Electric Potential Inside a Parallel-Plate Capacitor

1. Feb 27, 2009

k3N70n

Thank you for taking the time to look. I think I get the basic idea here but I'm must be missing something important. Any help is greatly appreciated.
1. The problem statement, all variables and given/known data

Two 1.4g beads, each charged to 5.1nC , are 2.2cm apart. A 2.8g bead charged to -1.0nC is exactly halfway between them. The beads are released from rest.

What is the speed of the positive beads, in cm/s, when they are very far apart?

2. Relevant equations

$$U_{i}+K_{i}=K_{f}+U_{f}$$
$$U_{q_{1}+q_{2}}=\frac{1}{4\pi\epsilon_{0}}$$

3. The attempt at a solution

$$m_{1}=0.0014 kg$$
$$m_{2}=0.0014 kg$$
$$m_{3}=0.0028 kg$$
$$q_{1}=5.1*10^{-9} C$$
$$q_{2}=5.1*10^{-9} C$$
$$q_{3}=-1*10^{-9} C$$
$$r_{12}=0.022 m$$
$$r_{13}=0.011 m$$

$$K_{i}=0$$ because initial velocity of all three is 0
$$U_{f}=0$$ because they end far apart --> r->$$\infty$$

So $$U_{i}=K_{f}$$
$$\Longrightarrow\frac{1}{4\pi\epsilon_{0}}\{\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}\}=\frac{m_{1}v_{f}^{2}}{2}$$

$$\Longrightarrow\sqrt\frac{2}{4\pi\epsilon_{0}m_{1}}\{\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}\}}$$ (the sqrt sgn should extend to the end of the expression but I suck as LaTeX)
=0.096m/s=9.6cm/s

2. Feb 27, 2009

got it now.