By denoting the source coordinates with prime, I get flux through a closed surface:

$$\Phi= \displaystyle\oint_{A} \mathbf{E}(x,y,z) \cdot \mathbf{\hat{n}}\ dA =q (x',y',z')$$

And now using the formal definition of divergence:

$$\displaystyle \nabla \cdot \mathbf{E}(x,y,z)=\dfrac{d\Phi}{dV} = \dfrac{d^3\ q (x',y',z')}{dx\ dy\ dz}=0 \tag1$$

In order to correctly get $\nabla \cdot \mathbf{E}(x,y,z)=\rho (x,y,z)$, we should be having:

$$\nabla \cdot \mathbf{E}(x,y,z)=\dfrac{d^3\ q(x,y,z)}{dx\ dy\ dz} \tag2$$

instead of equation $(1)$. Where am I going wrong?

EDIT: A simpler way of asking the same question:

By divergence theorem:

\begin{align} \iiint \nabla \cdot \mathbf{E}\ dV &= \unicode{x222F} \mathbf{E} \cdot \hat{\mathbf{n}}\ dS \\ &=q\ (x',y',z') \\ &=\iiint \rho (x',y',z')\ dV' \end{align}

Now how is the cancellation of triple integral in LHS and RHS justifiable when LHS has integral with respect to $V$ while RHS has integral with respect to $V'$?

  • $\begingroup$ q' must be the enclosed charged by the closed surface. $\endgroup$
    – Mohammad M
    May 18, 2019 at 14:54
  • $\begingroup$ Yes...... apparently $\endgroup$
    – N.G.Tyson
    May 18, 2019 at 14:55
  • $\begingroup$ The integrals are true for any arbitrary surface/volume. $\endgroup$ May 18, 2019 at 15:22
  • $\begingroup$ Maybe I'm misunderstanding the issue you are having. Can you explicitly say what the issue is? $\endgroup$ May 18, 2019 at 15:33
  • 1
    $\begingroup$ I think you're just confused by (arbitrary) labels - prime versus unprime $\endgroup$
    – nox
    May 18, 2019 at 16:00

3 Answers 3


Just because the two integrals over the same volume are equal, that does not imply that the integrands are equal. However, if the integral relation

$$ \int _V \nabla \cdot \mathbf{E} \ dV = \int _V \rho \ dV$$

holds for any volume, then it can be applied to an infinitesimally small volume, $V_{i}$, for which $\rho$ is a constant. Then you get the relation

$$ \int _{V_i} \nabla \cdot \mathbf{E} - \rho \ dV = 0$$

This can be applied everywhere, pointwise, and yields the differential form.

  • $\begingroup$ According to my post (please have a look at the edit), the RHS of your first equation should be $\int_V \rho (x',y',z') dV'$ and this is what is creating problems to me. $\endgroup$
    – N.G.Tyson
    May 18, 2019 at 18:05
  • $\begingroup$ I was just trying to shift focus from the prime notation. Go ahead and keep the right hand side you have. Apply the integral relation to a sufficiently small volume, for which, the charge density is constant. If the integrand on the right hand side is constant, then you can drop the prime. $\endgroup$
    – Daddyo
    May 18, 2019 at 18:27
  • $\begingroup$ But I do not see how $\rho (x,y,z)$ makes sense. If this is true then $\displaystyle\mathbf{E}=\rho(x,y,z) \int_{V'} \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV'$ and we get strange results. $\endgroup$
    – N.G.Tyson
    May 18, 2019 at 18:35

It looks like you are using the convention where unprimed coordinates represents The point of space you are "interested in" and primed coordinates refer to the coordinates of the charge density. So, for example, Coulomb's law becomes $$\mathbf E(\mathbf r)=\iiint\frac{\rho(\mathbf r')(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}\ \text dV'$$

However, for what you are trying to do the distinction between primed and unprimed coordinates is irrelevant. This is because for Gauss's law in integral form that you start with we don't need to distinguish between these two spatial coordinates. You're defining a surface and looking at the field on that surface and the charge contained within that surface. Therefore you only have one set of spatial coordinates to worry about, not two.

It might be better for you to consider a single set of coordinates for something like Coulomb's law anyway. We specify a fixed point in space $\mathbf r_0= x_0\hat x+y_0\hat y+z_0\hat z$ and then integrate over the spatial coordinates containing our charge distribution:

$$\mathbf E(\mathbf r_0)=\iiint\frac{\rho(\mathbf r)(\mathbf r_0-\mathbf r)}{|\mathbf r_0-\mathbf r|^3}\ \text dV$$

We are really only using one set of coordinates here along with a specified member of these coordinates.

So, putting together your whole derivation here, we start with Gauss's law in integral form: $$\displaystyle\oint\mathbf E(\mathbf r)\cdot\text d\mathbf A=q_{enc}$$ where we are integrating over an arbitrary surface on the left and the right is how much charge in contained within this surface. Note that the enclosed charge is not a function, it is a single value based on the surface we have chosen. Also notice that the infinitesimal area vector $\text d\mathbf A$ and the vector $\mathbf r$ refer to the same spatial coordinates.

Then we use the divergence theorem on the left and the definition of charge density on the right: $$\iiint\nabla\cdot\mathbf E(\mathbf r)\ \text dV=\iiint\rho(\mathbf r)\ \text dV$$

Notice that these two integrals are over the same volume, which is the volume contained by the surface we chose earlier. As has been mentioned in other answers, since the surface we chose was arbitrary we can say these integrands are equivalent.

  • $\begingroup$ The charge contained within that surface is $q(x',y',z')$ while the rest are functions of $x,y,z$. So I think we have two sets of spatial coordinates to worry about. Am I correct? $\endgroup$
    – N.G.Tyson
    May 18, 2019 at 22:16
  • $\begingroup$ @N.G.Tyson See my edit $\endgroup$ May 19, 2019 at 1:44

Note that $q$, in general, is not a function of space coordinates. The density $\rho$ is. So the RHS of the Maxwell equation really is $$ \int \rho(x, y, z) dV $$

  • $\begingroup$ How is this possible? How can $\rho$ be a function of $x,y,z$? $\endgroup$
    – N.G.Tyson
    May 18, 2019 at 17:03
  • $\begingroup$ Please can you reply. $\endgroup$
    – N.G.Tyson
    May 18, 2019 at 18:07

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