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Re: dissectible capacitor



On Fri, 2 Jun 2000, Carl E. Mungan wrote:

Bill wrote:

Even if the charge was immobilised upon the
dielectric surfaces, the metal plates would be charged electrophorus-like
by induction, and would act just like a conventional charged capacitor
after reassembly.

In order to generate a net charge, the electrophorus requires that
you touch a grounded wire to the outer surface of the metal plate
after contacting it with the dielectric.

Certainly. And when you do this, you produce a spark... AS IF the
neutral plate was already charged.

Here we are not doing this. Hence there is no net charge on the metal
plate. True there is a charge separation between the two surfaces.
But in the capacitor equation V=Q/C,

I think this equation is a bit too simple to describe the situation.

Q is the net charge on the
plate. As I see it, charge separation is not good enough.

As far as any external circuits are concerned, charge separation is
everything: by touching the electrophorus plate to ground we PRODUCED A
SPARK.

It's not wise to focus on theory while ignoring realworld phenomenon. A
very large electrophorus could kill you with its initial "charge
separation" spark to ground, and you would be just as dead as if you had
touched a quantity of net charge.


So in Leigh's experiment, just as net charge is transferred from the
plate to the glass during the charging (But why does this occur?), so
net charge must be transferred back during the reassembly. (Again,
why?)

It looks to me as if net charge is transferred to the glass during the
charging, but might not be transferred back again. If a neutral plate is
placed against the charged glass, then the neutral plate will develop some
separated charge by induction, and it can create sparks when touched with
a wire. I guess we could call this a "double sided electrophorus" since a
second metal plate is present.

My earlier "spray combs" idea was apparantly way off, and Leigh finds that
the charge-spray effect occurs during charging. Here's a way to visualize
it: there are three capacitors in series. Leigh's capacitor has some air
between the glass and the metal plates, and if this air should break down,
a plane of charge can be deposited upon the glass surface. If we see the
surface of the glass as being a capacitor plate, we have:

|
|
-------------
AAAAAAAAAAAAA Air
-------------
GGGGGGGGGGGGG
GGGGGGGGGGGGG
GGGGGGGGGGGGG
GGGGGGGGGGGGG
GGGGGGGGGGGGG Glass
GGGGGGGGGGGGG
GGGGGGGGGGGGG
GGGGGGGGGGGGG
GGGGGGGGGGGGG
-------------
AAAAAAAAAAAAA Air
-------------
|



...or better yet, this:


|
|
------------- C1 = air dielectric
-------------
|
|
------------- C2 = glass plate
GGGGGGGGGGGGG
-------------
|
|
------------- C3 = air dielectric
-------------
|
|


C1 and C3 both have very small gaps. When the composite capacitor is
connected to a power supply, C1 and C3 suffer breakdowns, and charge is
deposited on the plates of C2. (Question: do the two surfaces of the
glass plate receive charges which are approximately equal and opposite, or
is there a significant net charge on the glass.)

If the metal plates are then removed, C2 remains, and is still charged.
If the metal plates are discharged to ground and then replaced, C1 and C3
would each have zero potential, which means that a large potential would
appear across the external terminals. Hey, this suggests that during any
subsequent shorting of the external terminals, C1 and C3 might break down
again. If Leigh could look into the edge of the glass plate, he might see
a flash of light from breakdown of the air layers during discharge. If
this occurs, then the two glass surfaces might end up neutral at the end
of the demonstration.


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William J. Beaty SCIENCE HOBBYIST website
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