10. Let the inclination angle of line $AB$ be $\angle A F_{1} F_{2}=\theta$, and the right focus of the ellipse be $F_{2}$, as shown in Figure 2.
Then $\angle B F_{1} F_{2}=\pi-\theta$.
In $\triangle B F_{1} F_{2}$, let $\left|B F_{1}\right|=x$. Then
$$
\left|B F_{2}\right|=2 a-x=6-x \text {. }
$$

By the cosine rule,
$$
\begin{array}{l}
\left|B F_{2}\right|^{2}=\left|B F_{1}\right|^{2}+\left|F_{1} F_{2}\right|^{2}- \\
2\left|B F_{1}\right|\left|F_{1} F_{2}\right| \cos (\pi-\theta) \\
\Rightarrow(6-x)^{2}=x^{2}+4^{2}+8 x \cos \theta \\
\Rightarrow x=\left|B F_{1}\right|=\frac{5}{3+2 \cos \theta} .
\end{array}
$$

In the right triangle $\triangle Q B F_{1}$,
$$
\left|B F_{1}\right|=\left|Q F_{1}\right| \cos \theta=\frac{5}{2} \cos \theta \text {. }
$$

Thus, $\frac{5}{3+2 \cos \theta}=\frac{5}{2} \cos \theta$
$$
\begin{array}{l}
\Rightarrow 2 \cos ^{2} \theta+3 \cos \theta-2=0 \\
\Rightarrow(2 \cos \theta-1)(\cos \theta+2)=0 .
\end{array}
$$

Since $\cos \theta+2 \neq 0$, we have $\cos \theta=\frac{1}{2}$.
Given $0<\theta<\pi$, hence $\theta=\frac{\pi}{3}$.
Therefore, the slope of line $l$ is $\sqrt{3}$.
10. (20 points) Given the ellipse $E: \frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ with its left focus at $F_{1}$, a line $l$ passing through $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$. Let $Q\left(-\frac{9}{2}, 0\right)$. If $\overrightarrow{Q B} \perp \overrightarrow{A B}$, find the slope of the line $l$.