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Re: Real data from real experiments.
Tesla List wrote:
>
> >From dbell-at-baygate.bayarea-dot-netFri Nov 8 22:13:56 1996
> Date: Thu, 7 Nov 1996 22:31:30 -0800 (PST)
> From: Dave Bell <dbell-at-baygate.bayarea-dot-net>
> To: Tesla List <tesla-at-poodle.pupman-dot-com>
> Subject: Re: Real data from real experiments.
>
> Brent Turner asked:
>
> > two *identical value* caps in series yields a total capacitance
> > of 1/2 of the value. So, for a 1 uF cap, that gives us a total
> > value of 0.5 uF.
> >
> > I guess at this point, I can ask a related question:
> >
> > "When two identical capacitors are hooked in series, where does 1/2
> > of the capacitance go?"
> >
>
> Same place that "lost" potential energy went - absolutely nowhere!
> Not withstanding Jack C's attempts to explain away the experimental
> results with "spark losses", etc. the lost energy is a red herring.
> The POTENTIAL energy is merely redistributed over both caps, becoming
> less useful to perform work. You could also say the fully charged
> cap performed work in charging the second cap, storing the energy
> in the elusive field...
>
> I think the whole issue is sort of like the bellhop's "extra"
> dollar, if you all remember that old puzzle.
>
> On the two caps in series? Seriously, if you will, visualize
> it first as two caps connected by a length of wire; next with
> the two connected plates and wire replaced by a single plate
> in the middle, with the original dielectrics on each side of it;
> finally, slide out the (zero-thickness) "extra" plate. You
> have the two outer plates, separated by a double-thick dielectric
> layer. Half the capacitance...
>
> Dave
Thanks Dave. Actually, the question was rhetorical...I just wanted
to give another point of view to the original discussion. It's amazing
how we tend to give complicated answers to simple problems, isn't it?
The best analogy I can think of relating to the 'missing 1/2 of the
charge' puzzle is that of water stored behind a dam:
if the water is x-high, then you get y-amount of hydraulic or head
pressure. Using only the current amount of water in the dam, now
fill, as best as you can, *two* dams. Both dams now have 1/2 the
amount of water.
Since we now have 1/2 the water height (in a highly simple analogy)
we should get about 1/2y worth of head pressure, assuming a linear
function. The total amount of water hasn't changed. The total amount
of *potential* energy has though.
Tounge-in-cheek-wise, I suppose that Jack could do an analysis and tell
us that the pressure isn't 1/2, due to non-linearities in the resivour(?)
and such.... ;-)
- Brent