Time Delay for DC Voltage to Appear at End of 3×10¹⁰ Meter Wire with Negligible Resistance

I have a question about voltage.Suppose that there is 3*10^10 meters of long wire with negligible resistance. If I apply constant DC voltage at one end of the wire when would I see the voltage at the other end?

The speed of light is 3x10**8 m/sec and the speed of propagation in wire is about 1/10th as fast, ignoring the oddities of relativity. (If you're in a spaceship traveling at the speed of light and have a flashlight pointing in the direction you're going does anything come out? - of course it does, but what would a stationary observer see?)
Of course, if you have a load on the end of the wire things may be different, specifically if the load has capacitance, as virtually everything does, to one degree or another. The electrons would get there at the same rate no matter the load but it would take awhile for enough to accumulate to be observed.

For most electronic purposes the speed could be considered instantaneous. However, I do remember seeing a tech creating delays of a few nanoseconds with loops of wire.

Your wire is pretty long, about the distance to the moon, so It would take about 1000 seconds.

Well I would disagree with one of the answers as it assumes a velocity factor of 0.1. The cable would need a high dielectric constant or to have a spiral nature like delay coax for that to happen. However it does raise the question as to the positioning of the wire in its surroundings. If it were in space and a step response pulse was applied, and the velocity factor is 0.9 (more likely) the time would be 11.1 seconds. Now if such a wire were to exist, an interesting question may well be,how much of the sent energy reach the other end? If the wire had no DC losses, it would still have radiation resistance. I would suggest that almost all the signal would be radiated,. The wire being that long would exhibit an almost pure resistance (due to radiation) and its value will depend on its location with respect to other conductors. Hope this helps.
In disagreeing with previous arguments, I realize that I have stuck my neck out. and someone may chop it off. Ouch. It is all with good intention.

Oops The time should have been 111 seconds, not 11.1.

I will answer your question indirectly (think how electrons flow in the conduction bands in metals example).

When an electron flows it creates a current. This flow takes time because it is associated with a mass, no matter even if this current is the result of an e/h field.

The fact it takes time implies you will not see the effect instantly, i.e., the bulb will not light up instantly. This part is true.
So you simply need to look into how moving electrons in a wire propagate because this will tell you how long you'll have to wait for the bulb to light up.

FYI, it is not the electron in question here, it is the field. Example a CRT shoots electrons across a distance. It would take all the energy contained in the known universe to accelerate an electron to the speed of light so, you have bounded the min round trip time of at least 2 seconds, correct? Oh just a favor, don't use up all the universe's energy testing this theory, or at least wait till basket ball season has concluded.

And I have read that it's around 5% less than the speed of light -- assuming the wire is strait and is acting like an antenna.

With the comments I got here I am getting some picture of what will be happening in this case. I will summarize my understanding. Please comment if I am wrong
When the DC voltage is applied at time t=0 there is an electrical disturbance which causes a change in the electric field at one end of wire. This change of electric filed is propagated along the wire with a speed = speed of electromagnetic disturbance. Hence the speed of the electromagnetic disturbance in the medium of wire is taken into account and the time it takes has been calculated.
I hope this is what that happens when such a situation arises.

From http://en.wikipedia.org/wiki/Propagation_delay:

"Propagation delay is equal to d / s where d is the distance and s is the wave propagation speed. In wireless communication, s=c, i.e. the speed of light. In copper wire, the speed s generally ranges from .59c to .77c."

"Wires have an approximate propagation delay of 1 ns for every 6 inches (15 cm) of length."

Taking the last figure, which works out to 0.5c:

c = 3*10^8m/s
so propagation speed would be = 1.5*10^8m/s
so to travel 3*10^10 meters, the signal would take 200 seconds.

I dunno... I find it hard to believe that the propagation delay in copper can be as low as 0.5c or even 0.77c.

0.95c I would believe.

But, I'm no expert

Well, there's obviously only one thing for it now guys, prove it by doing the experi- . . .

Hang on a minute, where are we going to get a bit of wire that long ?

You are the only one who is actually closest to the actual answer, and it starts with how fields are propagated in a conductor, so you get three starts.

And no to the idea that electricity starts out down a wire, burns up in resistive losses then stops like a truck running out of gas, no, no, no. If the current starts, it will reach the end, what ever you define as the end.

Also as best we know once an e/h field is produced it literally goes on forever in terms of distance. It does not gust get to some point and then just stop. Anyone ever here of the big bang (no not the TV show)?

What books are you people reading? Answer the wrong ones.

Again all you need to do is review how electrons propagate in a conductor with respect to time. A 2nd year EE course covers this. The answer is in the book, just go get it and read it. Also good question while you are at it, this will help, what is the difference between a piece of wire connecting up your car's starter motor and your FM radio's antenna, both say copper, what is the difference? Answer, there is no difference. This is a clue.

Actually, he doesn't say it's a complete circuit, i.e. that it has a return path of "negligible resistance"). A voltage can be applied to one end of a wire, and if the other end of the wire is not connected to anything but open air (or a vacuum, etc.) then something different will happen. An instantaneous current will flow at the end that the voltage is applied to, but will gradually decrease along the length of the wire until there is no current at all at the other end of the wire. This is because the open end of the wire behaves like a capacitor and the current decreases because the capacitance is absorbing charge, and there is a transfer of charge motion into charge density. The increased charge density causes an increase of voltage -- a voltage gradient along the wire with the a peak voltage at the open end of the wire.

The open end of the wire behaves like a capacitor and the end that the voltage was applied to acts like an inductor. This forms a resonant circuit (or tank circuit). Where ever current was flowing in the wire, will cause a magnetic field to develop around the wire which will begin to collapse as the current decays. This collapsing magnetic filed will induce a current that will charge the capacitance of the wire to a potential that is higher than the potential applied to the wire. Once the magnetic field collapses, the higher potential at the open end of the wire will induce a current in the opposite direction. This will be a smaller current than what flowed originally, due to resistive loses. Again a magnetic field will form and then collapse. Now, depending on the impedance of the applied voltage source, will determine if this oscillation will dampen right away or continue for a few more cycles. Low impedance = dampened and fewer cycles (and exactly one cycle if the source impedance matches the impedance of the wire at the end where the source is connected). High impedance = more cycles.