Re: Error in Wikipedia article: Faraday's law of induction



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"Szczepan Bia³ek" <sz.bialek@xxxxx> wrote in message
news:g5mr11$ea$1@xxxxxxxxxxxxxxxxxxxxxx

<phil-news-nospam@xxxxxxxx> wrote news:g5mchp117hq@xxxxxxxxxxxxxxxxxxxx
On Wed, 16 Jul 2008 09:51:51 +0200 "Szczepan Bia?ek" <sz.bialek@xxxxx>
wrote:
|
| "Don Kelly" <dhky@xxxxxxx> wrote
news:OTefk.110294$gc5.47073@xxxxxxxxxxxx
|> >
|> It's a bit more than this simplistic and superficial approach.
|> To get a high voltage between two points, it is NECESSARY that there
is a
|> charge separation.
|
| No. Assume that you have very small cloud composed of 10 small charged
| drops. Between the cloud and the Earth is some voltage. If the drops
join
| together the voltagr rise.

Same total charge in Coulombs. Same distance in meters. Same potentin
in Volts.
What is it that you are changing to raise the voltage?

The ratio Q/C
In physics is the equation: C = Q/V. So we have:
1. For 10 drops: Let assume that the voltage on each is 1 V and Q is 1
electron and C is 1,
2. For one large drop: V = Q/C but now Q is 10 electrons and C is
proportional to the new radius of large drop. The new radius is 10^-3
bigger so the C also. In rsult we have: V = !0/10^-3 = 100V. I have used
the theoretical equations. In reality is 10V.
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Consider the 10 original drops having a radius of 1mm and spaced equally
around a sphere of 10cm radius. Modifying an expression for bundled
conductors of power lines, the equivalent radius of the whole becomes about
6 cm That is, a virtual conductor of 6cm cm radius carrying ten
electrons. At distances over a few m from the charge center, it is almost
impossible to tell the difference between the virtual sphere and the actual
charge distribution. For purposes of calculation of capacitance to ground
there is no measurable difference.
Now doing your math correctly the radius cubed should be the cube root of
10 times as large as the original droplet for the same total volume. The
single droplet (of the same total volume) that you propose is not 10^-3 (as
stated) or even 10^3 times as large but will have a radius that is 2.15 mm .
So, the group of droplets has a greater capacitance to ground than your
single large drop.

The original total capacitance of the separated drops depends on their
positions- with respect to each other and with respect to ground or whatever
is being used as a reference- this is something that you have entirely
ignored. You are trying to compare a single large drop with 10 electrons to
a single small drop of 1 electron. and haven't even got the math right.
Meaningless!



--------------



|>This separation of charge
|
| In contemporary science the separation of charge is used in place of
| build/lowering of voltage. Current flow from high voltage to lower. Not
from
| more charge to less.

The voltage is a difference between two points.

The second is the Earth. But you can have many charged particles in the
air with the same voltage. If the voltage is diferent current or sparks
start between them.
-----------
No kidding. Then each of these charges contribute to the field and if you
have a lot of similar charges- then one must consider the total effect. The
multiple individual charges in a cloud are individually negligable but the
total can amount to many Coulombs and can be treated as a charged ball. What
becomes important in breakdown (, is not the total voltage to ground but the
high fields in the vicinity of this large charge.
---------


|> is the cause of the voltage difference- look at the definition of
|> potential difference. Updrafts, wide temperature ranges, and charge
|> separation due to changes in temperature followed by freezing are some
of
|> the factors.
|> Suffice it to say that there are parts of a cloud that are negative
with
|> respect to earth and parts which are positive due to charge buildup.

It is impossible. All parts are negative but the voltage may be diffrent.
---------
You say it is not possible but it has been found by measurements that there
are regions of positive charge not just different amounts of negative
charges. There is theory backed by experiment that gives at least one
mechanism for this. In addition, there is solid data that some lightning
strokes originate in regions of positive charge. I'll stick with what is
known rather than what you say is not possible.
---------
|
| Are you talking about voltage?
| All parts of clouds have excess of electrons.

Maybe the whole planet has an excess.

Without any doubts.
Or the whole solar system. Or ...

| Meteorology is a new science. Physics is older. Lightnings folows the
| physics laws. Meteorologist should read physics.

They do.
But they use their own terminology. In physics negatively charged means
excess of electrons. Positively deficit. In a cloud no drops with deficit
of electrons.
S*
-------------
The references that I have given are physics and engineering references- not
meteorologists ' terminology. Consider a droplet that is being swept up
and cooled - there is a difference in mobility between the H+ ions and the
OH- ions in the drop. Suffice it to say that a supercooled droplet may
suddenly freeze (same as water in a supercooled stream can suddenly freeze
from the bottom) and the outer and lighter shards are + and are swept
upward faster and further than the heavier -ve core. Whether this is the
actual mechanism or some other mechanism occurs - there do exist regions of
positive charge (not less negative charge) - as you would know if you sat
down and read some of the technical literature with regard to lightning. A
review of electrostatics might also help. In other words take time to learn
more about the subject before making patently incorrect statements.

--

Don Kelly dhky@xxxxxxxxxxxx
remove the X to answer





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