# I'm confused about electric flux?

In a previous physics class, I learned that the electric flux was $\vec{E}\cdot\vec{A}$ (dot product), and hence the unit is $Nm^2/C$. But in my electromagnetics book, it says the unit is Coulomb, and that flux density is $C/m^2$. I'm really lost.

• Unfortunately, the book might be using cgs units. cgs is a system where they put in random factors of $c$ and $4\pi$ for no reason, and also change the units of every quantity for fun. Sep 18, 2018 at 22:05
• In SI units, the electric displacement $\vec{D}$ (the vector sum of the electric field and the electric polarization of a medium, which is sometimes called electric flux density) has units of $C/m^2$, which would mean that $D\cdot A$ would have units of $C$. Given that some textbooks consider $\vec{D}$ to be more fundamental, since Maxwell's equations in terms of $\vec{D}$ and $\vec{H}$ are simpler, it wouldn't be surprising if this is what they're referring to. Sep 18, 2018 at 22:26
• What is the book you're using for E&M? Sep 18, 2018 at 22:38
• engineering electromagnetics by hayt and buck Sep 18, 2018 at 22:39
• @probably_someone is right. I looked at the text. Hayt and Buck are defining electric flux as $\oint_S \vec{D} \cdot d\vec{S}$. So Gauss' law is $\oint_S \vec{D} \cdot d\vec{S} = Q_{\textrm{enclosed}}.$ Sep 18, 2018 at 22:53

In your previous physics class, electric flux was defined as

$$\Phi = \oint_S \vec{E} \cdot d\vec{S},$$

which has SI units of V-m or $\textrm{N-m}^2\textrm{-C}^{-1}$. Hayt and Buck are defining flux in terms of the electric flux density $\vec{D}$, given by $\vec{D} = \epsilon_{\textrm{o}}\vec{E}$ with the SI unit being $\textrm{C-m}^2$:

$$\Psi = \oint_S \vec{D} \cdot d\vec{S},$$

which has SI units of C. So $\Psi = \epsilon_{\textrm{o}}\Phi$, and Gauss' law can be written as

$$\Psi = \oint_S \vec{D} \cdot d\vec{S} = Q,$$

where $Q$ is the charge enclosed by the surface $S$.

• in my other book it was just E dot dA Sep 18, 2018 at 23:19
• The flux through the $d\vec{A}$ surface element is $\vec{E} \cdot d\vec{A}$, so we would write $d\Phi = \vec{E} \cdot d\vec{A}$. The integral gives the total flux $\Phi$. Sep 18, 2018 at 23:42
• so the flux has different definitions depending on the book? Sep 19, 2018 at 2:23
• @David it depends on the flux you are looking at. If you are looking at the electric field flux, the units are $Nm^2/C$. If you are looking at the flux of the electric displacement (not electric field), the units are $C$, giving the density as $C/m^2$. If you are looking at magnetic field flux the units are $Tm^2$. "Flux" isn't a single thing. It depends on what is "flowing" through the area you are calculating the flux through, not on the book (as long as everyone is using SI units here). Sep 19, 2018 at 2:51
• @David the general definition is consistent for everyone. Flux of "something" is "something"$\cdot$"area". Your book is using a different "something" ($D$) than what you are used to ($E$). But they are both consistent with what flux is, and they still agree with each other. Sep 19, 2018 at 2:57