Re: Electrostatic Field Question

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?
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.

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!

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.

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.

Thanks for any help or suggestions.

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

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