Re: TC Electrostatics (fwd)

To: tesla-at-pupman-dot-com
Subject: Re: TC Electrostatics (fwd)
From: Ed Harris <harris-at-parkave-dot-net>
Date: Mon, 21 Oct 1996 23:59:51 -0400
Subscriber: harris-at-parkave-dot-net Wed Jan 1 21:51:17 1997
    [The following text is in the "ISO-8859-1" character set]
    [Your display is set for the "US-ASCII" character set]
    [Some characters may be displayed incorrectly]



> Subscriber: hullr-at-whitlock-dot-com Tue Dec 31 22:31:30 1996
> Date: Tue, 31 Dec 1996 11:40:38 -0800
> From: Richard Hull <hullr-at-whitlock-dot-com>
> To: tesla-at-pupman-dot-com
> Cc: caydsi-at-aol-dot-com
> Subject: Re: TC Electrostatics (fwd)
> 
snip a bunch

> I have come to understand there is a vast difference between true 
> POTENTIAL (electrostatic only), static charge and forces as compared to 
> the voltage "pressure" found in a dynamic battery within a circuit.  Also

> the coulomb of the AMpere definition has no relation (not even the most 
> tenuous) to the coulomb of electrostatic charge!!  The difference is one 
> of current flow and magnetic field generation and the two seem unrelated 
> in spite of attempts to link them.  Yet, the temptation to link them 
> remains even in my mind.  I can find no electric field about a current 
> carryng wire.

Hi Richard,

There seems to be some fundamental difference in our perspectives.
Everything
you have reported about your experiments seems to be ok. But our 
interpertations is completely different. Your results are perfectly
consistent with 
my "old fashioned" understanding of electromagnetism. 

1 ampere of current though a wire will generate a magnetic field outside of
that wire, but any electric fields will only be parallel to the axis of the
wire
(since the chareges follow the electric field by ohms law). Now Richard is
asking why does this 1 coulomb per second of charge NOT generate a 
perceptible electric field around and perpendiclar to the wire axis just
as an "electrostatic" charge would.  Is does not in this case because no
NET charge is gained by the wire as in the electrostaic case. 1 coulomb per
second goes into the wire as 1 coulomb per second leaves the wire. 
Therefore there is no NET charge on the wire.

When the circuit is disconnected so that no current flows, the charges
can build up on opposite sides of the power supply just like if you
were charging  a capacitor. An electric field will also appear (just
as in the capacitor) between the oppositely charged wires. 

In both of these cases, the charges are just ordinary electrons. In one
cae the electrons move but the number of electrons in the wire
is exactly balanced by the number of stationary copper ions so there
is not net external electric field  - -- in the other case the chargind
current
of electrons is transient and simply leaves an excess of electrons as
compared to the copper ions in the (-) conductor and a deficit of electrons
on the (+) conductor. The final static condition has an electric field 
directed from the + to -  charges.

Bottom line is: A coulomb is a coulomb no matter if its moving
or not.

You are correct that there is subtle difference in types of "voltmeters". 
A "perfect" electrometer does not depend on a continuous transfer of
charge (infinite imperdance) so it can actually measure a voltage or
electric potential such as the built-in voltage across a diode. But
ordinary
DMMs for instance are really measuring what we physics types call 
"chemical-potential-difference" which requires charge transfer. This
kind of voltmeter has finite impedance and cannot measure a diode's 
built-in potential difference. Of course, the are different degrees of
voltmeter impedance so the line is a bit blurred as the impedances
go toward 10^15 ohm say.

 -Ed Harris