Are the Maxwell's equations enough to derive the law of Coulomb?

Are the 8 Maxwell's equations enough to derive the formula for the electromagnetic field created by a stationary point charge, which is the same as the law of Coulomb?

If I am not mistaken, due to the fact that Maxwell's equations are differential equations, their general solution must contain arbitrary constants. Aren't some boundary conditions and initial conditions needed to have a unique solution. How is it possible to say without these conditions, that a stationary point charge does not generate magnetic field, and the electric scalar potential is equal to

$$\Phi(\mathbf{r})=\frac{e}{r}.$$

If the conditions are needed, what kind of conditions are they for the situation described above (the field of stationary point charge)?

-
For essentially the opposite question(v3), see this Phys.SE post. –  Qmechanic Nov 17 '12 at 11:27
Yes, I have already read this post, but my question is quite different. –  achatrch Nov 17 '12 at 11:54
Wouldn't it be trivial to apply the divergence theorem to Gauss' law to get it in its integral form. From here it seems easy enough to use the usual tricks to find the electric field of a point charge, and then multiply by some charge to get your force. Surely this is Coulomb's force law? –  Daniel Blay Nov 17 '12 at 12:31
Why 8 Maxwell's equations and not 4? I am missing something? –  DaniH Nov 17 '12 at 12:35
@DanilH: I meant 8 scalar equations. From the 4 Maxwell's equations two are vector equations. –  achatrch Nov 17 '12 at 12:45

The short answer is yes, and in fact you only need one single Maxwell equation, Gauss's law, together with the Lorentz force, to get Coulomb's law.

More specifically, you need Gauss's law in its integral form, which is equivalent to the differential form for well-behaved fields because of Gauss's theorem. Thus, you use the law $$\nabla\cdot\mathbf{E}=\rho/\epsilon_0\quad\Leftrightarrow\quad \oint_S\mathbf{E}\cdot\mathrm{d}\mathbf{a}=Q/\epsilon_0,$$ where $Q$ is the total charge enclosed by the (arbitrary) surface $S$.

To derive Coulomb's law, consider the electric field of a single point particle, with nothing else in the universe. Because of isotropy (which must be added as an additional postulate), the electric field at a sphere of radius $r$ centred on the charge must be radial and with the same magnitude throughout. That means the integral is trivial and the electric field must be $$\mathbf{E}=\frac{Q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}.$$

Coupled with Lorentz's force law at zero velocity for the test particle (since Coulomb's law only holds in electrostatics) this yields Coulomb's law.

It is not obvious that this highly symmetric situation can give the general electrostatic force for multiple particles. This follows from the superposition principle, which is very much at the heart of classical electrodynamics, and which can be obtained from the linearity of Maxwell's equations. This gives you the field for a single source; add the fields for all the individual sources and you'll get the field for the collection of sources.

-
$$\stackrel{\tiny div}{\vec{\nabla}}\cdot\vec{\mathbf{E}}=\frac{1}{\epsilon_0}\rho \,\,\,\,\,\,\,\,\,\,\,\Big/\iiint\limits_V\,d^3\vec{r}$$ Then switch to integration over a closed surface, and also note that total charge inside this volume is Q: $$\iiint\limits_V\stackrel{\tiny div}{\vec{\nabla}}\cdot\vec{\mathbf{E}}\,\,d^3\vec{r}=\oint\vec{\mathbf{E}}\cdot d\vec\sigma=\iiint\limits_V\frac{1}{\epsilon_0}\rho\,\,d^3\vec{r}=\frac{Q}{\epsilon_0}$$ Now you need to note that the volume of integration is quite arbitrary and so is the surface, so we will use a sphere. You can describe the integral over a sphere using: $$\frac{Q}{\epsilon_0}=\oint\vec{\mathbf{E}}\cdot d\vec\sigma=\int\limits_{\phi=0}^{\phi=2\pi}\int\limits_{\theta=0}^{\theta=\pi}\mathbf{E}\hat{\vec{n}}\cdot\hat{\vec{n}}\,R\,d\phi\,R\,d\theta=4\pi R^2\mathbf{E}\,\,\,\,\,\,\Big/\frac{1}{4\pi R^2}$$ And so you obtain: $$\mathbf{E}=\frac{Q}{4\pi \epsilon_0 R^2}$$ It should be: $$\vec{\mathbf{E}}=\frac{Q}{4\pi\epsilon_0 R^2}\hat{\mathbf{r}}$$ But I lost the normal vector along the way (I hope that someone can correct this and edit this post).
Now you use the Lorentz Force law (where $\vec{\mathbf{B}}=\vec 0$): $$\vec{\mathbf{F}}_{lor}=q \vec{\mathbf{E}}+q \vec{\mathbf{V}}\times\vec{\mathbf{B}}=\frac{q\,Q}{4\pi\epsilon_0 R^2}\hat{\mathbf{r}}$$ And so you obtain the Coulomb force law.
thanks, I was hoping for a more strict (mathematically speaking) method of recovering the unit vector. It is true that charge is not a vector, so it is impossible to obtain a unit vector in this calculation. But simply "adding" (writing) it after derivation is complete is not what I prefer. Maybe we can do it strictly mathematically if we use the $\hat{\vec{z}}$ axis which currently is used only to define the $\theta$ angle inside the integral. –  cosurgi Jun 25 at 16:04