= Solution
Let $m$ be the sliding <point charge>'s <mass>. Expanding the <axial field of a uniformly charged annulus> for $|x|\ll a,b$ gives
$$
E_x=\frac{\sigma}{2\epsilon_0}\left(\frac1a-\frac1b\right)x+O(x^3)
=\frac{Q}{2\pi\epsilon_0ab(a+b)}x+O(x^3).
$$
Its <electric charge> is $-Q$, so its <electric force> points toward the centre. <Newton's second law> becomes
$$
m\ddot x=-\frac{Q^2}{2\pi\epsilon_0ab(a+b)}x+O(x^3).
$$
To leading order this is <simple harmonic motion>, with
$$
\boxed{\omega_{\rm osc}^2=\frac{Q^2}{2\pi\epsilon_0mab(a+b)},\qquad
T=2\pi\sqrt{\frac{2\pi\epsilon_0mab(a+b)}{Q^2}}.}
$$
The <mass> is an additional parameter, since the source supplies no numerical value. The constraint is essential: <stability> along the axis does not imply unconstrained three-dimensional trapping by a static <electric field>.
Back to article page
