Question regrading the meaning behind flux of Electric Field

Question

Consider a small area $dA$ ,now the normal vector to the area $dA$ is $\hat{n}$ ,now $E .\hat{n}$ gives us the flux along the direction of our area ,now my question is why multiply with dA ,(since dA is infinitesimal ,we can take it to be 0.0001 ) ,now multiplying with dA reduces our $\vec{E}.\vec{n}$ value ,does this mean that $E$ is somehow defined per unit area ? (If we intend to find the mass of a 2d shape ,density($\dfrac{mass}{area}$) *( some area $<1m²$ ) ,gives us a mass less than the density value) .

Consider the velocity field of Water $\vec{v}$,we are trying to find the amount of water exiting a closed surface ,now this is given by $\int_s \vec{v} .\hat{n}dA$.

The term $\vec{v}.\hat{n}dA$,captures the amount of water that is in the direction of the plane of area and multiply it to get the volume of water exiting the surface in a second,is there any good explanation like this for Gauss's Law

Answer 1

Within a uniform electric field $\vec{E}$, flux $\Phi_E$, through area $\vec{A}$ is found as a dot product

$$\Phi_E=\vec{E}\space\cdot\space\vec{A}=EA\cos(\theta)$$

where when $\vec{E}$ and $\vec{A}$ are perpendicular ($\theta = 0$), the resulting flux is simply the field strength multiplied by the area being measured. Thus the flux has units which depend on area.

Now, when the angle between the field is is parallel ($\theta = 90$), there is no flux measured. But what happens if the electric field is not uniform (varies in strength or direction across area $\vec{A}$)?

This is where breaking up area $\vec{A}$ into infinitesimal parts $d\vec{A}$ becomes useful, as we can then find the flux by integrating the expression of the varying electric field with every individual $d\vec{A}$. That is:

$$d\Phi_E=\vec{E} \space \cdot \space d\vec{A} \space\space\space\rightarrow\space\space\space \Phi_E = \int \vec{E}\space\cdot\space d\vec{A}$$

It is the same for closed surfaces where we can compare flux going in to flux exiting the surface of the volume, and find positive flux if there is more flux exiting the surface than entering (and vise versa).

$$\Phi_E = \oint \vec{E}\space\cdot\space d\vec{A}$$