= Electrostatic potential of a uniformly charged disk
{title2=$\psi(0,z)=\frac{Q}{2\pi\epsilon_0a^2}(\sqrt{a^2+z^2}-|z|)$}
A thin disk of radius $a>0$ in $z=0$ carries uniform <surface charge density> $\sigma=Q/(\pi a^2)$. Summing the <Coulomb potential> of concentric rings gives the displayed axis potential, with zero reference at infinity. Put $A=\sigma/(2\epsilon_0)$, use <spherical coordinates> $z=r\cos\theta$, and let $P_2(t)=(3t^2-1)/2$. The near-center expansion, separately harmonic above and below the charged sheet, is
$$
\psi(r,\theta)=A\left[a-r|\cos\theta|+\frac{r^2}{2a}P_2(\cos\theta)+O(r^4/a^3)\right].
$$
The absolute value represents the jump in normal <electric field> across the sheet; a single regular <solid harmonic> series cannot describe both half-spaces simultaneously. The constant and quadratic coefficients follow from the positive-axis Taylor series, and reflection symmetry supplies the other half-space. The far-field <multipole expansion> is
$$
\psi(r,\theta)=\frac{Q}{4\pi\epsilon_0}\left[\frac1r-\frac{a^2}{4r^3}P_2(\cos\theta)+O(a^4/r^5)\right].
$$
The monopole is the total charge, the dipole vanishes by reflection symmetry, and the next term is the axisymmetric <electric quadrupole>. These expansions converge for $r<a$ within each source-free half-space and for $r>a$ in the exterior, respectively; at a point of the sheet their potentials have continuous limiting values but their normal derivatives differ.
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