My question is more about how Gauss' Law can be used to find the strength of the electric field inside an insulator when the Gaussian surface used only contains a portion of the total charge.
From my understanding, the reason Gauss' Law works is because the field from a charge located outside the Gaussian surface fluxes into (negative flux) and out of (positive flux) the surface chosen, and as a result it contributes nothing to the total flux.
In cases like finding the field inside a ball (Note: not a shell) as the distance (r) from the center varies, all of the solutions I see involve choosing a small spherical surface of radius r concentric to the ball's surface to apply Gauss' Law on, and using $EA=4\pi kQ$ to find E.
Since the charge outside the Gaussian surface is essentially "invisible" to Gauss' Law, aren't these solutions leaving out the portion of the total field created by this charge? As you move out from the center of the ball, doesn't the charge "still ahead of you" exert a field in the same way that the charge "behind you" does? Or am I missing something?
An example of one of these solutions can be found here: