Let's say we have a 120 Volt AC source

[Drakkith]

Jared I don't think you understand what I meant. If you cut the high voltage wire carrying the power, you won't suddenly have the air conducting the electricity. The resistance in the air is simply too great. If you connect a smaller diameter cable, which has more resistance than the main one, current flows through both cables. There is simply a greater current in the cable with less resistance.

Evilbunny, lightning is similar to one of the poles of your generator. As the difference in electric potential gets higher and higher, it finally reaches a point where the difference is so great that the actual air ionizes and is stripped of electrons. When this happens the newly formed ions conduct electricity very well and the potential suddenly equalizes between the two points, causing a lightning bolt.

 

[Drakkith]

A battery is a voltage source. Is lightning similar to a battery?

Honest question...

A battery not connected to anything is like the conditions right before lightning forms. The difference is that in a battery the electric potential difference is not high enough to overcome the resistance of the internal parts of the battery. When you complete the circuit the electrons now have a low resistance path to get to the positive side of the battery and so you have current flowing. Once a battery is dead the positive side has received enough electrons to neutralize most of the positive ions there while the negative side has lost enough electrons from its negative ions to neutralize most of them as well. In a battery the ions gain and lose electrons through chemical reactions with their respective electrodes.

 

[JaredJames]

Jared I don't think you understand what I meant. If you cut the high voltage wire carrying the power, you won't suddenly have the air conducting the electricity. The resistance in the air is simply too great. If you connect a smaller diameter cable, which has more resistance than the main one, current flows through both cables. There is simply a greater current in the cable with less resistance.

Ah I'm with you.

I was looking at it from the potential difference stance. For a given PD there will be a 'maximum' resistance through which it will induce a current to flow. It won't flow through the air because the resistance is too high, but if you increase the PD then it can flow, eventually (lightning). Of course, the distance between 'poles' also comes into play.

In the case of the wire as per your example, it's still the path of least resistance (in comparison to all other paths) but it is also at a resistance that will allow the current to flow.

 

[Evil Bunny]

I think we are all in agreement that electrons will never ever want to go into the ground unless the ground ends up in the circuit as a return path back to the source.

We don't agree with that. You only have to look at lightning...

As Drakkith pointed out... Your lightning example has the ground in the circuit. In fact, it is actually the source in that particular circuit. One of the two poles of the source to be precise. The cloud would be the other pole.

So do you agree with it now?

And Drakkith, thank you for your help in clarifying things... I do still have a question for you on that arc. Are you saying that the arc itself is supplying the electrons needed to equalize the circuit? Or are you saying that the arc is a path leading from one pole to the other? The arc still needs to "touch" both poles in order for things to equalize, correct?

 

[JaredJames]

As Drakkith pointed out... Your lightning example has the ground in the circuit. In fact, it is actually the source in that particular circuit. One of the two poles of the source to be precise. The cloud would be the other pole.

So do you agree with it now?

My disagreement was with the end part what you said.

Where is the "return path" when it comes to lightning?

 

[Evil Bunny]

My disagreement was with the end part what you said.

Where is the "return path" when it comes to lightning?

The air is the return path. The big bright lightning bolt that flashes in the air is where the current is traveling.

The source, in this case, is the combination of the cloud and the earth. It's traveling from one pole (the cloud) of the source to the other pole (the earth) of the source. It is traveling in a circuit.

A really big loud one.

 

[Per Oni]

Okay... I'll rephrase that then...

Electrons acting on forces derived from an AC or DC voltage source with two poles will never ever want to go into the ground unless the ground ends up in the circuit as a return path back to the source.

Do you agree with that?

No I don’t and I’ll try to explain why. I’m not good at explaining but I’ll try my best.

As has been suggested before replace your battery with a charged capacitor. The advantage is that we need things to be as simple as possible to go to the hart of this problem. Also place this capacitor in a vacuum. Therefore we get rid of chemical processes and air. This way these 2 plates replace the terminals of your power supply. These plates are normally large in surface area and placed very close together. One is charged +ve, the second has an equal –ve charge to a voltage of (say)12 Volts. To a first approximation none of the electrons are attracted to the Earth or any other conductor, since all are attracted to the opposite +ve plate. However this situation alters dramatically if you start separating the plates further apart (having the power source disconnected). What will happen is that the voltage of this capacitor will increase since you need force to pull them apart and therefore you are supplying energy in this system. If you pull them so far apart that their distance is many times larger then the length and width of the plates then the final voltage is dependent on the size of these plates but that is not important here. What is important is that you have now –ve isolated charges with at first glance no +ve charges to be seen. This is the equivalent of the charged thunderclouds which have been discussed here.

Charges are surrounded by electrical fields lines and each line has to start and end on a charge. Since the +ve plate is far away the nearest conductor is (say) the earth. Most of the lines are now going to Earth and the electrons are attracted that way. If the new voltage stays below a certain value and ignoring very small leakage currents then the electrons cannot go to earth, except when we connect this plate with a conductor to ground. So you see these electrons will travel to Earth even though there’s no a return path.

 

[JaredJames]

...

Not bad at all.

 

[Drakkith]

Evilbunny, don't confuse lightning with an electric circuit. They aren't the same thing. Lightning is merely a very dramatic effect of static electricity. There is no circuit. There is only a difference in electric potential between two points.

The arc is the path through the air which has been ionized and has current passing through it. When the electrons are stripped from their atoms they create a plasma. Instead of all these neutral molecules that resist the movement of electrons, you now have a section of air that has free electrons that can be easily moved. The insulating air has been turned into a conductor. And yes, the arc would have to touch both poles.

 

[Evil Bunny]

These plates are normally large in surface area and placed very close together. One is charged +ve, the second has an equal –ve charge to a voltage of (say)12 Volts. To a first approximation none of the electrons are attracted to the Earth or any other conductor, since all are attracted to the opposite +ve plate. However this situation alters dramatically if you start separating the plates further apart (having the power source disconnected). What will happen is that the voltage of this capacitor will increase since you need force to pull them apart and therefore you are supplying energy in this system. If you pull them so far apart that their distance is many times larger then the length and width of the plates then the final voltage is dependent on the size of these plates but that is not important here. What is important is that you have now –ve isolated charges with at first glance no +ve charges to be seen. This is the equivalent of the charged thunderclouds which have been discussed here.

Interesting post...

Capacitance is the ability of something to hold a charge. The equation for capacitance is basically the surface area of the plates divided by the distance between the plates. Seems to me that your capacitance will be rapidly approaching zero as you increase the distance between these plates... Why wouldn't you lose your ability to hold a charge in these plates as you separate them? Maybe this equation doesn't apply if the distance gets too big?

...and if you could charge a capacitor, rip it in half, and walk away with a charged plate, why couldn't you do the same with a wire that was attached to a battery? Connect a wire to the battery, disconnect it, and walk away with a charged wire. Does it work like that?

Am I getting something mixed up here?

 

[Studiot]

...and if you could charge a capacitor, rip it in half, and walk away with a charged plate, why couldn't you do the same with a wire that was attached to a battery? Connect a wire to the battery, disconnect it, and walk away with a charged wire. Does it work like that?

I think you are confusing potential (voltage) and charge here. They are different concepts which is why we have both terms.

Capacitance is the ability of something to hold a charge.

Is it? I think that a positive charge will hold onto a negative one very well indeed if allowed to.
Does that make a positive charge a capacitor?

 

[Evil Bunny]

I think you are confusing potential (voltage) and charge here. They are different concepts which is why we have both terms.

I am in a constant state of confusion... You're absolutely right about that. I think he (or she, no offense) said we could rip a cap in half that had 12 volts across it and walk around with a charged plate. Why doesn't this work with the wire on the battery? Or does it? Is there no charge on the pole of a battery?

 

[Studiot]

This calls for one of my lightning sketches.



In the top sketch I have shown what happens if you connect one terminal only of a resistor, or a capacitor or a wire to one pole of a battery.
The second terminal is not connected in each case.
I have done this via a switch and numbered the terminals as shown.

The battery voltage is V_b

Before we make the connection the voltage at terminals 1, 2, 3, 4, 5 and 6 are indeterminate.

In the first position the switch connects to the resistor leaving the cap and wire not connected,
So there is no change at the cap and wire ie their voltage remains indeterminate.
At the resistor, terminals 1 and 4 take on the battery voltage V_b.

When we connect the switch to the capacitor, the resistor looses its connenction to the battery and thus its voltage so terminals 1 and 4 are now indeterminate again.
Terminal 2 on the capacitor takes on the battery voltage V_b. It has no charge at this stage just voltage.However the capacitor blocks terminal 5 from the battery so terminal 5 voltage remains indeterminate.

When we switch to the wire (terminal 3) terminal 2 on the capacitor looses its connection and its voltage becomes indeterminate. It has no charge to loose.
The whole of the wire including terminals 3 and 6 takes on the battery voltage V_b.

However it only the voltage changes in all the above. No charge moves, no current flows and in particular the capacitor is uncharged.

In the bottom sketch we repeat the experiment, but this time all the second terminals are connected to the other pole of the battery.

When we now set the switch to connect the resistor terminal 1 of the resistor takes on the battery voltage V_b. Terminal 3 on the resistor is connected to the other pole of the battery so charge moves and a current flows through the resistor. The voltage across the resistor is equal to the battery voltage V_b.
The charge flows continuously and evenly in time so there is no build up (or deficit) of charge anywhere.

Because there is no buildup of charge in the resistor, the voltage at terminal 1 becomes the same as the voltage at terminal 4 as soon as we disconnect by switching to the capacitor. There is no charge stored in the resistor to cut off and take away.

This time when we connect the switch to terminal 2 on the capacitor charge does move, but unevenly in time. The charge flows quickly at first and then ever more slowly collecting on the one plate of the capacitor since it is blocked from passing through the capacitor.
The accumulation of charge 'induces' a similar accumulation of opposite charge on the other plate of the capacitor. We say that charge has been separated and the capacitor is charged.
You can indeed now disconnect the capacitor and walk away with it and even cut it in half.

Finally if we switch to just the wire we have a short circuit which get as hot as my keyboard after typing all this.

 

[Evil Bunny]

Studiot,

I agree with everything you posted. Here is where everything gets a little fuzzy... At this point we have the battery terminals connected across the capacitor:

This time when we connect the switch to terminal 2 on the capacitor charge does move, but unevenly in time. The charge flows quickly at first and then ever more slowly collecting on the one plate of the capacitor since it is blocked from passing through the capacitor.
The accumulation of charge 'induces' a similar accumulation of opposite charge on the other plate of the capacitor. We say that charge has been separated and the capacitor is charged.
You can indeed now disconnect the capacitor and walk away with it and even cut it in half.

Let's disconnect the charged capacitor, cut it in half and walk out into the back yard with one of the halves (plates)... Will current flow between this plate and the Earth if we stick it in the ground outside?

 

[Studiot]

Sure it will, although it is better to say charge will flow. It will not matter which plate (- or +) that just determines the direction of flow.

However the flow rate will diminish rapidly with time. Time is always an important variable where capacitors are concerned.

I agree with everything you posted.

That's kind of you. The question is did you understand it?

go well

 

[Studiot]

I should add, before signing off for tonight, that you should never consider Earth as 'completing a circuit'.

The action of an Earth is entirely different, and does not require the completion of a circuit.

 

[Evil Bunny]

Sure it will, although it is better to say charge will flow. It will not matter which plate (- or +) that just determines the direction of flow.

Okay, so what if we didn't rip the capacitor in half... If we just stuck one of the cap leads into the Earth (and let's just say that we taped up the other cap lead really well and it was isolated), would current flow?

If your answer is yes, then it contradicts some of the posts in this thread, doesn't it?

If your answer is no, what is different about the scenario of us cutting it in half?

The question is did you understand it?

I understood everything except the part we're talking about now.

 

[rcgldr]

Okay, so what if we didn't rip the capacitor in half... If we just stuck one of the cap leads into the earth.

Each plate of a capacitor has a some amount of negative or positive charge. Connecting either plate to the Earth or any large conducting mass with zero charge would involve a temporary current flow until the charge on the plate and whatever the plate was connected to became uniformly distributed.

This is a bit different than the case with a battery, where the amount of charge per unit mass of the terminals is very low, and significant and sustained current relies on the internal transfer of electrons within the battery related to chemical reactions.

 

[Evil Bunny]

So which is it?

Are you familiar with a capacitor?, where equal charge is stored on either side of, say, parallel plates?...it's the electrical equivalent of what you example. Connecting either side to ground, individually, does NOT remove any charge from either side...electrons on the negative side (ungrounded) will oppose any more electrons being added if the positive side is grounded...in other words, the potential (voltage difference) won't change until a closed circuit is available...

Each plate of a capacitor has a some amount of negative or positive charge. Connecting either plate to the Earth or any large conducting mass with zero charge would involve a temporary current flow until the charge on the plate and whatever the plate was connected to became uniformly distributed.

 

[Studiot]

The answer, my friend, is blowing in the wind.

Time is always an important variable where capacitors are concerned.

In time if you connect either plate to ground, whilst insulating the other one, the charge from that plate will leak away.

 

[quietrain]

hey, i asked about something like this too, but from what i understand thus far, when you connect 1 side of the generator to the earth, there will be a very quick balancing of potentials so you would see a very quick surge of PD on your multimeter and then 0 after equalization of potential takes place.

i think that's how it works?

 

[Studiot]

Hey quietrain I was answering a question about charge, not potential

From earlier in this thread.

I think you are confusing potential (voltage) and charge here. They are different concepts which is why we have both terms.

 

[cabraham]

Let's say that V_A is the electric potential at pole A, V_B at pole B, and V_Earth of the earth, all measured relative to some common reference point.

So you're saying that

V_A - V_B = 120 V

but

V_A - V_Earth = 0 V

AND (somehow!)

V_B - V_Earth = 0V

You see that this is impossible, right? If pole A and pole B are not at the same potential, then they can't BOTH have the same potential relative to Earth.

If, for the sake of argument, V_B = V_Earth, then it must follow that V_A - V_Earth = 120 V (NOT 0 V)

So, in this example, if you connected pole A to Earth, current would flow, since pole A has a positive potential relative to Earth.

Instead of stating "Va - Vearth = 0V, etc.", let's look at it as follows.

Vae = small, Vbe = small, Ve1e2 = small, & Vab = 120 V. No paradox there. The "Ve1e2" is the voltage across 2 Earth points.

In reality, the 120V ac source mains do have an impedance to ground. There is some stray capacitance between each line & ground, namely hot-ground, & neutral-ground, Chg & Cng resp.

A small capacitance results in a high reactance resulting in little current. Also, the Earth has impedance. Soil is not a good conductor, especially when dry.

Measuring the line to Earth voltage can be tricky. Let's say the DMM has an ac impedance of 5 Mohm. If the impedance of the lines to Earth is the same order of magnitude of the DMM, or higher, than the DMM impedance skews the readings.

Let's say the hot-earth capacitive reactance is 20 Mohm, as well as the neutral-earth reactance. A voltage divider results so that if the DMM is connected hot-earth, then the 120V divides across 5 Mohm||20 Mohm & 20 Mohm. The DMM will read (120V)*((5||20)/(20+(5||20))) which computes to a mere 20V.

But the neutral to ground voltage is the balance of 100V. When the meter is removed, both line to Earth voltages are 60V, since the impedances are equal. If the DMM is placed across the neutral-earth, it will read 20V. Kirchoff's voltage law seems to have been breached but it hasn't. The DMM impedance is not always a lot higher than the network being measured, so it can influence the reading.

In real life, any impedance significantly smaller than the stray capacitive reactance will affect the voltage division. Placing a low impedance across one line & Earth results in a smaller fraction of the 120V appearing on that node. Did I explain it well?

There is a potential from hot-ground, & neutral-ground, as well as across 2 differing points on the earth. Does this help?

Claude

 

[quietrain]

Hey quietrain I was answering a question about charge, not potential

From earlier in this thread.

ok, but i was referring to evil bunny first post, where he says he sees no pd on his multimeter.

 

[Evil Bunny]

talking about a capacitor in a vaccum..

These plates are normally large in surface area and placed very close together. One is charged +ve, the second has an equal –ve charge to a voltage of (say)12 Volts. To a first approximation none of the electrons are attracted to the Earth or any other conductor, since all are attracted to the opposite +ve plate. However this situation alters dramatically if you start separating the plates further apart (having the power source disconnected).

This is an interesting scenario that hasn't been fully explored... As the distance between the plates increases, the capacitance decreases. The only reason the charge was on the plate in the first place was because it was being held there by the (opposite) charge on the other plate. What happens when the capacitance disappears? The attraction that was holding the charge there is now gone, correct? It's no longer a capacitor... it's now just a metal plate. What happened to the charge? The ability to hold a charge (capacitance) is no longer there... I guess the electrons didn't just fall off and land on the earth... they must still be there? What's holding them there?

What will happen is that the voltage of this capacitor will increase since you need force to pull them apart and therefore you are supplying energy in this system. If you pull them so far apart that their distance is many times larger then the length and width of the plates then the final voltage is dependent on the size of these plates but that is not important here. .

This part I don't understand. If we charged a capacitor and then removed the voltage source and increased the distance between the plates, our voltage would rise? This doesn't sound right to me...

What is important is that you have now –ve isolated charges with at first glance no +ve charges to be seen. This is the equivalent of the charged thunderclouds which have been discussed here. .

So what we have accomplished here is using the energy from an electric circuit to create a build up of negative (or positive) charge and created a whole new "static" electrical potential... Are we sure everything would work here like it has been layed out?

 

[Per Oni]

No doubt you know the formula relating charge q, voltage U, and capacitance C. q=UC. This determines the amount of charge on the capacitor. Being charged and the supply removed we can increase or decrease C as we like. In our case as you point out C is decreasing almost disappearing. Above formula still holds: U=q/C so U is indeed increasing as I pointed out. However as C approaches zero it looks as if U becomes infinite. That is impossible since that would imply an infinite energy content. Where does the formula break down? Well in post #83 cabraham talks about “stray capacitance”. I do approach this problem very differently but stray capacitance or parasitic capacitance is one of the keys. Another is self capacitance.
From wiki:

Self-capacitance
In electrical circuits, the term capacitance is usually a shorthand for the mutual capacitance between two adjacent conductors, such as the two plates of a capacitor. There also exists a property called self-capacitance, which is the amount of electrical charge that must be added to an isolated conductor to raise its electrical potential by one volt. The reference point for this potential is a theoretical hollow conducting sphere, of infinite radius, centered on the conductor. Using this method, the self-capacitance of a conducting sphere of radius R is given by:[19]

Example values of self-capacitance are:
for the top "plate" of a van de Graaff generator, typically a sphere 20 cm in radius: 20 pF
the planet Earth: about 710 μF[20]

All this stems from the fact that charges are in principle repelled or attracted by all other charges in the universe. Coulombs law never breaks down. This is the same as for gravity. You are attracted to Earth but also to a planet far far away.
In our case it means that electrons on the –ve plate are attracted in principle to the +ve plate but they also are at all times attracted to charges on Earth or power lines etc. In fact to determine our original charge we should also have included all the other real but small capacitances. All these add up as in parallel circuits. In reality we never have to do this (unless you are working with high frequency signals). However when decreasing our original 2 plate capacitance, all the other capacitances are becoming more and more important so that U never reaches infinity.

Now you also ask the question why don’t the electrons not just fall off? Which is a good question, not many texts go there. However you did not ask that point when the plates were close together. At that point in time all the electrons were attracted to the opposite plate. What holds them there?

In mechanic courses on forces you see pictures with arrows one way and a counter arrows the opposite way so that every thing is balanced even under accelerations. Why is electricity not treated that way? Why (in electricity) are the arrows pointing only one way even when charges are clearly at rest on this plate?
Also: if charges are attracted to the inside opposite plates how come we can discharge with a wire connected to the outside of the 2 plates?
I hope some other people will come to our rescue, since I’m going to watch some rugby. If not I’ll write some more. As far as I’m concerned you are asking the right questions.

 

[Evil Bunny]

Now you also ask the question why don’t the electrons not just fall off? Which is a good question, not many texts go there. However you did not ask that point when the plates were close together. At that point in time all the electrons were attracted to the opposite plate. What holds them there?

Well... what holds them there is the attraction. Like putting a piece of paper on the refrigerator with a magnet. the magnet (one plate) stays in place while the paper (dielectric) is in between it and the refrigerator (other plate).

Also: if charges are attracted to the inside opposite plates how come we can discharge with a wire connected to the outside of the 2 plates?

The charges are trying to get to each other and equalize, right? If you touch a wire to the two plates, you just gave them a path to get together and celebrate and have a party and, most importantly, equalize.

In all seriousness, I think this capacitance equation only holds true if the size of the parallel plates are very large compared to the distance between them. I think we could set up an experiment with a makeshift capacitor and move the plates away from each other and see if the voltage increases... I don't think it would. But I could be wrong. Wouldn't be the first time.

 

[shoestring]

Now you also ask the question why don’t the electrons not just fall off? Which is a good question, not many texts go there. However you did not ask that point when the plates were close together. At that point in time all the electrons were attracted to the opposite plate. What holds them there?

In mechanic courses on forces you see pictures with arrows one way and a counter arrows the opposite way so that every thing is balanced even under accelerations. Why is electricity not treated that way? Why (in electricity) are the arrows pointing only one way even when charges are clearly at rest on this plate?
Also: if charges are attracted to the inside opposite plates how come we can discharge with a wire connected to the outside of the 2 plates?
I hope some other people will come to our rescue, since I’m going to watch some rugby. If not I’ll write some more. As far as I’m concerned you are asking the right questions.

It takes a certain energy to free an electron from the metal, that's why they stay (unless temperature or field gets really high) The photoelectric effect (for which Einstein got the Noble prize) is all about which minimum photon energy (i.e. frequency of the light) it takes to make the electrons go away from the metal, and it depends on the metal used.

So electrons are, under normal circumstances, bound to the metal, or if you want to see it the other way: the metal is bound to the electrons. (Sounds a bit weird, but so what.)

The electric field acts on the electrons, and this (outer, perpendicular) force (on the surface charge) deforms the metal a tiny, tiny bit, so that the force on the charges is perfectly balanced by the tension in the metal. It's as if you put a tiny weight on a rubber band: It stretches to the point where the tension balances gravity, only the deformation of the metal is so tiny that it isn't visible.

As a result, you will have two forces balancing each other, even for the surface charges on each capacitor plate.

Alternatively, if there's a dielectric inbetween, this could be squeezed by the attraction between the positive and negative plate, and respond with a balancing elastic force. As long as things don't move, the forces involved must counteract each other somehow.

I should warn you that I figured that out myself, the thing involving elasticity of the metal. I can't remember reading it in a book, and it may not be a standard textbook explanation. Others may have better explanations!

 

[Per Oni]

Well... what holds them there is the attraction. Like putting a piece of paper on the refrigerator with a magnet. the magnet (one plate) stays in place while the paper (dielectric) is in between it and the refrigerator (other plate).

Yes but our capacitor is still placed in a vacuum and there’s no dielectric in between. Disregard the weight of these electrons completely, that’s so insignificant it’s unreal.

In all seriousness, I think this capacitance equation only holds true if the size of the parallel plates are very large compared to the distance between them. I think we could set up an experiment with a makeshift capacitor and move the plates away from each other and see if the voltage increases... I don't think it would. But I could be wrong. Wouldn't be the first time

Your experiment would confirm what I’m saying but it requires a good voltmeter. Ideally one which draws almost no current such as an electroscope.

 

[Per Oni]

It takes a certain energy to free an electron from the metal, that's why they stay (unless temperature or field gets really high) The photoelectric effect (for which Einstein got the Noble prize) is all about which minimum photon energy (i.e. frequency of the light) it takes to make the electrons go away from the metal, and it depends on the metal used.

So electrons are, under normal circumstances, bound to the metal, or if you want to see it the other way: the metal is bound to the electrons. (Sounds a bit weird, but so what.)

That’s a good answer but it still leaves me a bit cheated. We are quick to draw a force arrow to the opposite plate and can explain this arrow but there must exist an arrow with the same (but opposite) magnitude drawing the electrons back. Has anybody ever seen such an arrow with a proper classical explanation? I don’t think we need to go into qm and can stay in electrostatics/classical physics.