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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.

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    $\begingroup$ 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. $\endgroup$
    – knzhou
    Sep 18, 2018 at 22:05
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    $\begingroup$ 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. $\endgroup$ Sep 18, 2018 at 22:26
  • $\begingroup$ What is the book you're using for E&M? $\endgroup$ Sep 18, 2018 at 22:38
  • $\begingroup$ engineering electromagnetics by hayt and buck $\endgroup$
    – David
    Sep 18, 2018 at 22:39
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    $\begingroup$ @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 $\oint_S \vec{D} \cdot d\vec{S} = Q_{\textrm{enclosed}}.$ $\endgroup$ Sep 18, 2018 at 22:53

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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 $S$.

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  • $\begingroup$ in my other book it was just E dot dA $\endgroup$
    – David
    Sep 18, 2018 at 23:19
  • $\begingroup$ 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 $\Phi$. $\endgroup$ Sep 18, 2018 at 23:42
  • $\begingroup$ so the flux has different definitions depending on the book? $\endgroup$
    – David
    Sep 19, 2018 at 2:23
  • $\begingroup$ @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). $\endgroup$ Sep 19, 2018 at 2:51
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    $\begingroup$ @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. $\endgroup$ Sep 19, 2018 at 2:57

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