I read in this paper (http://www.math.iitb.ac.in/~srg/preprints/Chandigarh.pdf) that the following set is an affine variety:
$V_f=\{(t_0,...,t_N)\in \mathbb{F}_p^{N+1} : f(t_0,...,t_N)=0 \}$ where $f$ is an irreducible polynomial.
According to the same paper, the following set would be a projective variety:
$V_g=\{(x,y,w,z)\in \mathbb{P}_{\mathbb{F}_p}^3 : xy-wz=0 \}$ where $\mathbb{P}_{\mathbb{F}_p}^3$ is a 3-dimensional projective space over $\mathbb{F}_p$.
What's the dimension and degree of $V_g$?
Okay, so $$xy - wz = 0$$ cuts a quadric surface -- that is, a two-dimensional variety of degree 2 -- out of $\mathbb{P}^3$.
In order to see the dimension you really have to work scheme-theoretically, so that you see all of the non-$\mathbb{F}_p$-rational points of the variety -- otherwise it just appears to be a finite set of points, which would be zero-dimensional.
However, assuming you're willing to concede that a single equation cuts a two-dimensional variety out of three-dimensional space, it's not hard to calculate the degree of the intersection. Since $3 = 2 + 1$, we can determine this by intersecting our surface with a general line in $\mathbb{P}^3$. Let's take, say, the line $x = y = 0$. Substituting this into the equation for our surface, we have $x = y = w z = 0$. There are two solutions to this system of equations, namely $[0:0:1:0]$ and $[0:0:0:1]$. (In principle we should worry that the line we chose was somehow exceptional, but for now let's not.) Since the surface meets a linear subspace of complimentary dimension in two points, it has degree 2.