What is the amplitude of the electric field in a laser? I'm looking for reliable informations about the amplitude (not the intensity), in volt/meter, of the electric field in a typical laser.
Or in other words :  what are the typical amplitudes of the monochromatic plane waves in an intense electromagnetic wave ?
EDIT :  I'm interested in the highest amplitudes values we could find in an intense electromagnetic wave.  Maybe not "typical waves", after all.
I know that the maximum theoretical value of an electric field (from quantum mechanics) is about $10^{18} \text{volt/meter}$.  But this is for static fields, not waves.  What about intense waves ?
I hope the question is clearer now.
 A: The electric field strength is related to the power of the laser by the Poynting vector. This is given by:
$$ \mathbf{S} = \mathbf{E} \times \mathbf{H} $$
and the magnitude of $\mathbf{S}$ is the power. Assuming we can treat your laser as a plane wave (which seems reasonable) then $\mathbf{E}$ and $\mathbf{H}$ are at right angles so the power is simply:
$$ P = EH $$
and $H = E/\eta $ so we end up with:
$$ P = \frac{E^2}{\eta} $$
In this expression $P$ is the peak power but what we really want is the average power, because that's what your laser spec will give. As it happens this just introduces a factor of a half:
$$ P_\text{av} = \frac{E^2}{2\eta} $$
Remember that $P_\text{av}$ is the power per unit area so you need to take the power of your laser and divide by the beam area. Then substitute in the equation above and solve for $E$.
A: When you say things like

My question is just about the typical amplitude value in an experiment with lasers. 

you run into trouble, because "typical experiments" use laser powers which cover many orders of magnitude.
In general, you find the electric field from the intensity $I$ of the light, which is equal to the amount of power per unit area transmitted by the light, using the formula
$$I=\frac{c\varepsilon_0}{2}E^2,$$
which you can invert as
$$E=\sqrt{\frac{2I}{c\varepsilon_0}}.$$
Now, to get the intensity, you need the laser power and the spot size, and here is where you run into trouble, because those can vary enormously depending on what you want to achieve with the experiment. However:


*

*The spot size is usually not much of a problem. For a regular laser pointer you can take it to be of the order of $1\:\mathrm{mm^2}$, but if you focus tightly then you can get down to focal spots just bigger than the wavelength, so for visible light on the order of $1\:\mathrm{\mu m}^2$.

*On the low end, many experiments run on the single-photon per shot regime, with maybe one shot every microsecond, which is overall a very small laser power.

*For a regular laser pointer, the power will be limited to about $1\:\mathrm{mW}$. (Any higher than that and it becomes dangerous.)

*On the high end, the most intense lasers currently in use produce electric fields so strong that when an electron oscillates driven by the laser electric field, (i) it reaches the relativistic regime, and (ii) its cycle-averaged energy of oscillation becomes several times greater than the electron rest energy $m_ec^2$. That means that the electron's motion will spontaneously produce electron-positron pairs, with the positron later on recombining and emitting gamma rays. These are not nice experiments to stand around.
In terms of intensity, though, this happens somewhere along the $10^{19}\:\mathrm{W/cm^2}$ mark, but the record is probably rather higher than $10^{22}\:\mathrm{W/cm^2}$ these days.
That's enough to calculate the quantities you need; I have purposefully left some boxes empty so you can flex your muscles a bit.
A: A simple rule: Electric field (in V/m) = 2745*sqrt(intensity), whereas the intensity is given in W/cm²
