To: | [email protected] |
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Subject: | LF: Re: Re: To Ponder over the weekend |
From: | "Johan Bodin" <[email protected]> |
Date: | Fri, 8 Nov 2002 13:36:55 +0100 |
References: | <[email protected]> <[email protected]> |
Reply-to: | [email protected] |
Sender: | <[email protected]> |
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 |
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