Electric Potential Inside a Parallel-Plate Capacitor

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SUMMARY

The discussion centers on calculating the final speed of two positively charged beads, each with a charge of 5.1 nC, when they are released from rest and move apart from a negatively charged bead of -1.0 nC positioned between them. The initial potential energy is converted into kinetic energy, leading to the formula derived: v_f = √(2/(4πε₀m₁) * (q₁q₂/r₁₂ + q₁q₃/r₁₃)). The final speed of the positive beads is determined to be 9.6 cm/s when they are very far apart.

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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.

Homework Statement



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?
Express your answer using two significant figures.

Homework Equations



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

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
 
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