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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


2 years ago

Susmita
425 Points
							E=2×105-4×104tdisplacement 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 isId=JdA=3.54×10-7×5×10-2=1.77×10-8Ampere.

2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions