"Falk Tannhäuser" wrote
...
Rodney Blackall schrieb:
In article , Szczepan Bialek
wrote:
Anyway, since cloud and ground are oppositely charged (otherwise you
would not get discharges in the form of C-G lightning),
and C-C lightning shows charge varies within and between clouds

Right, the lower part of the cloud is negatively charged, while the
(usually anvil-shaped) summit carries positive charge. A small zone of
positive charge is also often observed at the cloud base where the rain
falls out.

The ground and clouds are both negatively charged (exces of electrons).
Different is the voltage.
So where are the positive charges?

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.

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.

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.

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.
S*





Falk