A parallel-plate capacitor consists of circular plates of radius 0.30 m separate
ID: 2015998 • Letter: A
Question
A parallel-plate capacitor consists of circular plates of radius 0.30 m separated by a distance of 0.20 cm. The voltage applied to the capacitor is made to increase at a steady rate of 2.0x10^3 V/s. Assume that the electric charge distributes itself uniformly over the plates, and ignore the fringing effects.(a)What is the rate of increase of the electric field between the plates?
(b)What is the displacement current between the plates within a radius of 0.15 m?
(c)What is the magnetic field between the plates at a radius of 0.15 m? At 0.30 m?
Explanation / Answer
circular plates of radius = R = 0.30 m so, area of the plates : A = R2 = 3.14(0.32)= 0.2826 m2 separated by a distance of =d = 0.2 x10-2 m capacitence of the parallel plate capacitor : voltage applied to the capacitor is made to increase at a steady rate of = dV /dt = 2.0x103 V/s (a) Since , electric field E = V /d or V = Ed dV /dt = d E /dt x d where : dE /dt represents the rate of change of electric field bet ween the plates so, dE / dt =[ dV /dt ] / d = 2.0x103 V/s / 0.2 x10-2 = 106 V/ms (b) dispalcement current : I = Q / A = C = Ao /d = (0.2826)(8.854 x10-12)/(0.2 x10-2) = 12.51 x 10-10 farad but , d V /dt = 2 x 103 V/s but , Q = C V or dQ /dt = C dV /dt I = ( 12.51 x 10-10 )(2 x 103 ) = 25.02 x10-7 A or I = 2.502 A (c) B = o I / 2 R = (4x10-7)(25.02 x10-7 )/ 2(0.15) = 333.6 x10-14 T at R = 0.3 m B' = B /2 = 166.8 x10-14 TRelated Questions
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