SO how many amps will I get if I charge a ONE FARRAD cap up to 12 volts?

[IconicPCB]

charging with electrons...the more electrons the more electronegative the notionally unearthed terminal

The electron is impeached... freedom to the hole

[danadak]

If I have a C, both plates connected, then Q = C x V = 0 because V is 0, and V is 0 because no
charge differential exists. Note no voltage on the C, even though there could be an equal amount
of excess charge on both plates, or no excess charge. # protons = # electrons

If I then extract some amount of charge from one plate, the other plate unconnected, then I develop
a V = Q / C. The charge comes from wherever, from the universe of charge, not necessarily the other
plate. So now we have a differential amount of charge on C. Thereby exhibiting a V.

Am I missing something ?

Regards, Dana.

[Ratch]

charging with electrons...the more electrons the more electronegative the notionally unearthed terminal

The electron is impeached... freedom to the hole

I thought I made it clear.  The same number of electrons are removed from the second plate as are added to the first plate.  You cannot give a cap a net charge of electrons.

Ratch

[Ratch]

If I have a C, both plates connected, then Q = C x V = 0 because V is 0, and V is 0 because no
charge differential exists. Note no voltage on the C, even though there could be an equal amount
of excess charge on both plates, or no excess charge. # protons = # electrons

If I then extract some amount of charge from one plate, the other plate unconnected, then I develop
a V = Q / C. The charge comes from wherever, from the universe of charge, not necessarily the other
plate. So now we have a differential amount of charge on C. Thereby exhibiting a V.

Am I missing something ?

Regards, Dana.

Yes, you are missing an explanation.  How are you going to extract a charge from one plate without giving the same charge to to other plate?  Charges, like mass don't just appear and disappear.  They have to be accounted for and explained. Applying a voltage across a capacitor will shift some of the charge from one plate to the other for a net charge change of zero, and an energy change of 1/2 C V^2.  That operation is energizing a capacitor.

Ratch

[mikerj]

Not great at Physics, but is there not an equivalency between charge and potential,
one does not exist without the other. So if we charge a cap we can talk about it having
and equivalent coulomb charge, or potential, they are one and the same as related,
Q = C x V. And we can also talk about the energy stored, and its relationship to charge.

Like matter, E = m x c^2, they are inextricably linked to each other. One can talk
about either, both being correct observations.


Regards, Dana.

Let me ask a quick question.  When you speak about charging a capacitor, what are you charging it with?  Please don't say coulombs of charge, because I showed previously that applying a voltage across a capacitor does not increase its net charge.  Another question.  What do you define as "potential"?

Ratch

Applying a voltage (or more accurately a current)  moves charge.  This process is charging.

[Vtile]

Heaviside explained this displacement as a spring analogy and resisted the charge concept to the end.
No I have not read the Electromagnetism 1&2, these are from 2nd hand sources.

Interesting discussion shows how much I have forgotten from the little I knew.

[danadak]

Change charge on one and only one plate -

1) Apply heat to one plate.
2) Radiate/bombard one plate.
3) Photo electric effect applied to one plate.

Of course then we have the issue the plate much more massive than the electron, so what
is the required escape velocity of the electron to break free of plate entirely. After all there
is gravity to be considered. What is the relative force to break free vs force to extract electron
from atom, eg. its coulomb and gravitational and atomic forces. So many questions, so little time.....

And then we have to consider the electrons in the parallel brane universes, what is going
on with them. And can a brane be considered a dielectric ? I do not know. Are we all just
a giant capacitor in the scheme of things ?

https://en.wikipedia.org/wiki/Brane

Then there is the coupling of the Heisenberg uncertainty principle and the virtual particle froth in the
universe, where/when does the anti electron get created ? And does the Heisenberg coupling imply
electrons are in violation of signed Coulomb force, that is they are somehow attracted to each other.
Ga....h head spinning, do capacitors really work ?





Regards, Dana.

[Vtile]

Atleast it doesn't make it any more clear that the electrons do flow against (so called) the current and the positive voltage side of the resistor is the depletion side of the electrons as there is more negative particles on the negative side that in the positive side which have less negative particles while the current do flow upstream (by convention). 

It is as conventional practical as holding the hammer from the head and using the shaft driving the nails.

So we are kind of stupid. We should have left hand rule instead of right hand rule.

[Ratch]

Not great at Physics, but is there not an equivalency between charge and potential,
one does not exist without the other. So if we charge a cap we can talk about it having
and equivalent coulomb charge, or potential, they are one and the same as related,
Q = C x V. And we can also talk about the energy stored, and its relationship to charge.

Like matter, E = m x c^2, they are inextricably linked to each other. One can talk
about either, both being correct observations.

You can't move a charge electrically without a voltage.  With what are you "charging" the capacitor?"

Ratch


Regards, Dana.

Let me ask a quick question.  When you speak about charging a capacitor, what are you charging it with?  Please don't say coulombs of charge, because I showed previously that applying a voltage across a capacitor does not increase its net charge.  Another question.  What do you define as "potential"?

Ratch

Applying a voltage (or more accurately a current)  moves charge.  This process is charging.

[Ratch]

Change charge on one and only one plate -

1) Apply heat to one plate.
2) Radiate/bombard one plate.
3) Photo electric effect applied to one plate.

Of course then we have the issue the plate much more massive than the electron, so what
is the required escape velocity of the electron to break free of plate entirely. After all there
is gravity to be considered. What is the relative force to break free vs force to extract electron
from atom, eg. its coulomb and gravitational and atomic forces. So many questions, so little time.....

And then we have to consider the electrons in the parallel brane universes, what is going
on with them. And can a brane be considered a dielectric ? I do not know. Are we all just
a giant capacitor in the scheme of things ?

https://en.wikipedia.org/wiki/Brane

Then there is the coupling of the Heisenberg uncertainty principle and the virtual particle froth in the
universe, where/when does the anti electron get created ? And does the Heisenberg coupling imply
electrons are in violation of signed Coulomb force, that is they are somehow attracted to each other.
Ga....h head spinning, do capacitors really work ?

Regards, Dana.

Is that the way you would energize a capacitor capacitor in a circuit?

Ratch

[danadak]

No, but physical laws don't change. I am simply questioning the statement made
earlier that net charge on a cap does not change.

Again, just asking questions, not a physicist, if we can't answer these then we
have a deficit in our C working knowledge.


Regards, Dana.

[IconicPCB]

Holy Sheldontalk (TM) ,,,

What is it with You ... The OP asked a simple question out of ignorance... what current will he get from... And lo and behold a pissing contest starts.

Engineers do not get involved in esoteria... whether it is energy expanded in creating a displacement current or work done in separating "charged" plates against an electrostatic field.. it does not matter.
A flow of electrons will take place between two points with different charge density tending towards minimum energy condition along any conductive path which may exist between these two points.

How this minimum energy condition may be reached depends on the path. It may be a long or short process but it will certainly not be a steady state process. IT will be a transient process requiring an analytic solution to define it.

And in a way of a stright forward no BS response to the OP...current is defined as flow of charge past a point. Charge is carried by electrons in classic definition.. although in modern semiconductor theory electrons coexist with holes ( or absence of electrons).

In the old days the Ampere was defined as flow of one coulomb of charge per second.

One coulomb of charge was defined as 6.242 x e18 elementary units of charge.

An elementary unit of charge is the charge carried by an electron ( deemed to be negative ) or a proton ( deemed to be positive)

One elementary unit of charge is defined as 1.6021766208×e?19 coulomb.

In case of a capacitor under steady state conditions a plate become charged with excess electrons.

One plate becomes more electro negative  with respect to the other and conversely one the other plate becomes more electro positive with respect to the first .

The increase of electrons on one plate results in a time varying change in electric field. This is known as displacement current. It is NOT a current of moving electrical charges.

From late forties onwards an ampere was defined in terms of a force which exist between two current carrying conductors.

It is this  force which is employed in D'Arsonval meter movement to provide an indication of current flow .

BAck to OP s question. The capacitor will provide as much current as the path joining the two plates will carry.It will be a time varying current since the charge density on the plates of a capacitor will vary in time. In order to estimate the current path impedance and duration of current flow need to be specified.

[Vtile]

It were answered in three first answers. Rest is just for fun. 

Besides this is really poorly discussed topic atleast on the EE books I have opportunity to get to my hands. There is plates, here is electric field, here is volts between and another is positive another is negative. The type of the model is hardly ever mentioned, nor the application window where it is usefull etc. Even the truth about the plate sizes versus cycletime is forgotten to mention.

Then when you start to dig around of it, you discover it is actually a transmission line and everything were a halftruths, assumptions, simplifications.   

The damn foil roll must be the most misunderstood component of all of them in electric and electronic engineering.

[free_electron]

how long after charging ?
what is the leakage rate ?
what is the esr and esl of the cap
what is the resistance and inductance in the wires ?
what kind of capacitor ( is it made for pulse discharge ? ) is it fat and short or tall and skinny ?

so many factors.

[danadak]

What is it with You ... The OP asked a simple question out of ignorance... what current will he get from... And lo and behold a pissing contest starts.

Engineers do not get involved in esoteria... whether it is energy expanded in creating a displacement current or work done in separating "charged" plates against an electrostatic field.. it does not matter.

I agree OP probably does not need this. It simply cascaded because of a question raised earlier about net charge.
Which brings me to the "esoteria" thought. I had the opportunity to work twice in environments with a team of engineers
and PHds in Physics, and lunch time was full of esoteric in the quest for a deeper understanding of phenomena not
well understood. I loved these sessions. These discussions did matter. Even when i worked on bit slice stuff I went to
lunch with Mrazek (inventer of TriState, head of 2900 team) and my boss and heads of Hybrids (read analog experts)
and we discussed many out of the box kinds of ideas. So yes, if designing an electronics frypan that needs to get to
market right away, I support your point of view, except at lunch and outside work, where esoteric is a good thing.

Regards, Dana.

PS : Not a pissing contest yet, no name calling has occurred.

[IconicPCB]

Hello Dana,

I dont go for name calling either. I was just a little bit miffed by the swirls in the discussion.

[NMNeil]

Funny, in beekeeping circles we have a saying, "Ask 5 beekeepers the exact same question, and you'll get 6 different answers"
This seems to also apply to engineers 

[Vtile]

Funny, in beekeeping circles we have a saying, "Ask 5 beekeepers the exact same question, and you'll get 6 different answers"
This seems to also apply to engineers 

Well both handle matters that sting and do make a buzz sounds, what did you expect.

[NMNeil]

Funny, in beekeeping circles we have a saying, "Ask 5 beekeepers the exact same question, and you'll get 6 different answers"
This seems to also apply to engineers 

Well both handle matters that sting and do make a buzz sounds, what did you expect.

Don't forget the magic smoke, beekeepers and engineers both generate smoke. 

[IconicPCB]

I thought I made it clear.  The same number of electrons are removed from the second plate as are added to the first plate.  You cannot give a cap a net charge of electrons.

Ratch


OK .. i am now sufficiently agitated to offer Maxwel's work in defence of my position:

I refer the reader to the first and second paragraph of

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

for a statement on displacement current and te fact that it is not a flow of chraged particles such as the conventional current or the more correct electron flow.

[Ratch]

OK .. i am now sufficiently agitated to offer Maxwel's work in defence of my position:

I refer the reader to the first and second paragraph of

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

for a statement on displacement current and te fact that it is not a flow of chraged particles such as the conventional current or the more correct electron flow.

And, how does that factoid about displacement current change or contradict the statement that capacitors are energized when the charge on their plates are unbalanced?

Ratch

[rsjsouza]

The way I learned was that, strictly speaking, when capacitors were connected to a DC voltage source they were able to sustain the potential difference (we called d.d.p. or diferença de potencial) across its terminals in open circuit.

We used the term "charge" simply due to the factor a capacitor under a load displays a similar behaviour as a battery - the imbalance of potential causes current to flow to/from its terminals.

The lectures on power and distribution systems used to call circuits and its elements "energized".

[MrAl]

Need to test the balls factor on some parts but by way of scientific method. My brain broke down when converting coulums to zipzaps...
Scariest cap ever!

Your question does not make sense for the reasons others have already given.  Also, a capacitor does not "charge".  It energizes.  A capacitor energized to 100 volts has the same net charge on it as it did at 0 volts.  For every coulomb of charge added to one plate, one coulomb is deleted from the opposite plate.  Capacitors store energy, not coulombs.

Ratch

Hi there Ratch,

From a purely physical viewpoint that may be true, but that limited view will just confuse people who are talking about this from a purely electronic viewpoint.  Surely you must have suspected something isnt right by now since almost everyone here questions that kind of statement.

The difference in the two viewpoints, again, is PURELY PHYSICAL vs PURELY ELECTRONIC.
When we talk about the physical nature of the cap itself from an internal viewpoint, we may want to know that the charge is balanced internally.  BUT there is a big difference when we view that capacitor from the EXTERNAL world.  So here we can say there is an INTERNAL vs EXTERNAL viewpoint difference.

The internal view says that the charge is balanced, but the external view sees it as a differential charge, not an absolute charge.  What this means is that we could have 1/2 of the total charge on one plate while we have a deficit of 1/2 of the total charge on the other plate, yet when we view this from the external world we have a view of the total charge which comes out to:
qTotal=q/2-(-q/2)=q

(or we could just say that the total charge is accounted for on just one plate only)

This is similar to when we use a plus and minus power supply (plus Vcc and minus Vee) and measure the voltage from Vcc to Vee instead of from one or the other to ground.  We dont get zero volts, which would be incredibly ridiculous.  We get the difference:
V=Vcc-Vee

and since Vee is negative if Vcc=5 and Vee=-5 we get:
Vtotal=Vcc-Vee=5-(-5)=10 volts.

THAT is the way we MUST look at it in the electronics world from the viewpoint of an electrical circuit and it's electrical analysis.  If we want to know what the cap is doing alone, we might look at that alone, but in a circuit we MUST know ther total charge equivalent which we almost always just call simply, "The Charge".

I dont think i can make it any more clear than that although the more usual view is that we just have charge on one plate that is equivalent to the 'total charge'.

You may want to note that there are other times when we have to depart from the purely physical view in the electronics world.  We sometimes have to apply different attributes to the same components depending on the actual application.  in modern times this would be called "anisotronic" behavior, where we can not analyze the circuit (note this is particular to circuit analysis) without knowing the multifaceted behavior which depends on the particular theoretical application at hand.

[hamster_nz]

When you 'charge' a cap, you are moving charge carriers (electrons) from one of it's terminals to the other.

Imagine charging a cap from a battery with a current limting resistor in series. A meter reading amps in series with the positive terminal will read the same as another meter connected in series with the negative terminal.

It will start off with the cap appearing to be a dead short (the full battery voltage will  be measured over the current limiting resistor), and then quickly fall away as the voltage across across the cap increases.

Once fully charged 0 amps will be flowing, and so the current limiting resistor will have zero volts over it - (although you can theoretically never fully charge a cap to the match a voltage source, only get very, very close - the whole exponential decay thingy).

And it is possible to add charge to one terminal of a capacitor (e.g a static charge), but that sort of makes a second capacitor between the uncharged terminal and the external environment...

[Assafl]

charging with electrons...the more electrons the more electronegative the notionally unearthed terminal

The electron is impeached... freedom to the hole

I thought I made it clear.  The same number of electrons are removed from the second plate as are added to the first plate.  You cannot give a cap a net charge of electrons.

Ratch

It may appear to be that there are capacitors that can accept a net charge as well. Leyden Jars are good examples of that. The physical mechanisms are the same but diametrically the opposite in quantitative terms. Caps can be tiny, but due to the size of the plates, distance between them and the dielectric constant, can accept a fairly large charge to opposed a reasonable electric field.

A Leyden Jar, on the flip side - is tiny, requires a very large electric field to charge it, and seemingly doesn't have an opposing plate - hence the net charge "electrostatic" claim (of course it does - the air around it is charged - that is why the air around it lights florescent bulbs, and is positively dangerous)...

One of the rather amazing (if unpleasant) surprises to owning an electrometer is that you realize everything around you (with the exception of good conductors) - including us humans - is a capacitor.