Two metal balls in a capacitor 
The situation is: let's have a capacitor connected to a voltage souce (I guess it's safe to assume that the capacitor is very long, so we don't have to consider effects at its edges). And in this capacitor we place two metal (conductive) balls, but - at the beginning - they are connected with a conductive meta rod/cable/whatever. 
The question is: after the removal of the conductive connection, will the balls start to move? 
My approach to the problem goes as follows: there will be obviously an electric field in the capacitor, let's say - for simplicity - it's homogenous and denote it's value by $E$. If - as in the drawing - the "upper" plate of the capacitor is connected to the positive pole of voltage source, then the electric field will be directed exactly as in the drawing.
Then, as the balls are made of metal, there obviously will be induced charges in it. And I suppose, until we disconnect the balls, that the total charge in both balls will we equal to zero, that is, in one ball there will be charge $Q$, and in the other one $-Q$. 
But now, how do I determine exactly the relationship between $Q$ and $E$ (I guess we have that $Q=f(E)$). Because if I knew what charge is going to be in each ball, then I would easily check if the forces acting between the balls (due to the charges) are greater than the force acting on the balls (due to the electric field $E$).
Any hints?
 A: Assuming that the balls are uncharged initially when they are placed connected to one another into the electric field negative charges will be induced at the top end of the top ball and positive charges will be induced at the bottom end of the bottom ball.   
This would increase the electric field between the top of the top ball and the top plate and similarly between the bottom of the bottom ball and the bottom plate.
The electric field between the two balls would be very much smaller.  
In terms of electric lines there would be a greater density near the plates and many fewer between the balls.
Overall this means that the induced charges are attracted to the plates more than they are attracted to each other.
Suppose the balls were fixed in position, what would happen to the distribution of charges on the balls when the connected conducting rod is removed.
The answer is probably not a lot.
So the balls move towards the plates.
Or put another way when you set up the arrangement the connecting rod will be in a state of tension.
Cutting the rod will mean that the balls will move away from one another.
In terms of finding the forces and electric field it might be easier to replace your balls with metal plates which are parallel to the plates connected to the battery?
