Irodov 3.356: Charge Conservation Inside Maxwell's Equations

Problem Irodov 3.356

Show that the law of conservation of electric charge, $\nabla\cdot\vect j = -\partial\rho/\partial t$ , is a consequence of Maxwell’s equations.

Given

$\nabla\times\vect H = \vect j + \partial\vect D/\partial t$
$\nabla\cdot\vect D = \rho$

Find

show that

\nabla\cdot\vect j = -\partial\rho/\partial t

Solution

Step 1The divergence of the circulation law

The circulation law for $\vect H$ ties the curl of $\vect H$ to the conduction current and the displacement current:

\nabla\times\vect H = \vect j + \frac{\partial\vect D}{\partial t} \tag{1}

The divergence of a curl is zero for every smooth vector field, since the mixed second derivatives cancel in pairs:

\nabla\cdot(\nabla\times\vect H) = 0

Taking the divergence of both sides of (1), and interchanging the space and time derivatives in the last term:

0 = \nabla\cdot\vect j + \frac{\partial}{\partial t}\left(\nabla\cdot\vect D\right) \tag{2}

Step 2Gauss’s law closes it

The remaining divergence is fixed by the other source equation, Gauss’s law, which names the density of free charge:

\nabla\cdot\vect D = \rho

Putting it into (2) and moving the time derivative to the right:

\nabla\cdot\vect j = -\frac{\partial\rho}{\partial t}

(3)

This is the law of charge conservation in local form, derived from the two source equations alone.

Step 3What it says about a volume

To read the law, integrate (3) over a fixed volume $V$ bounded by a closed surface $S$ , and turn the volume integral of the divergence into the flux through $S$ (Gauss’s theorem):

A fixed closed surface S round a region holding the charge q, with current density j leaving through S. — A fixed closed surface $S$ round a region holding the charge $q$ , with current density $\vect j$ leaving through $S$ .

\oint_S \vect j\cdot d\vect S = \int_V \nabla\cdot\vect j\,dV = -\int_V \frac{\partial\rho}{\partial t}\,dV = -\frac{dq}{dt}

The current flowing out through $S$ equals the rate at which the charge $q$ inside $V$ decreases: charge is never created or destroyed, only carried.

Check

In a steady state $\partial\rho/\partial t = 0$ , and (3) reduces to $\nabla\cdot\vect j = 0$ : as much current enters any volume as leaves it, which is Kirchhoff’s junction rule. Units: $\nabla\cdot\vect j$ is $(\text{A/m}^2)/\text{m} = \text{C}/(\text{m}^3\cdot\text{s})$ , the unit of $\partial\rho/\partial t$ .

Remark

Without the displacement current the divergence of (1) would give $\nabla\cdot\vect j = 0$ everywhere and always, which is false at the plate of a capacitor being charged. The term $\partial\vect D/\partial t$ is exactly what makes the circulation law consistent with charge conservation; seen this way, Maxwell’s addition was forced rather than guessed.