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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

Bhuvanesh Kumar , 6 Years ago
Grade 12
anser 1 Answers
Susmita

Last Activity: 6 Years ago

E=2×105-4×104t
displacement current density
J_{d}=\epsilon_{0}\frac{\partial E }{\partial t}
      =8.854*10^{-12}*(-4*10^{4})=3.54*10^{-7}A/m^2
So magnitude of  the current is
Id=JdA=3.54×10-7×5×10-2=1.77×10-8Ampere.
 
 

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