Size of a glass capillary for noticable capillary action How big would a glass capillary have to be to have noticable capillary action?
Also, does capillary action happen in plastic tubes?
 A: It all depends on what you consider "noticeable". The change in height in a capillary comes about from the curvature of the liquid, and the resulting change in pressure.
Simplifying for a moment, the curvature of the surface (for a small capillary) can be approximated to a section of a sphere (the real math is much harder... but the principles are easy to understand). The thing we need to do is to calculate the curvature of that sphere - because the more curved the surface, the greater the pressure difference across it. This is because you can think of the curved surface as a piece of a balloon: the pressure on one side is different than on the other side.
For a spherical surface of radius $R$, with surface tension $\sigma$, we can compute the difference in pressure across the surface by imagining the spherical surface cut exactly in half: the length of the cut would be $2\pi R$ and the area of the cut is $\pi R^2$. Now if the pressure difference is $\Delta p$, the total force supported at this "seam" is $\Delta p \cdot \pi R^2$ and the force per unit length is of course $\sigma$ (surface tension has units of N/m) so the total force we can apply is $F = 2\pi R \sigma$. Setting these two equal to each other we can solve for pressure difference:
$$\Delta p\; \pi\; R^2 = 2 \pi\; R\; \sigma\\
\Delta p = \frac{2 \sigma}{R}$$
In the case of the capillary, we need a relationship between $R$, the curvature of the surface, and $r$, the radius of the tube. We can find this if we know the contact angle between the water and the surface.
What is the contact angle? Well - when a liquid touches a solid surface, there is a certain amount of attraction (or repulsion) between the molecules. If the liquid is attracted to the surface (we call that hydrophilic meaning "likes water"), then the water will want to spread out on the surface. If it's hydrophobic ("hates water"), the water will tend to form beads to minimize the contact surface. Picture (source):

This contact angle is a constant property of the liquid/solid interface: no matter how much water you put down, it will try to make the contact angle the same shape. That means that the surface of your capillary will look like this (for a typical hydrophilic surface):

A bit of simple geometry shows that the relationship between $R$ and $r$ is given by
$$\cos\theta = \frac{r}{R}\\
R = \frac{r}{\cos\theta}$$
We can substitute that in the relationship we found before, to get
$$\Delta p = \frac{2 \sigma \cos \theta}{r}$$
When this pressure is negative (concave surface) the liquid will rise; when it is positive, the level will fall. The pressure must just support the weight of the column of liquid per unit area:
$$\Delta p = \rho\; g\; h$$
Thus we find
$$h = \frac{2 \sigma \cos \theta}{r\;\rho \; g}$$
This tells us the magnitude of the rise of the mean level of the liquid: for a really accurate answer you will want to take into account the real shape of the liquid (not a simple section of a sphere when $r$ becomes large) and the fact that some of the surface is higher than other parts (meaning that the curvature changes with position, and that the "height" is not so easy to define).
Finally, you can see that the height depends on the contact angle, and therefore on the material of the tube. Many plastics are slightly hydrophobic, and may result in the water level in the capillary being lower than the level outside. From the same link as before, here is another drawing to illustrate that:

A: Here is another approach to complement the nice answer from Floris:
Water rises in the tube to minimize its energy. The relevant energies are the gravitational potential energy of the column of fluid and the surface energies of the interfaces between fluid, tube, and air.
Suppose the water rises by an amount $\mathrm{d}h$. This is equivalent to moving a bit of water from the reservoir to the top of the tube, so
$$\mathrm{d}E = g (\rho A \mathrm{d}h) h$$
with $A$ the cross sectional area of the tube.
We do not change the area of the fluid-air interface, but we add area $C \mathrm{d}h$ to the liquid-tube interface and subtract the same amount from the tube-air interface, with $C$ the circumference of the tube. If we represent the surface tension of the liquid-tube interface minus that of the air-tube interface by $\gamma$, we have
$$\mathrm{d}E = \gamma C \mathrm{d}h$$
At equilibrium these are equal, so
$$g \rho A h = \gamma C$$
or 
$$h = \frac{2 \gamma}{r g \rho}$$ 
A: The wikipedia article on capillary action gives the following formula for the height $h$ of the meniscus in a glass tube of internal diameter $r$ in "laboratory conditions" (which I presume refers mainly to the purity of the water and the glass surface) :
$$h_g \approx \frac{1.48 \times 10^{-5}}{r} m $$
The article goes on to state that a 40mm diameter tube is sufficiently "small" to give a discernible difference in height of 0.7mm, while a 0.4mm diameter would give a difference of 70mm.
Laboratory plastic tubing is typically made of PVC or PTFE for which the surface tension and contact angle are 37.9, 85.6 and 19.4, 109.2 measured in $mJ/m^2$ and $^{\circ}$ respectively. Densities are about 1.1 and 2.2 $g/cm^3$ respectively. Using the general formula given in the wikipedia article, these values give
$$h_{PVC} \approx \frac{5.4 \times 10^{-7}}{r}m$$  $$h_{PTFE} \approx \frac{-5.9 \times 10^{-7}}{r}m$$  
(The minus sign indicates that the meniscus would be lower inside the tube than outside.) 
So the meniscus rises or falls about 25x as much with glass tubing compared with plastic. A 40/25=1.6mm diameter PTFE tube would be required to lower the meniscus by 0.7mm. 
A: The height $h$ of a liquid column is given by,
$$h={{2 \gamma \cos{\theta}}\over{\rho g r}}$$
where $\gamma$  is the liquid-air surface tension, $\theta$ is the contact angle, $\rho$ is the density of liquid (mass/volume), $g$ is local acceleration due to gravity, and $r$ is radius of tube. Thus the thinner the space in which the water can travel, the further up it goes. And vice versa, the larger the radius the smaller the height, but it will never be zero (but will become unnoticeable if small enough).
The material of the tube will not change the height of the column. Thus, plastic will work too.
NOTE: I am disregarding the shape of the meniscus and considering it flat, for simplicity purposes. The material does have an effect, as some comments mention, this is considered in the factor $\cos \theta$, and $\theta$ depends on the material. However, this factor is small: $\cos \theta$ is between [-1,1], so it decides if the height is positive or negative, and also contributes to the height of the meniscus, but not to the height of the column as measured from the water surface to the meniscus bottom.
