[Physics] Electric flux of a closed surface, $\Psi = Q $ or $\Phi =\int\vec{E}\cdot d\vec{A}$

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I have problem with the equation of electric flux. I use one book of fundamental physics and another book of electromagnetic engineering; the two of them give different equations for electric flux.

In the physics book, the electric flux is defined as the dot product of electric field and the normal vector of a surface then for a surface

$$\Phi = \int \vec{E} \cdot d\vec{A}$$

where $\vec{E}$ is the electric field through the surface and $d\vec{A}$ is the normal are element of the surface.

Also, it use Gauss's Law to state that for an closed surface $\Phi=Q_{\text{enc}}/\epsilon_0$, where $Q_{\text{enc}}$ is the total charge inside the surface.

On the other hand, the engineering book defines, for a closed surface,

$$\Psi = Q \, .$$

For Gauss's Law it says that the electric flux through a closed surface is equal to $Q_{\text{enc}}$ not $Q_{\text{enc}} / \epsilon_0$

There is also a difference in the sign of integral for a closed surface that I don't understand:

$$\Phi = \oint\vec{E} \cdot d\vec{A}$$

is used in my book while on internet I saw

$$\Phi = \unicode{x222F} \vec{E} \cdot d\vec{S} \, .$$

I do not understand these differing notations.

Best Answer

I will begin from the end. In the book where you see $\Phi = \oint\vec{E} \cdot d\vec{A}$ the meaning is a closed surface. The same meaning has the symbol in Internet and the circle/ellipse on the integrals means closed, i.e. a closed surface. From the element of integration you understand that you have to do with a surface.

About the presence of the $\epsilon _0$ it's matter of conventions of dimensions. In some books the Coulomb law is $\frac {1}{4 \pi \epsilon _0} \frac {q^2}{r^2}$, in other books the factor $\frac {1}{4 \pi \epsilon _0}$ is missing. Now, as I checked the dimensions, in both the system CGS and MKgS the correct way to work is with the factor $\frac {1}{4 \pi \epsilon _0}$.

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