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Question
Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.
Solution
Using Gauss’ law, the electric flux ΦE of a parallel plate capacitor having an area A, and a total charge Q is
`phi_E=EA=1/in_0Q/AxxA`
`=Q/in_0`
Where electric field is
`E=Q/(Ain_0)`
As the charge Q on the capacitor plates changes with time, so current is given by
i = dQ/dt
`:.(dphi_E)/dt=d/dt(Q/in_0)=1/in_0(dQ)/dt`
`=>in_0(dphi_E)/dt=(dQ)/dt=i`
This is the missing term in Ampere’s circuital law.
So the total current through the conductor is
i = Conduction current (ic) + Displacement current (id)
`:.i=i_c+i_d=i_c+in_0(dphi_E)/dt`
As Ampere’s circuital law is given by
`:.ointvecB.vec(dl)=mu_0I`
After modification we have Ampere−Maxwell law is given as
`ointB.dl=mu_0i_c+mu_0in_0(dphi_E)/dt`
The total current passing through any surface, of which the closed loop is the perimeter, is the sum of the conduction and displacement current.
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