A parallel-plate capacitor of plate area A is being charged by a current I flowi
ID: 2015997 • Letter: A
Question
A parallel-plate capacitor of plate area A is being charged by a current I flowing into its plates via external wires. At one instant, the charge on the capacitor plates is Q.
(a)Assume that the electric field between the plates is uniform. Show that the electric field between the plates at this instant is E=Q/A.
(b)Show that the electric flux crossing the mathematical midplane between the plate surfaces is Q/.
(c)What is the displacement current of this capacitor?
(d)Show that this displacement current matches the ordinary current I flowing into the plates.
Explanation / Answer
(a) It is known by the formual for capacitence of the parallel plate capacitor is : C = Ao / d If V be the potential difference between the plates then , electric field between them is, E = V / d If Q be the charge accumulates on the plates we have a relation for Q as Q = C V = [Ao / d ] V but ,E = V/d or V = Ed thus, Q = [Ao / d ] Ed hence , E = Q / Ao (b ) By the Gauss law , electricla flux crossing the mathematical midplane between the plate surfaces is : electrical flux = E = E A so, electrical flux = E = Q / Ao x A =Q / o (c) Displacement current of the capacitor is given by : D = Q / A (charge per unit surface area of the plate) (d) per unit surface ,flow of the charges across the plates is nothing but , normal current propagationRelated Questions
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