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.

That "game" leads nowhere.
It implies that all capacitors are internally a series string of
infinitely small (physical) capacitors and infinitely large
capacitance for each.

This all reminds me of the 3 men rent a hotel room puzzle.
The clerk tells them it will be 30 dollars, so each hands
him a 10 dollar bill. Later, the clerk realizes he's made
an error, the rate is only $25. So he gives 5 ones to the
bellhop & instructs him to give the money back to the men.
The bellhop can't divide it equally, so he goes to their
room and hands each man a dollar, and pockets the other 2
bucks. Now, each man has gotten a dollar back, so each
has given a total of $9 to the clerk. That's $27, and the
bellhop has 2 bucks in his pocket.
Where is the other dollar?




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.

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

Ed


John

On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
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.

That "game" leads nowhere.
It implies that all capacitors are internally a series string of
infinitely small (physical) capacitors and infinitely large
capacitance for each.

This all reminds me of the 3 men rent a hotel room puzzle.
The clerk tells them it will be 30 dollars, so each hands
him a 10 dollar bill. Later, the clerk realizes he's made
an error, the rate is only $25. So he gives 5 ones to the
bellhop & instructs him to give the money back to the men.
The bellhop can't divide it equally, so he goes to their
room and hands each man a dollar, and pockets the other 2
bucks. Now, each man has gotten a dollar back, so each
has given a total of $9 to the clerk. That's $27, and the
bellhop has 2 bucks in his pocket.
Where is the other dollar?




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.

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

Ed


John


John "The Bloviator" Larkin obfuscates the obvious. There is no
inconsistency. There is no magic. TINSTAAFL!

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

On Wed, 21 Jul 2010 20:02:50 -0700, Jim Thompson wrote:

On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
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.

That "game" leads nowhere.
It implies that all capacitors are internally a series string of
infinitely small (physical) capacitors and infinitely large
capacitance for each.

This all reminds me of the 3 men rent a hotel room puzzle.
The clerk tells them it will be 30 dollars, so each hands
him a 10 dollar bill. Later, the clerk realizes he's made
an error, the rate is only $25. So he gives 5 ones to the
bellhop & instructs him to give the money back to the men.
The bellhop can't divide it equally, so he goes to their
room and hands each man a dollar, and pockets the other 2
bucks. Now, each man has gotten a dollar back, so each
has given a total of $9 to the clerk. That's $27, and the
bellhop has 2 bucks in his pocket.
Where is the other dollar?




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.

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

Ed


John


John "The Bloviator" Larkin obfuscates the obvious. There is no
inconsistency. There is no magic. TINSTAAFL!

TANSTAAFL! 'Aint', not 'isn't' ) I've read Larry Niven too.

Grant.
...Jim Thompson

On Thu, 22 Jul 2010 15:36:20 +1000, Grant wrote:

On Wed, 21 Jul 2010 20:02:50 -0700, Jim Thompson wrote:

On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
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.

That "game" leads nowhere.
It implies that all capacitors are internally a series string of
infinitely small (physical) capacitors and infinitely large
capacitance for each.

This all reminds me of the 3 men rent a hotel room puzzle.
The clerk tells them it will be 30 dollars, so each hands
him a 10 dollar bill. Later, the clerk realizes he's made
an error, the rate is only $25. So he gives 5 ones to the
bellhop & instructs him to give the money back to the men.
The bellhop can't divide it equally, so he goes to their
room and hands each man a dollar, and pockets the other 2
bucks. Now, each man has gotten a dollar back, so each
has given a total of $9 to the clerk. That's $27, and the
bellhop has 2 bucks in his pocket.
Where is the other dollar?




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.

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

Ed


John


John "The Bloviator" Larkin obfuscates the obvious. There is no
inconsistency. There is no magic. TINSTAAFL!

TANSTAAFL! 'Aint', not 'isn't' ) I've read Larry Niven too.

Grant.
...Jim Thompson

I was just being politically correct ;-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:

On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:

[snip]

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.

Ed


"stuff" is the source of the error.

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

Jim Thompson wrote:
On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:


On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:


[snip]

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.


Ed



"stuff" is the source of the error.

...Jim Thompson

Exactly. A coulomb is massless. It is not a fixed physical thing
that occupies some particular volume, which seems to be the
misconception at the heart of the so called puzzle.

Ed

On Thu, 22 Jul 2010 13:43:51 -0400, ehsjr
wrote:

Jim Thompson wrote:
On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:


On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:


[snip]

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.


Ed



"stuff" is the source of the error.

...Jim Thompson

Exactly. A coulomb is massless. It is not a fixed physical thing
that occupies some particular volume, which seems to be the
misconception at the heart of the so called puzzle.

Ed

I'm not puzzled, John "The Bloviator" Larkin is. He asserts...

[begin quoted message]
NNTP-Posting-Date: Wed, 07 Jul 2010 10:50:55 -0500
From: John Larkin
Newsgroups: sci.electronics.design
Subject: Inverse Marx generator
Date: Wed, 07 Jul 2010 08:50:50 -0700
Message-ID:
[snip]

Right. If you dump all the energy from one charged cap into another,
discharged, cap of a different value, and do it efficiently, charge is
not conserved.

John
[end quoted message]


...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

On Thu, 22 Jul 2010 15:04:36 -0500, flipper wrote:

On Thu, 22 Jul 2010 08:07:34 -0700, Jim Thompson
wrote:

On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:

On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:

[snip]

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.

Ed


"stuff" is the source of the error.

Which 'error' about which 'stuff' are we supposedly talking about
here?

Is it the series coulomb math error?

...Jim Thompson

I'm trying to write it up, both I keep getting interrupted by work
that pays $:-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

flipper wrote:
On Thu, 22 Jul 2010 13:43:51 -0400, ehsjr
wrote:


Jim Thompson wrote:

On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:



On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:


[snip]


Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.



Ed



"stuff" is the source of the error.

...Jim Thompson

Exactly. A coulomb is massless. It is not a fixed physical thing


An electrostatic field is massless. Are you trying to suggest an
electrostatic field is not 'real'?


that occupies some particular volume,


In reality it does.

Nope. I guess I didn't make my point clear.
You can charge 2 different size (physical volume)
caps with a coulomb. There is no particular volume
associated with "a coulomb".




which seems to be the
misconception at the heart of the so called puzzle.


If you are referring to the, so called, 'missing coulomb' there isn't
one.

Correct, there is no missing coulomb. I've already said that,
earlier in the discussion.

Ed

snip

On Thu, 22 Jul 2010 13:43:51 -0400, ehsjr
wrote:

Jim Thompson wrote:
On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:


On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:


[snip]

Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.


Ed



"stuff" is the source of the error.

...Jim Thompson

Exactly. A coulomb is massless. It is not a fixed physical thing
that occupies some particular volume, which seems to be the
misconception at the heart of the so called puzzle.

Ed

Good. Electron mobility in copper wire is the tinyest fraction of the
speed of light.

"Jim Thompson" wrote in
message ...
I'm trying to write it up, both I keep getting interrupted by work
that pays $:-)

...Jim Thompson

I'd be interested to see your write up, Jim. This conundrum was asked of me
in a pre-uni interview more than 40 years ago, I forget my real answer, I
think I doubted the assumption of zero resistance. There were some wacky
claims by (I think it was) Ivor Catt in a discussion on this in Wireless
World where he claimed it proved the existence of some wonderful new thingy
("displacement current"?). Much sound and fury, but little edification
ensued as I recall. I should see if I still have the letters.

It did cause me to be wary of "stated conditions" aka basic assumptions in
engineering problems in later life ;-)

Regards
Ian

flipper wrote:
On Fri, 23 Jul 2010 00:12:50 -0400, ehsjr
wrote:


flipper wrote:

On Thu, 22 Jul 2010 13:43:51 -0400, ehsjr
wrote:



Jim Thompson wrote:


On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:




On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:


[snip]



Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.




Ed



"stuff" is the source of the error.

...Jim Thompson

Exactly. A coulomb is massless. It is not a fixed physical thing


An electrostatic field is massless. Are you trying to suggest an
electrostatic field is not 'real'?



that occupies some particular volume,


In reality it does.

Nope. I guess I didn't make my point clear.
You can charge 2 different size (physical volume)
caps with a coulomb. There is no particular volume
associated with "a coulomb".


I don't understand what you are trying to get at.

Are you of the opinion that there is one particular volume,
and one particular volume only, that a 1 coulomb charge
will occupy? If so, what is that volume?
If not, then you and I have no disagreement on the point.

To go further. A coulomb is a numerical result of a mathematical
operation, not a physical object. There is no physical object
that is an "ampere second". A coulomb is massless, and volumeless.
The puzzle misdirection appeared to be treating the coulomb as an
object that could be divided by dividing that coulomb's container.



Area and spacing, which define a volume, the dielectric of the volume
and the applied volts determine the coulombs. It's not an 'arbitrary'
thing and, in that sense, yes it is 'fixed' (by the particulars).
Otherwise one would not be able to write an equation for it.

It's even more the case when simply alternating between parallel or
series connections as none of their parameters change, not even V
(hence Q), and that's about as 'fixed' as one can get.


which seems to be the
misconception at the heart of the so called puzzle.


If you are referring to the, so called, 'missing coulomb' there isn't
one.

Correct, there is no missing coulomb. I've already said that,
earlier in the discussion.


Ok. So I threaded back and, unless I misunderstood, you seem to be
suggesting there is some unexplained 'difference' between whatever you
deem a 'real' 1F capacitor and two 2s in series, despite their
identical behavior, so that two 2s in series are "not a 1F cap" and 1F
"doesn't matter."

Not what I was suggesting.
In the puzzle the 2 coulombs is a number that tells you how much
current got stuffed into a container and how long the "stuffing"
took. Magically dividing the container into two containers
(or N containers) does not change the amount of current got
stuffed in, nor the amount of time it took to do that stuffing.



There's no need for mysterious 'differences' as, assuming all else is
equal, a single 1F, two 2s in series, four 4s in series, eight 8s in
series, and so on (although equal values are not required, just that
the resulting series comes to 1F), are black box indistinguishable. At
1V overall they'll each and every one have 1 coulomb because they all
have the exact same charging current go through.


The puzzle's magic division creates the "mysterious differences"
(it results in Vcap = V/N for each cap), hiding the fact that
total ampere seconds is the same.

You may have missed the context of the whole starting from
the beginning, when a "puzzle" (which wasn't a puzzle at all) was
presented. And I don't think it matters - the puzzle was just "mental
gymnastics" that caused confusion instead of making a point, in my
opinion. Your paragraph above shows you understand the principle,
so the puzzle is irrelevant. And the op clarified his meaning:
"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." His "puzzle"
was misdirection to show that a coulomb was somehow lost. No
coulomb was lost.

Perhaps what I've posted isn't clear to you, but if you
think that no coulomb was lost in the puzzle, and that a
coulomb of charge does not require one and only one specific
volume, then we're in agreement.

Ed






Ed

snip

On Sat, 24 Jul 2010 20:18:53 -0500, flipper wrote:

On Sat, 24 Jul 2010 15:29:45 -0400, ehsjr
wrote:

flipper wrote:
On Fri, 23 Jul 2010 00:12:50 -0400, ehsjr
wrote:


flipper wrote:

On Thu, 22 Jul 2010 13:43:51 -0400, ehsjr
wrote:



Jim Thompson wrote:


On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:




On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:


[snip]



Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.




Ed



"stuff" is the source of the error.

...Jim Thompson

Exactly. A coulomb is massless. It is not a fixed physical thing


An electrostatic field is massless. Are you trying to suggest an
electrostatic field is not 'real'?



that occupies some particular volume,


In reality it does.

Nope. I guess I didn't make my point clear.
You can charge 2 different size (physical volume)
caps with a coulomb. There is no particular volume
associated with "a coulomb".


I don't understand what you are trying to get at.

Are you of the opinion that there is one particular volume,
and one particular volume only, that a 1 coulomb charge
will occupy? If so, what is that volume?
If not, then you and I have no disagreement on the point.

Simply repeating the same thing doesn't explain it any more than the
first time did.

Changing the area/spacing ratios would, indeed, change the capacitance
but it's irrelevant as the physical parameters are subsumed into the
given capacitances and further irrelevant as none of the parameters
change anyway.

To go further. A coulomb is a numerical result of a mathematical
operation, not a physical object. There is no physical object
that is an "ampere second". A coulomb is massless, and volumeless.
The puzzle misdirection appeared to be treating the coulomb as an
object that could be divided by dividing that coulomb's container.

It can.

As you yourself pointed out, the story might as well begin with two 2s
in parallel and skip the fluff of 'splitting' the 4 in half as the
result is the same and nothing mysterious happens. You have two 2F
capacitors each with a 'charge' (Q) of 1 coulomb.

And as I pointed out, the entire parallel step is entirely superfluous
as you have the same supposed 'mystery' if you just look at two 2's in
series. Each have 1 coulomb and the total has one coulomb.

Of course, as I also explained, there is no 'mystery' because coulombs
do not add in series. They each and all have the same Q as the whole
because the exact same current goes through each and all as the whole.
I.E. instead of the 'mystery' posited by the diversion 1 + 1 = 1, "huh
where's the missing coulomb?" the correct answer is 1 = 1 = 1, just as
it should.

Area and spacing, which define a volume, the dielectric of the volume
and the applied volts determine the coulombs. It's not an 'arbitrary'
thing and, in that sense, yes it is 'fixed' (by the particulars).
Otherwise one would not be able to write an equation for it.

It's even more the case when simply alternating between parallel or
series connections as none of their parameters change, not even V
(hence Q), and that's about as 'fixed' as one can get.


which seems to be the
misconception at the heart of the so called puzzle.


If you are referring to the, so called, 'missing coulomb' there isn't
one.

Correct, there is no missing coulomb. I've already said that,
earlier in the discussion.


Ok. So I threaded back and, unless I misunderstood, you seem to be
suggesting there is some unexplained 'difference' between whatever you
deem a 'real' 1F capacitor and two 2s in series, despite their
identical behavior, so that two 2s in series are "not a 1F cap" and 1F
"doesn't matter."

Not what I was suggesting.

Ok. Then what did the two 2s are "not a 1F cap" and 1F
"doesn't matter" mean?

In the puzzle the 2 coulombs is a number that tells you how much
current got stuffed into a container and how long the "stuffing"
took. Magically dividing the container into two containers
(or N containers) does not change the amount of current got
stuffed in, nor the amount of time it took to do that stuffing.

Exactly. So what is the machination on "volume" about?


There's no need for mysterious 'differences' as, assuming all else is
equal, a single 1F, two 2s in series, four 4s in series, eight 8s in
series, and so on (although equal values are not required, just that
the resulting series comes to 1F), are black box indistinguishable. At
1V overall they'll each and every one have 1 coulomb because they all
have the exact same charging current go through.


The puzzle's magic division creates the "mysterious differences"
(it results in Vcap = V/N for each cap),

Not so. Capacitors in parallel have the same V and still do when then
'disconnected' from each other (since they're 'ideal' with no
leakage)..

hiding the fact that
total ampere seconds is the same.

There's nothing 'hidden' about 2 parallel coulombs divided into two
being 1 coulomb each.

Albeit superfluous the parallel step is entirely accurate. You begin
with .5V on 4F for 2 coulombs and you end up with two 2s each with .5V
and 1 coulomb.

'Volume', dielectric, and all other physical parameters are irrelevant
as they are subsumed into the 'given' capacitances. It doesn't matter
how they are made as 2F is 2F. In parallel it's 4F and in series it's
1F. That's it.

You may have missed the context of the whole starting from
the beginning, when a "puzzle" (which wasn't a puzzle at all) was
presented.

Paul Hovnanian started the thread with a proposed 'solution' to the
'cap charging a cap' puzzle and mine was the first reply. All I did
was point out that the 'solution' was equivalent to the oft posited
voltage source charging an 'ideal' cap 'puzzle' and that substituting
a cap for the voltage source simply makes the puzzle appear more
mysterious than it is. (A sub thread then ensued about R=0)

You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it.

And I don't think it matters - the puzzle was just "mental
gymnastics" that caused confusion instead of making a point, in my
opinion.

The purpose of a 'puzzle' is to posit what appears to be a conundrum
for the reader to then figure out.

The 'mystery' of the original puzzle is in leaving series R out of the
model and the 'mystery' of the second is being diverted into thinking
coulombs should 'add' when, in series, they do not.

Your paragraph above shows you understand the principle,
so the puzzle is irrelevant. And the op clarified his meaning:
"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." His "puzzle"
was misdirection to show that a coulomb was somehow lost. No
coulomb was lost.

That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'.

His first entry was to me and I did not respond because he hacked my
words to imply I said things I did not and changed the model to boot
so it wasn't even relevant to the discussion.


Perhaps what I've posted isn't clear to you, but if you
think that no coulomb was lost in the puzzle, and that a
coulomb of charge does not require one and only one specific
volume, then we're in agreement.

You still seem to be positing some 'mysterious' thing about 'volume'
that has nothing to do with solving the puzzle. Maybe if you put it to
numbers I could gather what you mean.

Here's my math

Q(Coulombs) = CV, E(Joules) = .5QV (or, by substitution, .5CV^2)

Each: 2F, .5V, 1C, .25J
Parallel: 2F + 2F = 4F, .5V = .5V = .5V, 1C + 1C = 2C, .5J
Series: 1/(1/2F + 1/2F) = 1F, .5V + .5V = 1V, 1C = 1C = 1C, .5J

Note that 'Q per volume' (whatever it might be) of the 2F caps remains
'fixed' because neither Q nor 'volume' change and neither does V, or
anything else. Or at least we can chose to 'pick' that simpler case
because 'volume' is not mentioned anywhere since, being subsumed into
F, it's completely irrelevant.

Energy is conserved, 'charge' (Q) is conserved, and all is right with
the world.

There is no 'puzzle' unless one is fooled by the riddle into thinking
series coulombs should add.

Ed


Ed

snip

Unless your business is located in San Fransicko, Californica

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

flipper wrote:
On Sat, 24 Jul 2010 15:29:45 -0400, ehsjr
wrote:


flipper wrote:

On Fri, 23 Jul 2010 00:12:50 -0400, ehsjr
wrote:



flipper wrote:


On Thu, 22 Jul 2010 13:43:51 -0400, ehsjr
wrote:




Jim Thompson wrote:



On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:





On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:


[snip]




Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.





Ed



"stuff" is the source of the error.

...Jim Thompson

Exactly. A coulomb is massless. It is not a fixed physical thing


An electrostatic field is massless. Are you trying to suggest an
electrostatic field is not 'real'?




that occupies some particular volume,


In reality it does.

Nope. I guess I didn't make my point clear.
You can charge 2 different size (physical volume)
caps with a coulomb. There is no particular volume
associated with "a coulomb".


I don't understand what you are trying to get at.

Are you of the opinion that there is one particular volume,
and one particular volume only, that a 1 coulomb charge
will occupy? If so, what is that volume?
If not, then you and I have no disagreement on the point.


Simply repeating the same thing doesn't explain it any more than the
first time did.

You did not answer the question. Apparently you can't/won't
or just want to argue.

I'll try, one last time:
You don't seem to realize that a coulomb is not a physical object.
It is a number, the result of a mathematical operation on amperes
and seconds. Thus all "puzzles" that attempt to treat it as a
physical object with mass and volume are flawed at the outset.
I have been trying to point out that it seems that the puzzle
that JL posed uses the misdirection of trying to get the reader to
think that the "stuff" in the cap called a coulomb is an object,
a thing with mass and volume. Maybe that is not the intended
misdirection, but you and I can't get past the point of agreeing
on what volume means. A coulomb does not have the property of
volume or mass. Whatever object has been charged does but the
coulomb itself is just a number. I'm sorry if that is a repeat,
but it is a fact you don't agree to. We can't get any further.

I do need to correct 2 points you made below:

"You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it."

Do not lay that puzzle at my doorstep! :-)
(Well, unless you want to stomp on it. Even then attribute
"ownership" where it belongs. :-) )

John introduced the parallel to series puzzle elsewhere.
I attributed it to him in my post. And it's clear from
the context that I'm talking about what someone else posted.

"That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'."

He posted the 'parallel to series' puzzle on SED - obviously you
missed it.

In any event, it is not *my* puzzle, and I didn't pose it.

Ed



Changing the area/spacing ratios would, indeed, change the capacitance
but it's irrelevant as the physical parameters are subsumed into the
given capacitances and further irrelevant as none of the parameters
change anyway.


To go further. A coulomb is a numerical result of a mathematical
operation, not a physical object. There is no physical object
that is an "ampere second". A coulomb is massless, and volumeless.
The puzzle misdirection appeared to be treating the coulomb as an
object that could be divided by dividing that coulomb's container.


It can.

As you yourself pointed out, the story might as well begin with two 2s
in parallel and skip the fluff of 'splitting' the 4 in half as the
result is the same and nothing mysterious happens. You have two 2F
capacitors each with a 'charge' (Q) of 1 coulomb.

And as I pointed out, the entire parallel step is entirely superfluous
as you have the same supposed 'mystery' if you just look at two 2's in
series. Each have 1 coulomb and the total has one coulomb.

Of course, as I also explained, there is no 'mystery' because coulombs
do not add in series. They each and all have the same Q as the whole
because the exact same current goes through each and all as the whole.
I.E. instead of the 'mystery' posited by the diversion 1 + 1 = 1, "huh
where's the missing coulomb?" the correct answer is 1 = 1 = 1, just as
it should.


Area and spacing, which define a volume, the dielectric of the volume
and the applied volts determine the coulombs. It's not an 'arbitrary'
thing and, in that sense, yes it is 'fixed' (by the particulars).
Otherwise one would not be able to write an equation for it.

It's even more the case when simply alternating between parallel or
series connections as none of their parameters change, not even V
(hence Q), and that's about as 'fixed' as one can get.



which seems to be the
misconception at the heart of the so called puzzle.


If you are referring to the, so called, 'missing coulomb' there isn't
one.

Correct, there is no missing coulomb. I've already said that,
earlier in the discussion.


Ok. So I threaded back and, unless I misunderstood, you seem to be
suggesting there is some unexplained 'difference' between whatever you
deem a 'real' 1F capacitor and two 2s in series, despite their
identical behavior, so that two 2s in series are "not a 1F cap" and 1F
"doesn't matter."

Not what I was suggesting.


Ok. Then what did the two 2s are "not a 1F cap" and 1F
"doesn't matter" mean?


In the puzzle the 2 coulombs is a number that tells you how much
current got stuffed into a container and how long the "stuffing"
took. Magically dividing the container into two containers
(or N containers) does not change the amount of current got
stuffed in, nor the amount of time it took to do that stuffing.


Exactly. So what is the machination on "volume" about?



There's no need for mysterious 'differences' as, assuming all else is
equal, a single 1F, two 2s in series, four 4s in series, eight 8s in
series, and so on (although equal values are not required, just that
the resulting series comes to 1F), are black box indistinguishable. At
1V overall they'll each and every one have 1 coulomb because they all
have the exact same charging current go through.


The puzzle's magic division creates the "mysterious differences"
(it results in Vcap = V/N for each cap),


Not so. Capacitors in parallel have the same V and still do when then
'disconnected' from each other (since they're 'ideal' with no
leakage)..


hiding the fact that
total ampere seconds is the same.


There's nothing 'hidden' about 2 parallel coulombs divided into two
being 1 coulomb each.

Albeit superfluous the parallel step is entirely accurate. You begin
with .5V on 4F for 2 coulombs and you end up with two 2s each with .5V
and 1 coulomb.

'Volume', dielectric, and all other physical parameters are irrelevant
as they are subsumed into the 'given' capacitances. It doesn't matter
how they are made as 2F is 2F. In parallel it's 4F and in series it's
1F. That's it.


You may have missed the context of the whole starting from
the beginning, when a "puzzle" (which wasn't a puzzle at all) was
presented.


Paul Hovnanian started the thread with a proposed 'solution' to the
'cap charging a cap' puzzle and mine was the first reply. All I did
was point out that the 'solution' was equivalent to the oft posited
voltage source charging an 'ideal' cap 'puzzle' and that substituting
a cap for the voltage source simply makes the puzzle appear more
mysterious than it is. (A sub thread then ensued about R=0)

You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it.


And I don't think it matters - the puzzle was just "mental
gymnastics" that caused confusion instead of making a point, in my
opinion.


The purpose of a 'puzzle' is to posit what appears to be a conundrum
for the reader to then figure out.

The 'mystery' of the original puzzle is in leaving series R out of the
model and the 'mystery' of the second is being diverted into thinking
coulombs should 'add' when, in series, they do not.


Your paragraph above shows you understand the principle,
so the puzzle is irrelevant. And the op clarified his meaning:
"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." His "puzzle"
was misdirection to show that a coulomb was somehow lost. No
coulomb was lost.


That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'.

His first entry was to me and I did not respond because he hacked my
words to imply I said things I did not and changed the model to boot
so it wasn't even relevant to the discussion.



Perhaps what I've posted isn't clear to you, but if you
think that no coulomb was lost in the puzzle, and that a
coulomb of charge does not require one and only one specific
volume, then we're in agreement.


You still seem to be positing some 'mysterious' thing about 'volume'
that has nothing to do with solving the puzzle. Maybe if you put it to
numbers I could gather what you mean.

Here's my math

Q(Coulombs) = CV, E(Joules) = .5QV (or, by substitution, .5CV^2)

Each: 2F, .5V, 1C, .25J
Parallel: 2F + 2F = 4F, .5V = .5V = .5V, 1C + 1C = 2C, .5J
Series: 1/(1/2F + 1/2F) = 1F, .5V + .5V = 1V, 1C = 1C = 1C, .5J

Note that 'Q per volume' (whatever it might be) of the 2F caps remains
'fixed' because neither Q nor 'volume' change and neither does V, or
anything else. Or at least we can chose to 'pick' that simpler case
because 'volume' is not mentioned anywhere since, being subsumed into
F, it's completely irrelevant.

Energy is conserved, 'charge' (Q) is conserved, and all is right with
the world.

There is no 'puzzle' unless one is fooled by the riddle into thinking
series coulombs should add.


Ed


Ed

snip

On Sun, 25 Jul 2010 23:09:57 -0500, flipper wrote:

On Sat, 24 Jul 2010 17:31:57 +0100, "Ian"
wrote:

"Jim Thompson" wrote in
message ...
I'm trying to write it up, both I keep getting interrupted by work
that pays $:-)

...Jim Thompson

I'd be interested to see your write up, Jim. This conundrum was asked of me
in a pre-uni interview more than 40 years ago, I forget my real answer, I
think I doubted the assumption of zero resistance. There were some wacky
claims by (I think it was) Ivor Catt in a discussion on this in Wireless
World where he claimed it proved the existence of some wonderful new thingy
("displacement current"?). Much sound and fury, but little edification
ensued as I recall. I should see if I still have the letters.

My understanding is that Ivor Catt doesn't believe in displacement
current or much else of conventional electromagnetism.

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

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


It did cause me to be wary of "stated conditions" aka basic assumptions in
engineering problems in later life ;-)

A wise caution.


Regards
Ian


"Ivor Catt" sounds like I should know him... '50's, early '60's ???

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

On Mon, 26 Jul 2010 00:33:58 -0500, flipper wrote:

On Sun, 25 Jul 2010 21:31:45 -0700, Jim Thompson
wrote:

On Sun, 25 Jul 2010 23:09:57 -0500, flipper wrote:

On Sat, 24 Jul 2010 17:31:57 +0100, "Ian"
wrote:

"Jim Thompson" wrote in
message ...
I'm trying to write it up, both I keep getting interrupted by work
that pays $:-)

...Jim Thompson

I'd be interested to see your write up, Jim. This conundrum was asked of me
in a pre-uni interview more than 40 years ago, I forget my real answer, I
think I doubted the assumption of zero resistance. There were some wacky
claims by (I think it was) Ivor Catt in a discussion on this in Wireless
World where he claimed it proved the existence of some wonderful new thingy
("displacement current"?). Much sound and fury, but little edification
ensued as I recall. I should see if I still have the letters.

My understanding is that Ivor Catt doesn't believe in displacement
current or much else of conventional electromagnetism.

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

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


It did cause me to be wary of "stated conditions" aka basic assumptions in
engineering problems in later life ;-)

A wise caution.


Regards
Ian


"Ivor Catt" sounds like I should know him... '50's, early '60's ???

Well, the wikipedia article says he was born in 1935 so that would
make him around 75.


...Jim Thompson

I was talking of the year-range where I think I met him... possibly at
Motorola. With a name like that, one rarely forgets :-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

flipper wrote:

much snipped


Where you claim the 'puzzle' lies, 'splitting' the 4F into two 2s,
there's no puzzle at all.


That is not what I said at all. I have no problem with magically
cutting the 4F charged to .5 volts and Q=2cap into 2 2F caps
each charged to .5 volts with Q=1 for each cap, as the puzzle
proposes.


As I explained, you have the ENTIRE 'puzzle' if you just put two 2s in
series and charge them up to 1V.

Ok, _now_ I see what you are missing. That is most definitely NOT in
the puzzle. The caps were not *charged* in series. You missed what
the puzzle says.

Both 2F capacitors have 1 coulomb and
the whole 1F capacitor has one coulomb. 1 + 1 = 1. Where is the
missing coulomb?

I don't understand your point above - I'm guessing that you're
indicating that coulombs don't add. But you are referring
to the wrong scenario, in any event.


Not ONE thing 'changes'. They are completely static with not even a
connection change. Explain that one with your 'volumes'.

Don't be absurd. The puzzle stipulates changes.


The only misdirection is in thinking coulombs add in series when they
don't.

Here

http://farside.ph.utexas.edu/teachin...es/node46.html

Two capacitors connected in series.

"... the fact that the charge is common to all three capacitors. "



Your cite above refers to a non-isolated system. You jumped into
the middle of an ongoing metadiscussion that started on SED, so you
are not aware of what has gone before. There was a large amount of
discussion on conservation of charge, and it spanned more than
one subject. JL made the point that charge is sometimes not
conserved.

I'll cut and paste the puzzle as he wrote it, below my comments.
His puzzle takes an isolated system, that starts out *already* charged
with 2 coulombs, and monkeys around with the volumes inside the system.
Your cite above refers to a non isolated system, and is not applicable
to an isolated system.

If you don't like the use of the word volume, fine. Substitute whatever
term you want that includes the concept of the capacity of that volume
and the coordinates that describe the place where your chosen term
exists. Or use whatever term(s) that allow you to realize that
2 2F capacitors in series are NOT *a* 1F capacitor in the
puzzle, without getting sidetracked into thinking that
1/Ct = 1/C1+1/C2...1/Cn applies to the isolated system.

The law of conservation of charge applies as long as the system
remains isolated. The number of coulombs remains constant, regardless
of the volume (size, whatever your term is) of the pieces you magically
slice the original cap into, or how you connect them, in that isolated
system.

The puzzle ends up stating that after the slicing & serial connecting,
you have *a* 1F cap charged to 1 volt. That is *false*. You have
2 2F caps, each with a charge of 1 coulomb, connected in series.

Now, because Q = CV, and because charge is constant in an isolated
system and you started with 2 coulombs charge, if you magically merged
the two caps into a 1F cap, then Q=CV means that V would have to be 2,
not 1. Charge must be conserved.

As I used "volume", those two caps occupy different cordinates, they
are not the magical merger the puzzle implies into a 1F cap, and they
must be considered as such: 2 separate capacitors, not a single 1F
capacitor. Different volumes, no merger of coordinates, no merger of
capacity. There's no current in or out of either cap. It simply does
not matter, in the isolated system, that the total series capacitance
would equal 1F in a non isolated system.

The misdirection seems to me to be in the monkeying around with the
volumes. It states that you have a 1F cap. In fact you don't even have
the equivalent of a 1F cap. The fact that you have 2 2F caps with an
equivalent of a single 1F *if connected in series in a circuit*
becomes important when you make the system no longer isolated by
discharging. But the puzzle does *not* "un-isolate" the system. No
current flows. The serial caps are just two caps, each of which has one
of its leads connected to the other cap, and its other lead floating
free. There is no circuit. Those two caps might just as well be lying
in your parts drawer. There is no equivalent capacitance.


Ed

I can't begin to capture the whole metadiscussion, which spans
several subjects and 2 (or more?) newsgroups, but at least
I can quote the part where John posted the puzzle:

Quoting from John Larkin's post, 7/10/2010 at 11:16 AM
on SED. Subject: Win Hill: Inverse Marx Generator ??

To celebrate the 21st century, I have composed a new riddle:

Start with a 4 farad cap charged to 0.5 volts. Q = 2 coulombs.

Carefully saw it in half, without discharging it, such as to have two
caps, each 2 farads, each charged to 0.5 volts. The total charge of
the two caps remains 2 coulombs, whether you connect them in parallel
or consider them separately.

Now stack them in series. The result is a 1F cap charged to 1 volt.
That has a charge of 1 coulomb. Where did the other coulomb go?

I think this is a better riddle.

John



Maybe that is not the intended
misdirection, but you and I can't get past the point of agreeing
on what volume means.


Really. Where did I ever dispute the meaning of 'volume'?


A coulomb does not have the property of
volume or mass.


And volume or mass does not have the property of charge, spin, or
color. So what?

You act as if the only 'real' things in the universe are volume and
mass so that all else are "only numbers;" and mysterious magical ones
that invisibly 'change' when nothing else has.


Whatever object has been charged does but the
coulomb itself is just a number.


And volume is "just a number."

Volume is the product of measurable quantities. Coulombs is the
product of measurable quantities. We can measure volume. We can
measure coulombs.

There is no mysticism involved.


I'm sorry if that is a repeat,
but it is a fact you don't agree to. We can't get any further.


You can't go any further because you can't put any math to your
'volume' theory, which means you have not solved the 'puzzle' as arm
waving about unexplained 'volume effects' does not solve equations.



I do need to correct 2 points you made below:

"You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it."

Do not lay that puzzle at my doorstep! :-)
(Well, unless you want to stomp on it. Even then attribute
"ownership" where it belongs. :-) )


I didn't say you invented or 'owned' it. I said you introduced it
here.


John introduced the parallel to series puzzle elsewhere.


If you say so.


I attributed it to him in my post. And it's clear from
the context that I'm talking about what someone else posted.


Terrific. And thank you for introducing it to us.


"That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'."

He posted the 'parallel to series' puzzle on SED - obviously you
missed it.


This isn't SED, it's ABSE.


In any event, it is not *my* puzzle, and I didn't pose it.


I didn't say either. All I said is you introduced it.

Good Lord, you act as if mentioning it is a heinous crime you plead
not guilty to.



Ed



Changing the area/spacing ratios would, indeed, change the capacitance
but it's irrelevant as the physical parameters are subsumed into the
given capacitances and further irrelevant as none of the parameters
change anyway.



To go further. A coulomb is a numerical result of a mathematical
operation, not a physical object. There is no physical object
that is an "ampere second". A coulomb is massless, and volumeless.
The puzzle misdirection appeared to be treating the coulomb as an
object that could be divided by dividing that coulomb's container.


It can.

As you yourself pointed out, the story might as well begin with two 2s
in parallel and skip the fluff of 'splitting' the 4 in half as the
result is the same and nothing mysterious happens. You have two 2F
capacitors each with a 'charge' (Q) of 1 coulomb.

And as I pointed out, the entire parallel step is entirely superfluous
as you have the same supposed 'mystery' if you just look at two 2's in
series. Each have 1 coulomb and the total has one coulomb.

Of course, as I also explained, there is no 'mystery' because coulombs
do not add in series. They each and all have the same Q as the whole
because the exact same current goes through each and all as the whole.
I.E. instead of the 'mystery' posited by the diversion 1 + 1 = 1, "huh
where's the missing coulomb?" the correct answer is 1 = 1 = 1, just as
it should.



Area and spacing, which define a volume, the dielectric of the volume
and the applied volts determine the coulombs. It's not an 'arbitrary'
thing and, in that sense, yes it is 'fixed' (by the particulars).
Otherwise one would not be able to write an equation for it.

It's even more the case when simply alternating between parallel or
series connections as none of their parameters change, not even V
(hence Q), and that's about as 'fixed' as one can get.




which seems to be the
misconception at the heart of the so called puzzle.


If you are referring to the, so called, 'missing coulomb' there isn't
one.

Correct, there is no missing coulomb. I've already said that,
earlier in the discussion.


Ok. So I threaded back and, unless I misunderstood, you seem to be
suggesting there is some unexplained 'difference' between whatever you
deem a 'real' 1F capacitor and two 2s in series, despite their
identical behavior, so that two 2s in series are "not a 1F cap" and 1F
"doesn't matter."

Not what I was suggesting.


Ok. Then what did the two 2s are "not a 1F cap" and 1F
"doesn't matter" mean?



In the puzzle the 2 coulombs is a number that tells you how much
current got stuffed into a container and how long the "stuffing"
took. Magically dividing the container into two containers
(or N containers) does not change the amount of current got
stuffed in, nor the amount of time it took to do that stuffing.


Exactly. So what is the machination on "volume" about?




There's no need for mysterious 'differences' as, assuming all else is
equal, a single 1F, two 2s in series, four 4s in series, eight 8s in
series, and so on (although equal values are not required, just that
the resulting series comes to 1F), are black box indistinguishable. At
1V overall they'll each and every one have 1 coulomb because they all
have the exact same charging current go through.


The puzzle's magic division creates the "mysterious differences"
(it results in Vcap = V/N for each cap),


Not so. Capacitors in parallel have the same V and still do when then
'disconnected' from each other (since they're 'ideal' with no
leakage)..



hiding the fact that
total ampere seconds is the same.


There's nothing 'hidden' about 2 parallel coulombs divided into two
being 1 coulomb each.

Albeit superfluous the parallel step is entirely accurate. You begin
with .5V on 4F for 2 coulombs and you end up with two 2s each with .5V
and 1 coulomb.

'Volume', dielectric, and all other physical parameters are irrelevant
as they are subsumed into the 'given' capacitances. It doesn't matter
how they are made as 2F is 2F. In parallel it's 4F and in series it's
1F. That's it.



You may have missed the context of the whole starting from
the beginning, when a "puzzle" (which wasn't a puzzle at all) was
presented.


Paul Hovnanian started the thread with a proposed 'solution' to the
'cap charging a cap' puzzle and mine was the first reply. All I did
was point out that the 'solution' was equivalent to the oft posited
voltage source charging an 'ideal' cap 'puzzle' and that substituting
a cap for the voltage source simply makes the puzzle appear more
mysterious than it is. (A sub thread then ensued about R=0)

You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it.



And I don't think it matters - the puzzle was just "mental
gymnastics" that caused confusion instead of making a point, in my
opinion.


The purpose of a 'puzzle' is to posit what appears to be a conundrum
for the reader to then figure out.

The 'mystery' of the original puzzle is in leaving series R out of the
model and the 'mystery' of the second is being diverted into thinking
coulombs should 'add' when, in series, they do not.



Your paragraph above shows you understand the principle,
so the puzzle is irrelevant. And the op clarified his meaning:
"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." His "puzzle"
was misdirection to show that a coulomb was somehow lost. No
coulomb was lost.


That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'.

His first entry was to me and I did not respond because he hacked my
words to imply I said things I did not and changed the model to boot
so it wasn't even relevant to the discussion.




Perhaps what I've posted isn't clear to you, but if you
think that no coulomb was lost in the puzzle, and that a
coulomb of charge does not require one and only one specific
volume, then we're in agreement.


You still seem to be positing some 'mysterious' thing about 'volume'
that has nothing to do with solving the puzzle. Maybe if you put it to
numbers I could gather what you mean.

Here's my math

Q(Coulombs) = CV, E(Joules) = .5QV (or, by substitution, .5CV^2)

Each: 2F, .5V, 1C, .25J
Parallel: 2F + 2F = 4F, .5V = .5V = .5V, 1C + 1C = 2C, .5J
Series: 1/(1/2F + 1/2F) = 1F, .5V + .5V = 1V, 1C = 1C = 1C, .5J

Note that 'Q per volume' (whatever it might be) of the 2F caps remains
'fixed' because neither Q nor 'volume' change and neither does V, or
anything else. Or at least we can chose to 'pick' that simpler case
because 'volume' is not mentioned anywhere since, being subsumed into
F, it's completely irrelevant.

Energy is conserved, 'charge' (Q) is conserved, and all is right with
the world.

There is no 'puzzle' unless one is fooled by the riddle into thinking
series coulombs should add.



Ed



Ed

snip

On Mon, 26 Jul 2010 15:37:12 -0500, flipper wrote:

On Mon, 26 Jul 2010 07:57:11 -0700, Jim Thompson
wrote:

On Mon, 26 Jul 2010 00:33:58 -0500, flipper wrote:

On Sun, 25 Jul 2010 21:31:45 -0700, Jim Thompson
wrote:

On Sun, 25 Jul 2010 23:09:57 -0500, flipper wrote:

On Sat, 24 Jul 2010 17:31:57 +0100, "Ian"
wrote:

"Jim Thompson" wrote in
message ...
I'm trying to write it up, both I keep getting interrupted by work
that pays $:-)

...Jim Thompson

I'd be interested to see your write up, Jim. This conundrum was asked of me
in a pre-uni interview more than 40 years ago, I forget my real answer, I
think I doubted the assumption of zero resistance. There were some wacky
claims by (I think it was) Ivor Catt in a discussion on this in Wireless
World where he claimed it proved the existence of some wonderful new thingy
("displacement current"?). Much sound and fury, but little edification
ensued as I recall. I should see if I still have the letters.

My understanding is that Ivor Catt doesn't believe in displacement
current or much else of conventional electromagnetism.

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

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


It did cause me to be wary of "stated conditions" aka basic assumptions in
engineering problems in later life ;-)

A wise caution.


Regards
Ian


"Ivor Catt" sounds like I should know him... '50's, early '60's ???

Well, the wikipedia article says he was born in 1935 so that would
make him around 75.


...Jim Thompson

I was talking of the year-range where I think I met him... possibly at
Motorola. With a name like that, one rarely forgets :-)

Oh. Well, how in the world would I know THAT? LOL

I do see references to him at Motorola. One says 'mid' 1960s and the
other said 'late' 1960s. I also see 1964 and 1966 referenced in some
papers.


...Jim Thompson

OK. I was Moto from 6/62 to 11/70

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Only as good as the person behind the wheel.

On Mon, 26 Jul 2010 18:03:11 -0400, ehsjr
wrote:

[snip]

I can't begin to capture the whole metadiscussion, which spans
several subjects and 2 (or more?) newsgroups, but at least
I can quote the part where John posted the puzzle:

Quoting from John Larkin's post, 7/10/2010 at 11:16 AM
on SED. Subject: Win Hill: Inverse Marx Generator ??

To celebrate the 21st century, I have composed a new riddle:

Start with a 4 farad cap charged to 0.5 volts. Q = 2 coulombs.

Carefully saw it in half, without discharging it, such as to have two
caps, each 2 farads, each charged to 0.5 volts. The total charge of
the two caps remains 2 coulombs, whether you connect them in parallel
or consider them separately.

Now stack them in series. The result is a 1F cap charged to 1 volt.
That has a charge of 1 coulomb. Where did the other coulomb go?

I think this is a better riddle.

John


[snip]

No riddle at all for anyone who even doodles a single picture (after
reading the Wiki Conservation of Charge Page)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

SED Has Crumbled to Below SEB Status
Populated Only by Bloviators and Pompous PhD's

On Mon, 26 Jul 2010 15:59:32 -0700, Jim Thompson wrote:
....
Populated Only by Bloviators and Pompous PhD's

Here I was, thinking you made that word up, so google found for me:
"
According to WikiPedia, to bloviate is to "speak pompously and
excessively", or "to expound ridiculously". A colloquial verb
coined in the United States, it is commonly used with contempt
to describe the behavior of politicians, academics, pundits
or media "experts", sometimes called bloviators, who hold forth
on subjects in an arrogant, tiresome way.
" -- http://www.bloviators.net/

There, they have their very own website!

Grant.

On Tue, 27 Jul 2010 00:21:20 -0500, flipper wrote:

On Mon, 26 Jul 2010 18:03:11 -0400, ehsjr
wrote:

flipper wrote:

much snipped


Where you claim the 'puzzle' lies, 'splitting' the 4F into two 2s,
there's no puzzle at all.


That is not what I said at all. I have no problem with magically
cutting the 4F charged to .5 volts and Q=2cap into 2 2F caps
each charged to .5 volts with Q=1 for each cap, as the puzzle
proposes.

Terrific.

As I explained, you have the ENTIRE 'puzzle' if you just put two 2s in
series and charge them up to 1V.

Ok, _now_ I see what you are missing. That is most definitely NOT in
the puzzle. The caps were not *charged* in series. You missed what
the puzzle says.

No, I did not 'miss' what the puzzle says. It's irrelevant as it makes
NO difference if you 'charge' them in parallel, in series, or
separately. The math is IDENTICAL, which is why I explained that the
parallel step is superfluous and you might as well start with them in
series. You have EXACTLY the same situation.


Both 2F capacitors have 1 coulomb and
the whole 1F capacitor has one coulomb. 1 + 1 = 1. Where is the
missing coulomb?

I don't understand your point above - I'm guessing that you're
indicating that coulombs don't add.

That's right. Coulombs do not add in series.

But you are referring
to the wrong scenario, in any event.

No I am not. It is the exact same "scenario" as it does not matter
'how' 1 coulomb gets in the 2F cap. .5V on 2F is 1 coulomb is 1
coulomb is 1 coulomb.

Put the two 2s in parallel, charge them to .5V, and you have 1 coulomb
on each. Charge them in series to 1V overall and you have 1 coulomb on
each. Charge them separately to .5V and you have 1 coulomb on each.

It DOES NOT MATTER. 1 coulomb is 1 coulomb is 1 coulomb.

Not ONE thing 'changes'. They are completely static with not even a
connection change. Explain that one with your 'volumes'.

Don't be absurd. The puzzle stipulates changes.

Nope. The 2F caps have .5V for 1 coulomb throughout the entire
exercise. NOTHING about the capacitors and their charge changes.


The only misdirection is in thinking coulombs add in series when they
don't.

Here

http://farside.ph.utexas.edu/teachin...es/node46.html

Two capacitors connected in series.

"... the fact that the charge is common to all three capacitors. "



Your cite above refers to a non-isolated system.

Oh? And just what are the bare wires sticking out each end 'connected'
to?

Doesn't matter anyway. Volts are Volts and Farads are Farads.

You jumped into
the middle of an ongoing metadiscussion that started on SED, so you
are not aware of what has gone before. There was a large amount of
discussion on conservation of charge, and it spanned more than
one subject. JL made the point that charge is sometimes not
conserved.

Doesn't matter what went on before. Capacitors can't read nor do they
care what sense or nonsense someone wrote.


I'll cut and paste the puzzle as he wrote it, below my comments.
His puzzle takes an isolated system, that starts out *already* charged
with 2 coulombs, and monkeys around with the volumes inside the system.
Your cite above refers to a non isolated system, and is not applicable
to an isolated system.

You see any reference to 'volumes' in CV? No. It's irrelevant.

If you don't like the use of the word volume, fine. Substitute whatever
term you want that includes the concept of the capacity of that volume
and the coordinates that describe the place where your chosen term
exists. Or use whatever term(s) that allow you to realize that
2 2F capacitors in series are NOT *a* 1F capacitor in the
puzzle, without getting sidetracked into thinking that
1/Ct = 1/C1+1/C2...1/Cn applies to the isolated system.

You, again, try to make hay from nonsense. Do you see anything in the
equations that qualify 'how' a "1F capacitor" is 'made'? No. It
doesn't matter if I use two plates or a hundred little plates in
parallel or a dozen large plates in series or a combination. Farads
are Farads.

Two 2s in series ARE "a 1F capacitor" and 1/Ct = 1/C1+1/C2...1/Cn is
not a 'sidetrack', it's a measurable fact.

If it makes you feel any better then pot them in a tube with two leads
sticking out. It's 1F.


The law of conservation of charge applies as long as the system
remains isolated. The number of coulombs remains constant, regardless
of the volume (size, whatever your term is) of the pieces you magically
slice the original cap into, or how you connect them, in that isolated
system.

The puzzle ends up stating that after the slicing & serial connecting,
you have *a* 1F cap charged to 1 volt. That is *false*.

It is perfectly true. You DO have a 1F capacitor charged to 1V.

You have
2 2F caps, each with a charge of 1 coulomb, connected in series.

Correct, which is a 1F capacitor charged to 1V for 1 coulomb.

Repeat after me: coulombs in series are all the same as the whole. It
MUST be so because the same current flows through each as the whole.


Now, because Q = CV, and because charge is constant in an isolated
system and you started with 2 coulombs charge, if you magically merged
the two caps into a 1F cap, then Q=CV means that V would have to be 2,
not 1.

Doesn't bother you blowing conservation of energy to hell and back,
eh?

You've fallen for the 'add' diversion. Coulombs DO NOT ADD in series,
just as amp/hours of batteries do not add in series. Volts do.

It doesn't matter if you "magically merged" the two or connect them in
series. The result is 1V on 1F for 1 coulomb.

Repeat after me: coulombs in series are all the same as the whole. It
MUST be so because the same current flows through each as the whole.

Charge must be conserved.

You are confusing your 'charges' (which, you may have noticed, is why
I usually put the word in tic marks often with an 'explanation' like
(Q) immediately following). A capacitor stores energy.

When 'charging' (or 'discharging') the (ideal) capacitor an electron
enters one plate and another electron exits the other (required by,
guess what, conservation of charge). The 'charges' (electrons, holes)
are rearranged, with equal - and + 'charges' on opposing plates, but
the NET charge (0) is unchanged.

There is no 'problem' conserving 0 whether you 'add', or not, and 0 +
0 is pretty much the same as 0 = 0. Close enough for government work
anyway.

Energy (.5CV^2) is a different matter, of course, and total E = .5J
better be the case no matter how they're connected or else the
universe will fold up into a reverse black hole (joke). Which is a
serious problem for your two .5V 2F caps each with E = .25J, total
E=.5J, merging in series to a 2V 1F cap with E = 2J.

AAAAAaaaaeeeeee-------..... help me

Quick, quick. It's really, just like the equations say, 1V on 1F for 1
coulomb and E= .5J

......-------eeeeeeaaaaAAAAA that's better.


As I used "volume", those two caps occupy different cordinates,

You see anything about "coordinates" in CV? No. It's irrelevant.

they
are not the magical merger the puzzle implies into a 1F cap,

Yes they are. 1/Cs = 1/Ca + 1/Cb. It's a measurable fact. Live it and
love it.

and they
must be considered as such: 2 separate capacitors, not a single 1F
capacitor.

Nonsense.

I give you two leads. You stick a capacitance meter on them and it
reads 1F. Now you tell me how many capacitors and in what arrangement
I used to make up the 1F you just measured.

Hint: No matter what you pick I'll claim it's another arrangement
because it DOES NOT MATTER. Farads are Farads.

Different volumes, no merger of coordinates, no merger of
capacity. There's no current in or out of either cap. It simply does
not matter, in the isolated system, that the total series capacitance
would equal 1F in a non isolated system.

And what magical transformation do you imagine takes place with the
"not a 1F capacitor" when I stick a capacitance meter on them and
measure 1F?

The misdirection seems to me to be in the monkeying around with the
volumes.

Nope. The misdirection is in thinking coulombs 'must add' no matter
how the caps are connected.

Repeat after me: coulombs in series are all the same as the whole. It
MUST be so because the same current flows through each as the whole.

It states that you have a 1F cap.

You do. 1/Cs = 1/Ca + 1/Cb. It's a measurable fact. Live it and love
it.

In fact you don't even have
the equivalent of a 1F cap.

Then, if not 1F, show me your 'volume' equations for what they 'really
are' in series.

You can't because they ARE 1F in series.

The fact that you have 2 2F caps with an
equivalent of a single 1F *if connected in series in a circuit*
becomes important when you make the system no longer isolated by
discharging.

What 'magical transformation' are you expecting to happen? Do 'missing
coulombs' mysteriously pop in or out?

But the puzzle does *not* "un-isolate" the system. No
current flows.

Which is why they still have the charge.

The serial caps are just two caps, each of which has one
of its leads connected to the other cap, and its other lead floating
free. There is no circuit. Those two caps might just as well be lying
in your parts drawer. There is no equivalent capacitance.

And what 'magical' transformation are you expecting to happen when one
puts a capacitance meter on them and they measure 1F in series?


Ed

I can't begin to capture the whole metadiscussion, which spans
several subjects and 2 (or more?) newsgroups, but at least
I can quote the part where John posted the puzzle:

Quoting from John Larkin's post, 7/10/2010 at 11:16 AM
on SED. Subject: Win Hill: Inverse Marx Generator ??

To celebrate the 21st century, I have composed a new riddle:

Start with a 4 farad cap charged to 0.5 volts. Q = 2 coulombs.

Carefully saw it in half, without discharging it, such as to have two
caps, each 2 farads, each charged to 0.5 volts. The total charge of
the two caps remains 2 coulombs, whether you connect them in parallel
or consider them separately.

Now stack them in series. The result is a 1F cap charged to 1 volt.
That has a charge of 1 coulomb. Where did the other coulomb go?

I think this is a better riddle.

Your rendition seems perfectly fine to me and there is no different
mystery, or any mystery at all.

As I said, you fell for the 'add' diversion. You have two 2F
capacitors charged to .5V and take, hook, line and sinker, 1+1
coulombs as gospel no matter how they're connected so why do you not
think volts 'must add' no matter how they're connected? I.E. If you're
so perplexed by there being only 1 coulomb in the series 1F capacitor
then why are you not perplexed there is only .5V on the parallel 4F
capacitor (or that the 'sum' of the volts on the two 2s is twice what
was there in parallel)?

I'll wager it's because you know that volts do not add in parallel so
why is it so darn hard for you to grasp that coulombs do not add in
series?

There is no '1 + 1 missing coulomb' in series any more than there is a
'missing .5V' when connected in parallel, nor any magical
transformations needed to explain why nothings missing, because,
repeat after me: coulombs in series are all the same as the whole. It
MUST be so because the same current flows through each as the whole.

I gave you the equations and you simply refuse to accept simple math.

Coulombs is CV. You see anything other than C and V in that equation?
No.

.5V x 2F = 1 coulomb

Put two of them in series and you have

1/Cs = 1/Ca + 1/Cb: 1F

Vs = Va + Vb: 1V

1V x 1F = 1 coulomb.

Those are all measurable FACTS.

There is no "1 + 1" mystery because coulombs DO NOT ADD in series,
just as volts do not add in parallel.

Coulombs (current): series: same.
Volts (potential): parallel: same.
Get it?

Note that those equations do not contain, nor does it matter, HOW the
capacitors got there nor what 'volume' they are, nor how they're made,
nor how many plates they have, nor how many capacitors make them up,
nor what the dielectric is, nor how they were charged, nor what
'coordinates' you magically assign them, nor anything else:

Charge them in parallel, charge them in series, charge them singly,
charge them with a iPod, charge them with a lighting bolt from Thor
the thunder god. Leave them in circuit, take them out of circuit,
'isolate' them, or don't. It DOES NOT MATTER. .5V on 2F is 1 coulomb
and 1V on 1F is 1 coulomb; and that's true whether it's 'one'
capacitor, of any form, or any number of capacitors or plates in any
combination of series and parallel connections that comes to the
stated Volts and Farads.

That's all there is and there ain't no more.

It amazes me how much effort you take, to try and get that penny to
move just a little, for JF to realise his Aha! moment...

I don't have your patience at all.

Grant.


John



Maybe that is not the intended
misdirection, but you and I can't get past the point of agreeing
on what volume means.


Really. Where did I ever dispute the meaning of 'volume'?


A coulomb does not have the property of
volume or mass.


And volume or mass does not have the property of charge, spin, or
color. So what?

You act as if the only 'real' things in the universe are volume and
mass so that all else are "only numbers;" and mysterious magical ones
that invisibly 'change' when nothing else has.


Whatever object has been charged does but the
coulomb itself is just a number.


And volume is "just a number."

Volume is the product of measurable quantities. Coulombs is the
product of measurable quantities. We can measure volume. We can
measure coulombs.

There is no mysticism involved.


I'm sorry if that is a repeat,
but it is a fact you don't agree to. We can't get any further.


You can't go any further because you can't put any math to your
'volume' theory, which means you have not solved the 'puzzle' as arm
waving about unexplained 'volume effects' does not solve equations.



I do need to correct 2 points you made below:

"You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it."

Do not lay that puzzle at my doorstep! :-)
(Well, unless you want to stomp on it. Even then attribute
"ownership" where it belongs. :-) )


I didn't say you invented or 'owned' it. I said you introduced it
here.


John introduced the parallel to series puzzle elsewhere.


If you say so.


I attributed it to him in my post. And it's clear from
the context that I'm talking about what someone else posted.


Terrific. And thank you for introducing it to us.


"That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'."

He posted the 'parallel to series' puzzle on SED - obviously you
missed it.


This isn't SED, it's ABSE.


In any event, it is not *my* puzzle, and I didn't pose it.


I didn't say either. All I said is you introduced it.

Good Lord, you act as if mentioning it is a heinous crime you plead
not guilty to.



Ed



Changing the area/spacing ratios would, indeed, change the capacitance
but it's irrelevant as the physical parameters are subsumed into the
given capacitances and further irrelevant as none of the parameters
change anyway.



To go further. A coulomb is a numerical result of a mathematical
operation, not a physical object. There is no physical object
that is an "ampere second". A coulomb is massless, and volumeless.
The puzzle misdirection appeared to be treating the coulomb as an
object that could be divided by dividing that coulomb's container.


It can.

As you yourself pointed out, the story might as well begin with two 2s
in parallel and skip the fluff of 'splitting' the 4 in half as the
result is the same and nothing mysterious happens. You have two 2F
capacitors each with a 'charge' (Q) of 1 coulomb.

And as I pointed out, the entire parallel step is entirely superfluous
as you have the same supposed 'mystery' if you just look at two 2's in
series. Each have 1 coulomb and the total has one coulomb.

Of course, as I also explained, there is no 'mystery' because coulombs
do not add in series. They each and all have the same Q as the whole
because the exact same current goes through each and all as the whole.
I.E. instead of the 'mystery' posited by the diversion 1 + 1 = 1, "huh
where's the missing coulomb?" the correct answer is 1 = 1 = 1, just as
it should.



Area and spacing, which define a volume, the dielectric of the volume
and the applied volts determine the coulombs. It's not an 'arbitrary'
thing and, in that sense, yes it is 'fixed' (by the particulars).
Otherwise one would not be able to write an equation for it.

It's even more the case when simply alternating between parallel or
series connections as none of their parameters change, not even V
(hence Q), and that's about as 'fixed' as one can get.




which seems to be the
misconception at the heart of the so called puzzle.


If you are referring to the, so called, 'missing coulomb' there isn't
one.

Correct, there is no missing coulomb. I've already said that,
earlier in the discussion.


Ok. So I threaded back and, unless I misunderstood, you seem to be
suggesting there is some unexplained 'difference' between whatever you
deem a 'real' 1F capacitor and two 2s in series, despite their
identical behavior, so that two 2s in series are "not a 1F cap" and 1F
"doesn't matter."

Not what I was suggesting.


Ok. Then what did the two 2s are "not a 1F cap" and 1F
"doesn't matter" mean?



In the puzzle the 2 coulombs is a number that tells you how much
current got stuffed into a container and how long the "stuffing"
took. Magically dividing the container into two containers
(or N containers) does not change the amount of current got
stuffed in, nor the amount of time it took to do that stuffing.


Exactly. So what is the machination on "volume" about?




There's no need for mysterious 'differences' as, assuming all else is
equal, a single 1F, two 2s in series, four 4s in series, eight 8s in
series, and so on (although equal values are not required, just that
the resulting series comes to 1F), are black box indistinguishable. At
1V overall they'll each and every one have 1 coulomb because they all
have the exact same charging current go through.


The puzzle's magic division creates the "mysterious differences"
(it results in Vcap = V/N for each cap),


Not so. Capacitors in parallel have the same V and still do when then
'disconnected' from each other (since they're 'ideal' with no
leakage)..



hiding the fact that
total ampere seconds is the same.


There's nothing 'hidden' about 2 parallel coulombs divided into two
being 1 coulomb each.

Albeit superfluous the parallel step is entirely accurate. You begin
with .5V on 4F for 2 coulombs and you end up with two 2s each with .5V
and 1 coulomb.

'Volume', dielectric, and all other physical parameters are irrelevant
as they are subsumed into the 'given' capacitances. It doesn't matter
how they are made as 2F is 2F. In parallel it's 4F and in series it's
1F. That's it.



You may have missed the context of the whole starting from
the beginning, when a "puzzle" (which wasn't a puzzle at all) was
presented.


Paul Hovnanian started the thread with a proposed 'solution' to the
'cap charging a cap' puzzle and mine was the first reply. All I did
was point out that the 'solution' was equivalent to the oft posited
voltage source charging an 'ideal' cap 'puzzle' and that substituting
a cap for the voltage source simply makes the puzzle appear more
mysterious than it is. (A sub thread then ensued about R=0)

You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it.



And I don't think it matters - the puzzle was just "mental
gymnastics" that caused confusion instead of making a point, in my
opinion.


The purpose of a 'puzzle' is to posit what appears to be a conundrum
for the reader to then figure out.

The 'mystery' of the original puzzle is in leaving series R out of the
model and the 'mystery' of the second is being diverted into thinking
coulombs should 'add' when, in series, they do not.



Your paragraph above shows you understand the principle,
so the puzzle is irrelevant. And the op clarified his meaning:
"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." His "puzzle"
was misdirection to show that a coulomb was somehow lost. No
coulomb was lost.


That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'.

His first entry was to me and I did not respond because he hacked my
words to imply I said things I did not and changed the model to boot
so it wasn't even relevant to the discussion.




Perhaps what I've posted isn't clear to you, but if you
think that no coulomb was lost in the puzzle, and that a
coulomb of charge does not require one and only one specific
volume, then we're in agreement.


You still seem to be positing some 'mysterious' thing about 'volume'
that has nothing to do with solving the puzzle. Maybe if you put it to
numbers I could gather what you mean.

Here's my math

Q(Coulombs) = CV, E(Joules) = .5QV (or, by substitution, .5CV^2)

Each: 2F, .5V, 1C, .25J
Parallel: 2F + 2F = 4F, .5V = .5V = .5V, 1C + 1C = 2C, .5J
Series: 1/(1/2F + 1/2F) = 1F, .5V + .5V = 1V, 1C = 1C = 1C, .5J

Note that 'Q per volume' (whatever it might be) of the 2F caps remains
'fixed' because neither Q nor 'volume' change and neither does V, or
anything else. Or at least we can chose to 'pick' that simpler case
because 'volume' is not mentioned anywhere since, being subsumed into
F, it's completely irrelevant.

Energy is conserved, 'charge' (Q) is conserved, and all is right with
the world.

There is no 'puzzle' unless one is fooled by the riddle into thinking
series coulombs should add.



Ed



Ed

snip

On Tue, 27 Jul 2010 13:10:45 +1000, Grant wrote:

On Mon, 26 Jul 2010 15:59:32 -0700, Jim Thompson wrote:
...
Populated Only by Bloviators and Pompous PhD's

Here I was, thinking you made that word up, so google found for me:
"
According to WikiPedia, to bloviate is to "speak pompously and
excessively", or "to expound ridiculously". A colloquial verb
coined in the United States, it is commonly used with contempt
to describe the behavior of politicians, academics, pundits
or media "experts", sometimes called bloviators, who hold forth
on subjects in an arrogant, tiresome way.
" -- http://www.bloviators.net/

There, they have their very own website!

Grant.

The only words I "make up" are shortenings of long silent letter words
like changing "through" to "thru" and creating acoustic expletives
like "Naaaah" :-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Performance only as good as the person behind the wheel.

On Tue, 27 Jul 2010 00:22:36 -0500, flipper wrote:

On Mon, 26 Jul 2010 15:53:21 -0700, Jim Thompson
wrote:

On Mon, 26 Jul 2010 15:37:12 -0500, flipper wrote:

On Mon, 26 Jul 2010 07:57:11 -0700, Jim Thompson
wrote:

On Mon, 26 Jul 2010 00:33:58 -0500, flipper wrote:

On Sun, 25 Jul 2010 21:31:45 -0700, Jim Thompson
wrote:
[snip]
"Ivor Catt" sounds like I should know him... '50's, early '60's ???

Well, the wikipedia article says he was born in 1935 so that would
make him around 75.


...Jim Thompson

I was talking of the year-range where I think I met him... possibly at
Motorola. With a name like that, one rarely forgets :-)

Oh. Well, how in the world would I know THAT? LOL

I do see references to him at Motorola. One says 'mid' 1960s and the
other said 'late' 1960s. I also see 1964 and 1966 referenced in some
papers.


...Jim Thompson

OK. I was Moto from 6/62 to 11/70

Ah. Well, then you very well might have met the Ivor.


...Jim Thompson

Like I said, I'm sure I did... how could you possibly forget a name
like that ?:-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Spice is like a sports car...
Performance only as good as the person behind the wheel.

flipper wrote:

You have totally missed it, either because you just want to argue,
or because you don't recognize or don't understand conservation
of charge. The puzzle starts with a cap with a charge of 2 coulombs.
You claim the end result is one coulomb. Where is the external
connection, to remove charge? The puzzle does not mention one.

"The Law of Conservation of Charge
The net charge of an isolated system remains constant. The only way to
change the net charge of a system is to bring in charge from elsewhere,
or remove charge from the system. "
http://physics.bu.edu/~duffy/semeste...servation.html

Argue with them, if you want.

Ed

On Mon, 26 Jul 2010 18:03:11 -0400, ehsjr
wrote:


flipper wrote:

much snipped

Where you claim the 'puzzle' lies, 'splitting' the 4F into two 2s,
there's no puzzle at all.


That is not what I said at all. I have no problem with magically
cutting the 4F charged to .5 volts and Q=2cap into 2 2F caps
each charged to .5 volts with Q=1 for each cap, as the puzzle
proposes.


Terrific.


As I explained, you have the ENTIRE 'puzzle' if you just put two 2s in
series and charge them up to 1V.

Ok, _now_ I see what you are missing. That is most definitely NOT in
the puzzle. The caps were not *charged* in series. You missed what
the puzzle says.


No, I did not 'miss' what the puzzle says. It's irrelevant as it makes
NO difference if you 'charge' them in parallel, in series, or
separately. The math is IDENTICAL, which is why I explained that the
parallel step is superfluous and you might as well start with them in
series. You have EXACTLY the same situation.



Both 2F capacitors have 1 coulomb and
the whole 1F capacitor has one coulomb. 1 + 1 = 1. Where is the
missing coulomb?

I don't understand your point above - I'm guessing that you're
indicating that coulombs don't add.


That's right. Coulombs do not add in series.


But you are referring
to the wrong scenario, in any event.


No I am not. It is the exact same "scenario" as it does not matter
'how' 1 coulomb gets in the 2F cap. .5V on 2F is 1 coulomb is 1
coulomb is 1 coulomb.

Put the two 2s in parallel, charge them to .5V, and you have 1 coulomb
on each. Charge them in series to 1V overall and you have 1 coulomb on
each. Charge them separately to .5V and you have 1 coulomb on each.

It DOES NOT MATTER. 1 coulomb is 1 coulomb is 1 coulomb.


Not ONE thing 'changes'. They are completely static with not even a
connection change. Explain that one with your 'volumes'.

Don't be absurd. The puzzle stipulates changes.


Nope. The 2F caps have .5V for 1 coulomb throughout the entire
exercise. NOTHING about the capacitors and their charge changes.



The only misdirection is in thinking coulombs add in series when they
don't.

Here

http://farside.ph.utexas.edu/teachin...es/node46.html

Two capacitors connected in series.

"... the fact that the charge is common to all three capacitors. "



Your cite above refers to a non-isolated system.


Oh? And just what are the bare wires sticking out each end 'connected'
to?

Doesn't matter anyway. Volts are Volts and Farads are Farads.


You jumped into
the middle of an ongoing metadiscussion that started on SED, so you
are not aware of what has gone before. There was a large amount of
discussion on conservation of charge, and it spanned more than
one subject. JL made the point that charge is sometimes not
conserved.


Doesn't matter what went on before. Capacitors can't read nor do they
care what sense or nonsense someone wrote.



I'll cut and paste the puzzle as he wrote it, below my comments.
His puzzle takes an isolated system, that starts out *already* charged
with 2 coulombs, and monkeys around with the volumes inside the system.
Your cite above refers to a non isolated system, and is not applicable
to an isolated system.


You see any reference to 'volumes' in CV? No. It's irrelevant.


If you don't like the use of the word volume, fine. Substitute whatever
term you want that includes the concept of the capacity of that volume
and the coordinates that describe the place where your chosen term
exists. Or use whatever term(s) that allow you to realize that
2 2F capacitors in series are NOT *a* 1F capacitor in the
puzzle, without getting sidetracked into thinking that
1/Ct = 1/C1+1/C2...1/Cn applies to the isolated system.


You, again, try to make hay from nonsense. Do you see anything in the
equations that qualify 'how' a "1F capacitor" is 'made'? No. It
doesn't matter if I use two plates or a hundred little plates in
parallel or a dozen large plates in series or a combination. Farads
are Farads.

Two 2s in series ARE "a 1F capacitor" and 1/Ct = 1/C1+1/C2...1/Cn is
not a 'sidetrack', it's a measurable fact.

If it makes you feel any better then pot them in a tube with two leads
sticking out. It's 1F.



The law of conservation of charge applies as long as the system
remains isolated. The number of coulombs remains constant, regardless
of the volume (size, whatever your term is) of the pieces you magically
slice the original cap into, or how you connect them, in that isolated
system.

The puzzle ends up stating that after the slicing & serial connecting,
you have *a* 1F cap charged to 1 volt. That is *false*.


It is perfectly true. You DO have a 1F capacitor charged to 1V.


You have
2 2F caps, each with a charge of 1 coulomb, connected in series.


Correct, which is a 1F capacitor charged to 1V for 1 coulomb.

Repeat after me: coulombs in series are all the same as the whole. It
MUST be so because the same current flows through each as the whole.



Now, because Q = CV, and because charge is constant in an isolated
system and you started with 2 coulombs charge, if you magically merged
the two caps into a 1F cap, then Q=CV means that V would have to be 2,
not 1.


Doesn't bother you blowing conservation of energy to hell and back,
eh?

You've fallen for the 'add' diversion. Coulombs DO NOT ADD in series,
just as amp/hours of batteries do not add in series. Volts do.

It doesn't matter if you "magically merged" the two or connect them in
series. The result is 1V on 1F for 1 coulomb.

Repeat after me: coulombs in series are all the same as the whole. It
MUST be so because the same current flows through each as the whole.


Charge must be conserved.


You are confusing your 'charges' (which, you may have noticed, is why
I usually put the word in tic marks often with an 'explanation' like
(Q) immediately following). A capacitor stores energy.

When 'charging' (or 'discharging') the (ideal) capacitor an electron
enters one plate and another electron exits the other (required by,
guess what, conservation of charge). The 'charges' (electrons, holes)
are rearranged, with equal - and + 'charges' on opposing plates, but
the NET charge (0) is unchanged.

There is no 'problem' conserving 0 whether you 'add', or not, and 0 +
0 is pretty much the same as 0 = 0. Close enough for government work
anyway.

Energy (.5CV^2) is a different matter, of course, and total E = .5J
better be the case no matter how they're connected or else the
universe will fold up into a reverse black hole (joke). Which is a
serious problem for your two .5V 2F caps each with E = .25J, total
E=.5J, merging in series to a 2V 1F cap with E = 2J.

AAAAAaaaaeeeeee-------..... help me

Quick, quick. It's really, just like the equations say, 1V on 1F for 1
coulomb and E= .5J

......-------eeeeeeaaaaAAAAA that's better.



As I used "volume", those two caps occupy different cordinates,


You see anything about "coordinates" in CV? No. It's irrelevant.


they
are not the magical merger the puzzle implies into a 1F cap,


Yes they are. 1/Cs = 1/Ca + 1/Cb. It's a measurable fact. Live it and
love it.


and they
must be considered as such: 2 separate capacitors, not a single 1F
capacitor.


Nonsense.

I give you two leads. You stick a capacitance meter on them and it
reads 1F. Now you tell me how many capacitors and in what arrangement
I used to make up the 1F you just measured.

Hint: No matter what you pick I'll claim it's another arrangement
because it DOES NOT MATTER. Farads are Farads.


Different volumes, no merger of coordinates, no merger of
capacity. There's no current in or out of either cap. It simply does
not matter, in the isolated system, that the total series capacitance
would equal 1F in a non isolated system.


And what magical transformation do you imagine takes place with the
"not a 1F capacitor" when I stick a capacitance meter on them and
measure 1F?


The misdirection seems to me to be in the monkeying around with the
volumes.


Nope. The misdirection is in thinking coulombs 'must add' no matter
how the caps are connected.

Repeat after me: coulombs in series are all the same as the whole. It
MUST be so because the same current flows through each as the whole.


It states that you have a 1F cap.


You do. 1/Cs = 1/Ca + 1/Cb. It's a measurable fact. Live it and love
it.


In fact you don't even have
the equivalent of a 1F cap.


Then, if not 1F, show me your 'volume' equations for what they 'really
are' in series.

You can't because they ARE 1F in series.


The fact that you have 2 2F caps with an
equivalent of a single 1F *if connected in series in a circuit*
becomes important when you make the system no longer isolated by
discharging.


What 'magical transformation' are you expecting to happen? Do 'missing
coulombs' mysteriously pop in or out?


But the puzzle does *not* "un-isolate" the system. No
current flows.


Which is why they still have the charge.


The serial caps are just two caps, each of which has one
of its leads connected to the other cap, and its other lead floating
free. There is no circuit. Those two caps might just as well be lying
in your parts drawer. There is no equivalent capacitance.


And what 'magical' transformation are you expecting to happen when one
puts a capacitance meter on them and they measure 1F in series?



Ed

I can't begin to capture the whole metadiscussion, which spans
several subjects and 2 (or more?) newsgroups, but at least
I can quote the part where John posted the puzzle:

Quoting from John Larkin's post, 7/10/2010 at 11:16 AM
on SED. Subject: Win Hill: Inverse Marx Generator ??

To celebrate the 21st century, I have composed a new riddle:

Start with a 4 farad cap charged to 0.5 volts. Q = 2 coulombs.

Carefully saw it in half, without discharging it, such as to have two
caps, each 2 farads, each charged to 0.5 volts. The total charge of
the two caps remains 2 coulombs, whether you connect them in parallel
or consider them separately.

Now stack them in series. The result is a 1F cap charged to 1 volt.
That has a charge of 1 coulomb. Where did the other coulomb go?

I think this is a better riddle.


Your rendition seems perfectly fine to me and there is no different
mystery, or any mystery at all.

As I said, you fell for the 'add' diversion. You have two 2F
capacitors charged to .5V and take, hook, line and sinker, 1+1
coulombs as gospel no matter how they're connected so why do you not
think volts 'must add' no matter how they're connected? I.E. If you're
so perplexed by there being only 1 coulomb in the series 1F capacitor
then why are you not perplexed there is only .5V on the parallel 4F
capacitor (or that the 'sum' of the volts on the two 2s is twice what
was there in parallel)?

I'll wager it's because you know that volts do not add in parallel so
why is it so darn hard for you to grasp that coulombs do not add in
series?

There is no '1 + 1 missing coulomb' in series any more than there is a
'missing .5V' when connected in parallel, nor any magical
transformations needed to explain why nothings missing, because,
repeat after me: coulombs in series are all the same as the whole. It
MUST be so because the same current flows through each as the whole.

I gave you the equations and you simply refuse to accept simple math.

Coulombs is CV. You see anything other than C and V in that equation?
No.

.5V x 2F = 1 coulomb

Put two of them in series and you have

1/Cs = 1/Ca + 1/Cb: 1F

Vs = Va + Vb: 1V

1V x 1F = 1 coulomb.

Those are all measurable FACTS.

There is no "1 + 1" mystery because coulombs DO NOT ADD in series,
just as volts do not add in parallel.

Coulombs (current): series: same.
Volts (potential): parallel: same.
Get it?

Note that those equations do not contain, nor does it matter, HOW the
capacitors got there nor what 'volume' they are, nor how they're made,
nor how many plates they have, nor how many capacitors make them up,
nor what the dielectric is, nor how they were charged, nor what
'coordinates' you magically assign them, nor anything else:

Charge them in parallel, charge them in series, charge them singly,
charge them with a iPod, charge them with a lighting bolt from Thor
the thunder god. Leave them in circuit, take them out of circuit,
'isolate' them, or don't. It DOES NOT MATTER. .5V on 2F is 1 coulomb
and 1V on 1F is 1 coulomb; and that's true whether it's 'one'
capacitor, of any form, or any number of capacitors or plates in any
combination of series and parallel connections that comes to the
stated Volts and Farads.

That's all there is and there ain't no more.



John



Maybe that is not the intended
misdirection, but you and I can't get past the point of agreeing
on what volume means.


Really. Where did I ever dispute the meaning of 'volume'?



A coulomb does not have the property of
volume or mass.


And volume or mass does not have the property of charge, spin, or
color. So what?

You act as if the only 'real' things in the universe are volume and
mass so that all else are "only numbers;" and mysterious magical ones
that invisibly 'change' when nothing else has.



Whatever object has been charged does but the
coulomb itself is just a number.


And volume is "just a number."

Volume is the product of measurable quantities. Coulombs is the
product of measurable quantities. We can measure volume. We can
measure coulombs.

There is no mysticism involved.



I'm sorry if that is a repeat,
but it is a fact you don't agree to. We can't get any further.


You can't go any further because you can't put any math to your
'volume' theory, which means you have not solved the 'puzzle' as arm
waving about unexplained 'volume effects' does not solve equations.




I do need to correct 2 points you made below:

"You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it."

Do not lay that puzzle at my doorstep! :-)
(Well, unless you want to stomp on it. Even then attribute
"ownership" where it belongs. :-) )


I didn't say you invented or 'owned' it. I said you introduced it
here.



John introduced the parallel to series puzzle elsewhere.


If you say so.



I attributed it to him in my post. And it's clear from
the context that I'm talking about what someone else posted.


Terrific. And thank you for introducing it to us.



"That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'."

He posted the 'parallel to series' puzzle on SED - obviously you
missed it.


This isn't SED, it's ABSE.



In any event, it is not *my* puzzle, and I didn't pose it.


I didn't say either. All I said is you introduced it.

Good Lord, you act as if mentioning it is a heinous crime you plead
not guilty to.




Ed




Changing the area/spacing ratios would, indeed, change the capacitance
but it's irrelevant as the physical parameters are subsumed into the
given capacitances and further irrelevant as none of the parameters
change anyway.




To go further. A coulomb is a numerical result of a mathematical
operation, not a physical object. There is no physical object
that is an "ampere second". A coulomb is massless, and volumeless.
The puzzle misdirection appeared to be treating the coulomb as an
object that could be divided by dividing that coulomb's container.


It can.

As you yourself pointed out, the story might as well begin with two 2s
in parallel and skip the fluff of 'splitting' the 4 in half as the
result is the same and nothing mysterious happens. You have two 2F
capacitors each with a 'charge' (Q) of 1 coulomb.

And as I pointed out, the entire parallel step is entirely superfluous
as you have the same supposed 'mystery' if you just look at two 2's in
series. Each have 1 coulomb and the total has one coulomb.

Of course, as I also explained, there is no 'mystery' because coulombs
do not add in series. They each and all have the same Q as the whole
because the exact same current goes through each and all as the whole.
I.E. instead of the 'mystery' posited by the diversion 1 + 1 = 1, "huh
where's the missing coulomb?" the correct answer is 1 = 1 = 1, just as
it should.




Area and spacing, which define a volume, the dielectric of the volume
and the applied volts determine the coulombs. It's not an 'arbitrary'
thing and, in that sense, yes it is 'fixed' (by the particulars).
Otherwise one would not be able to write an equation for it.

It's even more the case when simply alternating between parallel or
series connections as none of their parameters change, not even V
(hence Q), and that's about as 'fixed' as one can get.





which seems to be the
misconception at the heart of the so called puzzle.


If you are referring to the, so called, 'missing coulomb' there isn't
one.

Correct, there is no missing coulomb. I've already said that,
earlier in the discussion.


Ok. So I threaded back and, unless I misunderstood, you seem to be
suggesting there is some unexplained 'difference' between whatever you
deem a 'real' 1F capacitor and two 2s in series, despite their
identical behavior, so that two 2s in series are "not a 1F cap" and 1F
"doesn't matter."

Not what I was suggesting.


Ok. Then what did the two 2s are "not a 1F cap" and 1F
"doesn't matter" mean?




In the puzzle the 2 coulombs is a number that tells you how much
current got stuffed into a container and how long the "stuffing"
took. Magically dividing the container into two containers
(or N containers) does not change the amount of current got
stuffed in, nor the amount of time it took to do that stuffing.


Exactly. So what is the machination on "volume" about?





There's no need for mysterious 'differences' as, assuming all else is
equal, a single 1F, two 2s in series, four 4s in series, eight 8s in
series, and so on (although equal values are not required, just that
the resulting series comes to 1F), are black box indistinguishable. At
1V overall they'll each and every one have 1 coulomb because they all
have the exact same charging current go through.


The puzzle's magic division creates the "mysterious differences"
(it results in Vcap = V/N for each cap),


Not so. Capacitors in parallel have the same V and still do when then
'disconnected' from each other (since they're 'ideal' with no
leakage)..




hiding the fact that
total ampere seconds is the same.


There's nothing 'hidden' about 2 parallel coulombs divided into two
being 1 coulomb each.

Albeit superfluous the parallel step is entirely accurate. You begin
with .5V on 4F for 2 coulombs and you end up with two 2s each with .5V
and 1 coulomb.

'Volume', dielectric, and all other physical parameters are irrelevant
as they are subsumed into the 'given' capacitances. It doesn't matter
how they are made as 2F is 2F. In parallel it's 4F and in series it's
1F. That's it.




You may have missed the context of the whole starting from
the beginning, when a "puzzle" (which wasn't a puzzle at all) was
presented.


Paul Hovnanian started the thread with a proposed 'solution' to the
'cap charging a cap' puzzle and mine was the first reply. All I did
was point out that the 'solution' was equivalent to the oft posited
voltage source charging an 'ideal' cap 'puzzle' and that substituting
a cap for the voltage source simply makes the puzzle appear more
mysterious than it is. (A sub thread then ensued about R=0)

You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it.




And I don't think it matters - the puzzle was just "mental
gymnastics" that caused confusion instead of making a point, in my
opinion.


The purpose of a 'puzzle' is to posit what appears to be a conundrum
for the reader to then figure out.

The 'mystery' of the original puzzle is in leaving series R out of the
model and the 'mystery' of the second is being diverted into thinking
coulombs should 'add' when, in series, they do not.




Your paragraph above shows you understand the principle,
so the puzzle is irrelevant. And the op clarified his meaning:
"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." His "puzzle"
was misdirection to show that a coulomb was somehow lost. No
coulomb was lost.


That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'.

His first entry was to me and I did not respond because he hacked my
words to imply I said things I did not and changed the model to boot
so it wasn't even relevant to the discussion.





Perhaps what I've posted isn't clear to you, but if you
think that no coulomb was lost in the puzzle, and that a
coulomb of charge does not require one and only one specific
volume, then we're in agreement.


You still seem to be positing some 'mysterious' thing about 'volume'
that has nothing to do with solving the puzzle. Maybe if you put it to
numbers I could gather what you mean.

Here's my math

Q(Coulombs) = CV, E(Joules) = .5QV (or, by substitution, .5CV^2)

Each: 2F, .5V, 1C, .25J
Parallel: 2F + 2F = 4F, .5V = .5V = .5V, 1C + 1C = 2C, .5J
Series: 1/(1/2F + 1/2F) = 1F, .5V + .5V = 1V, 1C = 1C = 1C, .5J

Note that 'Q per volume' (whatever it might be) of the 2F caps remains
'fixed' because neither Q nor 'volume' change and neither does V, or
anything else. Or at least we can chose to 'pick' that simpler case
because 'volume' is not mentioned anywhere since, being subsumed into
F, it's completely irrelevant.

Energy is conserved, 'charge' (Q) is conserved, and all is right with
the world.

There is no 'puzzle' unless one is fooled by the riddle into thinking
series coulombs should add.




Ed




Ed

snip

On Tue, 27 Jul 2010 11:28:14 -0400, ehsjr
wrote:

flipper wrote:

You have totally missed it, either because you just want to argue,
or because you don't recognize or don't understand conservation
of charge. The puzzle starts with a cap with a charge of 2 coulombs.
You claim the end result is one coulomb. Where is the external
connection, to remove charge? The puzzle does not mention one.

"The Law of Conservation of Charge
The net charge of an isolated system remains constant. The only way to
change the net charge of a system is to bring in charge from elsewhere,
or remove charge from the system. "
http://physics.bu.edu/~duffy/semeste...servation.html

Argue with them, if you want.

Ed

[snip]

Absolutely correct...

Only YOU can prevent forest fires. but only "The Bloviator" can
destroy charge or create non-conservation events :-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
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SED Has Crumbled to Below SEB Status
Populated Only by Bloviators and Pompous PhD's

On Tue, 27 Jul 2010 07:33:21 -0500, flipper wrote:

On Tue, 27 Jul 2010 19:50:51 +1000, Grant wrote:

On Tue, 27 Jul 2010 00:21:20 -0500, flipper wrote:

snip

It amazes me how much effort you take, to try and get that penny to
move just a little, for JF to realise his Aha! moment...

That really would make it worthwhile, wouldn't it? To help out an Aha!
moment.

Yes, it's truly wonderful to see the penny drop To help another
get there too, is good.

Besides, I had a bit of fun with magical coulombs and reverse black
holes

Adding a bit of fun can help at times too, maybe break that 'eyes
glazed over' look which I saw way too often when I was trying to
explain my ideas.

Grant.

flipper wrote:
On Sat, 24 Jul 2010 17:31:57 +0100, "Ian"
wrote:

"Jim Thompson" wrote in
message ...
I'm trying to write it up, both I keep getting interrupted by work
that pays $:-)

...Jim Thompson
I'd be interested to see your write up, Jim. This conundrum was asked of me
in a pre-uni interview more than 40 years ago, I forget my real answer, I
think I doubted the assumption of zero resistance. There were some wacky
claims by (I think it was) Ivor Catt in a discussion on this in Wireless
World where he claimed it proved the existence of some wonderful new thingy
("displacement current"?). Much sound and fury, but little edification
ensued as I recall. I should see if I still have the letters.

My understanding is that Ivor Catt doesn't believe in displacement
current or much else of conventional electromagnetism.

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

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

It did cause me to be wary of "stated conditions" aka basic assumptions in
engineering problems in later life ;-)

A wise caution.

Regards
Ian


Of course, if there's no displacement current, there are no
electromagnetic waves, proving that Ivor Catt (whoever he is) can't see
his own shadow.

Moving on.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net