Derivation: Displacement Current for the 4th Maxwell Equation

Here we want to derive the so-called displacement current $I_{\text e}$ and the displacement current density $\boldsymbol{j}_{\text e}$, which must appear in the fourth Maxwell equation of electrodynamics:

4. Maxwell equation of electrostatics (integral form)

$$ \begin{align} \oint_{L} \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{l} ~=~ \mu_0 \, \class{red}{I} \end{align} $$

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:

4. Maxwell equation of electrostatics (differential form)

$$ \begin{align} \nabla \times \class{violet}{\boldsymbol{B}} ~=~ \mu_0 \, \class{red}{\boldsymbol{j}} \end{align} $$

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 so-called 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:

$$ \begin{align} I_{\text e} ~&=~ \frac{\text{d}Q}{\text{d}t} \\\\
~&=~ \varepsilon_0 \, \frac{\text{d}}{\text{d}t} \oint_A \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

At the second equal sign, we inserted the first Maxwell equation in integral form.

We assumed here a plate capacitor in which the E-field is homogeneous. However, since the electric field $\boldsymbol{E}(t, x, y, z)$ can be inhomogeneous (i.e., location-dependent) in general, the time derivative is noted as the partial derivative $\partial$. Moreover, if the electric field is a continuous ("mathematically well-behaved") function, then we can, if we want, drag the time derivative into the integral:

Displacement current caused by the time change of the E-field

$$ \begin{align} I_{\text e} ~=~ \varepsilon_0 \, \oint_A \frac{\partial \boldsymbol{E}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

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 time-varying E-field:

Corrected 4th Maxwell equation with the displacement current

$$ \begin{align} \oint_{L} \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{l} ~&=~ \mu_0 \, \left( I ~+~ \varepsilon_0 \int_{A} \frac{\partial \boldsymbol{E}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a}\right) \\\\
~&=~ \mu_0 \, I ~+~ \mu_0 \, \varepsilon_0 \int_{A} \frac{\partial \boldsymbol{E}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

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 E-field.

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:

Displacement current density caused by the temporal change of the E-field

$$ \begin{align} \boldsymbol{j}_{\text e} ~=~ \varepsilon_0 \, \frac{\partial \boldsymbol{E} }{\partial t} \end{align} $$

This allows us to correct the differential form 2 of Maxwell's fourth equation:

Corrected 4th Maxwell equation with the displacement current density

$$ \begin{align} \nabla ~\times~ \class{violet}{\boldsymbol{B}} ~&=~ \mu_0 \, \left( \boldsymbol{j} ~+~ \varepsilon_0 \, \frac{\partial \boldsymbol{E}}{\partial t} \right) \\\\
~&=~ \mu_0 \, \boldsymbol{j} ~+~ \mu_0 \varepsilon_0 \, \frac{\partial \boldsymbol{E}}{\partial t} \end{align} $$

By introducing the displacement current not only the above contradiction with the capacitor is solved, but also the conservation of charge is preserved!

Video: Maxwell Equations. Here You Get the Deepest Intuition!