According to "Lectures on Quantum Mechanics" by Steven Weinberg, the formula of Coulomb potential is $$V(r) = - \frac{Z e^2}{r}.$$
But it this true? I calculated the integral $$V = - \int _\infty ^r \vec{E} \cdot d\vec{r} = \frac{q}{4 \pi \epsilon _0} \frac{1}{r}.$$
I don't know what $Z$ is but I'm unfamiliar with the formula in the book.
Yes, using the integral $V = -\int \mathbf{E} \cdot \mathbf{dr}$ to calculate the potential is correct, but the expression - $V(r) = \dfrac{-Ze^2}{r}$ is for the potential energy of an electron in Bohr's classical model of an atom. $Z$ is just the number of protons in the atom.
You could relate the Coulombic force with the centripetal force for an electron in a hydrogen atom, and get the relation,
$$ E_{kinetic} = \dfrac{1}2 mv^2 = \dfrac{1}2 \dfrac{kZe^2}{r}\tag*{(1)} $$ And by the Virial Theorem for a spherical system ($n = -1$), $$ 2\langle T \rangle = -1\langle U \rangle\tag*{(2)} $$ Where $\langle T \rangle$ and $\langle U \rangle$ are the total kinetic and potential energies of the system.
Therefore, substituting $(1)$ in $(2)$ , we have, $$ U = -\dfrac{kZe^2}{r} $$
The top equation is electric potential energy while the bottom is electric potential. They use that coulomb potential energy term for hydrogen like atoms in quantum mechanics where Z is the number of protons in the nucleus.
Usually when we see $Z$ in this equation, we are relating the potential of electrical forces between electron and protons. Hence, $Z$ is a scalar for the number of particles with $+e$, i.e. protons that interact with the $-e$ electron, rendering $Ze^2.$ This comes up in the Bohr model of the hydrogen atom, for example.
