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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.
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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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