A parallel-plate capacitor has a capacitance of 100pF, a plate areaof 135cm 2 ,

Question

A parallel-plate capacitor has a capacitance of 100pF, a plate areaof 135cm2, and a mica dielectric ( =5.4) completely filling the space between the plates. At 75Vpotential difference, calculate the following. (a) the electric field magnitude E inthe mica
(b) the amount of excess free charge on each plate
(c) the amount of the induced surface charge on the mica
(a) the electric field magnitude E inthe mica
(b) the amount of excess free charge on each plate
(c) the amount of the induced surface charge on the mica

Explanation / Answer

a) C = k0A/d =>d =k0A/C = 5.4*8.85*10-12*135*10-4 m2 / 100*10-12 F =6.45*10-3m the electric field E = V/d = 75V/6.45*10-3m =1.16*104 V/m b) charge Q = CV = 100*10-12F*75V =7.5*10-9 C = 7.5nC

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