Differential and integral forms of Gauss' Law leading to different answers

Question

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 Gauss' (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 Gauss' (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. 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 Gauss' 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.

Answer 1

It looks like you're confusing the Divergence operator $\nabla\cdot E$ with the derivative or gradient operator $\nabla$. These are two very different things (as the divergence operator applies to vector fields and returns a scalar while the gradient applies to scalars and returns a vector field).

Answer 2

Look up divergence operator in spherical coordinates. The radial part is not just the derivative of Er in respect to r. If you use the right expression for the Div operator you get the same answer by both methods.