# Why is Gauss' Law for Electric Fields still applicable when $\vec{E}=kr^3\hat{r}$?

Isn't Gauss' Law for electric fields $=\frac{q_{enc}}{\epsilon_0}$derived based on the assumption that the $r^2$ (one from $d\vec{S}_{spherical}$ and the other from $\vec{E}$) cancel, otherwise the flux would depend on the radius of the sphere and we would need to know the relation between $k$ and $q_{enc}$?

It is easiest to see that $\Phi_E = \oint \vec{E} \cdot d\vec{a} = \frac{q_{enc}}{\epsilon_0}$ for spherical volumes surrounding point charges, and that's generally how it's justified to students. But in fact, it holds for for *all* volumes and charge configurations (not just spherical volumes surrounding point charges). And if you buy that, you can then show that $\vec{\nabla} \cdot \vec{E} = \rho/\epsilon_0$; in other words, the two statements are equivalent. The way to show this (informally) is as follows:

Imagine a cuboid spanning the coordinates $[x, x+\Delta x]$, $[y, y+\Delta y]$, and $[z, z+\Delta z]$. We will assume that $\Delta x$, $\Delta y$, and $\Delta z$ are "small" in some appropriate sense. The net flux through the faces whose normals are in the $x$-direction is approximately $$ \Phi_{E,x} \approx [E_x(x+\Delta x, y, z) - E_x(x,y,z)]\Delta y \Delta z = \frac{E_x(x+\Delta x, y, z) - E_x(x,y,z)}{\Delta x} \Delta x \Delta y \Delta z \approx \frac{\partial E_x}{\partial x} \Delta V $$ where $\Delta V \equiv \Delta x \Delta y \Delta z$. Note that the fluxes have opposite signs because the face at $x$ has normal $-\hat{\imath}$ while the face at $x +\Delta x$ has normal $+\hat{\imath}$. Similarly, the net fluxes through the pairs of faces facing in the $y$ and $z$ directions are $$ \Phi_{E,y} \approx \frac{\partial E_y}{\partial y} \Delta V \qquad \Phi_{E,z} \approx \frac{\partial E_z}{\partial z} \Delta V $$ respectively. Thus, the total flux through these surfaces is approximately $$ \Phi_E = \left( \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} \right) \Delta V = (\vec{\nabla} \cdot \vec{E}) \Delta V. $$ And if Gauss's Law holds for this volume, which holds a charge $\Delta q$, then we have $$ (\vec{\nabla} \cdot \vec{E}) \Delta V \approx \Phi_E = \frac{\Delta q}{\epsilon_0} \quad \Rightarrow \quad \vec{\nabla} \cdot \vec{E} \approx \frac{1}{\epsilon_0} \frac{\Delta q}{\Delta V} \approx \frac{\rho}{\epsilon_0}. $$ All of the "approximately equals signs" above become exact equalities in the limit that $\Delta x, \Delta y, \Delta z \to 0$.