Displacement current

1. Maxwell equations:

\begin{align}
\nabla \times \pmb{H}&=\pmb{J}+\frac{\partial \pmb{D}}{\partial t} \\
\nabla \times \pmb{E}&=\frac{\partial \pmb{B}}{\partial t}
\end{align*}

Taking the divergence of the above two equations, the left-hand sides are 0(due to curl has no divergency)

\begin{align*}
0 = &\nabla \cdot \pmb{J}+\frac{\partial \pmb{\nabla \cdot D}}{\partial t}
  = \nabla \cdot \pmb{J}+\frac{\partial \pmb{\rho}}{\partial t} \\
\nabla \cdot & \pmb{J} = -\frac{\partial \pmb{\rho}}{\partial t}.......(3) \\
0= &  \frac{\partial ( \nabla \cdot \pmb{B})}{\partial t}\\ 
\nabla \cdot & \pmb{B}=0 .........(4)
\end{align*} 

Equ (1) shows two types of current:
1. conductor current--Jc
2. displacement current--Jd=dD/dt
In a conductor, Jc>>Jd. In a dielectric(insulator), Jd >>Jc.

ex. Assume a parallel plate capacitor connected to a source with conducting wires. The displacement current within capacitor

\pmb{J}_d=\frac{\partial \pmb{D}}{\partial t}=\varepsilon \frac{\partial \pmb{E}}{\partial t}\\
=-\varepsilon \frac{\partial \nabla V}{\partial t}=\varepsilon \frac{\partial  V}{d\partialt}\\
=\frac{\varepsilon }{d}\frac{dV}{dt}

Assume source Vs=V0sinwt, we have

\begin{align *}
\pmb{J}_d=\frac{\varepsilon }{d}\frac{dV}{dt}=\frac{\varepsilon }{d}V_0\omega cos(\omega t)\\
I_d=\int_S J_d ds= \frac{\varepsilon S}{d}V_0\omega cos(\omega t)=CV_0\omega cos(\omega t)
\end{align}

This is the same result frrom circuit theory,Ic=Cdv/dt