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Dimensions of Parallel Plate Capacitors

  1. Feb 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The electric field between two circular plates of a capacitor is changing at a rate of 1.5 x 10^6 V/m per second. If the displacement current at this instant is Id = 0.80 x 10^-8A, find the dimensions of the plate.

    2. Relevant equations

    Id = ΔQ/Δt = ε0(ΔΦE/Δt)
    ΦE = EA
    Q = CV
    C = ε0AE
    ε0 = 8.85 x 10^-12

    3. The attempt at a solution

    I am unsure how to go about this question. Any guidance is appreciated!
  2. jcsd
  3. Feb 9, 2015 #2


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    Homework Helper

    You're told Id and ΔE/Δt ... I'm pretty sure you can look up ε in the textbook front cover. So, how does A depend on radius?
    (btw, your Capacitance equation should be εA/d, where d is the gap distance.)
  4. Feb 9, 2015 #3
    So I made this question out to be harder it actually was. I was thinking about the changing electric field in the wrong way.

    So here's my work:

    Id = 0.80 x 10^-8 A
    dE/dt = 1.5 x 10^6 V/m
    ΦE = EA

    Id = ε0(ΔΦE/Δt)
    Id = ε0(A)(ΔE/Δt)
    A = Id/[ε0 x (ΔE/Δt)]
    A = 0.80 x 10^-8/(8.85 x 10^-12)(1.5 x 10^6)
    A = πr2 = 6.03 x 10^-4 m2
    r = 1.4 x 10^-2 m or 1.4 cm
  5. Feb 9, 2015 #4


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    Yes. You cannot actually determine the gap distance, because you're not told the voltage change rate.
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