10. On a sphere with a radius of $\frac{\sqrt{3}}{2}$, there are 4 points $A, B, C, D$, and $A, B, C, D$ form a regular tetrahedron. The maximum distance from the center of the sphere to any face of the tetrahedron $ABCD$ is $\qquad$
10. $\frac{\sqrt{3}}{6}$

Let $O$ be the center of the sphere. Since $A, B, C, D$ form a regular tetrahedron, by symmetry, $O A \perp$ plane $B C D$, and the distance from $O$ to each face is equal, denoted as $h$. Therefore, the volumes of the tetrahedra $O A B C, O A C D, O A B D, O B C D$ are equal, so,
$$
V_{A B C D}=\frac{1}{3}(A O+h) \cdot S_{\triangle B C D}=4 V_{C O C D}=\frac{1}{4} \cdot \frac{1}{3} h S_{\triangle B C D} .
$$

Thus, $h=\frac{1}{3} A O=\frac{\sqrt{3}}{6}$.