Message-ID: <003b01c28728$72240500$0504210a@c.scope>
From: "g6tmk" <g6tmk@thersgb.net>
To: rsgb_lf_group@blacksheep.org
References: <7D653C9C42F5D411A27C00508BF8803DCB9CBB@mail.dstl.gov.uk> <000701c28713$e15270c0$0504210a@c.scope> <005201c28723$9fbd8cc0$c12565d5@oemcomputer>
Date: Fri, 8 Nov 2002 13:11:39 -0000
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Subject: LF: Re: Re: Re: To Ponder over the weekend
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You can measure viscosity by measuing the time taken for oil to drain
through a tube ( at least I've always assumed that's why heating oil is
graded in "seconds")
I wonder if you could measure capacitance (or charge) be measuing the time
taken for oil to drain out from between the vanes of a capacitor ?
There are probably less messy ways of measuring it :-)
Hugh

PS Yes, we had a KiloVolt meter like a free-turning variable capacitor in
the school physics lab I remember ...

----- Original Message -----
From: "Johan Bodin" <sm6lkm.jbeab@swipnet.se>
To: <rsgb_lf_group@blacksheep.org>
Sent: Friday, November 08, 2002 12:36 PM
Subject: LF: Re: Re: To Ponder over the weekend


g6tmk wrote:

> My guess is the 10uF capacitor becomes 1uF and the voltage rises to 100V.

That's my guess too, but... W [stored energy] = V^2*C / 2:

With oil: V=10V, C=10u => W = 10*10*10u / 2 = 500 uJ
Without oil: V=100V, C=1u => W = 100*100*1u / 2 = 5000 uJ

Assuming that no charge goes away with the oil, the stored energy
will increase by a factor of ten. Hmm... Force * distance = energy.
When gravity pulls the oil away from the plates, the oil has to fight
against an electrostatic force thereby injecting energy into the
capacitor, I guess...

Will a charged air variable capacitor try to turn itself towards
the maximum setting? :-)

73
Johan SM6LKM