"Falk Tannhäuser" schrieb
...
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.

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.

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.

They dont use term "voltage". Each drop has volume and charge. V = Q/C
(Voltage = Charge/Capacitance). Normally each drop which hang has excess of
electrons and proper voltage. Nomally because during lighning (which is an
oscillate phenomenon) the deficit appears periodically. So positive C-G can
not 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!

82*10^6 near to ground level. The higher the smoler number.

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.

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

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". 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.
S*

"Rodney Blackall" wrote
...
In article ,
Szczepan Bia³ek wrote:

We must distinguish the charges from the charged bodies. Droplets are
bodies. In electrostatics the positive meens defficiency of electrons and
negative the exces of electrons.in a body. Droplets when hang in the air
have exces of electrons. If they loss them have to fall down (Millican).

Good, so you can prove this by experiment.

Get two similar metal plates, connect them to the terminals of a large
potential difference. When you have a natural fog, turn on the voltage
source and see which plate collects most water.

Now I know that in electrostatic experiment should be used such isolators
which are negatively charged, because they do not collect water on their
surface. As I have any opportunity to make experiments I will be waiting
for ready results, like this:
http://www.physik.fu-berlin.de/~stelmas/SHG.html

Falk Tannhauser has given you a very good answer. Note that if you start
calculating the fall speed of raindrops, viscosity is important.

Falk tried to calculate. It is now rather difficult. It should be measured.
S*

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

"Falk Tannhäuser" wrote
...
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)

All clouds have always excess of electrons. See what shrieb people from
Berlin: http://www.physik.fu-berlin.de/~stelmas/SHG.html
"(e.g. the droplets of thunderstorm clouds have around several 104
elementary charges on them, while the number of charges on non-thunderstorm
clouds is usually 102)"
In your links positive negative probably meens higher or lover voltage.


82*10^6 near to ground level. The higher the smoler number.

You mean because gravity weakens with height?

E weakens with height. At 12 km it is only 2% of that at the ground.



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.

In a cloud (in the same cloud region) the voltage is the same. It meens
that the ratio of mass to charge must obey the equation V = Q/C (C js
increasing with the radius while their mass inceases with their cube)

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!

It will be better to wait on the results from Berlin.

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),

It is impossible. In one region The voltage is almost the same. Here also
will be better to wait as somebody measure it. It will be in near future.
See: http://www.agu.org/pubs/crossref/200...JD004468.shtml


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!

I wrote to Rodney:
"All what I am talking about was explained in XIX century. In that time
Armstrong made the vapour generator (of high voltage) and Kelvin the drop
generator. The all was described by J. Frenkel in words: ".. Clouds are
electrogravitational generators in in continual run. In place of the
friction and the induction (as it take place in normal generators) are the
condensation and the droplets grow."
But the most important is to remember that the surface of the Earth has
ALWAYS excess of electrons. They are ewerywhere and in sunny days migrate
up. It is obvious that they must come back."
S*


Falk