Re: Capacitor charge, were is it?

To: tesla-at-pupman-dot-com
Subject: Re: Capacitor charge, were is it?
From: "Fred W. Bach, TRIUMF Operations" <music-at-triumf.ca>
Date: Fri, 08 Nov 1996 18:06:06 PST
Cc: music-at-triumf.ca
>Message-ID: <199611070541.WAA29812-at-poodle.pupman-dot-com>
>Date: Wed, 6 Nov 1996 22:41:55 -0700
>From: Tesla List <tesla-at-poodle.pupman-dot-com>
>To: Tesla-list-subscribers-at-poodle.pupman-dot-com
>Subject: Re: Capacitor charge, were is it?

>From hullr-at-whitlock-dot-comWed Nov  6 22:32:49 1996
>Date: Wed, 06 Nov 1996 16:17:27 -0800
>From: Richard Hull <hullr-at-whitlock-dot-com>
>To: tesla-at-pupman-dot-com
>Subject: Re: Capacitor charge, were is it?
>

>Fred ,
>
>I would agree that absolutely dry de-ionized air merely introduced or 
>mixed with the inside air would never discharge the electroscope or 
>dosimeter.  Too bad this stuff doesn't exist on planet earth without 
>man's intervention.  The charge within the electroscope or or the 
>dosimeter is retained by the air molecules only at the interfacial point 
>of the metal and the air.  The great volume of air in the vessel is not 
>charged!  Coulombic forces hold these interfacial molecules tightly to 
>the metal as separated charge is wont to do.  A lot of physical force or 
>electrical neutralization of the thin layer of air at this junction would 
>be required to dirupt or move this charge.  Very high speed air would do 
>this, dry or not.   All that would be necessary is to overcome the 
>coulombic force.  At molecular ranges this can be tremendous!

   Sounds good to me!  I feel better after this explanation.


>A neat experiment which is isultrative of this interfacial nature of 
>charge retention of dielectrics is to insert three pieces of mylar in a 
>disectable capacitor.  Charge it.  (I've done this by the way)
>
>Now disect the capacitor.  pass all of the sheets around the room.  roll 
>em up, stomp on them, etc.  Now gather them up.  Take the top sheet and 
>reassemble the capacitor with this sheet only.  Discharge it.  POW!  
>Now take the bottom sheet and do the same. POW!  Now the middle 
>sheet...Zippo, goose egg, zilch!  Charge is still 100% retained by the 
>dielectric but only at the outer layers where the differing dielectric 
>constants are found and not within the bulk dielectric.  The charge is 
>there regardless of the presence of a metal plate.

   What happens if people put their wet fingers around the charged
   piece (from one side to the other) and slide their fingers all up
   and down the mylar so that every square millimeter of the surface
   has been discharged by their fingers?  (Probably what I would have
   done if I were at your 'capacitor party'.)  Anyway it sounds like
   a wild idea for a party!


>  This is true in the vacuum of space, in liquids or solids.  The vacuum 
>seems to form the "tightest" interfacial bond. (though charge energy per 
>unit space is very low)  So tight, in fact, that it looks as if the metal 
>is indeed charged.

   [ snip ]

   Yes, I would like very much to explore this particular case because
   I believe there is a fine point to be learned here about vacuum
   capacitors being a special case which is bothering me about the
   electrons and what they are attracted/bound to.

   Earlier we agreed on the value of the electric field at the ends of
   a metal rod inside a chamber with voltage V across the chamber.  V
   is the applied voltage and D is the distance between the walls or
   plates.  There is a metal bar in the middle of the field but its
   ends do not touch either plate.  The field away from the rod is
   pretty well V / D in volts per meter, but the field at the rod ends
   is much higher because the rod effectively makes D shorter at that
   location (since there can be no voltage field across a conductor
   says Maxwell).

   Now what would happen if we cut that rod in two, put a very tiny
   microammeter between the two half-rods and connected the leads of
   the microammeter to the two half-rods such that the ammeter was now
   in series with the rods, and then we turned on the field V.  We
   agree that in the steady state there is no current flow in the rod. 
   But what would the meter read if we started with the field off and
   the we turned it on?  Would the meter ever read any current?  Would
   the current start and flow in one direction and stop, or would the
   current turn around and flow back the other way?  Would there be
   any net charge transferred through the microammeter if we left the
   voltage on?

   Regards,
 
 Fred W. Bach ,    Operations Group        | Internet: music-at-triumf.ca
 TRIUMF (TRI-University Meson Facility)    | Voice:  604-222-1047 loc 6327/7333
 4004 WESBROOK MALL, UBC CAMPUS            | FAX:    604-222-1074
 University of British Columbia, Vancouver, B.C., CANADA   V6T 2A3
 "Accuracy is important. Details can mean the difference between life & death."
 These are my opinions, which should ONLY make you read, think, and question.
 They do NOT necessarily reflect the views of my employer or fellow workers.