A parallel plate capacitor has two plates, each of area3.0 x 10 -4 m 2 , separat
ID: 1673571 • Letter: A
Question
A parallel plate capacitor has two plates, each of area3.0 x 10-4 m2, separatedby 3.5 x 10-4 m. The space betweenthe plates is filled with a dielectric. When the capacitor isconnected to a source of 120 V rms at 8.5kHz, an rms current of 1.6 x10-4 A is measured. (a) What is the capacitive reactance?(b) What is the dielectric constant of the material between theplates of the capacitor?
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(a) What is the capacitive reactance?
(b) What is the dielectric constant of the material between theplates of the capacitor?
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Explanation / Answer
Area of plates A = 3.0 x10-4 m2, distance between two plates d = 3.5 x10-4 m. capacitance of parallel plate capacitor with out dielectric Co = o A / d = (8.85*10-12 )(3.0 x 10-4m2,) / ( 3.5 x 10-4m. ) = 7.5 p F V r ms = 120 V I r m s = 1.6-*10-4 A capicitive capacitance X C = V r m s / I r ms = 120 /1.6 *10 -4 = 75*104 capacitance of the capacitor with dielectric C = 1 / 2 fX C = 1 / 2 *3.14 * 8.5*103 * 75*104 = 2.4466*10-11 The relation between C and Co is C = k Co dielectric constant k = C / Co = 24.55 / 7.5 = 3.23