= Axial field of a uniformly charged annulus
{title2=$E_x=\frac{\sigma x}{2\epsilon_0}[(x^2+a^2)^{-1/2}-(x^2+b^2)^{-1/2}]$}
For an <annulus> of inner radius $a>0$, outer radius $b>a$ and uniform <surface charge density> $\sigma$, integrate circular rings of <electric charge> to get $V(x)=\sigma[\sqrt{x^2+b^2}-\sqrt{x^2+a^2}]/(2\epsilon_0)$ with zero <electric potential> at infinity. Differentiation gives the displayed axial <electric field>. Its magnitude is greatest at $|x|=(ab)^{2/3}/\sqrt{a^{2/3}+b^{2/3}}$. Near the centre, $E_x=Qx/[2\pi\epsilon_0ab(a+b)]+O(x^3)$, where $Q=\pi\sigma(b^2-a^2)$; an oppositely charged particle constrained to the axis consequently undergoes small-amplitude <simple harmonic motion>.
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