= 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> $dq=2\pi\sigma r\,dr$. Its transverse <electric field> components cancel in pairs, and its <electric potential> is $dV=dq/[4\pi\epsilon_0\sqrt{x^2+r^2}]$. Setting $V(\infty)=0$ and integrating gives the <axial field of a uniformly charged annulus>:
$$
\boxed{V(x)=\frac{\sigma}{2\epsilon_0}\left(\sqrt{x^2+b^2}-\sqrt{x^2+a^2}\right),}
$$
$$
\boxed{\mathbf E(x)=-\frac{dV}{dx}\hat{\mathbf x}
=\frac{\sigma x}{2\epsilon_0}\left(\frac1{\sqrt{x^2+a^2}}-\frac1{\sqrt{x^2+b^2}}\right)\hat{\mathbf x}.}
$$
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>:
$$
\frac{dE_x}{dx}=\frac{\sigma}{2\epsilon_0}
\left[\frac{a^2}{(x^2+a^2)^{3/2}}-\frac{b^2}{(x^2+b^2)^{3/2}}\right].
$$
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 $B=b^{2/3}$ gives
$$
\boxed{x_{\max}^2=\frac{a^{4/3}b^{4/3}}{a^{2/3}+b^{2/3}},\qquad
x_{\max}=\frac{(ab)^{2/3}}{\sqrt{a^{2/3}+b^{2/3}}}.}
$$
The 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 $x_{\max}\to a/\sqrt2$.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-resit-annulus-fields.png]
{title=Axial potential and electric field of a uniformly charged annulus}
{description=Original dimensionless curves for $b=2a$ and positive <surface charge density>. The dotted line marks the calculated positive-side maximum of the <electric field>; the dashed tangent shows its linear behaviour near the centre.}
Back to article page
