This is just a quick question of a misconception I have. I've looked through books, and online pretty extensively, and I couldn't find the simple answer I was looking for, so I came here.
Gauss's Law is of the form: $$ \Phi_E=\oint \vec E\cdot ~\mathrm d\vec a = \frac {q_\textrm{enc}} {\varepsilon_0}$$
Say we have some charged sphere, with continuous charge distribution $ \rho $, and radius $R.$ We have to draw a Gaussian surface, of radius $r.$
My question is... how do we determine the volume parameters inside $q_\textrm{enc}$ I know it can be defined as $q_\textrm{enc} = \displaystyle \int\rho ~\mathrm dV $, and since the charge distribution is continuous we can pull it out, and integrate and get $q_\textrm{enc} = \rho V $. We also know what $\rho$ is, and that is $\rho = q/V.$ Depending on where the Gaussian surface is placed, these volumes can be different.
For example, if we are inside the sphere, we get $ q_\textrm{enc} = \rho V=\frac q {V_1}V_2=\frac {q} {\frac 4 3\pi R^3}(\frac 4 3\pi r^3) =\frac {qr^3} {R^3}$. This will then simplify to the proper electric field, once we set it divided by $\varepsilon_0$ equal to the flux, which is $\vec E =\frac 1 {4\pi \varepsilon_0} \frac {qr} {R^3}$.
How do we determine which volume is which, the volume the sphere encloses, or the volume the Gaussian surface encloses? Because when we examine the electric field outside the sphere, $V_1$ and $V_2$ equal each other and cancel, giving $\vec E = \frac 1 {4\pi \varepsilon_0} \frac q {r^2} \hat r\,.$ So those $V_1$ and $V_2$ must have some reasoning behind their choosing, or I'm interpreting something wrong, and everything I've written is invalid.