Displacement Current for the 4th Maxwell Equation
Here we want to derive the socalled displacement current \(I_{\text e}\) and the displacement current density \(\boldsymbol{j}_{\text e}\), which must appear in the fourth Maxwell equation of electrodynamics:
The Stokes integral theorem can be used to convert the integral form 1
into the differential form in which the current density \( \boldsymbol{j} \) occurs:
In the more general case, when the E and B fields are allowed to change in time, the integral form 1
and differential form 2
of Maxwell's fourth equation are not valid.
Consider a plate capacitor with electrode surface area \(A\) and charged with a current \(I\). An electric field \( \boldsymbol{E} \) builds up between the electrodes.

If a loop \(S_1\) is chosen according to Ampere's law
1
(see Illustration 2), then Ampere's law gives the current \(I\). 
If, on the other hand, the loop \(S_2\) is chosen so that it passes between the capacitor plates, then the included current obviously becomes zero because no current flows between the capacitor plates, of course.
A different choice of the areas enclosing the current will give a different result.... Not good!
To resolve this contradiction, the socalled displacement current \(I_{\text e}\) is introduced. When the current \(I\) flows through the wire, the charge \(Q\) on the electrodes of the capacitor changes (the capacitor is charged). This change of charge \(\dot{Q}\) in time means a change of the charge difference on the electrodes, i.e. a change of the electric field \( E \) between the electrodes:
~&=~ \varepsilon_0 \, \frac{\text{d}}{\text{d}t} \oint_A \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$
At the second equal sign, we inserted the first Maxwell equation in integral form.
We assumed here a plate capacitor in which the Efield is homogeneous. However, since the electric field \(\boldsymbol{E}(t, x, y, z)\) can be inhomogeneous (i.e., locationdependent) in general, the time derivative is noted as the partial derivative \(\partial\). Moreover, if the electric field is a continuous ("mathematically wellbehaved") function, then we can, if we want, drag the time derivative into the integral:
This allows us to correct the integral form 1
of Maxwell's fourth equation. Thus, the total current \(I\) in 1
generally does not correspond ONLY to a current caused by flowing charges, but it is composed of a current \(I\) caused by flowing charges AND a displacement current \(I_{\text e}\) caused by the timevarying Efield:
~&=~ \mu_0 \, I ~+~ \mu_0 \, \varepsilon_0 \int_{A} \frac{\partial \boldsymbol{E}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$
Here \(I\) is, as said, the part of the current caused by moving charges and no longer the total current as in 1
. We have chosen the same notation for it to avoid additional notation. We could also have denoted this current part as \(I_{\text c}\), so that the total current can be written as: \( I = I_{\text c} ~+~ I_{\text e} \).
The differential form 2
of Maxwell's fourth equation must also be modified. There we only have to adjust the total current density \(\boldsymbol{j}\) (current per area). It is now composed of the current density \(\boldsymbol{j}\) (same name) caused by moving charges and the displacement current density \(\boldsymbol{j}_{\text e}\) caused by the changing Efield.
The current is generally defined as the surface integral of the current density. If you compare the displacement current 4
with this definition, you will see that the displacement current density must correspond to the following expression:
This allows us to correct the differential form 2
of Maxwell's fourth equation:
~&=~ \mu_0 \, \boldsymbol{j} ~+~ \mu_0 \varepsilon_0 \, \frac{\partial \boldsymbol{E}}{\partial t} \end{align} $$
By introducing the displacement current not only the above contradiction with the capacitor is solved, but also the conservation of charge is preserved!