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Capacitance of Capacitor

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A capacitor is constructed from two square
    metal plates. A dielectric [itex]\kappa[/itex] = 4.61 fills the
    upper half of the capacitor and a dielectric
    [itex]\kappa[/itex] = 6.9 fills the lower half of the capacitor.
    Neglect edge effects. Calculate the capacitance C of the device.
    Answer in units of pF

    2. Relevant equations
    C[itex]_{TOP}[/itex] = [itex]\kappa[/itex][itex]\epsilon[/itex]A/d
    C[itex]_{bottom}[/itex] = [itex]\kappa[/itex][itex]\epsilon[/itex]A/d
    C[itex]_{total}[/itex] = C[itex]_{TOP}[/itex] + C[itex]_{BOTTOM}[/itex]

    3. The attempt at a solution
    C[itex]_{TOP}[/itex] = [itex]\kappa[/itex][itex]\epsilon[/itex]A/d
    = 4.61(8.85419*10[itex]^{-12}[/itex])(.12[itex]^{2}[/itex])/0.0005
    = 1.175553098*10[itex]^{-9}[/itex]
    C[itex]_{bottom}[/itex] = 6.9(8.85419*10[itex]^{-12}[/itex])(.12[itex]^{2}[/itex])/0.0005
    = 1.759504637*10[itex]^{-9}[/itex]
    C[itex]_{total}[/itex] = 2.935057735*10[itex]^{-9}[/itex]
    Is this correct?
  2. jcsd
  3. Mar 1, 2012 #2


    User Avatar

    Staff: Mentor

    Ummm. Plate size? Plate separation? Are the plates oriented vertically or horizontally?
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