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P: 92
 Quote by turin I think you accidently used the cylindrical version of the divergence. Why are you integrating?
I have used the cylindrical version, though this problem only depends on r so maybe use spherical? I integrated the partial differential with respect to r.

if I assume that Er is not a function of r but a constant then,

$$\frac{\partial (r E_{r})}{\partial r} = E_{r}$$

or if i use a radial divergence from the beginning:

$$\nabla . E = \frac{\partial (E)}{\partial r}$$

but by using that I get:

$$E = \frac{\rho r}{\epsilon_0}$$

i.e missing a factor of 3