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

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Susmita · Answer

E=2×10 5 -4×10 4 t displacement current density So magnitude of the current is I d =J d A=3.54×10 -7 ×5×10 -2 =1.77×10 -8 Ampere.

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the electric field between 2 plates of a parallel plate(area=510^-2 m^2) capacitor is given by E=2.010^5 -4*10^4 t) where t is in sec.What is the magnitude of displacement current between plates

the electric field between 2 plates of a parallel plate(area=510^-2 m^2) capacitor is given by E=2.010^5 -4*10^4 t) where t is in sec.What is the magnitude of displacement current between plates

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