Re: Capacitor charge, were is it?

To: tesla-at-pupman-dot-com
Subject: Re: Capacitor charge, were is it?
From: Richard Hull <hullr-at-whitlock-dot-com>
Date: Mon, 11 Nov 1996 16:22:04 -0800
Tesla List wrote:
> 
> >From ed-at-alumni.caltech.eduSun Nov 10 21:48:25 1996
> Date: Sun, 10 Nov 1996 19:27:19 -0800 (PST)
> From: "Edward V. Phillips" <ed-at-alumni.caltech.edu>
> To: tesla-at-pupman-dot-com
> Subject: Re: Capacitor charge, were is it?
> 
> "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?"
>         At the time the field was applied there would be a
> transient current through the microammeter, since the two
> ends of the rod to which it is connect are, in effect, small
> capacitors coupled to the two plates.  After the initial transient
> those two capacitors are charged, so a net current had to flow.
>         On turnoff the same thing will happen, with the direction
> of current flow reversed.
>         (Above assumes that that I read your statement right and
> this is not a trick question which fooled me!  I'm all pooed
> out tonight.
> Ed


Ed,

I agree, the two istropic capacitors would react as you say.  This is a 
metallic circuit, so current could flow from one metal rod to the 
other during the polarization process within the dielectric.

Richard Hull, TCBOR