Assuming the charge doesn't drain out with the oil, then, for the
relationship Q=CV to be maintained, the voltage must rise to compensate for
the loss of capacitance.
My guess is the 10uF capacitor becomes 1uF and the voltage rises to 100V.
I looked for the answer, upside down, at the bottom of the page ... but it
wasn't there...
73
Hugh M0WYE
----- Original Message -----
From: "Talbot Andrew" <[email protected]>
To: "LF Group (E-mail)" <[email protected]>
Sent: Friday, November 08, 2002 9:38 AM
Subject: LF: To Ponder over the weekend
This simple puzzle caused more discussion in my office this morning than
it
had any right to. Have a ponder while waiting for next QRSS symbol to
come
through.
Andy G4JNT
--------------------------------------------
>From Circuit Cellar magazine, Nov 2002
An oil filled capacitor consists of two plates separated by a dielectric
made of oil with a dielectric constant of 10. The capacitance is 10uF and
it is charged to 10V. The oil is drained, leaving air between the plates.
the capacitance falls to 1uF. Does the stored energy level change?. If
so,
where did the extra energy come from or go to?
--------------------------------------------
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