# What is electric flux density?

We say the electric flux is the number of field "lines", thus electric flux density is the number of field "lines" per a given area. However, let's say we had a point charge $$Q$$ centered at the origin and we were to enclose this charge with a surface of radius $$R$$. If we wanted to integrate the electric flux density over that surface, why do we get the total charge enclosed instead of the electric flux? I know this mathematically makes sense because we can write Gauss's law and multiply both sides by epsilon to yield this result, but I don't think this makes sense with what an electric flux density "is". Do I have a misunderstanding of what an electric flux density "is"? Please keep the math simple as I am only in first year university with limited knowledge of multivariate calculus.

\begin{align} \int \vec{E} \cdot ds &= Q / \varepsilon \rightarrow \varepsilon \int \vec E \cdot ds = Q_\text{enclosed} \\ \int \vec D \cdot ds &= Q_\text{enclosed} \qquad \vec D = \varepsilon \vec E \end{align}

• You don't. If you integrate the electric flux density over a given surface you get the electric flux. If the surface is closed, and you multiply the result with $\epsilon_0$, then that product happens to coincide, because of Gauss's law, with the charge enclosed by the surface. But the integral itself is the electric flux. Why do you think otherwise? Mar 2 '19 at 22:51
• @EmilioPisanty In my professor's note, he wrote that the contour integral of electric flux density dotted with a differential of surface is equal to the charge enclosed, which left me confused. Because I thought it should be Q/epsilon instead. Can I assume he made a typo? Mar 2 '19 at 22:56
• The notes are correct. You seem to be misunderstanding the notes (or there are mistakes in the text of the notes which you haven't provided). Mar 2 '19 at 23:14
• The 'electric flux' is the closed surface (gaussian) integral of electric field, which is Q/e_0, by gauss's law. This integral is quite clearly the gaussian integral of electric field multiplied by e_0, which is quite clearly the electric flux times e_0. This value is therefore Q. I'm not sure what else i can say
– user86425
Mar 2 '19 at 23:33
• yes, well. It would be the electric flux multiplied by epsilon through that surface. Gauss's law states that for a closed surface, this is Q. I suppose you can calculate flux for an open surface, so I would modify my last statement to say that. However Gauss's law applies to a closed surface
– user86425
Mar 2 '19 at 23:42

It seems that you're likely to be using non-standard terminology.

If I understand your comments correctly, you seem to be associating the name "electric flux density" with the vector field $$\vec D= \varepsilon \vec E$$. If that is the case, then this is definitely non-standard terminology. The $$\vec D$$ field certainly does suffer from a lack of a universally-accepted name (though the problem is less bad than in magnetism), but a better name is the one used in Wikipedia, "electric displacement field". Using the name "electric flux density" for $$\vec D$$ is acceptable when writing in a generic context, but it is vital that it be clearly identified as such when it is introduced.

To be clear: the electric flux across a surface $$S$$ is defined as $$\Phi(S) = \iint_S \vec E(\vec r) \cdot\mathrm d\vec S,$$ in terms of the electric field $$\vec E$$ itself. If the surface $$S$$ is closed, then the electric flux is still $$\Phi(S) = \oint_S \vec E(\vec r) \cdot\mathrm d\vec S$$ and it just happens, because of Gauss's law, to coincide with $$Q_S/\varepsilon_0$$, i.e. the total charge enclosed by $$S$$ divided by the vacuum permittivity. Moreover, if you're dealing with the electric field $$\vec E$$ produced by a free-charge distribution that's embedded in a homogeneous, isotropic linear dielectric with permittivity $$\varepsilon$$, then $$\Phi(S)$$ can also be seen to equal $$Q_{S,\mathrm{free}}/\varepsilon$$, where $$Q_{S,\mathrm{free}}$$ is the free charge enclosed by $$S$$.

Because of this, the direct understanding of the term "electric field density" is to assign that to the vector field $$\vec E$$ itself, since it is the vector field whose surface integrals give the electric flux.

It is possible to start re-defining those terms so that you have a bit more operational room in how you distinguish between $$\vec E$$ and $$\vec D$$, though you run the risk of creating an extremely confusing situation. It looks to me that what's happened is that your lecturer has chosen a confusing choice of terminology and that's led you into some conceptual contradictions which are impossible to clear up within that framework. However, without seeing the notes in full as provided directly by your lecturer, it's impossible to tell that for sure.

• @HarryHolmes Electric flux and radiated power are separate and different concepts, and saying "flux density" for the latter does not make sense. As this separately, or in Physics Chat. Apr 2 at 16:03
• Hmmm, I've seen the term 'photon flux density' used. Leave radiation; was my understanding of flux correct? Like, if I was talking about electric flux, would it be okay? Apr 2 at 16:22
• @HarryHolmes The comment thread of an unrelated old thread is not the place for this. Start a new thread or ask on chat. Apr 2 at 17:23
• I would, but my question's sort of a 'check-if-my-idea-is-right' type of query; asking it, I think, would be frowned upon here. Apr 3 at 1:50
• Then ask on chat. This is my last comment here. Apr 3 at 6:33

It's just proportional to the electric field in order to remove the epsilon constant. There cannot be any other 'intuition' that doesn't also apply to electric field because the constant of proportionally is somewhat arbitrary.

I think there's an issue with your question:

Why do we get charge enclosed 'instead of' electric flux?

Well Gauss's law says that the two are equal for a closed surface (by a constant factor). The explanation of Gauss's law has probably been covered on this website