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Hello all,

I have been boggled by this problem the whole day today. Lacking a charge electrometer, I am limited in the ways I can test my theories.

I'll explain the problem in small steps:

Suppose you have two capacitors, each with the same capacitance. Lets say, 1 farad each.

You apply 5 Volts to each capacitor individually, thus having them store 5 Coulombs of charge each. (Charge / Capacitance = Voltage).

From my best understanding of charge storage on capacitors, (based on wikipedia), 5 coulombs of charge is on one plate the capacitor and -5 coulombs of charge is on the other plate. This applies for both of the capacitors.

To make the explanation easier, label the first capacitor's positive plate A+, its negative plate A-. Label the second capacitor's positive plate B+, its negative plate B-.

If I connect A- and B+ to my precise voltmeter, I get 0 V.

While the voltmeter is connected, if I connect A+ to B-, then I get 10 V on my voltmeter.

In classical terms of electromagnetics, this makes perfect sense. By connecting A+ and B-, the two capacitors are essentially in series, both with +5 V. Thus you get +10 V on the voltmeter.

This is, however an issue that I cannot get around:

If you have the two isolated capacitors, and connect A- to B+ with a very low resistance wire, apparently (as far as I was able to tell) there is absolutely no charge transfer between the two points. Let me restate this: There is 5 Coulombs on one end of a wire and -5 Coulombs on the other end of the wire and there is no movement at all of these charges.

If the two charges were isolated from the capacitors and connected by the wire, they would neutralize.. probably with a loud bang because 5 Coulombs is a lot of charge.

At first I tried to argue against this logic with the fundamental idea of a capacitor:

A- is attracted to A+. B- is attracted to B+. In order for charge to both decrease in A- and B+, (by decrease I mean neutralize.. charge is conserved), it must decrease in A+ and B-. Otherwise there would be excess charge in A+ and B-.

This means that the force in the capacitors between A+ and A- is a LOT more powerful than A- and B+, even though the charge of A+ = B+.

The force between A+ and A- is based on the capacitance. (Distance between the plates and the area of the plates.) However what is the force between A- and B+? ...assume 0 resistance wire.

...Maybe it's not called "Force" but something else.. either way, the logic stays the same.

The "pull" between A+ and A- is apparently much stronger than the "pull" between A- and B+ because there is no interaction when B+ and A- are connected.

I would leave is problem to rest except that it shows a small issue:

Even if A+ and A- attract each other much more than A- and B+ does, you would expect that at least an extremely small amount of charge would move between A- and B+ when they are connected.

This should in theory make A+ have more charge than A-.

Likewise B- would have more charge than B+.

However I have never seen this in practice.

I have searched for a solution for some time, and remembered an interesting phenomena:

Emission mechanism (for example thermionic and cold emission) has negative charge basically coming out of the conductor, into the air (or wherever.. could be a vacuum). Apparently high voltage, or high temperature is enough to overcome the attraction between A+ and A-.

Well then... Why doesn't the attraction between A- and B+ not give any noticeable charge movement at all?? Even though the voltage is only 5 or 10 Volts, 5 and -5 Coulombs on both ends should amount for something, no?

In mathematical terms, what is the force between A- and B+? ..regardless of this force, why isn't there any noticeable charge movement at all?

Thank you very much for reading. I have given much thought to this problem.

I hope you have something useful to share.

(Note - For those of you looking for an error in what I have written, I have simplified the problem to make it easier to understand. The total "pull" is (A- and B+) vs (A+ and A-, and B+ and B-). Meaning 2 times the "pull" of (A+ and A-) vs (A- and B+). Either way there should still be a little charge transfer between A- and B+ in theory.. Unless I'm missing something.)

I have been boggled by this problem the whole day today. Lacking a charge electrometer, I am limited in the ways I can test my theories.

I'll explain the problem in small steps:

Suppose you have two capacitors, each with the same capacitance. Lets say, 1 farad each.

You apply 5 Volts to each capacitor individually, thus having them store 5 Coulombs of charge each. (Charge / Capacitance = Voltage).

From my best understanding of charge storage on capacitors, (based on wikipedia), 5 coulombs of charge is on one plate the capacitor and -5 coulombs of charge is on the other plate. This applies for both of the capacitors.

To make the explanation easier, label the first capacitor's positive plate A+, its negative plate A-. Label the second capacitor's positive plate B+, its negative plate B-.

If I connect A- and B+ to my precise voltmeter, I get 0 V.

While the voltmeter is connected, if I connect A+ to B-, then I get 10 V on my voltmeter.

In classical terms of electromagnetics, this makes perfect sense. By connecting A+ and B-, the two capacitors are essentially in series, both with +5 V. Thus you get +10 V on the voltmeter.

This is, however an issue that I cannot get around:

If you have the two isolated capacitors, and connect A- to B+ with a very low resistance wire, apparently (as far as I was able to tell) there is absolutely no charge transfer between the two points. Let me restate this: There is 5 Coulombs on one end of a wire and -5 Coulombs on the other end of the wire and there is no movement at all of these charges.

If the two charges were isolated from the capacitors and connected by the wire, they would neutralize.. probably with a loud bang because 5 Coulombs is a lot of charge.

At first I tried to argue against this logic with the fundamental idea of a capacitor:

A- is attracted to A+. B- is attracted to B+. In order for charge to both decrease in A- and B+, (by decrease I mean neutralize.. charge is conserved), it must decrease in A+ and B-. Otherwise there would be excess charge in A+ and B-.

This means that the force in the capacitors between A+ and A- is a LOT more powerful than A- and B+, even though the charge of A+ = B+.

The force between A+ and A- is based on the capacitance. (Distance between the plates and the area of the plates.) However what is the force between A- and B+? ...assume 0 resistance wire.

...Maybe it's not called "Force" but something else.. either way, the logic stays the same.

The "pull" between A+ and A- is apparently much stronger than the "pull" between A- and B+ because there is no interaction when B+ and A- are connected.

I would leave is problem to rest except that it shows a small issue:

Even if A+ and A- attract each other much more than A- and B+ does, you would expect that at least an extremely small amount of charge would move between A- and B+ when they are connected.

This should in theory make A+ have more charge than A-.

Likewise B- would have more charge than B+.

However I have never seen this in practice.

I have searched for a solution for some time, and remembered an interesting phenomena:

Emission mechanism (for example thermionic and cold emission) has negative charge basically coming out of the conductor, into the air (or wherever.. could be a vacuum). Apparently high voltage, or high temperature is enough to overcome the attraction between A+ and A-.

Well then... Why doesn't the attraction between A- and B+ not give any noticeable charge movement at all?? Even though the voltage is only 5 or 10 Volts, 5 and -5 Coulombs on both ends should amount for something, no?

In mathematical terms, what is the force between A- and B+? ..regardless of this force, why isn't there any noticeable charge movement at all?

Thank you very much for reading. I have given much thought to this problem.

I hope you have something useful to share.

(Note - For those of you looking for an error in what I have written, I have simplified the problem to make it easier to understand. The total "pull" is (A- and B+) vs (A+ and A-, and B+ and B-). Meaning 2 times the "pull" of (A+ and A-) vs (A- and B+). Either way there should still be a little charge transfer between A- and B+ in theory.. Unless I'm missing something.)

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