The square plates of a 2000-pF capacitor measure 20 mm by 20 mm and are separated by a dielectric that is 0.27?

The square plates of a 2000-pF capacitor measure 20 mm by 20 mm and are separated by a

dielectric that is 0.27 mm thick. The voltage rating of the capacitor is 500 V. The dielectric

constant of the dielectric is closest to:

A) 110 B) 150 C) 170 D) 140 E) 120

2 Answers

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  • 9 years ago
    Best Answer

    Given parameters of capacitor are:

    Capacitance C = 2000pF = 2000 x 10-12pF (1pF = 10-12F)

    Separation of plates = thickness of dielectric = 0.27mm = 27 x 10-5m

    Plates are square with dimension 20mm x 20mm

    Hence area of capacitor plates = 20mm x 20mm = 400mm2 = 400 x 10-6m2

    Voltage rating of capacitor = 500V

    Now the capacitance of a capacitor is given by the formula:

    C = Kє0 A/d

    Where C= capacitance of capacitor in farad

    K = dielectric constant

    є0 = permittivity of free space = 8.854 x 10-12 Farad/m

    d = separation of the capacitor plates

    In this problem, since we are looking for dielectric constant, we make it the subject of formula thus:

    K = Cd/ є0A

    putting all our known values we have:

    K = (2000 x 10-12 x 27 x 10-5)/(8.854 x 10-12 x 400 x 10-6) = 152.47 = 150 nearest approx

    Source(s): capacitance formula from: http://instrumenttoolbox.blogspot.com/search/label...
  • renzi
    Lv 4
    3 years ago

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