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Electric Potential Inside a Parallel-Plate Capacitor

  1. Feb 27, 2009 #1
    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?
    Express your answer using two significant figures.

    2. Relevant 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
     
  2. jcsd
  3. Feb 27, 2009 #2
    got it now.
     
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