Differential and integral forms of <PERSON>' Law leading to different answers
I came across an interesting problem in one of my exam preparation worksheets recently. To make absolutely sure that I do not omit any important information from the problem, I have included below a direct screenshot of its introduction.
I was able to to parts a.) - d.) without issue, but part e.) was where I got confused. It asks:
e.) Explain the steps that the student must follow in order to derive $\rho(r)$ from $E(r)$.
Here was my original solution:
By <PERSON>' (Integral) Law, $$\unicode{x222F} E(r) dA= 4 \pi r^2 E (r) = \frac{Q_{enc}}{\epsilon} = \frac{\int_{0}^{r}4 \pi r^2 \rho(r) dr}{\epsilon}.$$ Which leads to $$4 \pi \epsilon r^2 E(r) = \int_{0}^{r} 4 \pi r^2 \rho(r) dr.$$
Differentiating on both sides, $$4 \pi \epsilon (E'(r)r^2 + 2r E(r)) = 4 \pi r^2 \rho(r).$$
Solving for $\rho(r)$, $$ \rho(r) = \epsilon (E'(r) + \frac{2E(r)}{r}).$$
However, I've also been able to arrive at a different answer by <PERSON>' (Differential) Law. If you define the three axis of this space to be one along $r$ and the other two exactly orthogonal to each other and to $r$, then, unless I'm mistaken, the field will only have any strength in the $r$-dimension due to symmetry conditions. 474
The other two dimensions will be locked at zero for all points along the $r$-axis. However, if one travels slightly off this axis, then there will be a non-zero value again. This implies that the $r$-axis is a line of extrema, and therefore that the two non-$r$ derivatives must be 0 when evaluated on the $r$-axis. That would transform <PERSON>' Law from saying $$\nabla \bullet E(x,y,r) = \frac{p(x,y,r)}{\epsilon} \Rightarrow E'(r) = \frac{\rho(r)}{\epsilon}.$$
Solving for $\rho(r)$, $$\rho(r) = \epsilon E'(r).$$
These both can't be true, so I am wondering on which path, differential or integral, I made an error. I assume it's the differential path, but that's really nothing more than a hunch. 474