managed. A suggestion is made to draw a free body diagram to analyze the forces acting on the balls, including gravity, electrostatic forces, and string tension. Overall, the focus is on deriving the equations necessary to determine the balls' motion as they lose charge.
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Homework Statement


Assume each ball is loosing charge at a rate of 1.20nC/s, as what instantaneous relative speed do the balls approach each other initially.
m=11.2g L=122cm x=4.70cm

http://img40.imageshack.us/img40/9708/electballs.jpg


Homework Equations



I figured since the value given is a rate it would be the dq=-1.20nC/s, but I am not entirely sure what exactly this rate is referring to, I am assuming it would be derivative to charge q. But the units are C/s. I am a bit lost now, because I don't know where to start, if I use Coloumbs law with dF and find force and divide it by mass that is given then I can maybe find acceleration and continue from there, but somehow I doubt it.

I know that v=dx/dt, maybe I have to find a way to find these equations from the rate given.


Thank You
 
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Ok, I have missed one huge glaring detail, that x is defined as x=[q^2L/2(pi)mg)]^1/3, I guess taking the derivative of this will result in the answer I am still trying to work it out, but it is not a simple derivative.

If someone could please help a little. That would be great.

Thank You.
 
What are the forces on the balls? Gravity and electrostatic forces, and the string tension, right? Draw a free body diagram, sum the forces, and use that to calculate the motions. The fact that the charge is leaking away at some rate does complicate the electrostatic force components, but should still be manageable.
 
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