Both these problems are standard fare in 'older' physics texts- anyone
rember Resnick & Halliday, for example?
There is another more interesting side to Andy's problem as stated however.
Consider the A/d ratio for a practical air spaced capacitor of 1uF and
adequate voltage 'withstand' capability.
It's not exactly small!!! In practice the spacing would need to be somewhat
greater than the minimum purely for reasons of mechanical support/stability,
making the required plate area even greater!
OK - it was way OT but reason I raised it is I am currently looking at using
just such a capacitor arrangement (with many plates, not just a pair) and
some sort of oil (maybe peanut) to resonate my LF transmitting loop on 180
kHz.
I have been down this path before but never actually progressed the matter
till now--as I am thinking about raising the transmitter RF output power I
will have to rebuild the present arrangement which has a bunch of 4700 pf
500 volt polystyrenes in series parallel. The idea of just pumping
dielectric in and out to tune the thing has always had appeal but has been
offset with other possible practical issues-- eg. I'll be running this up in
the open air initially-just in case!
The neat thing with the oil dielectric is it is basically self healing ,
even if it does arc over- no permanent capacitor damage will result as long
as the power is shut off fast enough.. and that's the current issue-no pun
intended!
73
Dave
ZL3FJ
----- Original Message -----
From: "James Moritz" <[email protected]>
To: <[email protected]>
Sent: Saturday, November 09, 2002 12:51 AM
Subject: Re: LF: To Ponder over the weekend
Dear Andy, LF group,
At 09:38 08/11/2002 +0000, you wrote:
>This simple puzzle caused more discussion in my office this morning than
it
>had any right to....
Here is another conundrum along similar lines - you have two identical,
loss-free 1uF capacitors, one is charged to 10V, the other is discharged.
You then connect the two in parallel so the total C is now 2uF - charge
will flow from the charged capacitor into the discharged capacitor.
Assuming charge Q is conserved, and that Q=CV, the voltage must now be 5V.
But the stored energy in a capacitor =1/2CV^2, so with the single charged
capacitor, the stored energy is 1/2 x 1u x 100 = 50uJ, while with both
capacitors in parallel it is only 1/2 x 2u x 25 = 25uJ. So where has the
other 25uJ gone?
Cheers, Jim Moritz
73 de M0BMU
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