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Capacitor change in distance

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A parallel plate capacitor with plates of area [itex]460 cm^2[/itex] is charged to a potential difference [itex]V[/itex] then disconnected from the source of voltage. When the plates move [itex].3 cm[/itex] farther apart, the voltage between them increases by [itex]120 V[/itex]. What is the charge [itex]Q[/itex] on the positive plate of the capacitor? Answer in [itex]nC[/itex]


    2. Relevant equations
    [tex]Q=\frac{ε_0AV}{d}[/tex]


    3. The attempt at a solution
    [tex]V=\frac{Qd}{ε_0A}[/tex]
    [tex]Q=\frac{ε_0A(V+120)}{(d+.003)}[/tex]
    Inserting one into the other...
    [tex]Q=\frac{ε_0A}{(d+.003)}(\frac{Qd}{ε_0A}+120)[/tex]
    distribute and canceling yields...
    [tex]Q=\frac{Qd+120ε_0A}{(d+.003)}[/tex]
    Finally
    [tex]Q=\frac{120ε_0A}{.003}[/tex]

    So using [itex]A=.046 m^2[/itex], I keep getting [itex]16nC[/itex]

    Am I doing something wrong assuming [itex]Q[/itex] is constant?
     
  2. jcsd
  3. Sep 22, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Your assumption about Q being constant is correct, and I don't see any errors in your method. That leaves significant figures or application errors (the marking program has been given a wrong answer to look for) as suspects.
     
  4. Sep 23, 2013 #3
    Ok thanks for the feedback. This homework system has a reputation for making errors.
     
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