[SI-LIST] Re: FEM/BEM/MoM

Hi Abdul

My post seems to have been distorted somehow by the equals signs.  Try this.

A capacitor has one end connected to the ground end of a high impedance
voltmeter, and the voltmeter measures the other terminal. 
From my text-book on electricity and magnetism (Bleaney and Bleaney):
the capacitor charge corresponding to a voltage V is
Q equals C times V.
Energy equals integral from 0 to V(QdV) equals 0.5 * C* Vsquared joules.

For a capacitance of one farad, charged up to one volt, energy is 0.5 joule,
and charge is one coulomb.
A second capacitor is connected to the first but through a switch, which is
open for now, and the other end is grounded.
The second capacitor has no charge, voltage is zero.
Close the switch.  The charge distributes between the two C's. By
conservation of charge the charge on each is now 0.5 coulomb and the voltage
is 0.5V.
Therefore the energy on each is 0.5*1*0.5*0.5 equals 0.125 joule.
Total energy is now 0.25 joule.
Compared to the initial energy of 0.5 joule, 0.25 joule has been lost as
heat and electromagnetic radiation.

This example was just to show that a circuit diagram of a physical structure
doesn't tell you all the electrical properties of that structure.

Cheers
Geoff

> -----Original Message-----
> From: Abdulrahman Rafiq [mailto:arafi001@xxxxxxx]
> Sent: 28 August 2003 18:36
> To: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] FEM/BEM/MoM
> 
> 
> Geoff, 
> 
> Could you ellaborate on this a little more, as I am afraid i 
> can't quite see what happened to the 1/4 Joule of energy. 
> Perhaps if you could do a quick back of the envelope 
> calculation as an example. 
> 
> -----------------
> 
> Just a little diversion to show the error in a schematic:
> An old example often quoted is the problem of connecting two 
> one farad
> capacitors together by a switch; one is at 1V potential, the 
> other zero.
> The initial energy is 1/2 CV**2 =3D 0.5 joule.  After closing 
> the switch th=
> ere
> is charge distribution, and energy =3D 1/8 +1/8 =3D 0.25 
> joule.  Where did =
> the
> missing 0.25 joule go?  If you did an electrical degree, 
> you'd see the
> paradox.  If you studied high frequencies, you'd know the 
> answer.  (I did a
> Physics degree, then built RF circuits, so for me a capacitor 
> is not a
> capacitor.)
> 
> The answer is
> 1) it's not physically possible to put two capacitors 
> together at one point
> 2) therefore they are separated by a distance
> 3) therefore on closing the switch, the discharge current 
> travels a distance
> 4) the conductors have finite conductivity
> 5) therefore there is a varying electromagnetic field and 
> energy is
> dissipated and radiated
> 6) please don't talk about too much about inductance because 
> it's only an
> approximation.
> -------------------
> --------------------------------------- 
> Abdulrahman Rafiq
> Department of Physics
> University of California
> Riverside, Ca. 92521
> Email: arafi001@xxxxxxx
> URL: www.geocities.com/arafiq786
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