Re: Disectable Capacitor

[Sam Sampere <sampere@SUHEP.PHY.SYR.EDU>]

John S. Denker wrote:

> Hi Folks --
>
> Here's what may be going on in the Cenco dissectable capacitor.  (There's
> no money-back guarantee, since I've never seen the device in question, but
> this hypothesis is easy to check.)
>
>
>
> 3) In the dissectable capacitor, the relationship of the insulator to the
> conductors is, I believe, a distraction, so consider the following modified
> version:
>
>         P i Q   R--------R   Q i P
>         P i Q   R--------R   Q i P
>         P i Q   R--------R   Q i P
>         P i Q   R--------R   Q i P
>         P i Q   R--------R   Q i P
>         P i Q   R--------R   Q i P
>         P i Q   R--------R   Q i P
>         P i Q   RRRRRRRRRR   Q i P
>         P i Q                Q i P
>        P i QQQQQQQQQQQQQQQQQQ i P
>        P iiiiiiiiiiiiiiiiiiiiii P
>         PPPPPPPPPPPPPPPPPPPPPPPPPP
>
> where all the capital letters are metal, and (i) is the insulator.  The
> metal "liner" (Q) is something I've added to clarify the situation.
>
> So....
>
> a) Imagine that the inner plug (R) has been removed.
>
> b) Further imagine that the (PQ) capacitor has a large charge on it.  This
> is, ironically, in compliance with the Kirchhoff law that says all charges
> should be confined inside capacitors.  Therefore the usual checks for
> "static electricity" will indicate zero unbalanced charge.
>
> c) Replace the plug (R).  If the (QR) gap is small, or better yet if (Q)
> and (R) are in solid electrical contact, (R) now becomes one terminal of
> the capacitor.  If you short-circuit from (P) to (R), then a goodly WHAM is
> to be expected.

I'm confused.  I'm not sure what you just accomplished with this experiment.
If Q an R are not touching, then touching P to R will charge R. Even if there
are no charges on P, R will be charged by induction.  Since P has excess
charge, that charge will be shared with the uncharged R, even if Q in neutral.
So now P and Q are oppositely charged, touching P to R will certainly charge R
for the above reasons, induction and charge sharing, right? If R touches P, the
R IS P, it's part of the capacitor.

Later,

Sam

--
Sam Sampere
Syracuse University
Department of Physics
Syracuse, NY 13244
315-443-5999