How does Gauss's law account for charges both inside and outside the surface?

[gracy]

Electric flux through a closed surface S
= q/ε0
(1.31)
q = total charge enclosed by S
In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field [whose flux appears on the left side of Equation above is due to all the charges, both inside and outside S. The term q on the right side of Gauss’s law, however, represents only the total charge inside S.I want to know how that's work?

 

[Andrew Mason]

The flux from all charges not enclosed by the gaussian surface = 0. This is because the flux through the surface into the interior of the space enclosed by the Gaussian surface also passes through another part of the surface upon exiting the space so the net flux into the space = 0. Net flux through the surface is from enclosed charge only.

AM