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Electric Potential Inside a Parallel-Plate Capacitor

  • Thread starter k3N70n
  • Start date
67
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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.
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?
Express your answer using two significant figures.

2. Homework Equations

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

3. The Attempt at a Solution

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

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

So [tex]U_{i}=K_{f}[/tex]
[tex]\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}[/tex]

[tex]\Longrightarrow\sqrt\frac{2}{4\pi\epsilon_{0}m_{1}}\{\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}\}}[/tex] (the sqrt sgn should extend to the end of the expression but I suck as LaTeX)
=0.096m/s=9.6cm/s
 
67
0
got it now.
 

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