Write Maxwell's generalization of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is `I=varepsilon_0 (dphi_E)/dt,`where *Φ _{E}* is the electric flux produced during charging of the capacitor plates.

#### Solution

Maxwell's generalisation of Ampere's circuital law is given as follows:

`ointvecB.vec"dl"=mu_0(I+I_D)=mu_0(I+varepsilon_0 (dphi)/dt)`

Consider that a parallel capacitor *C *is charging in a circuit.

The magnitude of electric field between the two plates will be `E=q/(varepsilon_0 A)`and is perpendicular to the surface of the plate.

`phi_E=vecE.vecA=EA cos0=q/(varepsilon_0 A)xxA=q/varepsilon_0`

`=>(dphi_E)/dt=(d(q/varepsilon_0))/dt`

`=>(dq)/dt=varepsilon_0(dphi_E)/dt ("Here, dq/dt is rate of change of charge with time.")`

`=>I=varepsilon_0(dphi_E)/dt`