# Gauss's Law with electric flux density

Vyse007
I am taking a course in Electromagnetic Wave Theory, and the prescribed book for us is Engineering Electromagnetics by William Hayt. The book if excellent, till I reached the part about Gauss's Law. I will be describing my queries here, kindly help me out with them.

1) The book first states Faraday's experiments with charged sphere, and then says that "flux is proportional to charge. However, in SI units the constant of proportionality is 1, and hence flux equals charge." I didn't get this. The fact that charge is same as flux is then used throughout the chapter.

2) The book then introduces flux density (D), and proves the relation D=$$\epsilon$$E, where E is the electric field. The proof seems alright, but again thats assuming that charge equals flux.

3) Finally, coming to Gauss's Law, the book states that total flux through any closed surface is equal to the charge enclosed. Mathematically, integral of D.ds over a closed surface equals volume integral of rho.dv (sorry I tried latex but I get a weird formatting bug), where D is the flux density, rho is the volumetric charge density and ds, dv have their usual meanings. I just cant seem to relate this with the Gauss's Law that I have learned, saying that the flux that the flux is equal to the charge enclosed upon the permittivity, the flux being the areal integral of of electric field and elemental area. I don't quite see how its the same thing, and I am really confused.

Even Gauss's law in differential form has the same thing. Also, I tried looking these things up over the internet, but apparently no one uses flux density with Gauss's Law.

Any help would be highly appreciated.

Gold Member
It's all just the system of units used. You may prefer to view the relations using the MKS units which explicitly state all these proportionality constants. However, there are other systems of units. The other popular variant is the CGS which generally uses the Gaussian but there is also the Lorentz-Heaviside unit system.

So there are three main systems. MKS, which is SI units, Gaussian (CGS) and Lorentz-Heaviside (CGS). There is a further difference that can be applied called rationalization. Rationalized unit systems are not different systems of unit but rather they change where the factors of 4\pi are related. MKS is rationalized. So what does this mean? Well Gauss' Law can be expressed, using MKS, as

$$\nabla \cdot \mathbf{D} = \rho$$
$$\mathbf{D} = \epsilon \mathbf{E} = \epsilon_r \epsilon_0 \mathbf{E}$$

Heaviside-Lorentz, which is also rationalized, chooses a system of units so that the vacuum permittivity and permeability and speed of light (\epsilon_0, \mu_0 and c) are unity. Thus,

$$\nabla\cdot\mathbf{D} = \rho$$
$$\mathbf{D} = \epsilon \mathbf{E} = \epsilon_r \mathbf{E}$$

So in free-space, we say that D = E and that the divergence of the electric field is the same as the charge density.

Now, for Gaussian. Gaussian uses the same set of units as Heaviside-Lorentz so that \epsilon_0, \mu_0 and c are unity. But Gaussian is not rationalized and so the factor of 4\pi is moved off of Coloumb's Law onto Gauss' Law.

$$\nabla\cdot\mathbf{D} = 4\pi\rho$$
$$\mathbf{D} = \epsilon \mathbf{E} = \epsilon_r \mathbf{E}$$

Coloumb's Law in MKS (rationalized) is

$$F = \frac{1}{4\pi\epsilon_0} \frac{Q_1Q_2}{r^2}$$

But in a unrationalized system like Gaussian, Coulomb's Law becomes

$$F = \frac{Q_1Q_2}{r^2}$$

So we can see that in MKS, which I assume your book uses, the divergence of the electric flux density is equal to the charge density. Now, the electric flux is defined as

$$\Phi_E = \oint \mathbf{E}\cdot d\mathbf{S} = \frac{Q_{enclosed}}{\epsilon}$$

where we assume that the medium is homogeneous. So we can see that the electric flux is proportional to the charge enclosed in the volume. Now I have seen a very esoteric definition of electric flux. Here, it is defined as

$$\Phi_E = \oint \mathbf{D}\cdot d\mathbf{S} = Q_{enclosed}$$

This is applicable for an inhomogeneous volume and it is, in MKS, directly proportional to charge. So my guess is that your book is choosing to define the electric flux as the surface integral over the electric flux density. Oddly enough, this makes logical sense from the name for the D field, electric flux density, but I have rarely ever seen it defined as such. Most of the time when we talk of the electric flux we mean the surface integral over the E field, not the D. If we used Gaussian units then there would be an added factor of 4\pi in the proportionality. This sort of confusion between names and terms comes up a lot. You will find that the names for the D, B and H fields in your engineering text will most likely be at odds with any physics text. In physics, D is usually called the displacement field. B is the magnetic field and H is ummm... some other heathen name I guess. I can't remember.

Now on to your second point, the relationship between the E and D fields. This doesn't have anything to do with how they define the proportionality via rationalization. For all three systems E and D are related by the permittivity, \epsilon. The D field incorporates the polarization of the medium due to an applied electric field.

As for the third point, you can change between the differential form of Gauss' Law, which I showed in the answer to the first point above, and the integral form, shown when discussing the definition of electric flux above, by use of the Divergence Theorem.

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Vyse007
Wow that really helped a lot! I knew that maybe units had something to do with the whole mess, but I could never figure ALL THAT out!
Thanks a lot for you help. I guess I get it now. But still, to date I haven't come across a book that chooses to define Gauss's Law in terms of electric field density, until now that is!

Gold Member
Wow that really helped a lot! I knew that maybe units had something to do with the whole mess, but I could never figure ALL THAT out!
Thanks a lot for you help. I guess I get it now. But still, to date I haven't come across a book that chooses to define Gauss's Law in terms of electric field density, until now that is!

We can only explicitly state Gauss' Law using the electric field for a homogeneous medium. In that case, we can separate the \epsilon, which is now a constant, and move it to the right hand side where the charge density lies. But once we allow for an inhomogeneous medium, one where the \epsilon varies in space (like if we have a sphere of dielectric sitting in vacuum), then we cannot move the \epsilon out of the divergence operator (without expanding out the divergence operator using the chain rule). So expressing it with the electric field is really only a special case. The general case uses the D field which may be why you have not seen it before.

Vyse007
When you say that the general case uses the D field, you mean the flux density right? Also, you said that epsilon can't be taken out of the divergence operator. But how can that be explained in the integral form of Gauss's Law?

Gold Member
When you say that the general case uses the D field, you mean the flux density right? Also, you said that epsilon can't be taken out of the divergence operator. But how can that be explained in the integral form of Gauss's Law?

I mean in regards to expressing Gauss' Law. Sometimes you may see Gauss' Law expressed using E and \epsilon as opposed to D.

Guass' Law in its integral form, expanding the electric flux density into \epsilon and E, is

$$\oint \epsilon \mathbf{E}\cdot d\mathbf{S} = Q_{enclosed}$$

If \epsilon is a constant, as in the case where the volume enclosed within the Gaussian surface is a homogeneous medium, then we can move the \epsilon outside of the integral on the left and over to the right-hand side as I did so above. This is a common expression for Gauss' Law but again it is just a special case.

Vyse007
Ok I guess I get it all now. Thanks a lot for all your help. I can finally move ahead with the course!