Voltage as the cause of motion of charges?

[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