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the electric field between 2 plates of a parallel plate(area=5*10^-2 m^2) capacitor is given by E=2.0*10^5 -4*10^4 t) where t is in sec.What is the magnitude of displacement current between plates
E=2×105-4×104tdisplacement current density So magnitude of the current isId=JdA=3.54×10-7×5×10-2=1.77×10-8Ampere.
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