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 displacment current.

Explanation

Using Gauss' law, the electric flux `phi_(epsi)` of a parallel plate capacitor having an area A, and a total charge Q is
`phi_(epsi)=EA=(1)/(epsi_(0))(Q)/(A)xxA=(Q)/(epsi_(0))`
Where electric field is, `E=(Q)/(Aepsie_(0))`.
As the charge Q on the capacitor plates changes with time, so current is given by
`i=dQ//dt`
`(phi_(epsi))/(dt)=(d)/dt((Q)/(epsi_(0)))=(1)/(epsi_(0))(dQ)/(dt)`
`rArrepsi_(0)(depsi_(0))/(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)`
`therefore " "i=i_(c)+i_(d)=i_(c)+epsi(dphi_(E))/(dt)`
As Ampere's circuital law is given by
`therefore phivecB*vecdt=mu_(0)I`
After modification we have Ampere's Maxwell law is given as
`phiB.dt=mu_(0)i_(c)+mu_(0)epsi_(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 diplacement current.