10. (20 points) Given $O$ as the origin, $F$ as the right focus of the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, a line $l$ passing through point $F$ intersects the ellipse $C$ at points $A$ and $B$. Points $P$ and $Q$ on the ellipse satisfy
$$
\overrightarrow{O P}+\overrightarrow{O A}+\overrightarrow{O B}=\overrightarrow{O P}+\overrightarrow{O Q}=0 \text {, }
$$

and points $P, A, Q, B$ are concyclic. Find the eccentricity of the ellipse $C$.
10. First, prove a lemma.

Lemma: For the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, if two chords intersect at point $E$, then the necessary and sufficient condition for points $A, B, C, D$ to be concyclic is that the inclination angles of $AB$ and $CD$ are supplementary.

Proof: Let $E\left(x_{0}, y_{0}\right)$, and the inclination angles of $AB$ and $CD$ be $\alpha$ and $\beta$ respectively.
Then the parametric equations of $AB$ are $\left\{\begin{array}{l}x=x_{0}+t \cos \alpha, \\ y=y_{0}+t \sin \alpha\end{array}\right.$
Substituting these into the ellipse equation and simplifying, we get
$\left(b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha\right) t^{2}+\left(2 b^{2} x_{0} \cos \alpha+\right.$
$\left.2 a^{2} y_{0} \sin \alpha\right) t+b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}=0$.
Then $|A E||B E|=\left|t_{1} t_{2}\right|$
$$
=\frac{\left|b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha} \text {. }
$$

Similarly,
$$
|C E||D E|=\frac{\left|b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \beta+a^{2} \sin ^{2} \beta} \text {. }
$$

By the intersecting chords theorem,
$$
\begin{array}{l}
\frac{\left|b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}=\frac{\left|b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \beta+a^{2} \sin ^{2} \beta} \\
\Rightarrow\left(b^{2}-a^{2}\right)\left(\cos ^{2} \alpha-\cos ^{2} \beta\right)=0 .
\end{array}
$$

Noting that $\alpha$ and $\beta$ are inclination angles, then $\alpha+\beta=\pi$. The lemma is proved.
By the lemma, we know that the inclination angles of $AB$ and $PQ$ are supplementary.
Let the slope of $PQ$ be $k$. Then the slope of $AB$ is $-k$, and the equations of the two lines are
$$
l_{P Q}: y=k x, l_{A B}: y=-k(x-c),
$$

The intersection point of these two lines is $\left(\frac{c}{2}, \frac{k c}{2}\right)$.
Thus, $Q(c, k c)$.
Combining the equation of line $AB$ with the ellipse equation and simplifying, we get
$$
\begin{array}{l}
\left(a^{2} k^{2}+b^{2}\right) x^{2}-2 a^{2} k^{2} c x+a^{2} k^{2} c^{2}-a^{2} b^{2}=0 \\
\Rightarrow x_{1}+x_{2}=\frac{2 a^{2} k^{2} c}{a^{2} k^{2}+b^{2}}=c \\
\Rightarrow b^{2}=a^{2} k^{2} .
\end{array}
$$

Since point $Q$ is on the ellipse, we have
$$
\frac{c^{2}}{a^{2}}=\frac{1}{2} \Rightarrow e=\frac{c}{a}=\frac{\sqrt{2}}{2} \text {. }
$$