[Prev][Next][Index][Thread]
Re: 50%
Tesla List wrote:
>
> >From kg7bz-at-whitemtns-dot-comSat Nov 9 21:23:06 1996
> Date: Fri, 08 Nov 1996 19:35:47 -0700
> From: "August H. Johnson" <kg7bz-at-whitemtns-dot-com>
> To: Tesla List <tesla-at-pupman-dot-com>
> Subject: Re: 50%
>
> I've been reading this thread about the 50% energy loss and it reminded
> me of a experiment I'd done about 15-20 years ago. I'd just been given
> a neat 5,000pF 10,000V Jennings variable vacuum capacitor. I wish I
> still had it now! I'd made a calibrated turns counting dial on it so I
> could adjust it to whatever capacitance I wanted.
>
> I set it to 5,000pF and charged it to 5,000 volts. Then I cranked it
> out to 2,500pF and used one of those neat 1000:1 40KV Tektronix scope
> probes to measure the voltage. It was about 10,000 volts. The
> measurement was a bit hard to make since the time constant was only a
> couple seconds when I connected the probe to measure the voltage but a
> storage scope helped. I also did the reverse of this, charge the
> 2,500pF capacitor to 10,000 volts and then adjust it in to 5,000pF. It
> then measured about 5,000 volts. My readings weren't the most accurate
> due to things like leakage on the glass and likely some corona discharge
> in the air but I got in the ballpark. Now do the math on the two:
>
> (5000V)**2 x (5000pF/2) = 0.0626j
> (10000V)**2 x (2500pF/2) = 0.125j
>
> Same charge, but the capacitance of the capacitor was changed. Now,
> if the mechanics were frictionless I could have extracted mechanical
> energy when I let the plates of the capacitor move closer together
> (5000pF) since the plates have opposite charge and are attracted to each
> other, and it would have taken mechanical energy to move them farther
> apart (2500pF) for the same reason. Is this mechanical energy equal to
> the energy change from 0.0625j to 0.125j? I've often wondered about
> this but don't know how to do the conversions.
>
> Now that I think about it, due to the actual mechanical construction
> of that capacitor (many concentric cylinders with alternate ones
> connected together) I don't think the actual attraction between the
> plates is all that well translated to the actuator.
>
> August
> --
> August Johnson KG7BZ AMI 733 http://www.whitemtns-dot-com/~kg7bz
> P.O. Box 795
> Pinetop, AZ 85935
August,
>From a theoretical standpoint, your reasoning is absolutely correct!
Work is required to further separate the charged plates, resulting in
increased potential. When moving charges, the amount of work can be
determined from:
Work = 0.5*QV (Joules/Coulomb)
However, Q in a capacitor Q = CV. Substituting for Q:
Work = 0.5*CV^2 (Joules)
You've done work by overcoming the attractive force separating the
plates, and moving them until the potential was doubled. The amount of
work required to increase the potential by 5000 Volts is directly
converted to an increase in potential energy (or voltage) across the
capacitor. Going back the other way would have given you back the same
amount of mechanical energy. In fact, an electrostatic motor can be
fabricated which uses this principle to provide useful (low power)
mechanical work.
However, from a practical standpoint, you're also correct in assuming
that the mechanical linkage in your vacuum cap may not accurately
translate this force to the actuator.
Safe coilin' to ya!
-- Bert --