#### 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 *Φ*_{E} 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 (i_{c}) + Displacement current (i_{d})

`:.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.