Re: Electrostatic Induction

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"Vince Morgan" <vinharAtHereoptusnet.com.au> wrote in message
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"Don Kelly" <dhky@xxxxxxx> wrote in message
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"Vince Morgan" <vinharAtHereoptusnet.com.au> wrote in message
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As for your oscillating charged body - why do you think that this will
produce a current in the wire? --

Hi Don,
As an experiement I recently made a flat plate capacitor
from and old cd and aluminium foil. Passing a mildly statically charged
plastic tube (bic pen body) back and forth over the upper capacitor plate
induced an AC flow through an air cored inductor connected in series to
the
plates. There is no exchange of energy as the pen's static charge decayed
(apparently) as it would at rest on an insulated surface. I believe the
source of the energy is the field ascociated with the charge on the pen,
which if insulated sufficiently should decay at a rate that has little
relationship, if any at all, to it's deployment.

From what little I do know of the science of electrostatics, I believed
that
my result was to be expected.
A charged body will induce a charge seperation on a metal object, some
field
lines from the charged body will fall on those sepperated charges and in
doing so create an electrical potential.

Perhaps I am misunderstanding something here. That wouldn't be anything
new
;)
Highest regards,
Vince Morgan

Try it without the inductor but with a closed circuit-just a wire- see if
there is any current. Alternatively leave it "open circuit" but with a very
high impedance voltmeter across the capacitor - I would be curious about
your results. You are unlikely to make the rod move fast enough to make
the inductance a factor so its presence should not be necessary Small C and
small L means that with anything you can move manually, you are a long way
from any conditions where the inductance will be important.
Yes, as you predicted, you will have a change in the charge balance and a
current will flow in the external loop (which has a low impedance, quite
different from a large air gap in the device mentioned before -in which the
current in the air would be extremely small-possibly not measurable).

Another way of looking at it is as a two capacitor system in which one
capacitor is the rod-plate capacitor which you are varying. Using quite
standard energy balance analysis techniques it should be fairly simple to
show what is going on. Note that force in a capacitive system can be due to
changing charge on a fixed capacitor or changing capacitance with a fixed
charge. Output energy will be lost in resistance in the wire. Input energy
comes from you. The electrostatic system is simply a mechanical/electrical
conversion device supplying no energy from the charges. The force that you
apply to move the pen has a component due to the acceleration of the mass of
the rod and also a component in that you are modifying the electrostatic
conditions in the system. The magnetic equivalent is moving a magnet
through a coil- no net energy is removed from the magnetic field which
simply acts as a transfer mechanism. You could also get an effect with a
parallel plate air core capacitor and physically flexing it to change the
gap between plates. This will not take energy from the electrostatic
field -and is at the heart of some electrostatic transducers.

The Swiss device looks interesting but there is the same problem that occurs
with a lot of magnetic "free energy" machines -inadequate and inappropriate
instrumentation as proper instrumentation requires a good knowledge of what
is being measured and equipment well beyond that available at the nearest
Radio Shack. Essentially this machine appears to have a moving charge
situation and a charge in motion is a current which produces a varying
magnetic field coupled to a coil- presto -Faraday's Law. Move the charge
fast enough and n appreciable voltage can be induced in the coil. In
addition, as with most of these devices, there is a high frequency component
(rectified or not) and often resonance. Resonance seems great but the high
currents and or voltages do not translate into high power. The better the
resonance, the less the average input and output power (which is what is of
importance) involved.

Don Kelly dhky@xxxxxxxxxxxx
remove the X to answer





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