 Solution

Write $A=Q/(2\pi {ϵ}_0{a}^2)$. The Taylor expansion $\sqrt{a^2+z^2}=a+z^2/(2a)+O(z^4/a^3)$ gives the near-axis resultTo extend off the axis, work separately above and below the charged plane. In either source-free half-space the regular axisymmetric solid harmonics are $r^n{P}_n(\cos \theta )$. Matching the positive-axis coefficients, using $P_n(1)=1$, and then reflecting in the disk givesHere $P_2(t)=(3t^2-1)/2$ is a Legendre polynomial. The quadratic term is $A(z^2-{\rho }^2/2)/(2a)$. The linear term is $-Az$ above and $+Az$ below; it represents the discontinuity of the normal electric field. A single regular spherical expansion across the entire charged sheet would incorrectly remove that jump.