Question

A parallel plate air-filled capacitor is being charged as inthefigure above.The circular plates have a radius of 4.00cm, and at a particular instant the conduction current in the wiresis 0.280A.

A. What is the displacement current density (j_D) in theair space between the plates?

B. What is the rate at which the electric field between the platesis changing?

C.What is the induced magnetic field between the plates at adistance of 2.00 cm from the axis?

D.What is the induced magnetic field between the plates at adistance of 1.00 cm form the axis?

Answer 1

A capacitor is charged by a current i c has a displacement current equal to i c between the plates

∴displacement current i_D =i _c=0.280 Aradius of the circular plate r = 4 cm=0.04 m

( A ) . displacement current density j D = i_D / ( π r ² )=0.28 / ( 3.14 * 0.04 ² )=55.73 A / m ²

( B ) . displacement current density j D = ε( dE / dt )the rate at which the electric field between the plates is changing( dE / dt ) = j_D/ εdE/ dt ) = 55.73/ 8.85 * 10 ^-12=6.3*10 ¹² N / C - s

( C ) . the induced magnetic field between theplates B = ( μ * r / 2π R ²) * i _{c ----( 1 )}whereR= 2 cm=0.02 mr= 4 cm=0.04 mμ=permeability of free space=4π* 10 ^-7 H

( D )substitute R = 1 cm = 0.01 m inequation ( 1 ),we get answer