A parallel plate capacitor has two plates, each of area3.0 x 10 -4 m 2 , separat

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?
=

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

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