# Voltage as the "cause of motion" of charges?

by gralla55
Tags: cause of motion, charges, voltage
 P: 57 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.
 P: 853 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.
 P: 57 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?
P: 440
 Quote by DaleSpam 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.
Mentor
P: 16,472
 Quote by 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.
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.
 P: 57 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.
Mentor
P: 16,472
 Quote by gralla55 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.
P: 370
 Quote by 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 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.
 P: 57 "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.
 P: 370 Yes of course. Did anyone claim something different?
 P: 57 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?
 P: 370 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
Mentor
P: 16,472
 Quote by gralla55 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.
Mentor
P: 16,472
 Quote by gralla55 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.
 P: 57 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?
Mentor
P: 16,472
 Quote by gralla55 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.
 P: 997 Voltage does not "drive current". Claude

 Related Discussions Engineering Systems & Design 1 Computers 14 Electrical Engineering 3 Introductory Physics Homework 1 Classical Physics 1