Re: Electrostatic Field Question
- From: John C. Polasek <jpolasek@xxxxxxxxxx>
- Date: Wed, 05 Jul 2006 21:42:15 GMT
On Wed, 05 Jul 2006 15:18:27 -0400, chuck <nospam@xxxxxxxxxx> wrote:
John C. Polasek wrote:
On Tue, 04 Jul 2006 10:01:24 -0400, chuck <nospam@xxxxxxxxxx> 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.John Polasek
Chuck
http://www.dualspace.net
----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
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
.
- References:
- Electrostatic Field Question
- From: chuck
- Re: Electrostatic Field Question
- From: John C . Polasek
- Re: Electrostatic Field Question
- From: chuck
- Electrostatic Field Question
- Prev by Date: Re: Electrostatic Field Question
- Next by Date: Re: Electrostatic Field Question
- Previous by thread: Re: Electrostatic Field Question
- Next by thread: Radiation pressure -photon momentum
- Index(es):
Relevant Pages
|
