# Electric Potential Inside a Parallel-Plate Capacitor

#### 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. 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?

2. Homework 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

Related Introductory Physics Homework Help News on Phys.org

got it now.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving