Let $p$ be a prime number and let $k$ be a field with $char(k)\neq p$ such that all finite extensions have degree coprime to $p$. (For example, we can take $k=\mathbb{R}$ and $p\neq 2$ or let $k$ the union of $\mathbb{F}_{l^{(p^n)}}$ for all $n\geq 1$.)

Let $C$ be a regular affine curve over $k$ with (regular) projective compactification $\widetilde{C}$.

Then my question is as follows:

Is $\operatorname{Pic}(C)$ a $p$-divisible group? In other words, is every divisor on $C$ linearly equivalent to $pD$ for a divisor $D$ on $C$?

In a result I have proven related to strong approximation over function fields, the $p$-divisibilty is one of the necessary and sufficient conditions, so a positive answer here would allow me to simplify the statement.

One of the reasons why I suspect the above statement could be true is that the statement is true if $k$ is algebraically closed and the assumption on $k$ implies that $l^\times$ is a $p$-divisible group for all finite extensions $l/k$. If the statement is false, then I would be very interested in a counterexample and under what additional assumptions the statement is true.

**My attempt thus far**:
If $\widetilde{C}\setminus C$ contains a rational point, or more generally, there exists a divisor of degree $1$ supported in $\widetilde{C}\setminus C$, then the restriction map $\operatorname{Pic}^0(\widetilde{C})\rightarrow \operatorname{Pic}(C)$ is surjective so under this assumption we can reduce the question to

Is $\operatorname{Pic}^0(\widetilde{C})$ a $p$-divisible group? In other words, is every divisor on $C$ of degree $0$ linearly equivalent to $pD$ for a divisor $D$ on $\widetilde{C}$?

If $\widetilde{C}$ is smooth (automatic if $k$ is perfect) and has a rational point, then $\operatorname{Pic}^0(\widetilde{C})$ are the $k$-points of the Jacobian of $\widetilde{C}$. This may allow us to use tools from the theory of abelian varieties. Maybe $A(k)$ is $p$-divisible for any (principally polarized) abelian variety $A$?