Voltage as the cause of motion of charges?

[gralla55]

Voltage as the "cause of motion" of charges?

I know voltage is defined as the potential energy difference per unit charge between two points A and B. In textbooks they often describe this potential difference as the "cause" of current through a wire. Further, if this potential difference gets high enough, the charges does not even require a wire to get from point A to point B (like lightning).

This is what I fail to understand. Shouldn't the coulomb force still be the "cause" of motion for charges? If you increase the distance between two opposite charges, you also increase the voltage between them, but then the coulomb force is reduced, making their pull towards each other less.

My question is basically, is a high potential difference enough to cause lightning, and if not, is it really correct to label voltage as the cause of current through a wire?

 

[Dale]

The Coulomb force is the gradient of the voltage. The two are completely equivalent.

 

[cabraham]

In general voltage is not the cause of current. The 2 are related through Ohm's law or non-linear relation such as diode equation, light bulb, etc. But energy conversion such as hydroelectric generator, coal burning/steam engine, chemical redox in battery, etc., is the "cause" of current, as well as of voltage. To get voltage you have to move charge, but that movement is indeed current. It takes current to produce voltage, and once the charges are separated, they can influence other charges to move.

It's chickens and eggs.

Claude

 

[gralla55]

DaleSpam:

You are right of course! Voltage actually "decreases" with distance, so that makes a lot more sense. I think what confused me was that I thought in terms of gravitational potential energy near the surface of the Earth (= mgh), which increases with distance.

 

[gralla55]

Perhaps I'm just really stupid... but I thought some more about this, and now it does not make any sense again. The electric field is the "negative" of the gradient of the voltage, so the voltage does increase with distance (though not in magnitude).

 

[gralla55]

Now I'm spamming my own tread, but say you have a charged plate with infinite area. The force is the same everywhere, while the potential is E times the perpendicular distance away from the plane. So even if you increase the voltage by moving away, it doesn't increase the strength of the electric field, and shouldn't cause any more movement of charges...

 

[rcgldr]

Now I'm spamming my own thread, but say you have a charged plate with infinite area. The force is the same everywhere, while the potential is E times the perpendicular distance away from the plane.

That would be true for a negatively charged plate. The sign convention for a field and potential assumes a positive charged source. For a positively charged plate, the voltage is at a maximum at the surface of the plate and decreases with distance. Assuming a positively charged particle in the field, accelerating away from the positively charged plate, the potential energy of the charged particle decreases and the kinetic energy of the charged particle increases. I'm not sure if or how the field generated from the accelerating charged particle affects this.

I'm not sure of a convention for reference voltage in such a situation. The voltage could be defined as zero at the surface of a positively charge plate, and approach -∞ as the distance approaches ∞. For a capacitor, voltage could be defined as zero at the surface of the negatively charged plate, linearly increasing with distance from the negative plate to a maximum at the positively charged plate.

 

[jartsa]

Now I'm spamming my own tread, but say you have a charged plate with infinite area. The force is the same everywhere, while the potential is E times the perpendicular distance away from the plane. So even if you increase the voltage by moving away, it doesn't increase the strength of the electric field, and shouldn't cause any more movement of charges...

In this case a water hose analogy works perfectly: We have a vertical elastic water tube filled with water, at the low end of the tube there will be some hydrostatic pressure. Then we stretch the tube and observe an increase of the hydrostatic pressure. Every water molecule weighs the same as before though.
Let's consider some charges inside an elastic wire, placed in a homogeneous electric field. There are some charges side by side (parallel), and some charges one after the other (in series).

When we stretch this wire, charges organize themselves so that they are more in series, and less parallel.

Million point charges one after the other, each feeling a coulomb force of one pico Newtons --> Large voltage
Million point charges side by side, each feeling a coulomb force of one pico Newtons --> Small voltage

 

[gralla55]

You're right, I meant for a negatively charged plate. It doesn't really matter where I define the voltage to be zero, the difference in voltage between two points will come out the same.

Anyway, the point of my question was to understand why higher voltage equals higher current. I get that you need a voltage difference for there to even be a current, but it seems to me that a higher voltage does not necessarily equal a higher coulomb force between the two points.

 

[Crazymechanic]

Higher voltage is more charges per given area.
more alike charges on a given area means stronger electric field there which means more potential to do work if a current path is formed.More charges flow.

About the lightning and breakdown, well a very high PD across two points of say 1 metre apart have a very strong electric field between those points.since we have air around us and many dielectrics like wood and plastic the field polarizes the stuff wwe normally don't call conductors and if the field is strong enough these materials can start to conduct and become conductors, like the breakdown of air when it has happened the current forms a path called an arc.resistance drops in this arc and it become a conductor like a wire in the air.

Vacuum is the best insulator , mainly because it is almoust empty of particles and matter so there is not much for the field to polarize so a very low probability of a breakdown and a very high PD required to achieve one.

 

[jartsa]

You're right, I meant for a negatively charged plate. It doesn't really matter where I define the voltage to be zero, the difference in voltage between two points will come out the same.

Anyway, the point of my question was to understand why higher voltage equals higher current. I get that you need a voltage difference for there to even be a current, but it seems to me that a higher voltage does not necessarily equal a higher coulomb force between the two points.

There was some problem regarding two charged plates moved away from each other, and some charges between the plates. What was the problem? Are the charges in a wire? Does the wire move when the plates are moving?

Oh yes, one more thing that is unclear: Who were you replying to?

 

[gralla55]

jartsa:

I didn't notice your reply, so I wasn't replying to you! What you wrote made sense though. The initial problem was that I saw voltage described as the driving force for current in several places. I then pictured some cases where higher voltage meant lower electric field, or the same electric field, which should mean lower current in spite of higher voltage. This in turn would mean that voltage can't really be seen as the driving force for current.

For the uniform eletric field case:

V = Ed means E = V/d

so the voltage is proportional to the electric field, but given a very high voltage, the electric field could still be small if d is even higher.

But this just looks like some stripped down version of ohms law. I imagine the length of the wire is part of what makes up resistance? If so, higher voltage would drive a larger current, with everything else being equal, but the electric field itself is what drives the current, and that depends on more than voltage.

 

[Dale]

DaleSpam:

You are right of course! Voltage actually "decreases" with distance, so that makes a lot more sense. I think what confused me was that I thought in terms of gravitational potential energy near the surface of the Earth (= mgh), which increases with distance.

Perhaps I'm just really stupid... but I thought some more about this, and now it does not make any sense again. The electric field is the "negative" of the gradient of the voltage, so the voltage does increase with distance (though not in magnitude).

Now I'm spamming my own tread, but say you have a charged plate with infinite area. The force is the same everywhere, while the potential is E times the perpendicular distance away from the plane. So even if you increase the voltage by moving away, it doesn't increase the strength of the electric field, and shouldn't cause any more movement of charges...

Sorry I missed this earlier. You are correct, the E field is the negative gradient of the voltage. Sorry for being sloppy. It depends on the sign of the charge whether voltage increases or decreases with distance.

However, it seems like you still are confused. For a uniform E field the negative gradient of the voltage is uniform. The voltage increases with distance, but it's negative gradient is constant. Recall from your own OP that it is voltage differences which drive current. The gradient operator gives the local "defferences" of a scalar field.

 

[gralla55]

I'll admit to still being a little confused about this. The whole point of the thread was that I couldn't understand "why" voltage is the supposed driving force of current, as it is the electric field which asserts a force on a charge (and thereby causes it to move).

As you pointed out, the voltage can increase without an increase in electric field.

 

[Dale]

I'll admit to still being a little confused about this. The whole point of the thread was that I couldn't understand "why" voltage is the supposed driving force of current, as it is the electric field which asserts a force on a charge (and thereby causes it to move).

As you pointed out, the voltage can increase without an increase in electric field.

Voltage doesn't drive current, voltage differences do.

Think of the water analogy. Pressure doesn't drive water through pipes, pressure differences do.

As you said in your OP:

In textbooks they often describe this potential difference as the "cause" of current through a wire.

There is mathematically no difference between saying that the E field drives it and saying voltage differences drive it. They are mathematically the same thing: the E field is the negative gradient of the voltage in electrostatics.

 

[jartsa]

This in turn would mean that voltage can't really be seen as the driving force for current.

Yeah. Point charge does not know about different voltage differences. That's why charges stay in a million volt power line, the charges flow over voltage difference of one volt and distance of ten meters along the line, instead of flowing over distance of ten meters and voltage difference of million volts to the ground.

The long cylinder shaped charged object inside the wire may be thought to be siphoned from high potential to low potential, maybe. (The electron gas is the long object)

 

[gralla55]

Voltage = potential difference.

As for the water analogy, you could say that pressure difference drives the water, but it is the gravitational field thas is responsible for the pressure in the first place.
I don't think the analogy is perfect either, as the significant gravitational field is generated by the earth, not from the water molecules themselves.

"There is mathematically no difference between saying that the E field drives it and saying voltage differences drive it. They are mathematically the same thing: the E field is the negative gradient of the voltage in electrostatics. "

They are related mathematically, but the slope of a function between two points is not "the same" as the differences in function value between those two points.

For linear functions of one variable, there are two ways to increase the difference f(b) - f(a):

1) Increase the difference between b and a.
2) Increase the slope

I understand how 2), which corresponds to increasing the electric field, would increase a current, at the same time as it increases the voltage. 1) on the other hand, should as far as I can tell, have no effect on the current.

 

[Dale]

They are related mathematically, but the slope of a function between two points is not "the same" as the differences in function value between those two points.

For linear functions of one variable, there are two ways to increase the difference f(b) - f(a):

1) Increase the difference between b and a.
2) Increase the slope

I understand how 2), which corresponds to increasing the electric field, would increase a current, at the same time as it increases the voltage. 1) on the other hand, should as far as I can tell, have no effect on the current.

I don't think that this distinction is a real distinction for this particular application.

If you are talking about the "driving" of the current at a point then you have either the E-field or the negative gradient of the voltage. If you are talking about the "driving" of the current over some path between two points then you either have to integrate the E-field along the path or take the difference of the voltage at the end points.

Either way is mathematically equivalent for either case.

Basically, if you say "the E-field drives the current at point a" then it must also be true that "the gradient of the voltage drives the current at point a" since $E=-\nabla V$. And if you say "the voltage difference drives the current from a to b" then it must also be true that "the E-field along the path drives the current from a to b" since $$-\int_a^b E\cdot ds = \Delta V$$

 

[gralla55]

I think the distinction still applies. I imagine a conductor of some length l, placed in a uniform electric field.

If you take three points a, b and c on the conductor, where v(a) < V(b) < v(c), then V(c) - V(a) > V(b) - V(a).

If the current is larger at point c, then it really is the same thing as you're saying. However, isn't current in a conductor always constant? In that case, larger voltage does not equal larger current, unless the larger voltage is due to a larger eletric field.

 

[Crazymechanic]

I see what you're thinking.

First of all , how can there be any current in a wire that just hang in mid air , isn't attached to any circuit or anything? There is and cannot be any current in such wire.
You can place the wire in a uniform or non uniform electric field , all that will happen is the wire will polarize , which means that as you implied different parts of the wire will have varying strengths of charge , if you would apply a varying electric field to the wire and the attach the wire to a circuit , then you would get current induced because now the charge would flow as a PD would be formed.

And yes once you have a current path or a circuit the current is constant , before that it itsn't but that's because there is no current at all before that.It's like a river that as been blocked by a dam. If the river isn't flowing we don't call it a river, maybe a lake.So the water has accumulated in this one area and has a potential to do something but only when you let it flow towards the lower place , lower potential only then we say that water flows. So as current.You can have a wire in a high electric field etc but if that wire isn't atached to anything no current can flow and when current does start to flow it is the same everywhere.

 

[gralla55]

Well you're right, there can't be any prolonged current in such a wire, but I just imagined it in the first nanosecond as the charges started to rearrange themselves to oppose the outside field.

For some reason I always find the most basic things to be the hardest to understand. I can do problems involving differential forms of maxwell equations without much trouble, but thinking about something like the meaning behind ohms law kills my brain.

How about the lighting example again. You have a cloud which has a lot of negative charge, but not enough for lightning to strike. If you could raise the cloud another mile, it would increase the potential difference between the cloud and the ground, but would that increase or decrease the chance of discharge?

 

[jartsa]

There is mathematically no difference between saying that the E field drives it and saying voltage differences drive it. They are mathematically the same thing: the E field is the negative gradient of the voltage in electrostatics.

I don't know anything, but this sounds correct:
Electric field is another name for voltage gradient.

This sounds wrong:
Voltage difference is another name for electric field.

 

[Dale]

I think the distinction still applies. I imagine a conductor of some length l, placed in a uniform electric field.

If you take three points a, b and c on the conductor, where v(a) < V(b) < v(c), then V(c) - V(a) > V(b) - V(a).

If the current is larger at point c, then it really is the same thing as you're saying. However, isn't current in a conductor always constant? In that case, larger voltage does not equal larger current, unless the larger voltage is due to a larger eletric field.

Adding a 3rd point doesn't change anything I said in post 18. Frankly, I cannot figure out why you think it would. $E=-\nabla V$ and $\int E\cdot ds=\Delta V$. It doesn't matter how many points you add.

 

[gralla55]

Adding the third point was only to illustrate that a larger voltage does not automatically imply a stronger electric field.

Another example: If all you know is that person X drove 100 miles, and that person Y drove 50 miles, it does not say anything about their velocities, even though velocity is the derivative of distance.

 

[Dale]

Adding the third point was only to illustrate that a larger voltage does not automatically imply a stronger electric field.

For a conservative field a larger potential drop along a path does, in fact, automatically imply a larger field along the path. That is precisely what $-\int E\cdot ds=\Delta V$ states.

I think that the problem you are having is that you are trying to compare a voltage drop along a path with an E-field at a point. You compare E-fields at a point with the gradient of the voltage at that same point. You compare E-fields along a path with the difference in the voltage at the end points of the same path.

 

[DrZoidberg]

I think the distinction still applies. I imagine a conductor of some length l, placed in a uniform electric field.

If you take three points a, b and c on the conductor, where v(a) < V(b) < v(c), then V(c) - V(a) > V(b) - V(a).

If the current is larger at point c, then it really is the same thing as you're saying. However, isn't current in a conductor always constant? In that case, larger voltage does not equal larger current, unless the larger voltage is due to a larger electric field.

You should realize that many laws are just approximations and can't necessarily be applied in all situations. Current is only constant in a dc circuit. In a high frequency ac circuit however the current can be completely different at different points of the same wire.
You could take a wire with very high resistivity - so it will take a few seconds for the charges to rearrange - and put that wire in a uniform electric field. Since the field inside the wire is uniform at the beginning the current will be uniform too. But after a moment the top and bottom ends will start to gain a charge which then leads to a non uniform field and therefore a different current in different parts of the wire. In the middle of the wire the current will be strongest.

 

[gralla55]

"For a conservative field a larger potential drop along a path does, in fact, automatically imply a larger field along the path."

Yes, if it is the same path. But if you have the same potential difference between the endpoints of a wire with length L, and a wire with length 2L, shouldn't the strength of the eletric field be half inside the second wire? The integral says that the work the field does along the path is the same. If the path is twice as long, the field would have to be twice as weak to get the same work.

 

[DrZoidberg]

Yes of course. Did anyone claim something different?

 

[gralla55]

Well, that's all I've been trying to point out for the latter half of this thread. A voltage "by itself" is not enough to determine the strength of an electric field between two points, you have to also know the distance between the two points in question.

But anyway, I think I understand this whole thing much better know. Potential differences imply there is an electric field, and if you increase the potential difference without increasing the distance between the points of high and low potential, you also increase the electric field, which in a circuit means increasing the current.

One more question just popped into my head. The electric field inside a wire is as far as I know, always parallel to the wire. This would of course imply that the orientation of the wire itself is inconsequential to the strength of the current. How can one explain this on a micro-scale?

 

[DrZoidberg]

Basically it's because charges on the surface of the wire always arrange themselves such that the field inside is parallel.
Take a look at this
http://www.phy-astr.gsu.edu/cymbalyuk/Lecture16.pdf

 

[Dale]

Yes, if it is the same path. But if you have the same potential difference between the endpoints of a wire with length L, and a wire with length 2L, shouldn't the strength of the eletric field be half inside the second wire? The integral says that the work the field does along the path is the same. If the path is twice as long, the field would have to be twice as weak to get the same work.

Yes. If you use half the force over twice the distance then you have the same amount of work. In the equation $-\int E\cdot ds=\Delta V$ if s increases then E must decrease (all other things the same) for ΔV to stay the same.

 

[Dale]

A voltage "by itself" is not enough to determine the strength of an electric field between two points, you have to also know the distance between the two points in question.

There is no such thing as "the strength of an electric field between two points", as far as I know. There is the strength of the E field at a point, $|E|$, and there is the E field along a path, $\int E\cdot ds$. Each is related to the voltage appropriately.

Suppose that you have a very strong uniform E field and a path perpendicular to the E field. Despite being strong, the E field does not drive any current along the path because $E \cdot ds=0$. It is only the E field along the path which drives current along the path.

 

[gralla55]

DaleSpam:

Strictly speaking, you are correct of course. However, if the magnitude of the electric field X at every point between point A and B were larger than the magnitude of the electric field Y on every point between point C and D, it's common to refer to field X as the "stronger electric field", even though it is unpresise as a mathematical description.

DrZoidberg:

Thanks a lot for the link! Great diagrams and drawings. One thing that's still not clear to me, is if the magnitude of an electric field inside a wire with DC current is taken to be constant everywhere in the wire?

 

[Dale]

if the magnitude of the electric field X at every point between point A and B were larger than the magnitude of the electric field Y on every point between point C and D, it's common to refer to field X as the "stronger electric field", even though it is unpresise as a mathematical description.

Yes, but as I pointed out above, if this is what you mean by the term then the strength of the electric field does not drive current along a path.

In the end, your textbook is correct, the voltage does drive it. You can also say the E field drives it, because of how the E field and the voltage are related.

 

[cabraham]

Voltage does not "drive current".

Claude