I'm trying to rigorously derive the integral form of Gauss's law from Coulomb's law and the divergence theorem. Arrive at $$ \oint\limits_{\partial V} E\cdot da = \begin{cases} \frac q\epsilon_{o} & \text{if $\partial V$ encloses $q$ } \\ 0 & \text{if $\partial V$ does not enclose $q$ } \end{cases} $$ using $ E=\frac{q\hat r}{4\pi\epsilon_{o}r^2} $ and $\oint \limits_{\partial V} E\cdot da = \int \limits_V \nabla \cdot E \space d\tau $

Proof attempt: For the case where closed surface $\partial V $ doesn't enclose any charge, I'm trying to show that $\int \limits_V \nabla \cdot E d\tau $ is 0 and then use divergence theorem to complete the proof for this case. Used the fact that $\nabla \cdot E$ is 0 everywhere except at $r=0$, since I have $\frac1{r^2}$ radial field but how do I show that $\nabla \cdot E $ is 0 through out the entire volume $V$ ?

I know this very basic, but please point out what's it, that I'm missing.

EDIT: Now I understand that $\nabla\cdot E$ is 0 as $V$ doesn't contain $r=0$(the location of charge). But why does it turn out so, even though the field lines appear to "diverge" at each point. I'm looking for some kind of interpretation for this. Is it because "if we take a region small enough within which $E$ is almost constant, $\nabla\cdot E$ is 0".

  • $\begingroup$ Why would it not be zero if $V$ does not have the origin as an element? $\endgroup$ Commented Oct 8, 2014 at 16:03
  • $\begingroup$ Related: physics.stackexchange.com/q/38404/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Oct 8, 2014 at 18:13
  • $\begingroup$ @Mateus Sampaio. Thanks, I added another question in the EDIT. $\endgroup$
    – levitt
    Commented Oct 9, 2014 at 3:20
  • $\begingroup$ The name "divergence" of the differential operator $\nabla\cdot$ should not be taken to literal. It may be the case that lines "diverge" in some sense but the divergence of the field is null, as is the case. $\endgroup$ Commented Oct 9, 2014 at 3:24
  • 1
    $\begingroup$ Because in spherical coordinates, the divergence of a radial field is given by $\frac1{r^2}\partial_r(r^2 E)$. Take as an example a field $\mathbf{E}=(x,0,0)$. Even though the lines do not diverge, the divergece is not null. $\endgroup$ Commented Oct 9, 2014 at 3:54

1 Answer 1


$\oint E\cdot dS = \frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV $

if $\frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV$ then $\frac{\rho}{\epsilon} = \nabla \cdot E$

if $\rho=0$ then $\frac{\rho}{\epsilon} = 0 = \nabla \cdot E$

Is this what your looking for?

$\rho$ would be zero say, outside of a point charge.

One of the paradoxes you'll find when considering a point charge is that the divergence is zero for the field created a point charge, except at the origin in which case it is undefined. This paradox is resolved using the dirac delta function like in this excellent website which I recommend. http://farside.ph.utexas.edu/teaching/em/lectures/node30.html

In case you didn't know the divergence theorem comes from summing up the flux in infinitesimally small cubes that constitute a closed surface. The only fluxes that "survive" are the exterior ones. If $\nabla \cdot E = 0$ then this means that the flux on an infinitesimally small cube is zero. Which means that if the field lines were an actual fluid then the density of that small cube is not changing. (Because the same water flowing in flows out).

  • $\begingroup$ Ill add an answer to your second part shortly. $\endgroup$
    – DLV
    Commented Oct 9, 2014 at 4:04
  • 1
    $\begingroup$ I've added it now. $\endgroup$
    – DLV
    Commented Oct 9, 2014 at 4:09

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