Solution
= Solution
At vertical displacement $x$ from the axis, the gap is
$$
h(x)=h_0+a-\sqrt{a^2-x^2}.
$$
For $|x|\ll a$,
$$
\sqrt{a^2-x^2}=a-\frac{x^2}{2a}+O(x^4/a^3),
$$
so
$$
\boxed{h(x)\simeq h_0+\frac{x^2}{2a}
=h_0\left(1+\frac{x^2}{2ah_0}\right).}
$$
The narrow-gap length scale is therefore $(ah_0)^{1/2}$.