Does Coulomb's law hold as long as $\dot{\rho} = 0$?

Question

Does Coulomb's law,

$$\textbf{E}\left(\textbf{r}\right) = \frac{1}{4\pi\epsilon_0}\int \rho\left(\textbf{r}'\right)\frac{\textbf{r} - \textbf{r}'}{\left|\textbf{r} - \textbf{r}'\right|^3}dV',$$

hold always when $\dot{\rho} = 0$? Even if $\dot{\textbf{J}} \neq \boldsymbol{0}$, as in, for example, a scenario with a spinning ring of charge that keeps speeding up? It seems to me that Gauss's law implies that Coulomb's law always holds when $\dot{\rho} = 0$, but $\dot{\textbf{J}} \neq \boldsymbol{0}$ implies a time-varying magnetic field which alarmingly means that $\textbf{E}$ has a nonzero curl.

Answer 1

Gauss's law does not imply Coulomb's law, not even when $\dot\rho=0$.

Two of Maxwell's equations are $$ \nabla\cdot\mathbf{E}(\mathbf{x})\propto\rho(\mathbf{x}) \tag{1} $$ and $$ \nabla\times\mathbf{E}(\mathbf{x})\propto\mathbf{\dot B}(\mathbf{x}). \tag{2} $$ Equation (1) can only determine one component of the vector $\mathbf{E}$ per point in space, because it is a scalar equation. The other two components of $\mathbf{E}$ are governed by equation (2).

This is easier to see after taking a Fourier transform with respect to the spatial coordinates, so that equations (1)-(2) become $$ \mathbf{p}\cdot\mathbf{E}(\mathbf{p})\propto\rho(\mathbf{p}) \tag{3} $$ and $$ \mathbf{p}\times\mathbf{E}(\mathbf{p})\propto\mathbf{\dot B}(\mathbf{p}). \tag{4} $$ Equation (3) only determines the component of $\mathbf{E}$ that is parallel to $\mathbf{p}$. Equation (4) governs the components that are orthogonal to $\mathbf{p}$.

For example, a propagating EM wave has a non-zero $\mathbf{E}$ even when the charge and current densities are both zero. Equation (3) only says that the longitudinal component of the electric field must be zero in this case.

Answer 2

A theorem due to Helmholtz ( https://en.wikipedia.org/wiki/Helmholtz_decomposition ) states that there is a unique smooth vector field if we know its divergence and curl, provided it vanishes at infinity.

So, we have to know both $\nabla {\bf E} $ and $\nabla \times{\bf E} $ to uniquely characterize a vector field. In the case of the electric field its divergence is always given by the charge density $\rho$ (irrespectively from its time dependence). The key point for answering the question is if $\dot \rho = 0$ implies the vanishing of the rotational part of the electric field. It is not so because the condition of a stationary charge density implies a condition only on the divergence of the current density vector (via continuity equation), leaving full freedom on the curl of $\bf j $ and then it still allows the presence of a magnetic field whose variation with time controls $\nabla \times{\bf E} $.

Summarizing, your question has a negative answer for any field, if only its divergence is known.

Answer 3

Strictly speaking, Coulomb’s law holds for electrostatics - no motion of charges - but in practice it also holds very well for quasistatic situations, where the charges move with velocities $v$ much less than the speed of light, i.e. $v/c\ll 1$. When this no longer holds one needs the full machinery of electrodynamics.

Answer 4

As a counter exemple, you can consider the induction heating: Ideally, an infinite metallic cylinder into an infinite solenoid (the two with the same axe to be simple and symetric). Volume charge density in the conductor is $0$ and the electric field is non zero : an orthoradial electric field which rotate around the magnetic field lines.