 Solution

Write the uniform surface charge density on the annulus as $\sigma =Q/[\pi (b^2-a^2)]$, and let $x$ be signed distance along its axis. A ring of radius $r$ and width $dr$ contains electric charge $dq=2\pi \sigma rdr$. Its transverse electric field components cancel in pairs, and its electric potential is $dV=dq/[4\pi {ϵ}_0\sqrt{x^2+r^2}]$. Setting $V(\infty )=0$ and integrating gives the axial field of a uniformly charged annulus:At $x=0$, each element's electric field cancels that of the diametrically opposite element, so $\mathbf{E}=0$. There is no surface charge at the centre because $a>0$; the axis electric field is continuous there.

For positive $Q$ and $x>0$, differentiate the signed axial electric field:Its stationary condition is $a^{4/3}(x^2+b^2)=b^{4/3}(x^2+a^2)$. Factoring with $A=a^{2/3}$ and $B=b^{2/3}$ givesThe derivative starts positive, has exactly one positive zero and is negative at large $x$, so this is the positive-side maximum. For either sign of $Q$, the electric field magnitude has equal maxima at $x=\pm x_{\max }$. As $b\to a$, the thin-ring limit is $x_{\max }\to a/\sqrt{2}$.