# Tubelights+power lines pictures?

I've come across many pictures like these, sometimes in chain emails reporting the dangers of power lines.

Another claim is that they run on "wasted" energy.

The explanations given are that the wires set up a field, and thus a potential difference. This p.d. gives current in the wires.

Now, I can't help but feeling skeptical. I can't see how the tubes would stay lit for the time it takes to set up the whole grid. Here's my logic:

If current flows through the tubelight, there will be a net charge distribution on the caps. At one point, the charge distribution will be such that it creates its own potential, halting the flow of light. This works even if one end is grounded. An easier way to analyse this is a capacitor (GP: G for ground, P for power lines), with a bulb inside it. Each terminal of the bulb is connected to a plate (A and B). There's an optional grounding wire W. In this situation, it's obvious that eventually the plates AB will get enough charge (same charge density as PG incidentally) to create zero p.d. Current should be there for a brief moment in time.

I understand that this is a DC situation, but for power lines, I think that something similar should happen. Or maybe not. I'm clear that the potential difference across the live and neutral switches polarities, but I'm not sure how the potential field is outside the wire. In fact, I feel that there shouldn't be a V/E field, only a magnetic field.

If it's true that the tubelights stay lit, then how is the energy "wasted energy"? I feel that even if the tubelight can stay lit, the energy is being drawn from the power lines. This looks just like how one can naively say that there is energy wasted in an open socket; when in reality plugging something in draws more energy from the system (and can dim other appliances).

Update: So my questions are thus:

1. I'd like a clearer explanation of the potential and E field outside an AC power line.
2. Is the energy "wasted energy"?
• Note that power-lines are AC. How does that change your analysis? Commented Feb 25, 2012 at 16:48
• @dmckee I have written that (right below the diagram). I'm not exactly clear about the field and potential outside an AC wire. I'd like aclearer explanation than my muddled thoughts. Commented Feb 25, 2012 at 16:59
• Imagine adding a switch between the battery and the rest of your circuit and turning it on and off really fast. The current is there for a brief moment every time you turn it on, and flows in the opposite direction when you turn it off. AC is basically like this, and if this thing with the tube lights works, then I imagine that's how. Commented Feb 26, 2012 at 14:33
• Yeah, I kinda understood that, but I'd like a more detailed explanation, especially of the "power loss" Commented Feb 26, 2012 at 14:33