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Electrostatic Field Question
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j.alexander.hawkins@gmail
science forum beginner


Joined: 28 Jun 2006
Posts: 5

PostPosted: Sat Jul 08, 2006 2:36 am    Post subject: Re: Electrostatic Field Question Reply with quote

http://www.ee.nmt.edu/~langmuir/E100/E100.html
chuck wrote:
Quote:
John, do a search on "electrometer" or
"fieldmeter" or "E-field probe" or
"field mill" and I think you'll find a
lot of non-mechanical measurement
techniques. The field mill uses a
rotating "chopper" but that is not
really a transducer, but rather a device
to convert a DC voltage to AC for
"easier amplification".


Chuck

chuck,

quite right. a very clever trick. i saw some meteorologists (?) sight,
they gave their setup and schematic for there meausement... check out.
http://www.ee.nmt.edu/~langmuir/E100/E100.html. the AC part is
neccesary i think though.

john
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Joseph S. Powell, III
science forum addict


Joined: 23 May 2005
Posts: 82

PostPosted: Fri Jul 07, 2006 4:20 pm    Post subject: Re: Electrostatic Field Question Reply with quote

Quote:
A very high impedance electrometer will
indicate this voltage, but presumably, a
10 Mohm DMM will not due to excessive
loading.
I did a, (by no means exhaustive) search on it over google, and all of
the ones ive seen use force, displacement etc, mechanical
meausremtns...and transducers etc. to give you a voltage. all of them
ive seen have to measure some mechanical displacement, and then give
you a voltage.
snip


John, do a search on "electrometer" or
"fieldmeter" or "E-field probe" or
"field mill" and I think you'll find a
lot of non-mechanical measurement
techniques. The field mill uses a
rotating "chopper" but that is not
really a transducer, but rather a device
to convert a DC voltage to AC for
"easier amplification".


Chuck

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John C. Polasek
science forum Guru


Joined: 30 Apr 2005
Posts: 321

PostPosted: Wed Jul 05, 2006 9:42 pm    Post subject: Re: Electrostatic Field Question Reply with quote

On Wed, 05 Jul 2006 15:18:27 -0400, chuck <nospam@nospam.net> wrote:

Quote:
John C. Polasek wrote:
On Tue, 04 Jul 2006 10:01:24 -0400, chuck <nospam@nospam.net> wrote:

My question has to do with actual
measurements of a homogeneous
electrostatic field of E V/m in air.

A voltage is observed between two
parallel plates in the field. Assume the
capacitance of these plates is 1 pF and
their separation is 0.1 m. I expect the
voltage to be 0.1V.

You have a lot to learn about capacitors and you won't learn it in cgs
electrostatics with charges on plates. By what equation would you
expect the voltage to be 0.1V?

Is not the very meaning of an electric
field of 1 volt per meter that the
voltage between two conductors in that
field is 1 volt x their separation in
meters?

One volt/meter is 0.1 volt/0.1 meter by
definition, is it not?

The data of 1pF and .1m allows one only to calculate the area of the
plates. Nothing allows you to guess voltage.

The voltage was not guessed, of course.
The exercise here is to be able to
estimate the error in measuring electric
field intensity using a voltmeter of
some internal resistance Rm.

I have assumed the area of the plates is
such that with a separation of 0.1
meter, their capacitance is 1 pF.

The real action in capacitors is that the dielectric even vacuum is
stressed and polarized so that Q = DxA.

A very high impedance electrometer will
indicate this voltage, but presumably, a
10 Mohm DMM will not due to excessive
loading. While this idea appears often,
the only numbers usually presented are
the 10(12) ohm input impedance of the
electrometer and the 10(7) ohm input
impedance of the DMM.

Your 1pF and.1V equals 6 million electrons. That's nearly nothing!


What difference does it make? Would you
be happier with 10 pF, or 100? Would the
theory care?

My question is this: how do I determine
the "internal resistance" of the two
plates in the electric field? How can I
know the quantitative effect of
different voltmeter loadings on the
measurement (other than "less loading is
obviously better")?

There is no internal resistance of the plates. When you connect a
resistive load like the DMM, you have a time constant RC. The DMM will
give you a 10 usec time constant; in 10 usec the reading will have
dropped to 37% or the true value. V = V1*e^-t/RC.

Yes, I did that calculation. And
"internal resistance" is in quotes. But
the plates are being "charged" (also in
quotes) by the electric field. Else,
what is the source of the energy in the
capacitor?

If a capacitor is connected directly to
a constant voltage source, you really
can't compute a time constant. If the
two plates in my example are being
"charged" by the electric field, then is
the time constant you calculate that of
the plates in isolation or "in parallel
with" the electric field that is
"charging" them.


Knowing the charge on the plates, it
seems that the DMM will discharge the
capacitor in microseconds. What is the
time rate at which the electric field
replenishes the charge on the plates?

What electric field?

The electric field that is providing
charge to the capacitor.

The field and charge are proportional.

Of course.

Youhave to learn about capacitors. The charge out of the battery is
stored in the polarization in the medium. The plates are merely
electric handles that touch the whole face of the dielectric.

Maybe the question can be stated
differently. When a voltmeter (Rm
internal resistance) is connected to two
plates in an electric field, electric
energy from the field is converted to
heat energy in the voltmeter. What
determines the rate at which energy can
be delivered from the field to the
voltmeter?

Obviously, the more intense the electric
field, and the greater the area of the
plates, the greater the rate at which
energy can be delivered.

The rate at which a capacitor in
isolation can be discharged into the
meter is known. What is the rate at
which the field is "charging" that
capacitor? As the voltmeter is
discharging the capacitor, the field is
simultaneously "charging" it.

The field of interest, FWIW is the
earth's fairweather field of
approximately 100 V/m. Surely the
physics will apply to a field of 1
volt/meter.

I appreciate your efforts to enlighten
me. ;-)

Chuck

Incidentally the Wikipedia article I saw is couched in cgs terms and
you will learn nothing from it.
Use a text that uses SI units and recognizes eps0 as the real
permittivity of the vacuum.

Thanks for any help or suggestions.

Chuck
John Polasek
http://www.dualspace.net

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Why did I go to all that work when you were holding out on me? Now you
say the earth has an innate field of 100V/m and you are going to stick
two plates out, connect a voltmeter and you want to know what's going
to happen. I don't know.
The best way to find a potential without draindown is to keep trying
out different voltages which you repeatedly connect to the plates
through a very sensitive current detector and find one where the
deflection is zero.
You have some research to do.

John P
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Joseph S. Powell, III
science forum addict


Joined: 23 May 2005
Posts: 82

PostPosted: Wed Jul 05, 2006 7:18 pm    Post subject: Re: Electrostatic Field Question Reply with quote

John C. Polasek wrote:
Quote:
On Tue, 04 Jul 2006 10:01:24 -0400, chuck <nospam@nospam.net> wrote:

My question has to do with actual
measurements of a homogeneous
electrostatic field of E V/m in air.

A voltage is observed between two
parallel plates in the field. Assume the
capacitance of these plates is 1 pF and
their separation is 0.1 m. I expect the
voltage to be 0.1V.

You have a lot to learn about capacitors and you won't learn it in cgs
electrostatics with charges on plates. By what equation would you
expect the voltage to be 0.1V?

Is not the very meaning of an electric
field of 1 volt per meter that the
voltage between two conductors in that
field is 1 volt x their separation in
meters?

One volt/meter is 0.1 volt/0.1 meter by
definition, is it not?

Quote:
The data of 1pF and .1m allows one only to calculate the area of the
plates. Nothing allows you to guess voltage.

The voltage was not guessed, of course.
The exercise here is to be able to
estimate the error in measuring electric
field intensity using a voltmeter of
some internal resistance Rm.

I have assumed the area of the plates is
such that with a separation of 0.1
meter, their capacitance is 1 pF.

Quote:
The real action in capacitors is that the dielectric even vacuum is
stressed and polarized so that Q = DxA.

A very high impedance electrometer will
indicate this voltage, but presumably, a
10 Mohm DMM will not due to excessive
loading. While this idea appears often,
the only numbers usually presented are
the 10(12) ohm input impedance of the
electrometer and the 10(7) ohm input
impedance of the DMM.

Your 1pF and.1V equals 6 million electrons. That's nearly nothing!


What difference does it make? Would you
be happier with 10 pF, or 100? Would the
theory care?

Quote:
My question is this: how do I determine
the "internal resistance" of the two
plates in the electric field? How can I
know the quantitative effect of
different voltmeter loadings on the
measurement (other than "less loading is
obviously better")?

There is no internal resistance of the plates. When you connect a
resistive load like the DMM, you have a time constant RC. The DMM will
give you a 10 usec time constant; in 10 usec the reading will have
dropped to 37% or the true value. V = V1*e^-t/RC.

Yes, I did that calculation. And
"internal resistance" is in quotes. But
the plates are being "charged" (also in
quotes) by the electric field. Else,
what is the source of the energy in the
capacitor?

If a capacitor is connected directly to
a constant voltage source, you really
can't compute a time constant. If the
two plates in my example are being
"charged" by the electric field, then is
the time constant you calculate that of
the plates in isolation or "in parallel
with" the electric field that is
"charging" them.

Quote:

Knowing the charge on the plates, it
seems that the DMM will discharge the
capacitor in microseconds. What is the
time rate at which the electric field
replenishes the charge on the plates?

What electric field?

The electric field that is providing
charge to the capacitor.

The field and charge are proportional.

Of course.

Quote:
Youhave to learn about capacitors. The charge out of the battery is
stored in the polarization in the medium. The plates are merely
electric handles that touch the whole face of the dielectric.

Maybe the question can be stated
differently. When a voltmeter (Rm
internal resistance) is connected to two
plates in an electric field, electric
energy from the field is converted to
heat energy in the voltmeter. What
determines the rate at which energy can
be delivered from the field to the
voltmeter?

Obviously, the more intense the electric
field, and the greater the area of the
plates, the greater the rate at which
energy can be delivered.

The rate at which a capacitor in
isolation can be discharged into the
meter is known. What is the rate at
which the field is "charging" that
capacitor? As the voltmeter is
discharging the capacitor, the field is
simultaneously "charging" it.

The field of interest, FWIW is the
earth's fairweather field of
approximately 100 V/m. Surely the
physics will apply to a field of 1
volt/meter.

I appreciate your efforts to enlighten
me. ;-)

Chuck

Quote:
Incidentally the Wikipedia article I saw is couched in cgs terms and
you will learn nothing from it.
Use a text that uses SI units and recognizes eps0 as the real
permittivity of the vacuum.

Thanks for any help or suggestions.

Chuck
John Polasek
http://www.dualspace.net

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John C. Polasek
science forum Guru


Joined: 30 Apr 2005
Posts: 321

PostPosted: Wed Jul 05, 2006 3:57 pm    Post subject: Re: Electrostatic Field Question Reply with quote

On Tue, 04 Jul 2006 10:01:24 -0400, chuck <nospam@nospam.net> wrote:

Quote:
My question has to do with actual
measurements of a homogeneous
electrostatic field of E V/m in air.

A voltage is observed between two
parallel plates in the field. Assume the
capacitance of these plates is 1 pF and
their separation is 0.1 m. I expect the
voltage to be 0.1V.

You have a lot to learn about capacitors and you won't learn it in cgs
electrostatics with charges on plates. By what equation would you
expect the voltage to be 0.1V?
The data of 1pF and .1m allows one only to calculate the area of the
plates. Nothing allows you to guess voltage.
The real action in capacitors is that the dielectric even vacuum is
stressed and polarized so that Q = DxA.

Quote:
A very high impedance electrometer will
indicate this voltage, but presumably, a
10 Mohm DMM will not due to excessive
loading. While this idea appears often,
the only numbers usually presented are
the 10(12) ohm input impedance of the
electrometer and the 10(7) ohm input
impedance of the DMM.

Your 1pF and.1V equals 6 million electrons. That's nearly nothing!

Quote:
My question is this: how do I determine
the "internal resistance" of the two
plates in the electric field? How can I
know the quantitative effect of
different voltmeter loadings on the
measurement (other than "less loading is
obviously better")?

There is no internal resistance of the plates. When you connect a
resistive load like the DMM, you have a time constant RC. The DMM will
give you a 10 usec time constant; in 10 usec the reading will have
dropped to 37% or the true value. V = V1*e^-t/RC.

Quote:
Knowing the charge on the plates, it
seems that the DMM will discharge the
capacitor in microseconds. What is the
time rate at which the electric field
replenishes the charge on the plates?

What electric field? The field and charge are proportional.
Youhave to learn about capacitors. The charge out of the battery is
stored in the polarization in the medium. The plates are merely
electric handles that touch the whole face of the dielectric.
Incidentally the Wikipedia article I saw is couched in cgs terms and
you will learn nothing from it.
Use a text that uses SI units and recognizes eps0 as the real
permittivity of the vacuum.

Quote:
Thanks for any help or suggestions.

Chuck
John Polasek

http://www.dualspace.net
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j.alexander.hawkins@gmail
science forum beginner


Joined: 28 Jun 2006
Posts: 5

PostPosted: Wed Jul 05, 2006 6:12 am    Post subject: Re: Electrostatic Field Question Reply with quote

chuck,
i think you got the DMM thing wrong. you cant measure a static E field
with a DMM
chuck wrote:
Quote:
My question has to do with actual
measurements of a homogeneous
electrostatic field of E V/m in air.

A voltage is observed between two
parallel plates in the field. Assume the
capacitance of these plates is 1 pF and
their separation is 0.1 m. I expect the
voltage to be 0.1V.

yes, but i dont think you can read the voltage with a DMM... to use a

dc circuit analogy, look at it this way....
________________
| == Cupper
| |
+ |
V === C test R load in paralell
- |
| |
| === Clower
| |
|________________|

you switched Ctest in with say V(0)=0, the dc analogy kind of gets
more convoluted, but in any event, no charge can be supplied to your
test capacitor by the generator...
the only charge moving was to cancel the E field inside the conductor.
but no net charge is delivered by the generator...(the source that
generates your electrostatic field)
- - - - - -
\\\\\\\\\\\ E=0
++++++


- - - - - - -
\\\\\\\\\\\\\ E=0
++++++

but that it. you have potential to do mechanical work, the plates want
to move to each other, etc. but since no net charge is supplied to the
cap, it can drive anything, no matter the resistance, i.e. a DMM.
Quote:
A very high impedance electrometer will
indicate this voltage, but presumably, a
10 Mohm DMM will not due to excessive
loading.
I did a, (by no means exhaustive) search on it over google, and all of

the ones ive seen use force, displacement etc, mechanical
meausremtns...and transducers etc. to give you a voltage. all of them
ive seen have to measure some mechanical displacement, and then give
you a voltage.
<snip>
Quote:
What is the
time rate at which the electric field
replenishes the charge on the plates?

that my big point/problem with the claim a better DMM can measure the

Efield over the cap... it cant replenish charge..
Quote:
Thanks for any help or suggestions.

Chuck
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Don Kelly
science forum Guru Wannabe


Joined: 30 Apr 2005
Posts: 166

PostPosted: Wed Jul 05, 2006 4:18 am    Post subject: Re: Electrostatic Field Question Reply with quote

"chuck" <nospam@nospam.net> wrote in message
news:1152022590_3907@sp6iad.superfeed.net...
Quote:
chuck wrote:
My question has to do with actual measurements of a homogeneous
electrostatic field of E V/m in air.

Sorry. I left out my assumption that E= 1 V/m.


A voltage is observed between two parallel plates in the field. Assume
the capacitance of these plates is 1 pF and their separation is 0.1 m. I
expect the voltage to be 0.1V.

A very high impedance electrometer will indicate this voltage, but
presumably, a 10 Mohm DMM will not due to excessive loading. While this
idea appears often, the only numbers usually presented are the 10(12) ohm
input impedance of the electrometer and the 10(7) ohm input impedance of
the DMM.

My question is this: how do I determine the "internal resistance" of the
two plates in the electric field? How can I know the quantitative effect
of different voltmeter loadings on the measurement (other than "less
loading is obviously better")?

Knowing the charge on the plates, it seems that the DMM will discharge
the capacitor in microseconds. What is the time rate at which the
electric field replenishes the charge on the plates?

Thanks for any help or suggestions.

Chuck

Consider the extremes. If you have no voltmeter and the field is uniform as

you say,so the plates are on equipotential curves and do not affect the
field, then the voltage will be 0.1V. Now short the plates (voltmeter
resistance =0) Voltage between plates is 0 , but the field external to the
plates, with the original source, must now be somewhat greater than 1V/m.
With a real voltmeter you will be somewhere in between- external field rises
and field between plates drops. The internal resistance of the plates
doesn't appear to be a factor. You could consider that there is a
capacitive divider with one of the capacitors in parallel with a resistive
load.
--

Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
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Joseph S. Powell, III
science forum addict


Joined: 23 May 2005
Posts: 82

PostPosted: Tue Jul 04, 2006 2:24 pm    Post subject: Re: Electrostatic Field Question Reply with quote

chuck wrote:
Quote:
My question has to do with actual measurements of a homogeneous
electrostatic field of E V/m in air.

Sorry. I left out my assumption that E=
1 V/m.

Quote:

A voltage is observed between two parallel plates in the field. Assume
the capacitance of these plates is 1 pF and their separation is 0.1 m. I
expect the voltage to be 0.1V.

A very high impedance electrometer will indicate this voltage, but
presumably, a 10 Mohm DMM will not due to excessive loading. While this
idea appears often, the only numbers usually presented are the 10(12)
ohm input impedance of the electrometer and the 10(7) ohm input
impedance of the DMM.

My question is this: how do I determine the "internal resistance" of the
two plates in the electric field? How can I know the quantitative effect
of different voltmeter loadings on the measurement (other than "less
loading is obviously better")?

Knowing the charge on the plates, it seems that the DMM will discharge
the capacitor in microseconds. What is the time rate at which the
electric field replenishes the charge on the plates?

Thanks for any help or suggestions.

Chuck
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Joseph S. Powell, III
science forum addict


Joined: 23 May 2005
Posts: 82

PostPosted: Tue Jul 04, 2006 2:01 pm    Post subject: Electrostatic Field Question Reply with quote

My question has to do with actual
measurements of a homogeneous
electrostatic field of E V/m in air.

A voltage is observed between two
parallel plates in the field. Assume the
capacitance of these plates is 1 pF and
their separation is 0.1 m. I expect the
voltage to be 0.1V.

A very high impedance electrometer will
indicate this voltage, but presumably, a
10 Mohm DMM will not due to excessive
loading. While this idea appears often,
the only numbers usually presented are
the 10(12) ohm input impedance of the
electrometer and the 10(7) ohm input
impedance of the DMM.

My question is this: how do I determine
the "internal resistance" of the two
plates in the electric field? How can I
know the quantitative effect of
different voltmeter loadings on the
measurement (other than "less loading is
obviously better")?

Knowing the charge on the plates, it
seems that the DMM will discharge the
capacitor in microseconds. What is the
time rate at which the electric field
replenishes the charge on the plates?

Thanks for any help or suggestions.

Chuck
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