Coulomb’s Law and divergence
This is answered directly by the differential form of Gauss's Law

$$\vec\nabla\cdot\vec{E}=\frac{\rho}{\epsilon}$$

In regions where there is no charge, the divergence is zero.

In regions where there is a positive charge, the divergence is positive.

In regions where there is a negative charge, the divergence is negative.

In the case of an ideal line charge, the charge density is infinite along the line, so the divergence of \$\vec{E}\$ will also be infinite (with sign depending on the sign of the charge). To avoid working with the infinities, it will usually be easier to work with the integral form of Gauss's Law.