Szczepan Bialek schrieb:
"Falk Tannhäuser" wrote
...
Due to electrostatic induction, the area on the ground lying directly
below the negatively-charged cloud gets positively charged, leading to a
reversal of the usual fair-weather field.

Are you sure? It would mean that the technical grounding is not zero under a
cloud. When electrons are in clouds the field is reversal.

Yep - that's http://en.wikipedia.org/wiki/Electrostatic_induction. It
occurs because the Earth is quite a good conductor.

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."
__________________________________________________ ______________________

Hence the negative C-G lightning actually increases the net negative
charge of the earth. Thunderstorms effectively act as generators - without
them, fair-weather current would soon make disappear the difference of
potential between ground and atmosphere.
Note that positive C-G lightning also occurs, but is considerably rarer
than negative one.

I have read that they start from place where the normal lightning has stroke
(in the same moment) . So they are C-C.

C-C between the positive anvil and negative cloud centre as well as the
negative cloud centre and the positive rain base do happen, of course.
However, positive C-G can occur independently.

Typically it originates from the cloud's anvil and strikes a place on the
ground that is peripheral to the thunderstorm (and thus negatively
charged). It is more often found during dissipating storms (where the lower
cloud parts often disappear first) or in winter thunderstorms (when the
cloud summits are lower).
Typical field strengths are on the order of magnitude of E = 10 kV/m
between ground and cloud, and 100 kV/m within the cloud. For comparison,
fair-weather field strength is about 0.15 kV/m near to ground level.

Concerning gravitational and electrostatic forces:
Consider a spheric rain droplet of a mass of m = 1 mg.
(It has a volume of V = 1 mm^3 and hence a diameter of 1.24 mm, since V =
4/3*pi*r^3 - not an unreasonable size).
Its weight (force exercised by gravitation) is m*g = 9.81*10^-6 N
(with g = 9.81 m/s^2).
The electrostatic force equals q*E where q is the charge of the droplet.
If electrostatic force is supposed to prevent our droplet from falling
down, it has to compensate the gravitational force. Then we can calculate
the charge needed for this. If we set E = 100 kV/m = 100 kN/C,
we obtain that our droplet has to have a charge of 9.81*10^-11 C.
This would mean 1.02*10^10 such droplets (corresponding to 10.2 m^3 of
water) would carry an aggregate charge of 1 Coulomb.
Now we let's consider that we may find about 100000 m^3 of water in a
small thunderstorm cloud (just to get an idea of the order of magnitude -
this would correspond to 10 mm of precipitation over 10 km^2, note however
that only a part of the water in the cloud finally makes it to the earth
as precipitation). The aggregate charge of this mass of water would then
equal to 10000 C - a value that seems much to high to me! Average
lightnings transport a charge of less than 10 C - furthermore, a punctual
charge of Q = 10000 C would produce a field of about
E = 10 MV/m at a distance of d = 3 km
(E = Q / (d^2 * 4*pi*eps_0)
eps_0 being the vacuum permittivity of 8.8541878176*10^-12 F/m)
- which is stronger by a factor of 100 than the values actually observed
in thunderstorm clouds.

Excelent job. Calculate now how many of the water particles (H2O) can one
electron lift when E = 0.15 kV/m. It will be something as cross-examining.
Not the all electrons fall down in form of lightnings. The most as the
normal electric current.

Water has a molar mass of 18 g/mol and with the Avogadro constant of
6.022*10^23/mol we obtain the molecule's mass of 2.99*10^-26 kg and a
weight of 2.93*10^-25 N.
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!

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

As a conclusion, I believe that electrostatic force can be neglected when
compared to gravity, and even more the vertical winds in a cumulonimbus,
where updrafts commonly reach 30 m/s and more.

Here is not place for "believe". The calculations should be done.

Well, when I wrote "believe", it was because I based myself on
simplifying assumptions (punctual charge at a distance of 3 km, instead
of charge continuously distributed within the cloud as in reality) and
guesstimating (quantity of water in the cloud). And of course, all
clouds are different! A better model for charge distribution between the
negative cloud centre and the positive anvil would perhaps have been a
plate capacitor. However, as the text form the website I cited above
shows, my estimation was not that far from the truth - typical charges
in a cumulonimbus are of an order of magnitude closer to 100 C rather
than 10000 C. 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.