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 $$ F~=~k_e \frac{q_1q_2}{r^2}~? $$ 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


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

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    $\begingroup$ For essentially the opposite question(v3), see this Phys.SE post. $\endgroup$
    – Qmechanic
    Nov 17, 2012 at 11:27
  • $\begingroup$ Yes, I have already read this post, but my question is quite different. $\endgroup$
    – achatrch
    Nov 17, 2012 at 11:54
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    $\begingroup$ 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? $\endgroup$ Nov 17, 2012 at 12:31
  • $\begingroup$ Why 8 Maxwell's equations and not 4? I am missing something? $\endgroup$
    – Dani
    Nov 17, 2012 at 12:35
  • $\begingroup$ @DanilH: I meant 8 scalar equations. From the 4 Maxwell's equations two are vector equations. $\endgroup$
    – achatrch
    Nov 17, 2012 at 12:45

4 Answers 4


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.


The exact derivation goes as follows. You start from Gauss' Law, integrate on both sides over some volume V:

$$ \stackrel{\tiny div}{\vec{\nabla}}\cdot\vec{\mathbf{E}}=\frac{1}{\epsilon_0}\rho \,\,\,\,\,\,\,\,\,\,\,\Big/\iiint\limits_V\,d^3\vec{r} $$ 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.

  • $\begingroup$ No you didn't lost the unit vector. You calculated the charge, which is not a vector. It's a geometric argument(already given in the previous answer physics.stackexchange.com/a/44423/16689 : the electric field at a sphere of radius r centred on the charge must be radial and with the same magnitude throughout) that allow you to recover the vector form of the electric field. $\endgroup$
    – FraSchelle
    Jun 25, 2013 at 15:42
  • $\begingroup$ 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. $\endgroup$ Jun 25, 2013 at 16:04

If you are asking about the Coloumb's law for the electric fields, yes, which you can see others' answers.

If you are asking about the Coloumb's law for the electric force,


Maxwell equations do NOT tell you about how the force acting on the charges $q$ or currents $\vec{J}$.

Simply speaking, to FULLY understand classical E&M (i.e. one can determine the physics from a initial value problem to determine all its consequences - physics is all about to determine/predict the future), you need BOTH:

(1) Maxwell's equations

(2) Lorentz force law (Newtonian mechanics, E&M equivalence of Newtonian gravitation force.)

Punch line I: (1) and (2) are absolute different things.

Lagrangian and variational principle E.O.M. viewpoint

However, if you start from a Lagrangian viewpoint, writing down the action: $$ S=\int (-\frac{1}{2} |f|^2 + A \wedge *J)=\int d^3xdt (-\frac{1}{4} f_{\mu\nu} f^{\mu\nu} + A_\mu J^\mu) $$ with the 2-form field strength $f_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ gives you E and M fields, you can determine Maxwell's equation from the equations of motions(E.O.M.) by doing the variation principle on the 1=form gauge field $A$. The source is a 1-form current $J=(\rho,\vec{J})$.

Maxwell's equations: E.O.M. respect to varying 1-form gauge field $A$

You obtain Maxwell equations by varying $A$: $$ d*F=J \;\;\;\text{-Gauss law for electricity, Maxwell-Ampere's law} $$ and $$ dF=d^2A=0 \;\;\;\text{-Gauss's law for magnetism, Maxwell–Faraday equation} $$

How about the Lorentz force law? You can do variation respect to the spatial coordinate $x^\mu=(t,x)$, and you need to specify which massive particle with mass $m$ experiencing the force $F$, which is $F=m \ddot{\vec{x}}$ by Newtonian mechanics. To specify massive particle in the Lagrangian/action, you just need to add its kinetic energy $\frac{1}{2}m \dot{\vec{x}}^2$.

Lorentz force law: E.O.M. respect to varying spacetime coordinates $x^\mu=(t,x)$

$$ S=\int d^3xdt (-\frac{1}{4} f_{\mu\nu} f^{\mu\nu} + A_\mu \wedge J^\mu+\frac{1}{2}m \dot{\vec{x}}^2) \to \int d^3xdt (+ q\Phi - q \dot{\vec{x}} \cdot \vec{A}+\frac{1}{2}m \dot{\vec{x}}^2) $$

you will derive Lorentz force law $$ m \ddot{\vec{x}}=q\vec{E}+q \dot{\vec{x}} \times \vec{B} $$

Punch line II: The action and Variational principles are very powerful to unite (1) Maxwell's equations and (2) Lorentz force law, in the same framework.

  • $\begingroup$ The above sketch of derivation of force on point particles appears in good books, but has a fundamental flaw for point particles. For such particles, the interaction field term has no value, since the field is singular at the particle, and the pure field term is infinite, which invalidates any formal differentiation. The result, although it seems correct, is wrong: as vector potential, the total electric and magnetic fields are not defined at the place of the particle $\vec{x}$. $\endgroup$ Jan 22, 2014 at 5:08
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    $\begingroup$ @ Ján Lalinský, thanks but you miss my point. The field is NOT from the singular-size point particle, but from the external source surrounding it. Such as external current, electric planes, etc. The derivation is carried out the external EM effects on the point particle. $\endgroup$
    – wonderich
    Jan 22, 2014 at 5:29
  • $\begingroup$ So your comment is actually NOT to my point. $\endgroup$
    – wonderich
    Jan 22, 2014 at 5:32
  • $\begingroup$ In the first part of your derivation, you talk about deriving Maxwell equations, so there $f_{\mu\nu}, A_\mu$ are total fields produced by prescribed sources. In the second part after "How about the Lorentz force law? ", if by $f_{\mu\nu}, A_\mu$ you refer to external fields, the derivation is correct, but please consider making this clear in your text: due to the first part and even without it, when most people see these symbols, they assume they refer to the total field. $\endgroup$ Jan 22, 2014 at 8:37

Given the Gauss law AND the Lorentz force, yes it's possible to derive Coulomb's law as it already been answered. So I think the question is if it's possible to derive it given ONLY the four Maxwell equations (and not the Lorentz force). The answer is still yes since the Lorentz force is equivalent to the Faraday law and it can be derive from it. The relation between Faraday law and Lorentz force is not trivial in 3D since the Faraday paradox arises (check wiki). On the other hand when electromagnetism is expressed in the covariant formulation no such paradox exists.

  • $\begingroup$ Lorentz force formula does not follow from Maxwell's equations, nor from the Faraday law $\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}$, which is part of Maxwell's equations. $\endgroup$ Jan 16, 2014 at 1:22

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