A dielectric circular cylinder used between the plates of a capacitor has a thickness of 0.2 mm and a radius of 1.4 cm. The dielectric properties are ER = 400 and σ = 10-5 S/m. a) Calculate C: C = EREOS d (400) (8.854 x 10-12) (1.4 x 10-2)2 2 x 10-4 = 1.09 x 10-8 = 10.9 nF b) Find the quality factor Qor (QQF = wRC) of the capacitor at f = 10 kHz: Use the relation RC = e/o to write
A dielectric circular cylinder used between the plates of a capacitor has a thickness of 0.2 mm and a radius of 1.4 cm. The dielectric properties are ER = 400 and σ = 10-5 S/m. a) Calculate C: C = EREOS d (400) (8.854 x 10-12) (1.4 x 10-2)2 2 x 10-4 = 1.09 x 10-8 = 10.9 nF b) Find the quality factor Qor (QQF = wRC) of the capacitor at f = 10 kHz: Use the relation RC = e/o to write
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A dielectric circular cylinder used between the plates of a capacitor has a thickness of 0.2 mm and a
radius of 1.4 cm. The dielectric properties are ER = 400 and σ =
10-5 S/m.
a) Calculate C:
C =
EREOS (400) (8.854 x 10-12) (1.4 x 10-2)²
2 x 10-4
d
= 1.09 x 10-8
QQF = WRC =
=
b) Find the quality factor Qor (QoF=wRC) of the capacitor at f = 10 kHz: Use the relation
RC = e/o to write
2лfe (2 x 104) (400) (8.854 x 10-12)
10-5
0
10.9 nF
= 22.3
c) If the maximum field strength permitted in the dielectric is 2 MV/m, what is the maximum per-
missible voltage across the capacitor? Vmax = EBDd = (2x 106) (2 x 10-4)= 400 V.
d) What energy is stored when this voltage is applied?
1
1
We,max =
= CV ax =
(10.9 x
(10.9 × 10−9)(400) = 8.7 x 10−4 = 0.87 m
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