John Larkin wrote:

On Sat, 17 Jul 2010 02:42:20 -0400, ehsjr
wrote:

Paul Hovnanian P.E. wrote:
My solution for the missing energy.


But John's puzzle stated that energy *is* conserved, so there is no
missing energy in his puzzle.

He said _charge_ was not conserved, in some cases.

Or maybe it's a different puzzle? What I recall was a 4F cap
that was charged to .5 volts, therefore Q=2 coulombs. He magically
cut in half. (I don't know why he didn't start with 2 caps, 2F
each, in parallel.)

The resultant 2F caps each retained the .5V charge
for Q=1 coulomb per cap.

He then puts them in series resulting in
*a 1F cap charged to 1 volt* which is Q=1 coulomb, and claims there's 1
coulomb missing.

That's the fallacy. The result is NOT a 1F cap.
The result is 2 2F caps, in series. Each cap
has Q=1 coulomb - there is no missing coulomb.
Doesn't matter that total circuit C = 1 F.

Ed

But every 1F cap is, internally, two 2F caps in series. Just draw a
dotted line midway through the dielectric. Play that game all you
want.

After the two 1F caps are separated and rearranged in series, if you
discharge them through a resistor, you will recover 1 coulomb.

My point all along is that, in actual circuit design, the generalism
"charge is conserved" is dangerous, given that "charge" is
ampere-seconds that you can actually measure and use.

John

The original fallacy is cutting the cap in half and putting the two
halves in series. From Wikipedia:

"Coefficients of potential

"C = QV does not apply when there are more than two charged plates, or
when the net charge on the two plates is non-zero. To handle this case,
Maxwell introduced his "coefficients of potential". If three plates are
given charges Q1,Q2,Q3, then the voltage of plate 1 is given by

V1 = p11Q1 + p12Q2 + p13Q3 ,

"and similarly for the other voltages. Maxwell showed that the
coefficients of potential are symmetric, so that p12 = p21, etc."

http://en.wikipedia.org/wiki/Capacitance

I think there is a more fundamental problem that everyone is missing.

When you charge a capacitor, electrons flow from the negative supply into
the capacitor lead. An equal number of electrons from the positive lead
flow into the positive terminal of the power supply.

After the capacitor is charge, you can disconnect the power supply. You
can also remove the dielectic. The capacitor plates will have a small
residual charge, but most of the energy is in the polarized atoms in the
dielectric. You can recover this energy by putting the dielectric back
between the electrodes.

The question is what happened to the electrons that entered the negative
lead of the capacitor while it was being charged?

They cannot flow through the dielectric - it is an insulator. They do not
stay on the electrode - you can remove the dielectric and find a minor
charge on the electrode. Most of the energy is in the dielectric, but the
electrons did not go there. They simply disappeared.

Similarly, where did the electrons come from on the positive electrode?
They seem to suddenly appear from nothing.

The puzzle seems to come from the displacement current term that Maxwell
added to Ampère's equation:

"The idea was conceived by Maxwell in his 1861 paper On Physical Lines of
Force in connection with the displacement of electric particles in a
dielectric medium. Maxwell added displacement current to the electric
current term in Ampère's Circuital Law. In his 1865 paper A Dynamical
Theory of the Electromagnetic Field Maxwell used this amended version of
Ampère's Circuital Law to derive the electromagnetic wave equation. This
derivation is now generally accepted as an historical landmark in physics
by virtue of uniting electricity, magnetism and optics into one single
unified theory. The displacement current term is now seen as a crucial
addition that completed Maxwell's equations and is necessary to explain
many phenomena, most particularly the existence of electromagnetic waves.

http://en.wikipedia.org/wiki/Displacement_current

The displacement current is obviously real, and can easily be measured.
But how do the electrons disappear at the negative terminal, and suddenly
reappear at the positive terminal?

The puzzle becomes more perplexing when you consider a capacitor with a
vacuum dielectric. These are often used in precision rf applications, or
high-powered transmitters.

In a vacuum capacitor, there are no atoms in a physical dielectric to
stress. But the electrons still do their magical disappearing and
reappearing act. And they obviously do not flow through the vacuum
between the electrodes.

But how do they get from one electrode to the other?

Mike

Mike wrote:
"Coefficients of potential

"C = QV does not apply when there are more than two charged plates,

Typo. Should read:

C = Q/V does not apply when there are more than two charged plates,

On Tue, 27 Jul 2010 21:09:52 GMT, Mike wrote:

Mike wrote:
"Coefficients of potential

"C = QV does not apply when there are more than two charged plates,

Typo. Should read:

C = Q/V does not apply when there are more than two charged plates,


I suspect the real subtlety is the part about "net charge is not
zero".

...Jim Thompson
--
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| Analog Innovations, Inc. | et |
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Spice is like a sports car...
Performance only as good as the person behind the wheel.

On Tue, 27 Jul 2010 21:01:19 GMT, Mike wrote:

John Larkin wrote:

On Sat, 17 Jul 2010 02:42:20 -0400, ehsjr
wrote:

Paul Hovnanian P.E. wrote:
My solution for the missing energy.


But John's puzzle stated that energy *is* conserved, so there is no
missing energy in his puzzle.

He said _charge_ was not conserved, in some cases.

Or maybe it's a different puzzle? What I recall was a 4F cap
that was charged to .5 volts, therefore Q=2 coulombs. He magically
cut in half. (I don't know why he didn't start with 2 caps, 2F
each, in parallel.)

The resultant 2F caps each retained the .5V charge
for Q=1 coulomb per cap.

He then puts them in series resulting in
*a 1F cap charged to 1 volt* which is Q=1 coulomb, and claims there's 1
coulomb missing.

That's the fallacy. The result is NOT a 1F cap.
The result is 2 2F caps, in series. Each cap
has Q=1 coulomb - there is no missing coulomb.
Doesn't matter that total circuit C = 1 F.

Ed

But every 1F cap is, internally, two 2F caps in series. Just draw a
dotted line midway through the dielectric. Play that game all you
want.

After the two 1F caps are separated and rearranged in series, if you
discharge them through a resistor, you will recover 1 coulomb.

My point all along is that, in actual circuit design, the generalism
"charge is conserved" is dangerous, given that "charge" is
ampere-seconds that you can actually measure and use.

John

The original fallacy is cutting the cap in half and putting the two
halves in series. From Wikipedia:

"Coefficients of potential

"C = QV does not apply when there are more than two charged plates, or
when the net charge on the two plates is non-zero. To handle this case,
Maxwell introduced his "coefficients of potential". If three plates are
given charges Q1,Q2,Q3, then the voltage of plate 1 is given by

V1 = p11Q1 + p12Q2 + p13Q3 ,

"and similarly for the other voltages. Maxwell showed that the
coefficients of potential are symmetric, so that p12 = p21, etc."

http://en.wikipedia.org/wiki/Capacitance

I think there is a more fundamental problem that everyone is missing.

When you charge a capacitor, electrons flow from the negative supply into
the capacitor lead. An equal number of electrons from the positive lead
flow into the positive terminal of the power supply.

After the capacitor is charge, you can disconnect the power supply. You
can also remove the dielectic. The capacitor plates will have a small
residual charge, but most of the energy is in the polarized atoms in the
dielectric. You can recover this energy by putting the dielectric back
between the electrodes.

The question is what happened to the electrons that entered the negative
lead of the capacitor while it was being charged?

They cannot flow through the dielectric - it is an insulator. They do not
stay on the electrode - you can remove the dielectric and find a minor
charge on the electrode. Most of the energy is in the dielectric, but the
electrons did not go there. They simply disappeared.

Similarly, where did the electrons come from on the positive electrode?
They seem to suddenly appear from nothing.

The puzzle seems to come from the displacement current term that Maxwell
added to Ampère's equation:

"The idea was conceived by Maxwell in his 1861 paper On Physical Lines of
Force in connection with the displacement of electric particles in a
dielectric medium. Maxwell added displacement current to the electric
current term in Ampère's Circuital Law. In his 1865 paper A Dynamical
Theory of the Electromagnetic Field Maxwell used this amended version of
Ampère's Circuital Law to derive the electromagnetic wave equation. This
derivation is now generally accepted as an historical landmark in physics
by virtue of uniting electricity, magnetism and optics into one single
unified theory. The displacement current term is now seen as a crucial
addition that completed Maxwell's equations and is necessary to explain
many phenomena, most particularly the existence of electromagnetic waves.

http://en.wikipedia.org/wiki/Displacement_current

The displacement current is obviously real, and can easily be measured.
But how do the electrons disappear at the negative terminal, and suddenly
reappear at the positive terminal?

The puzzle becomes more perplexing when you consider a capacitor with a
vacuum dielectric. These are often used in precision rf applications, or
high-powered transmitters.

In a vacuum capacitor, there are no atoms in a physical dielectric to
stress. But the electrons still do their magical disappearing and
reappearing act. And they obviously do not flow through the vacuum
between the electrodes.

But how do they get from one electrode to the other?

Mike

They don't. If you apply current to a vacuum cap, electrons enter one
plate and pile up there. Electrons leave the other plate and it's
short a bunch. No electrons travel through the vacuum.

Try a 1-plate capacitor, like for example a conductive sphere floating
in space, equal numbers of electrons and protons. Now fire a stream of
electrons into it. You now have a current flowing into the sphere, the
voltage is ramping negative, no problem.

John