Example 4 The center of the ellipse is the origin $O$, its minor axis length is $2 \sqrt{2}$, and the directrix $l$ corresponding to the focus $F(c, 0)(c>0)$ intersects the $x$-axis at point $A, |O F|=2|F A|$, a line passing through point $A$ intersects the ellipse at points $P$ and $Q$.
(1) Find the equation of the ellipse and its eccentricity;
(2) If $\overrightarrow{O P} \cdot \overrightarrow{O Q}=0$, find the equation of the line $P Q$.
Solution: (1) According to the problem, we can set the equation of the ellipse as $\frac{x^{2}}{a^{2}}+$ $a=\sqrt{6}, c=2$.
$\therefore$ The equation of the required ellipse is $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$, and the eccentricity is $e=\frac{\sqrt{6}}{3}$.
(2) From (1), we get $A(3,0)$.
Let the equation of line $P Q$ be $y=k(x-3)$,
From the system of equations $\left\{\begin{array}{l}\frac{x^{2}}{6}+\frac{y^{2}}{2}=1, \\ y=k(x-3),\end{array}\right.$ we get $\left(3 k^{2}+1\right) x^{2}-$ $18 k^{2} x+27 k^{2}-6=0$,

According to the problem: $\Delta=12\left(2-3 k^{2}\right)>0$, we get $-\frac{\sqrt{6}}{3}<k<$ $\frac{\sqrt{6}}{3}$
Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$,
Then $x_{1}+x_{2}=\frac{18 k^{2}}{3 k^{2}+1} \cdots \cdots \cdots$ (1),
$$
x_{1} \cdot x_{2}=\frac{27 k^{2}-6}{3 k^{2}+1} \ldots \ldots \ldots \ldots \text { (2), }
$$

From the equation of line $P Q$, we get $y_{1}=k\left(x_{1}-3\right), y_{2}=$ $k\left(x_{2}-3\right)$,

Thus, $y_{1} y_{2}=k^{2}\left(x_{1}-3\right)\left(x_{2}-3\right)=k^{2}\left[x_{1} x_{2}\right.$ $\left.-3\left(x_{1}+x_{2}\right)+9\right] \cdots \cdots \cdots \cdots$. 3,
$\because \overrightarrow{O P} \cdot \overrightarrow{O Q}=0$,
$\therefore x_{1} x_{2}+y_{1} y_{2}=0$
From (1)(2) (3)(4) we get $5 k^{2}=1$, thus $k= \pm \frac{\sqrt{5}}{5} \in$ $\left(-\frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{3}\right)$,
$\therefore$ The equation of line $P Q$ is $x-\sqrt{5} y-3=0$ or $x$ $+\sqrt{5} y-3=0$.