Szczepan Bialek schrieb:
"Falk Tannhäuser" schrieb
[Electrostatic induction]
It should be measured and calculated. I bet that under a cloud is more
electrons (on the surface of the ground) then under the clear sky.

It has been. Have you checked the links below? Under positively charged
cloud regions (rain base, anvil) there surely is a surplus of electrons
(resulting in an electric field vector directed downwards), while under
negatively charged cloud parts there is a lack of electrons on the
ground (resulting in an electric field vector directed upwards, which
will accelerate negatively charged particles downwards).

I found some interesting web sites about thunderstorm charge distribution:
http://www.britannica.com/eb/art-19731/Electrical-charge-distribution-in-a-thunderstorm-When-the-electrical-charge
or http://minilien.fr/a0khcv (one can see that ground charge is negative
under the small centre of positive charge at the rain cloud base and
positive under the (negatively charged) remaining part of the cloud base -
the negative ground charge in fair-weather conditions is not depicted.
http://scf-cfs.rncan-nrcan.gc.ca/index/lightning-faq/3 shows a similar
picture and even gives examples of observed electrical charges:
__________________________________________________ ______________________
"The three centres of accumulated charge are commonly labeled p, N, and P.
The upper positive centre, P, occupies the top half of the cloud. The
negative charge region, N, is located in the middle of the cloud. The
lowest centre, p, is a weak, positively charged center at the cloud base.
The N and the P regions have approximately the same charge, creating the
positive dipole. Malan (1963) documented charges and altitudes above
ground level for the p, N, and P regions of a typical South African
thundercloud (1.8 km above sea level) as +10 coulombs (C) at 2 km, -40 C
at 5 km, and +40 C at 10 km. These are representative of values that can
vary considerably with geography and from cloud to cloud."
__________________________________________________ ______________________

[...]
So positive C-G can not occur independently.

It does. See for example
http://scf-cfs.rncan-nrcan.gc.ca/index/lightning-faq/5.

[...]
OTOH an electron has a charge of 1.602*10^-19 C and experiences an
electrostatic force of 2.403*10^-17 N in a field of 150 V/m. That's the
weight of 82 million water molecules!
82*10^6 near to ground level. The higher the smoler number.

You mean because gravity weakens with height? Well, it is inversely
proportional to the square of distance to the centre of gravity (Earth's
centre). So you have

g(h) = g(0) * R^2 / (R + h)^2

With the Earth's average radius of R = 6370 km and h = 10 km (a typical
height of a cumulonimbus summit) we obtain g(10 km) = 0.9969*g(0)
With g(0) = 9.81 m/s^2 (which is an average value, as g varies a little
bit with latitude, too) we have g(10 km) = 9.78 m/s^2 - this difference
is peanuts!

However, 100 C gives only 6.24*10^20 electrons while in 100000 m^3 of
water there are 3.34*10^33 molecules - a ratio of 5.35*10^12
[...]
typical charges in a cumulonimbus are of an
order of magnitude closer to 100 C rather than 10000 C.

The estimate of the total charge in a cloud is impossible.

It *is* possible. See the text from
http://scf-cfs.rncan-nrcan.gc.ca/index/lightning-faq/3 that I cited
above. Of course, it is just an estimation and the values will vary from
cloud to cloud - but the electrostatic fields and the charges per mass
of water will not deviate by a factor of 100.

The single drop should be observed like in Millican's experiment.

- Split this charge among droplets (or calculate the ratio
of water molecules per electron), and you'll see that resulting
electrostatic force is pretty weak compared to gravity.

Yes. For this reason large drops fall down (but the fine hang pretty well).

As long as the ratio of mass to charge (or water molecules to electrons)
is the same, drop size doesn't matter. Electrostatic force is q*E,
gravity force is m*g. If a big and a small drop are in the same cloud
region, the both experience the same electric field E and gravity g.

Of course, for very small droplets, the quantisation of electric charge
has to be taken into account. As I wrote above, a charge of 100 C in
100000 m^3 of water is 5.35*10^12 water molecules per electron, or
1.602*10^-13 kg of water per electron.
A droplet of a diameter D = 1 micrometer (10^-6 m, a typical droplet
size observed in Millikan's experiment) has a volume of V = pi/6*D^3 =
5.236*10^-19 m^3 and a mass of 5.236*10^-16 kg. I means that the average
charge per droplet would be 0.00327 electrons - hence most such droplets
are not charged, but one out of 306 carries a charge of one electron
(and thus has a charge 306 times higher than average).
For droplets of 10 micrometer, average charge is 3.27 electrons. The
actual distribution of electrons per droplets follows the Poisson
distribution with lambda = 3.27: P(k) = e^-lambda * lambda^k / k!

k (electrons/droplet): 0 1 2 3 4 5 6 7 8 9 10
P(k) (% of droplets): 3.8 12.4 20.3 22.2 18.1 11.8 6.4 3.0 1.2 0.4 0.1

This is the reason why Millikan's experiment works. Note however that
while the electric field typically used for this experiment is of
comparable strength to the one found within a thunderstorm cloud (about
100 kV/m), the average charge of the oil droplets is considerably higher
- hence the observed droplets' sizes are around 1 micrometer, rather
than 10 micrometer.
OTOH water droplet size in clouds is about 1 to 15 micrometer.

Dave " ..did some searching and it is said that warm rising air keeps clouds
up. Is it possible the static
charge in the clouds could also have an effect?"
I have never seen textbook on meteorology. There should be something about
it. In textbook on electrostatic are two pages about Absolute Earth
Potential and Atmospheric Electricity. In the Fluid Dynamics by Prantl are
many pages about atmosphere but nothing about charged fluids.
Weatherlawyer wrote: "It is a mystery why the clouds of vapour don't
condense though. It is as if some magical force is holding them apart" and
"It doesn't make sense that they don't fall to earth".

Small droplets fall slower than large ones because of aerodynamic drag
(increasing with the square of their diameter while their mass inceases
with their cube) - they reach a terminal velocity of a just a few
centimetres per second. No "magical force" and no electrostatic force is
involved here.
Bigger droplets of course fall faster and reach the ground as drizzle or
rain (100 micrometer to 5 mm), drops bigger than 6 mm usually break up
sooner or later.

This "magical force"
works in each atmospherical conditions. We have XXI century. In my opinion
in XIX century people have known what that was. In the XX all worked on
details and forgot about fundamentals.

Aerodynamic drag was known back in 19th century - guys like Reynolds,
Navier and Stokes did some research on fluid dynamics back then. I doubt
their results were forgotten during 20th century - Millikan and
Cunningham used them!

Falk