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
- 10 years agoFavorite 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 approxSource(s): capacitance formula from: http://instrumenttoolbox.blogspot.com/search/label...
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