voltage gradient in the atmosphere
Vincent Canter
Someone on another group claimed that there is an increase in voltage as
you go upward in the atmosphere of about 100 Volts per foot. Is that way
off?
Vince
James Logajan
Vincent Canter (cvca...@gsbux1.uchicago.edu) wrote:
: Someone on another group claimed that there is an increase in voltage as
: you go upward in the atmosphere of about 100 Volts per foot. Is that way
: off?
It is about 100 volts/meter. See chapter 9 "Electricity in the Atmosphere" in
volume II of The Feynman Lectures on Physics for a nice discussion.
The Feynman lectures reference "Atmospheric Electricity" by Chalmers, J. Alan,
Pergamon Press, London, 1957.
Kent Nielsen
James Logajan <jam...@netcom.com> skrev i artiklen
<jameslEr...@netcom.com>...
I'm close to 2 m in height - should I feel pain?
/Kent Nielsen
James Logajan
Kent Nielsen (kent.n...@get2net.dk) wrote:
: James Logajan <jam...@netcom.com> skrev i artiklen
: <jameslEr...@netcom.com>...
[On voltage gradient...]
: > It is about 100 volts/meter. See chapter 9 "Electricity in the
: Atmosphere" in
: > volume II of The Feynman Lectures on Physics for a nice discussion.
: > The Feynman lectures reference "Atmospheric Electricity" by Chalmers, J.
: Alan,
: > Pergamon Press, London, 1957.
: I'm close to 2 m in height - should I feel pain?
Only if you forget to duck when trying to walk through 1 m high doors.
Feynman answers that question right away in the lecture I referenced:
"Or you might wonder: 'If there is really a potential difference of 200
volts between my nose and my feet, why is it I don't get a shock when I go
out into the street?'"
...
"Your body is a relatively good conductor. If you are in contact with the
ground, you and the ground will tend to make one equipotential surface."
And another bit of info:
"The total potential difference from the surface of the earth to the top of
the atmosphere is about 400 000 volts."
The lecture has quite a bit of fascinating bits of information; you might
want to check it out.
Ray Tomes
In article <jameslEs...@netcom.com>, jam...@netcom.com (James
Logajan) wrote:
>Feynman answers that question right away in the lecture I referenced:
>"Or you might wonder: 'If there is really a potential difference of 200
>volts between my nose and my feet, why is it I don't get a shock when I go
>out into the street?'"
>...
>"Your body is a relatively good conductor. If you are in contact with the
>ground, you and the ground will tend to make one equipotential surface."
>And another bit of info:
>"The total potential difference from the surface of the earth to the top of
>the atmosphere is about 400 000 volts."
>The lecture has quite a bit of fascinating bits of information; you might
>want to check it out.
I found this interesting because a friend, Thomas Peterson, asked his
teachers how they knew that the ground was 0 volts and was never
satisfied with the answer. In effect it was defined as 0. So Thomas
made a device to measure the true potential of the ground and found that
it varied from about 300,000 to 600,000 volts.
What I didn't know before was that this was addressed by Feynman,
thanks.
-- Ray Tomes -- http://www.kcbbs.gen.nz/users/rtomes/rt-home.htm --
Cycles email list -- http://www.kcbbs.gen.nz/users/af/cyc.htm
Boundaries of Science http://www.kcbbs.gen.nz/users/af/scienceb.htm
John M. Feiereisen
In <jameslEs...@netcom.com>, jam...@netcom.com (James Logajan)
wrote:
>Feynman answers that question right away in the lecture I referenced:
>"Or you might wonder: 'If there is really a potential difference of 200
>volts between my nose and my feet, why is it I don't get a shock when I go
>out into the street?'"
>...
>"Your body is a relatively good conductor. If you are in contact with the
>ground, you and the ground will tend to make one equipotential surface."
This I don't understand. Doesn't this mean that if I take my
multimeter and hold the two probes out in the air separated by 1 m
vertical distance, it should read 100 volts? The probes themselves
are pretty well insulated from me and from each other.
--
John
Harry H Conover
John M. Feiereisen (feie...@utrc.utc.com) wrote:
: In <jameslEs...@netcom.com>, jam...@netcom.com (James Logajan)
If you have a sufficiently high-impedance voltmeter, and the area
you hold it up in has nothing else in the nearby vicinity to result
in leakage to ground, yes.
You can easily demonstrate the presence of the atmospheric potential
gradient if you have have a television antenna on top of your home.
For the purpose of the experiment, it needs to be ungrounded, and
have an insulated lead-in cable running down to your point of
measurement.
Now, take one of those small, neon-bulb type voltage testers and
connect it between the antenna lead-in wire and ground, and it
will light quite brightly. (Indicating that more than 90-Volts
or so is present, but not providing a quantitative measurment.)
The fact that only one element of the lamp is lit indicates that
what you are seeing is d.c. (It's dim, but bright enough to be
seen in a dim basement.)
To measure the actual voltage, you require one of those very
high-impedance electrostative measurement type voltmeters, since
the atmospheric impedance is on the order of hundreds of
Meg-Ohms.
I discovered this accidently while in high-school (working part-time
installing TV antennas). I kept getting little shocks from the
lead-in, and wondered why, so I hooked up the voltage tester and
verified that there was a good bit of voltage present.
It was years later, half-way through a degree in physics,
that I learned why.
By the way, the reason that you can feel a slight shock is due to
the capacity of the antenna and lead-in, probably only a couple of
hundred pico-farads, but charged to several hundred volts delivers
a brief tingle when discharged though your body and fingers.
Harry C.
Peter Berdeklis
On Sat, 25 Apr 1998, Vincent Canter wrote:
> Someone on another group claimed that there is an increase in voltage as
> you go upward in the atmosphere of about 100 Volts per foot. Is that way
> off?
In clear skies the "fair-weather field" is approx. 100 - 200 V/m at the
ground, but the field drops off non-linearly with altitude. The field is
created by the potential difference between the ionosphere (pos. charged)
and the ground (neg. charged), which is maintained by global thunderstorm
activity.
Pete
---------------
Peter Berdeklis
Dept. of Physics, Univ. of Toronto
Rick Baartman
>>>>> "feierijm" == John M Feiereisen <feie...@utrc.utc.com>
>>>>> wrote the following on Tue, 28 Apr 1998 11:40:12 GMT
feierijm> In <jameslEs...@netcom.com>, jam...@netcom.com (James
feierijm> Logajan) wrote:
>> Feynman answers that question right away in the lecture I referenced:
>> "Or you might wonder: 'If there is really a potential difference of
>> 200 volts between my nose and my feet, why is it I don't get a shock
>> when I go out into the street?'" ... "Your body is a relatively good
>> conductor. If you are in contact with the ground, you and the ground
>> will tend to make one equipotential surface."
feierijm> This I don't understand. Doesn't this mean that if I take my
feierijm> multimeter and hold the two probes out in the air separated by
feierijm> 1 m vertical distance, it should read 100 volts? The probes
feierijm> themselves are pretty well insulated from me and from each
feierijm> other.
Well, yes it does mean that. But "well-insulated" means the
resistance of the insulation around the leads and the meter itself
must be very large compared with the resistance of a similar configuration
of air. This is extremely difficult to achieve. Feynman suggests better
ways of measuring it.
--
rick baartman
(delete the extra 'a' when replying)
Ken Fischer
Harry H Conover wrote;
: You can easily demonstrate the presence of the atmospheric potential
: gradient if you have have a television antenna on top of your home.
: For the purpose of the experiment, it needs to be ungrounded, and
: have an insulated lead-in cable running down to your point of
: measurement.
:
: Now, take one of those small, neon-bulb type voltage testers and
: connect it between the antenna lead-in wire and ground, and it
: will light quite brightly. (Indicating that more than 90-Volts
: or so is present, but not providing a quantitative measurment.)
: The fact that only one element of the lamp is lit indicates that
: what you are seeing is d.c. (It's dim, but bright enough to be
: seen in a dim basement.)
: I discovered this accidently while in high-school (working part-time
: installing TV antennas). I kept getting little shocks from the
: lead-in, and wondered why, so I hooked up the voltage tester and
: verified that there was a good bit of voltage present.
Harry, isn't some of this due to TV transmitters?
Ken Fischer
---
Peter Berdeklis
On Tue, 28 Apr 1998, Ray Tomes wrote:
> In article <jameslEs...@netcom.com>, jam...@netcom.com (James
> Logajan) wrote:
>
> >Feynman answers that question right away in the lecture I referenced:
>
> >"Or you might wonder: 'If there is really a potential difference of 200
> >volts between my nose and my feet, why is it I don't get a shock when I go
> >out into the street?'"
> >...
> >"Your body is a relatively good conductor. If you are in contact with the
> >ground, you and the ground will tend to make one equipotential surface."
Maybe because the spark would occur only at the breakdown field
of air, which is 3,000,000 V/m at sea level. 100 V/m is a pretty piddly
little field, and 100 V with no current to speak of is pretty paultry too!
Feynman wasn't right about everything just because he was Feynman you
know.
Harry H Conover
Ken Fischer (kefi...@iglou.com) wrote:
:
: Harry, isn't some of this due to TV transmitters?
Ken, we lived something like 30 miles from the nearest tv station.
At that distance, the signal from the tranmitter is measured in
in microvolts. If you live within earshot of a tv transmitter
tower, that's another issue entirely.
Also, you can eliminate the possible induction from nearby power
lines, since only one pole of the neon bulb lights, indicating that
it's d.c.
Harry C.
James Logajan
Peter Berdeklis (pe...@atmosp.physics.utoronto.ca) wrote:
: Maybe because the spark would occur only at the breakdown field
: of air, which is 3,000,000 V/m at sea level. 100 V/m is a pretty piddly
: little field, and 100 V with no current to speak of is pretty paultry too!
I thought it was closer to 1,000,000 V/m.... Of course, the breakdown
field strength isn't the same as the resistance.
: Feynman wasn't right about everything just because he was Feynman you
: know.
Too many snips...in my original post, I pointed out that in fact that
Feynman lecture gives a references to some-one elses work. I assume that
is where Feynman, Leighton or whoever actually gave the lecture got the
information.
What can I say? Someone asks what the voltage gradient of the atmosphere
is, I happen to remember reading about it in the Feynman lectures and
so I give an answer and a reference. After that, anyone who reads my
response is on their own. No warranties as to accuracy of information
supplied is implied or guaranteed. This _is_ Usenet after all!
P.S. In this case, I think Feynman is right.
John Popelish
John M. Feiereisen wrote:
> This I don't understand. Doesn't this mean that if I take my
> vertical distance, it should read 100 volts? The probes themselves
> are pretty well insulated from me and from each other.
>
> --
> John
This works as long as the resistance of the meter mechanism is very much
higher than the resistance of the air between the two probe tips.
otherwise, the meter shorts out the potential difference. Hint: your
meter looks like a nearly perfect short circuit compared to 2 meters of
air even if it does have a 10 megohm impedance.
John Popelish
Peter R Newman
> >"The total potential difference from the surface of the earth to the top of
> >the atmosphere is about 400 000 volts."
>
> >The lecture has quite a bit of fascinating bits of information; you might
> >want to check it out.
>
> I found this interesting because a friend, Thomas Peterson, asked his
> teachers how they knew that the ground was 0 volts and was never
> satisfied with the answer. In effect it was defined as 0. So Thomas
> made a device to measure the true potential of the ground and found that
> it varied from about 300,000 to 600,000 volts.
With respect to what? Voltage is a measure of potential *difference*
(PD). PD is not an absolute.
Pete
--
http://sa1.star.uclan.ac.uk/~prn
Happy is he who has been able to learn the causes of things - Virgil
Peter R Newman
John M. Feiereisen wrote about potential gradients in the atmosphere:
>
> This I don't understand. Doesn't this mean that if I take my
> multimeter and hold the two probes out in the air separated by 1 m
> vertical distance, it should read 100 volts? The probes themselves
> are pretty well insulated from me and from each other.
If you had a multimeter with an internal resistance comparable to the
resistance of the air, then yes, otherwise Ohm's law defeats your
measurement. A gold-leaf electroscope or its modern equivalent should
show it at a guess (although I've not done the experiment).
Peter R Newman
Harry H Conover wrote:
> You can easily demonstrate the presence of the atmospheric potential
> gradient if you have have a television antenna on top of your home.
> For the purpose of the experiment, it needs to be ungrounded, and
> have an insulated lead-in cable running down to your point of
> measurement.
>
> Now, take one of those small, neon-bulb type voltage testers and
> connect it between the antenna lead-in wire and ground, and it
> will light quite brightly. (Indicating that more than 90-Volts
> or so is present, but not providing a quantitative measurment.)
> The fact that only one element of the lamp is lit indicates that
> what you are seeing is d.c. (It's dim, but bright enough to be
> seen in a dim basement.)
Neat! I must try this tonight when I get home. I think there's a neon
bulb in one of my screwdrivers (and a resistor to be removed)...
This reminds me of some film I saw on television years ago of somebody
holding a domestic fluorescent light bulb, vertically, under overhead
high-voltage cables, at night. They were grounding one end with their
hand, and the bulb was glowing brightly. 300kV at 30m above ground
gives 10kV/m, considerably larger than the natural potential gradient.
Cheers,
b.h.j...@hw.ac.uk
On 1998-04-28 rto...@kcbbs.gen.nz(RayTomes) said:
Newsgroups: sci.physics
In article <jameslEs...@netcom.com , jam...@netcom.com (James
Logajan) wrote:
Feynman answers that question right away in the lecture I
referenced:
"Or you might wonder: 'If there is really a potential difference
of 200 volts between my nose and my feet, why is it I don't get a
shock when I go out into the street?'"
...
(snip)
It's less than a year ago I read in a monthly electronics mag how to
construct a gadget to monitor this potential.
The difficult bit was to ensure a resistance to earth of giga-ohms.
Otherwise the plate which you basically stuck above your roof, to charge
up to the potential at that level, would merely pull it down to (almost)
zero.
Could it have been the Maplin electronics magazine? I think so. They are
hot on interesting gadgets if they can sell you the parts.
If interested, ask
(not me!)
============ ===== ===== BILL J. ===== ===== ============
GM8APX, qthr Edinburgh, Scotland, UK
Silent enim leges sunt inter arma
Net-Tamer V 1.11 - Registered
b.h.j...@hw.ac.uk
On 1998-04-28 kefi...@iglou.com(KenFischer) said:
Harry H Conover wrote;
: You can easily demonstrate the presence of the atmospheric
potential : gradient if you have have a television antenna on top
of your home. : For the purpose of the experiment, it needs to be
ungrounded, and : have an insulated lead-in cable running down to
your point of : measurement.
:
: Now, take one of those small, neon-bulb type voltage testers and
: connect it between the antenna lead-in wire and ground, and it
: will light quite brightly. (Indicating that more than 90-Volts
: or so is present, but not providing a quantitative measurment.)
: The fact that only one element of the lamp is lit indicates that
: what you are seeing is d.c. (It's dim, but bright enough to be
: seen in a dim basement.)
:
: I discovered this accidently while in high-school (working
part-time : installing TV antennas). I kept getting little shocks
from the : lead-in, and wondered why, so I hooked up the voltage
tester and : verified that there was a good bit of voltage present.
Harry, isn't some of this due to TV transmitters?
Ken Fischer
---
No. TV watchers are very lucky if they get a few millivolts of signal.
============ ===== ===== BILL J. ===== ===== ============
GM8APX, qthr Edinburgh, Scotland, UK
Si vales, bene est
Dag ]gren FYSI
Peter Berdeklis (pe...@atmosp.physics.utoronto.ca) wrote:
> > In article <jameslEs...@netcom.com>, jam...@netcom.com (James
> > Logajan) wrote:
> > >Feynman answers that question right away in the lecture I referenced:
> > >"Or you might wonder: 'If there is really a potential difference of 200
> > >volts between my nose and my feet, why is it I don't get a shock when I go
> > >out into the street?'"
> > >...
> > >ground, you and the ground will tend to make one equipotential surface."
> of air, which is 3,000,000 V/m at sea level. 100 V/m is a pretty piddly
> little field, and 100 V with no current to speak of is pretty paultry too!
> know.
However, nobody was talking about sparks.
--
I)/\(, - Dag Agren - dag...@abo.fi - Goaway on IRC
Please don't go to http://www.abo.fi/~dagren/
-> Legalize oregano! <-
Dag ]gren FYSI
Rick Baartman (baaa...@alph04.triumf.ca) wrote:
> >>>>> "feierijm" == John M Feiereisen <feie...@utrc.utc.com>
> >>>>> wrote the following on Tue, 28 Apr 1998 11:40:12 GMT
> feierijm> multimeter and hold the two probes out in the air separated by
> feierijm> 1 m vertical distance, it should read 100 volts? The probes
> feierijm> themselves are pretty well insulated from me and from each
> feierijm> other.
> Well, yes it does mean that. But "well-insulated" means the
> resistance of the insulation around the leads and the meter itself
> must be very large compared with the resistance of a similar configuration
> of air. This is extremely difficult to achieve. Feynman suggests better
> ways of measuring it.
Could somebody explain these methods?
I'm interested in measuring the atmospheric electirc field, especially
during a thunderstorm (so the antenna idea is not that great... :).
Doesn't have to be anything accurate, as long as it's pretty simple.
Richard Herring
Harry H Conover (con...@tiac.net) wrote:
> Ken Fischer (kefi...@iglou.com) wrote:
> :
> : Harry, isn't some of this due to TV transmitters?
> Ken, we lived something like 30 miles from the nearest tv station.
> At that distance, the signal from the tranmitter is measured in
> in microvolts.
An acceptable UHF TV signal is something like 3 millivolts/metre.
--
Richard Herring | richard...@gecm.com | Speaking for myself
GEC-Marconi Research Centre |
Peter Berdeklis
On Tue, 28 Apr 1998, James Logajan wrote:
> Peter Berdeklis (pe...@atmosp.physics.utoronto.ca) wrote:
> : Feynman wasn't right about everything just because he was Feynman you
> : know.
>
> What can I say? Someone asks what the voltage gradient of the atmosphere
> is, I happen to remember reading about it in the Feynman lectures and
> so I give an answer and a reference. After that, anyone who reads my
> response is on their own. No warranties as to accuracy of information
> supplied is implied or guaranteed. This _is_ Usenet after all!
Sorry, I didn't mean to sound so snarfy. I forgot to add the emoticon. :)
Peter Berdeklis
On 29 Apr 1998, Dag ]gren FYSI wrote:
>
> I'm interested in measuring the atmospheric electirc field, especially
> during a thunderstorm (so the antenna idea is not that great... :).
>
> Doesn't have to be anything accurate, as long as it's pretty simple.
Look up the book,
Uman, M. A., 1969: Lightning. McGraw-Hill.
Chapter 3 describes the techniques for measuring the electric and magnetic
fields in thunderstorms. Note that this is much easier than measuring the
fair-weather field, because the field at the ground under a thunderstorm
peaks at about 8 kV/m, as opposed to 100 V/m.
The instrument is called a field mill. Basically, it measures the voltage
between an antenna and ground, as was suggested by another poster.
However, because the field is so much stronger the antenna can be quite
short (a couple of feet high in research units).
You can also look up:
Krehbiel, Brook and McCrory: An analysis of the charge structure of
lightning discharges to ground. J. of Geophys. Res., vol 84, C5, 2432 -
2456.
which includes a photo and circuit diagram of a field mill.
Fairly simple instrument.
Ray Tomes
In article <3545BA...@web.pages>, Peter R Newman <see...@web.pages>
wrote:
>Ray Tomes wrote:
>> I found this interesting because a friend, Thomas Peterson, asked his
>> teachers how they knew that the ground was 0 volts and was never
>> satisfied with the answer. In effect it was defined as 0. So Thomas
>> made a device to measure the true potential of the ground and found that
>> it varied from about 300,000 to 600,000 volts.
>With respect to what? Voltage is a measure of potential *difference*
>(PD). PD is not an absolute.
If you think about charges then there must be such a thing as a zero
potential. Not zero potential difference, but zero potential.
In another post you wrote:
>If you had a multimeter with an internal resistance comparable to the
>resistance of the air, then yes, otherwise Ohm's law defeats your
>measurement. A gold-leaf electroscope or its modern equivalent should
>show it at a guess (although I've not done the experiment).
which is pretty much on the right track.
Tom Potter
Peter R Newman <see...@web.pages> wrote in article
<3545C5...@web.pages>...
> John M. Feiereisen wrote about potential gradients in the atmosphere:
> >
> > vertical distance, it should read 100 volts? The probes themselves
> > are pretty well insulated from me and from each other.
>
> If you had a multimeter with an internal resistance comparable to the
> resistance of the air, then yes, otherwise Ohm's law defeats your
> measurement. A gold-leaf electroscope or its modern equivalent should
> show it at a guess (although I've not done the experiment).
Although I have been out of instrumentation
for over thirty years, back in the "old days"
these kinds of measurements were made with
"electrometers" and "electro-static voltmeters."
Both had extremely high input impedance, on the
order of 10^14 ohms as I recall. Electrometers
use field effect transistors ( FET ) for the input.
Of course, ordinary voltmeters have an
input impedance of about 20,000 ohms per volt,
and electronic voltmeters typically have
input impedance's of 1,000,000 ohms.
It is good for a voltmeter to have a high
input impedance so that it does not
load the circuit under test, but if the
input impedance is extremely high,
it is sometimes hard to know when the
meter is being affected by ambient electrostatic
fields, rather than a particular circuit point.
Alternating current ( A.C. ) fields are ( Were )
measured with tuned, calibrated, receivers
whose input impedance was matched to the
transmission line and the antenna, and the
measurements were made in terms of
microvolts per megacycle per meter,
and were calibrated in terms of
average, RMS, peak and quasi peak values.
Of course an oscilloscope could be used to
display the instantaneous values within its
bandwidth and sensitivity range.
I might mention that once when I was working
at the Bendix Research Labs, we kept
getting unreliable readings from
a Keithley electrometer and eventually
we discovered that the input impedance
had been corrupted ( Smoking, aging, etc. )
and the input impedance was
less than 10^9 ohms ( Instead of 10^14 ).
--
Tom Potter http://home.earthlink.net/~tdp
Peter R Newman
Ray Tomes wrote:
> that I wrote:
> >that he wrote:
> [snip]
> >> So Thomas
> >> made a device to measure the true potential of the ground and found that
> >> it varied from about 300,000 to 600,000 volts.
>
> >With respect to what? Voltage is a measure of potential *difference*
> >(PD). PD is not an absolute.
>
> If you think about charges then there must be such a thing as a zero
> potential. Not zero potential difference, but zero potential.
Then how do you get a result in volts, which are a measure of potential
difference?
> In another post you wrote:
> >If you had a multimeter with an internal resistance comparable to the
> >resistance of the air, then yes, otherwise Ohm's law defeats your
> >measurement. A gold-leaf electroscope or its modern equivalent should
> >show it at a guess (although I've not done the experiment).
>
> which is pretty much on the right track.
Are you saying 'zero potential' equals neutral charge (i.e. equal
positive and negative ions)? You seem to be implying that an uncharged
electroscope, when connected to the ground, should show a charge. Is
this the case? I'd be fascinated to see more details on Thomas's
experiment.
me...@cars3.uchicago.edu
In article <35564b90...@aklobs.org.nz>, rto...@kcbbs.gen.nz (Ray Tomes) writes:
>In article <3545BA...@web.pages>, Peter R Newman <see...@web.pages>
>wrote:
>
>>Ray Tomes wrote:
>
>>> I found this interesting because a friend, Thomas Peterson, asked his
>>> teachers how they knew that the ground was 0 volts and was never
>>> made a device to measure the true potential of the ground and found that
>>> it varied from about 300,000 to 600,000 volts.
>
>>With respect to what? Voltage is a measure of potential *difference*
>>(PD). PD is not an absolute.
>
>If you think about charges then there must be such a thing as a zero
>potential. Not zero potential difference, but zero potential.
>
which is arbitrary. Note that EM is fully summed up in Maxwell's
equations where the potential itself deoesn't appear, only its
gradient does.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Ray Tomes
In article <35498D...@web.pages>, Peter R Newman <see...@web.pages>
wrote:
>Are you saying 'zero potential' equals neutral charge (i.e. equal
>positive and negative ions)?
Yes, if the distribution of charge about a point is exactly balanced
(e.g all made of of non-ionised atoms) then the potential at that point
must be zero, mustn't it?
>You seem to be implying that an uncharged
>electroscope, when connected to the ground, should show a charge. Is
>this the case? I'd be fascinated to see more details on Thomas's
>experiment.
I suggest that you send him an email at tfp...@aol.com (Thomas
Peterson).
me...@cars3.uchicago.edu
In article <3556bcc9...@aklobs.org.nz>, rto...@kcbbs.gen.nz (Ray Tomes) writes:
>In article <35498D...@web.pages>, Peter R Newman <see...@web.pages>
>wrote:
>
>>Are you saying 'zero potential' equals neutral charge (i.e. equal
>>positive and negative ions)?
>
>Yes, if the distribution of charge about a point is exactly balanced
>(e.g all made of of non-ionised atoms) then the potential at that point
>must be zero, mustn't it?
>
Tom Potter
me...@cars3.uchicago.edu wrote in article
<EsDD0...@midway.uchicago.edu>...
> In article <3556bcc9...@aklobs.org.nz>, rto...@kcbbs.gen.nz (Ray
Tomes) writes:
> >In article <35498D...@web.pages>, Peter R Newman <see...@web.pages>
> >wrote:
> >
> >>Are you saying 'zero potential' equals neutral charge (i.e. equal
> >>positive and negative ions)?
> >
> >Yes, if the distribution of charge about a point is exactly balanced
> >(e.g all made of of non-ionised atoms) then the potential at that point
> >must be zero, mustn't it?
> >
> No.
If you plot the ionization potentials of atoms in various states,
you will see that the ionization potentials do not converge
on zero, but on negative values.
This is known as "electro-negativity".
me...@cars3.uchicago.edu
In article <01bd76ee$40882480$fe7a0b26@ELN/tdp>, "Tom Potter" <t...@earthlink.net> writes:
>
>me...@cars3.uchicago.edu wrote in article
><EsDD0...@midway.uchicago.edu>...
>> In article <3556bcc9...@aklobs.org.nz>, rto...@kcbbs.gen.nz (Ray
>Tomes) writes:
>> >In article <35498D...@web.pages>, Peter R Newman <see...@web.pages>
>> >wrote:
>> >
>> >>Are you saying 'zero potential' equals neutral charge (i.e. equal
>> >>positive and negative ions)?
>> >
>> >Yes, if the distribution of charge about a point is exactly balanced
>> >(e.g all made of of non-ionised atoms) then the potential at that point
>> >must be zero, mustn't it?
>> >
>> No.
>
>If you plot the ionization potentials of atoms in various states,
>you will see that the ionization potentials do not converge
>on zero, but on negative values.
>
potential ain't a potential".
An electrostatic potential is simply defined as the scalar function
the gradient of which is the electric field. Thus it is defined only
up to a constant. It is a standard and convenient choice to fix this
constant so that the value of the potential at infinity converges to
zero. But that's just a matter of convenience, not of actual physical
content.
b.h.j...@hw.ac.uk
On 1998-05-04 me...@cars3.uchicago.edu said:
In article <01bd76ee$40882480$fe7a0b26@ELN/tdp , "Tom Potter"
<t...@earthlink.net writes:
me...@cars3.uchicago.edu wrote in article
<EsDD0...@midway.uchicago.edu ...
In article <3556bcc9...@aklobs.org.nz , rto...@kcbbs.gen.
nz (Ray Tomes) writes:
In article <35498D...@web.pages , Peter R Newman <see.
m...@web.pages wrote:
Are you saying 'zero potential' equals neutral charge (i.e.
equal positive and negative ions)?
Yes, if the distribution of charge about a point is exactly
balanced (e.g all made of of non-ionised atoms) then the
potential at that point must be zero, mustn't it?
No.
If you plot the ionization potentials of atoms in various states,
you will see that the ionization potentials do not converge
on zero, but on negative values.
You've a confusion of terminology. Simply speaking, "an ionization
potential ain't a potential".
An electrostatic potential is simply defined as the scalar function
the gradient of which is the electric field. Thus it is defined
only up to a constant. It is a standard and convenient choice to
fix this constant so that the value of the potential at infinity
converges to zero. But that's just a matter of convenience, not of
actual physical content.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the
same"
Potential at a point, measured in volts, is defined as the work needed
to bring unit + charge (1 coulomb) to that point from infinity.
Infinity I like, because it is the same whatever your frame of
reference. Which makes it unnecessary to alarm students with frames of
reference. Nor even to mention scalars.
I got that from McKenzie's 2nd Course of Electricity, Cambridge
University Press, 3rd edition, 1973. It's pretty clear on many similar
topics, and does not use spurious words to reduce clarity.
I KNOW no-one is going to go around carrying a coulomb and measuring the
work needed to move it into our bathroom from infinity; but it does
provide the starting point for a fairly easy way of approaching
practical situations.
============ ===== ===== BILL J. ===== ===== ============
GM8APX, qthr Edinburgh, Scotland, UK
Silent enim leges sunt inter arma
Net-Tamer V 1.11 - Registered
t...@earthlink.net
In article <EsF2q...@midway.uchicago.edu>#1/1,
me...@cars3.uchicago.edu wrote:
>
> In article <01bd76ee$40882480$fe7a0b26@ELN/tdp>, "Tom Potter" <t...@earthlink.net> writes:
> >
> >me...@cars3.uchicago.edu wrote in article
> ><EsDD0...@midway.uchicago.edu>...
> >Tomes) writes:
> >> >In article <35498D...@web.pages>, Peter R Newman <see...@web.pages>
> >> >wrote:
> >> >
> >> >>Are you saying 'zero potential' equals neutral charge (i.e. equal
> >> >>positive and negative ions)?
> >> >
> >> >Yes, if the distribution of charge about a point is exactly balanced
> >> >(e.g all made of of non-ionised atoms) then the potential at that point
> >> >must be zero, mustn't it?
> >> >
> >> No.
> >
> >If you plot the ionization potentials of atoms in various states,
> >you will see that the ionization potentials do not converge
> >on zero, but on negative values.
> >
> You've a confusion of terminology. Simply speaking, "an ionization
> potential ain't a potential".
>
> An electrostatic potential is simply defined as the scalar function
> the gradient of which is the electric field. Thus it is defined only
> up to a constant. It is a standard and convenient choice to fix this
> constant so that the value of the potential at infinity converges to
> zero. But that's just a matter of convenience, not of actual physical
> content.
Of course, the point I was trying to make
was that even if you "balance" all the
quantum ( Integer ) charges about
some point, that the "potential"
will still not be zero.
How would you answer the question
posed by the previous poster,
considering that charges come only
in integer amounts, and that space
is more than one dimension?
Tom Potter
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
http://www.dejanews.com/ Now offering spam-free web-based newsreading
me...@cars3.uchicago.edu
In article <6ikktq$794$1...@cn1.hw.ac.uk>, b.h.j...@hw.ac.uk writes:
>
>On 1998-05-04 me...@cars3.uchicago.edu said:
> An electrostatic potential is simply defined as the scalar function
> the gradient of which is the electric field. Thus it is defined
> only up to a constant. It is a standard and convenient choice to
> fix this constant so that the value of the potential at infinity
> converges to zero. But that's just a matter of convenience, not of
> actual physical content.
> Mati Meron | "When you argue with a fool,
> me...@cars.uchicago.edu | chances are he is doing just the
> same"
>
>Potential at a point, measured in volts, is defined as the work needed
>to bring unit + charge (1 coulomb) to that point from infinity.
>
Actually, no. It is the work needed to bring unit charge to that
point from a chosen reference point. Said reference point is usually
chosen as infinity, but it doesn't have to be. It is all a matter of
convenience.
>Infinity I like, because it is the same whatever your frame of
>reference. Which makes it unnecessary to alarm students with frames of
>reference. Nor even to mention scalars.
>
Indeed. That's why infinity is customarily used, when it can be used.
Now, here is a little electrostatics problem for you. Given an
infinite, uniformly charged plane (I trust you know that this simply
yields a uniform field), find the potential in the vicinity of the
plane based on a reference point at infinity.
me...@cars3.uchicago.edu
In article <6ikvtj$d0s$1...@nnrp1.dejanews.com>, t...@earthlink.net writes:
>In article <EsF2q...@midway.uchicago.edu>#1/1,
> me...@cars3.uchicago.edu wrote:
>>
>> In article <01bd76ee$40882480$fe7a0b26@ELN/tdp>, "Tom Potter" <t...@earthlink.net> writes:
>> >
>> >If you plot the ionization potentials of atoms in various states,
>> >you will see that the ionization potentials do not converge
>> >on zero, but on negative values.
>> >
>> You've a confusion of terminology. Simply speaking, "an ionization
>> potential ain't a potential".
>>
>> the gradient of which is the electric field. Thus it is defined only
>> up to a constant. It is a standard and convenient choice to fix this
>> constant so that the value of the potential at infinity converges to
>> zero. But that's just a matter of convenience, not of actual physical
>> content.
>
>was that even if you "balance" all the
>quantum ( Integer ) charges about
>some point, that the "potential"
>will still not be zero.
OK, one small clarification, then I'll get to the point. The actual
value of the potential is of no consequence, only the gradient is.
So what we really ask is not whether the potential is zero but whether
it is constant.
Now to the point (sorry about the nitpicking). What you say (as far
as I get it) is that charges are quantized (the "integer" bit isn't
important here) one cannot, using a finite number of charges, generate
a constant potential, i.e. a zero E field, throughout space. Yes, I
fully agree. You can mathematically prove that using a finite number
of discrete charges you can have zero E field only in a finite number
of points.
>
>How would you answer the question
>posed by the previous poster,
>considering that charges come only
>in integer amounts, and that space
>is more than one dimension?
>
Oh, the above is true even in one dimension. Even along a line there
is still an infinity of points and the zero field condition can only
be satisfied on a finite number of points (given a finite number of
charges).
b.h.j...@hw.ac.uk
On 1998-05-04 me...@cars3.uchicago.edu said:
In article <6ikktq$794$1...@cn1.hw.ac.uk , b.h.j...@hw.ac.uk writes:
On 1998-05-04 me...@cars3.uchicago.edu said:
An electrostatic potential is simply defined as the scalar
function the gradient of which is the electric field. Thus it
is defined only up to a constant. It is a standard and
convenient choice to fix this constant so that the value of
the potential at infinity converges to zero. But that's just
a matter of convenience, not of actual physical content.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing
just the same"
Potential at a point, measured in volts, is defined as the work
needed to bring unit + charge (1 coulomb) to that point from
infinity.
Actually, no. It is the work needed to bring unit charge to that
point from a chosen reference point. Said reference point is
usually chosen as infinity, but it doesn't have to be. It is all a
matter of convenience.
Infinity I like, because it is the same whatever your frame of
reference. Which makes it unnecessary to alarm students with
frames of reference. Nor even to mention scalars.
Indeed. That's why infinity is customarily used, when it can be
used. Now, here is a little electrostatics problem for you. Given
an infinite, uniformly charged plane (I trust you know that this
simply yields a uniform field), find the potential in the vicinity
of the plane based on a reference point at infinity.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the
same"
Er, I don't _always_ fall down the same hole twice!
============ ===== ===== BILL J. ===== ===== ============
GM8APX, qthr Edinburgh, Scotland, UK
Memoria minuitur, nisi eam exerceas
me...@cars3.uchicago.edu
In article <6ilbhg$fe0$1...@cn1.hw.ac.uk>, b.h.j...@hw.ac.uk writes:
>
>On 1998-05-04 me...@cars3.uchicago.edu said:
> Indeed. That's why infinity is customarily used, when it can be
> used. Now, here is a little electrostatics problem for you. Given
> an infinite, uniformly charged plane (I trust you know that this
> simply yields a uniform field), find the potential in the vicinity
> of the plane based on a reference point at infinity.
>
>
most possible holes, once, and remembers them well enough not to do it
again. You're doing fine.
Ray Tomes
In article <EsDD0...@midway.uchicago.edu>, me...@cars3.uchicago.edu
wrote:
>rto...@kcbbs.gen.nz (Ray Tomes) writes:
>>.. if the distribution of charge about a point is exactly balanced
>>(e.g all made of of non-ionised atoms) then the potential at that point
>>must be zero, mustn't it?
>No.
Counterexample please.
[Please send reply by email also, as I will be away for the next 2 weeks
and will probably miss it otherwise]
me...@cars3.uchicago.edu
In article <354e53f9...@aklobs.org.nz>, rto...@kcbbs.gen.nz (Ray Tomes) writes:
>In article <EsDD0...@midway.uchicago.edu>, me...@cars3.uchicago.edu
>wrote:
>
>>rto...@kcbbs.gen.nz (Ray Tomes) writes:
>>>.. if the distribution of charge about a point is exactly balanced
>>>(e.g all made of of non-ionised atoms) then the potential at that point
>>>must be zero, mustn't it?
>
>>No.
>
>Counterexample please.
>
constant, say 42 (in whatever units). Is it consistent with Maxwell's
equations? Yes.
James Logajan
me...@cars3.uchicago.edu wrote:
: In article <354e53f9...@aklobs.org.nz>, rto...@kcbbs.gen.nz (Ray Tomes) writes:
: >In article <EsDD0...@midway.uchicago.edu>, me...@cars3.uchicago.edu
: >wrote:
: >
: >>rto...@kcbbs.gen.nz (Ray Tomes) writes:
: >>>.. if the distribution of charge about a point is exactly balanced
: >>>(e.g all made of of non-ionised atoms) then the potential at that point
: >>>must be zero, mustn't it?
: >
: >>No.
: >
: >Counterexample please.
: >
: Take a space with no charges at all. Define the potential there as a
: constant, say 42 (in whatever units). Is it consistent with Maxwell's
: equations? Yes.
But not consistent with QM. The constant becomes a phase difference. While
that is normally not observable, one can devise an experiment to detect the
phase difference.
See for example the classic experiment "Significance of Electromagnetic
Potentials in the Quantum Theory" by Y. Aharonov and D. Bohm, Phys. Rev. 115,
485-491 (1959).
Ray Tomes
In article <EsG5z...@midway.uchicago.edu>, me...@cars3.uchicago.edu
wrote:
>In article <6ikktq$794$1...@cn1.hw.ac.uk>, b.h.j...@hw.ac.uk writes:
>>Potential at a point, measured in volts, is defined as the work needed
>>to bring unit + charge (1 coulomb) to that point from infinity.
Sounds good to me.
>Actually, no. It is the work needed to bring unit charge to that
>point from a chosen reference point. Said reference point is usually
>chosen as infinity, but it doesn't have to be. It is all a matter of
>convenience.
It isn't all a matter of convenience. If you apply accepted theory to
many situations it is quite possible to determine an absolute potential.
Your repeated statement that "potential" means "potential difference" is
contradicted by the need to change the words. It may be inconvenient to
establish absolute potentials, but it can be done and leads to
interesting results.
Tom Potter
Ray Tomes <rto...@kcbbs.gen.nz> wrote in article
<3555efe3...@aklobs.org.nz>...
> In article <EsG5z...@midway.uchicago.edu>, me...@cars3.uchicago.edu
> wrote:
>
> >In article <6ikktq$794$1...@cn1.hw.ac.uk>, b.h.j...@hw.ac.uk writes:
> >>Potential at a point, measured in volts, is defined as the work needed
> >>to bring unit + charge (1 coulomb) to that point from infinity.
>
> Sounds good to me.
>
> >Actually, no. It is the work needed to bring unit charge to that
> >point from a chosen reference point. Said reference point is usually
> >chosen as infinity, but it doesn't have to be. It is all a matter of
> >convenience.
>
> It isn't all a matter of convenience. If you apply accepted theory to
> many situations it is quite possible to determine an absolute potential.
> Your repeated statement that "potential" means "potential difference" is
> contradicted by the need to change the words. It may be inconvenient to
> establish absolute potentials, but it can be done and leads to
> interesting results.
Also if you apply "accepted theory",
the universe is finite,
and folds back onto itself,
so how does infinity get in here?
me...@cars3.uchicago.edu
In article <3555efe3...@aklobs.org.nz>, rto...@kcbbs.gen.nz (Ray Tomes) writes:
>In article <EsG5z...@midway.uchicago.edu>, me...@cars3.uchicago.edu
>wrote:
>
>>In article <6ikktq$794$1...@cn1.hw.ac.uk>, b.h.j...@hw.ac.uk writes:
>>>Potential at a point, measured in volts, is defined as the work needed
>>>to bring unit + charge (1 coulomb) to that point from infinity.
>
>Sounds good to me.
>
>>Actually, no. It is the work needed to bring unit charge to that
>>point from a chosen reference point. Said reference point is usually
>>chosen as infinity, but it doesn't have to be. It is all a matter of
>>convenience.
>
>It isn't all a matter of convenience. If you apply accepted theory to
>many situations it is quite possible to determine an absolute potential.
All you say is that it is possible to pick a single reference point
and relate everything to it. It is also possible to pick any other
reference point, though. Why won't you start with Maxwell's equations
and see what they say on the topic.
Niels M. Møller
On Tue, 28 Apr 1998 14:51:10 GMT, Peter Berdeklis
<pe...@atmosp.physics.utoronto.ca> wrote:
>In clear skies the "fair-weather field" is approx. 100 - 200 V/m at the
>ground, but the field drops off non-linearly with altitude. The field is
>created by the potential difference between the ionosphere (pos. charged)
>and the ground (neg. charged), which is maintained by global thunderstorm
>activity.
I've known about this effect for some years, but I've always wondered,
if this really means, that taking an insulated wire, with de-insulated
ends, of say 10 meters and sticking it up in the sky, would produce a
difference in voltage of 1000 V, that can be used to produce a spark
in contact with the ground down where you're holding it?
/Greetings
Niels M. Møller, 1.g stud. art, Denmark
(niels_...@bigfoot.com)
Niels M. Møller
On Tue, 19 May 1998 15:07:21 GMT, niels_...@bigfoot.com (Niels M.
Møller) wrote:
Hmm...I kinda feel quite silly, posting a follow-up-message to my own
mail... ;-)
>I've known about this effect for some years, but I've always wondered,
>if this really means, that taking an insulated wire, with de-insulated
>ends, of say 10 meters and sticking it up in the sky, would produce a
>difference in voltage of 1000 V, that can be used to produce a spark
>in contact with the ground down where you're holding it?
(Warning. Here comes a pretty sick experiment of thought...)
...and does this mean, that a person who is say around 2 m tall, if he
has got an open wound - or even better a nail or acupuncture needle -
in his feet and head (to insure proper galvanic conact to inner body
fluids), will feel a faint current, resulting from the delta-U of 200
V, running trough him??
(And even sicker...)
Can you electricute yourself by securing a grounded wire to your hand
and jump around a bit?