= Solution
Integrating the local electric-field <divergence> equation and applying the <divergence theorem> gives <Gauss's law>:
$$
\boxed{Q=\int_V\rho\,dV=\varepsilon_0\int_V\nabla\cdot\mathbf E\,dV
=\varepsilon_0\int_{\partial V}\mathbf E\cdot d\mathbf S.}
$$
For a <point charge> at the origin, <spherical symmetry> requires $\mathbf E=E_r(r)\mathbf e_r$. A sphere of radius $r>0$ encloses charge $q$, so its electric flux is $4\pi r^2E_r(r)$. Thus
$$
\boxed{\mathbf E(\mathbf x)=\frac{q}{4\pi\varepsilon_0r^2}\mathbf e_r
=\frac{q\mathbf x}{4\pi\varepsilon_0r^3}.}
$$
The <electrostatic potential> satisfies $E_r=-d\phi/dr$. Integrating and choosing the reference value $\phi(\infty)=0$ yields
$$
\boxed{\phi(\mathbf x)=\frac{q}{4\pi\varepsilon_0r}\quad(r>0).}
$$
An arbitrary additive constant gives any other choice of potential reference. The singularity at the origin is represented by the <Dirac delta distribution>; it is not a point where the ordinary derivatives of this field are finite.
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