A parallel plate capacitor is made up of two flat conducting plates, each of are

Question

A parallel plate capacitor is made up of two flat conducting plates, each of area 30 cm2 , separated by a distance of 0.3 mm. The region between the plates is filled by stacking two dielectric sheets, each of thickness 0.15 mm, on top of each other. If the dielectric constants of the two sheets are 3 and 4, respectively, find the capacitance of the capacitor

*Solve this question from first principles by placing charges ±Q on the conductors, finding the fields and thus the voltage drop ... (i.e. without using the series capacitor circuit formula

)

Explanation / Answer

capacitance of capacitor= 8.85 x 10-12 ( 30 x 10^-4)/ ( 0.3 x 10^-3)=8.85 x 10^-11 F

when dielectrics are stacked,

C= 2 ( k1 K2)/ ( K1 +K2) x8.85 x 10^-11

C = 30.34 x 10^-11 F or 303. 4 pF

E(electric field) = charge density/ permitivity

charge density = Q/A

E =  Q/( A permitivity)

Voltage drop = Ed

C = Q/.V= Q/ Ed = QA permivity / d Q = A(permitiity)/ d

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