Why do charges outside surface create no net flux?

[yosimba2000]

Why do charges outside the surface contribute 0 net flux? The book I'm reading says it's because the flux entering the surface cancels out with the flux exiting the surface. But that means E dot Area must be exactly the same magnitude when entering and exiting to cancel out.

But we know E-field decreases by factor of R², so as you increase in distance, area must increase by R² in order to maintain the same magnitude of flux.

But what about in this situation? https://imgur.com/a/c2KS8

The areas penetrated by the E-Field have the same magnitude, but the E-fields are of different strength because E-Fields decrease by R². So E dot A is different at each area, so flux in this situation cannot be 0.

 

[Dale]

But what about in this situation? https://imgur.com/a/c2KS8

Flux also leaves the sides.

Sometimes thinking about lines of flux helps. If there is no charge in the surface then no lines start or stop inside the surface. Therefore, every line which goes in one side must come out somewhere else.

 

[NFuller]

But that means E dot Area must be exactly the same magnitude when entering and exiting to cancel out.

It means that the integral of $\mathbf{E}\cdot\mathbf{a}$ must be zero
$$\oint\mathbf{E}\cdot d\mathbf{a}=0$$

But we know E-field decreases by factor of R2, so as you increase in distance, area must increase by R2 in order to maintain the same magnitude of flux.

The size of the Gaussian surface can be considered constant. The Gaussian surface itself is arbitrary and how it is drawn and is not constrained by the behavior of the electric field. Although picking a surface with identical symmetry to the field can drastically simplify the problem.

The areas penetrated by the E-Field have the same magnitude, but the E-fields are of different strength because E-Fields decrease by R2. So E dot A is different at each area, so flux in this situation cannot be 0.

You forgot about the flux through the sides and the flux through the ends is not correct. The total flux is zero if you find it correctly.

 

[yosimba2000]

Flux also leaves the sides.

Sometimes thinking about lines of flux helps. If there is no charge in the surface then no lines start or stop inside the surface. Therefore, every line which goes in one side must come out somewhere else.

But the field strength when the line enters is stronger than when the field line leaves. So still, the areas must somehow compensate for this, right?

 

[NFuller]

But the field strength when the line enters is stronger than when the field line leaves. So still, the areas must somehow compensate for this, right?

Yes, but there is more area over which the field lines are leaving.

 

[Dale]

But the field strength when the line enters is stronger than when the field line leaves. So still, the areas must somehow compensate for this, right?

The field strength is proportional to the spacing of the lines. So it automatically ensures that as the field gets weaker it covers a larger area.