How do the number of plane curves over a finite field of a fixed genus increase with the degree? Let $k$ be a finite field. We define $N(d,g)$ to be the number of plane curves $f(x,y)$ defined over $k$ of degree $d$ with (geometric) genus $g$.  If $D(d) := (d-1)(d-2)/2$ (the maximum possible genus), I would expect that as $d$ goes to infinity that the proportion of curves of degree $d$ with genus $D(d)$ would go to 1.  If, on the other hand, we're interested in curves of a fixed small genus (say $g=0$ or 1), I would expect that $N(d,g)$ would still approach infinity, albeit at a much slower rate.  The question that I have is how do $N(d,0)$ or $N(d,1)$ approach infinity?  Is it a polynomial in $\log d$, faster, slower?
I realize that there has been a lot done with enumerative geometry (Gromon-Witten, Caporaso-Harris), but that seems a bit different, since it always works over an algebraically closed field, and classifies curves by having them pass through some set of generic points, and possibly prescribing, tangency, etc.
 A: The set of singular curves of degree $d$ contains the curves with $f(0,0)=f_x(0,0)=f_y(0,0)=0$. The chance that each of these values ($f(0,0),\ldots$) is zero is $1/q$ over the field of $q$ elements, so the proportion of such curves among all curves of degree $d$ is $1/q^3$, hence the proportion of smooth curves is at most $1-1/q^3$ regardless  of $d$ and doesn't go to $1$ as $d$ goes to infinity. (I believe this observation is due to Poonen).
The set of curves of degree $d$ and genus zero is the set of parametrized curves $(P_0(t):P_1(t):P_2(t))$, $P_i$ polynomials of degree $d$, modulo the action of $PGL_2(\mathbb{F}_q)$ on the variable $t$. So there are about $q^{3d-4}$ such curves. That's your $N(d,0)$. You should be able to do $N(d,1)$ using similar ideas.
A: Fix $g$, the genus, and $q$, the order of $k$.
$N(d,g)$ should be $\approx C q^{3d}$, where $C$ is some constant dependent on $q$ and $g$. (Note that my $C$ has absorbed the $q^{-4}$ in Felipe's answer.) There are some nonrigorous details here.
There are finitely many isomorphism classes of pair $(X, L)$ where $X$ is a genus $g$ curve over $k$ and $L$ is a degree $d$ line bundle. I claim that, for each such pair, there are roughly $C(X,L,q) q^{3d}$ degree $d$ curves in $\mathbb{P}^2$ such that the curve is isomorphic to $X$ and the pullback of $\mathcal{O}(1)$ is $L$, and $C(X,L,q)$ is some constant dependent only on $q$, $X$ and $L$.
Recall that such curves, more or less, come from linear maps $H^0(X, L) \to k^3$, modulo rescaling on the image. The number of such maps is $q^{3 \dim H^0(X, L)}$. By Riemman-Roch, for $d$ larger than $2g$, we have $\dim H^0(X, L) = d-g+1$. So this is where I get $q^{3d}$ from, absorbing everything else into the constant. 
To be more precise, we need to (1) discard the maps that have base points (2) discard the maps that correspond to branched covers rather than immersions (3) if $(X, L)$ has nontrivial automorphism group, count orbits under that group.
I claim that (1) reduces our count by a factor of about $Z_X(3)$, where $Z_X$ is the zeta function of $X$. (2) should be negligible for large $d$ -- I get that there are about $q^{2d}$ maps coming from $d$-fold covers of a line, and fewer for every other case. And (3) I would guess just divides by $|\mathrm{Aut}(X,L)|$ -- there probably aren't a lot of cases with nontrivial stabilizers. I haven't checked the claims in this paragraph carefully, but that is what I'd expect the counts to be.
