Electrons do "fill up your body" when you jump up and hit a high voltage wire - there is a property called the capacitance of the body that determines how much the voltage increases when you add a certain amount of charge - mathematically, $C = \frac{Q}{V}$.
But it's not charge that kills you, it is current: charge flowing per unit time. Since it takes relatively few electrons to bring the body up to 30,000 V or so, there is not much charge flowing and nobody gets killed. But you may have noticed a static "shock" when (especially in winter) you walked across a carpet, then touched a metal door and got a shock. As you walked across the carpet you built up static charge (with an associated potential that could reach several 10's of kV); and all that charge "leaks away" when you touch a grounded (conducting) surface. But while you can "feel" the current it's not enough to kill you.
So how much charge is there on your body when you are charged to 30,000 V? It's a bit hard to estimate the capacitance of a human body, so we'll use the physicist's trick of the "spherical cow": we approximate the human body as a sphere with 1 m diameter. The capacitance of a sphere is given by
$$C_{sphere} = 4\pi\epsilon_0 R = 0.11 nF$$
At 30 kV, that gives a charge of 3.3 µA; if that charge comes out of your body in 1 µs*, it would result in a peak current of 3.3 A which is why it feels like quite a jolt; however, the total amount of energy is only $\frac12 C V^2 = 0.05 J$ - and that is not enough to kill you. It's enough to kill sensitive electronic circuits, which is why you have to be careful how you handle "bare" electronics, especially in winter (low humidity = build up of static electricity as conductivity of air is lower).
EDIT
- if the current flows in 1 $\mu$s, that suggests that the time constant of body capacitance and skin resistance should be on that order. Since time constant is RC, solving for R gives about 10 kOhm. That’s a rather low resistance: skin resistance is higher, so peak current will be lower.
I'm assuming that you're imagining a long, skinny, series circuit with three simple resistive lamps, like this:
switch A B C
__/ _____________^v^v^v_________________^v^v^v_________________^v^v^v________
| |
= battery short |
|_____________________________________________________________________________|
(Sorry for the terrible ASCII diagram.)
The story we tell children about electric currents --- that energy in electric circuits is carried by moving electric charges --- is somewhere between an oversimplification and a fiction.
This is a transmission line problem. The bulbs illuminate in the order $A\to B\to C$, but reflections of the signal in the transmission line complicate the issue.
The speed of a signal in a transmission line is governed by the inductance and capacitance $L,C$ between the conductors, which depend in turn on their geometry and the materials in their vicinity. For a transmission line made from coaxial cables or adjacent parallel wires, typical signal speeds are $c/2$, where $c=30\rm\,cm/ns=1\rm\,foot/nanosecond$ is the vacuum speed of light.
So let's imagine that, instead of closing the switch at $x=0$ and leaving it closed, we close the switch for ten nanoseconds and open it again. (This is not hard to do with switching transistors, and not hard to measure using a good oscilloscope.) We've created a pulse on the transmission line which is about 1.5 meters long, or 5% of the distance between the switch and $A$. The pulse reaches $A$ about $200\rm\,ns$ after the switch is closed and illuminates $A$ for $10\rm\,ns$; it reaches $B$ about $400\rm\,ns$ after the switch is closed, and $C$ at $600\rm\,ns$.
When the pulse reaches the short at the $100\rm\,m$ mark, about $670\rm\,ns$ after the switch was closed, you get a constraint that's missing from the rest of the transmission line: the potential difference between the two conductors at the short must be zero. The electromagnetic field conspires to obey this boundary condition by creating a leftward-moving pulse with the same sign and the opposite polarity: a reflection. Assuming your lamps are bidirectional (unlike, say, LEDs which conduct only in one direction) they'll light up again as the reflected pulse passes them: $C$ at $730\rm\,ns$, $B$ at $930\rm\,ns$, $A$ at $1130\rm\,ns$.
You get an additional reflection from the open switch, where the current must be zero; I'll let you figure out the polarity of the second rightward-moving pulse, but the lamps will light again at $A, 1530\,\mathrm{ns}; B, 1730\,\mathrm{ns}; C, 1930\,\mathrm{ns}$.
(Unless you take care to change your cable geometry at the lamps, you'll also get reflections from the impedance changes every time a pulse passes through $A$, $B$, or $C$; those reflections will interfere with each other in a complicated way.)
How do we extend this analysis to your question, where we close the switch and leave it closed? By extending the duration of the pulse. If the pulse is more than $1330\rm\,ns$ long, reflections approaching the switch see a constant-voltage boundary condition rather than a zero-current condition; adapting the current output to maintain constant voltage is how the battery eventually fills the circuit with steady-state direct current.
Note that if your circuit is not long and skinny but has some other geometry, then transmission-line approximation of constant $L,C$ per unit length doesn't hold and one of your other answers may occur.
Best Answer
Hanging from a power line you should be as safe as a bird.
The voltage difference is between the lines (e.g. in a 3-phase system) and between the line and ground. This voltage difference exists across the insulators and pole, as well as through the air to ground. These voltage differences are obviously small enough to avoid striking an arc, hence no current flows between the lines or between line and ground. If you are hanging from one line, there is no change in the separation between the lines (unless you are swinging wildly) and hence again no current flows between the lines. As the distance between the lines will usually be smaller than that between your feet and the ground, again no current will flow, and you will be safe. Note that, if this distance were too small, you would not be safe standing under the line either!
Your real problem will be to get down from the line. Unless someone can switch off the power, or you are an acrobat who can jump from the lines to the pole, you will need to touch both the line and the pole simultaneously. If you are hanging from high voltage lines (tens of thousands, not hundreds), touching even a wood pole at the same time as the power line may kill you, unless the wood is extremely dry. Even though wood is considered a poor conductor, when it gets damp its conductivity increases dramatically.