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

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.

• As far as I know, Coulomb's law gives the electrostatic field. If you want the full electric field, you have to add up the non-conservative part. – FGSUZ Nov 28 '18 at 0:58
• You are right, there will be more to the field than just the static electric field. (@FGSUZ beat me to it while I was typing this up!) – BioPhysicist Nov 28 '18 at 0:59

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.

• "Gauss's law does not imply Coulomb's law" This isn't true in electrostatics right? If you apply Gauss's law to a point charge you get Coulomb's law. – BioPhysicist Nov 28 '18 at 2:17
• @AaronStevens Yes, I think we agree. In the context of electrostatics, meaning that nothing is changing in time, then the combination of equations (1) and (2) does imply Coulomb's law, aside from a possible background electric field with uniform magnitude and direction. Since the question mentioned a time-dependent current and time-dependent magnetic field, I inferred that the context was not meant to be restricted to electrostatics, but maybe my inference was incorrect. (Wouldn't be the first time I've made an incorrect inference.) – Chiral Anomaly Nov 28 '18 at 4:18
• I appreciated the suggestion to Fourier transform the Maxwell eqs, and for a physical insight I offer this: it amounts to saying 'let's try plane wave solutions' followed by 'and any solution can be written in terms of those'. – Andrew Steane Nov 28 '18 at 8:35

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.

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.

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.