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"?
  • $\begingroup$ Note that power-lines are AC. How does that change your analysis? $\endgroup$ Commented Feb 25, 2012 at 16:48
  • $\begingroup$ @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. $\endgroup$ Commented Feb 25, 2012 at 16:59
  • $\begingroup$ 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. $\endgroup$
    – N. Virgo
    Commented Feb 26, 2012 at 14:33
  • $\begingroup$ Yeah, I kinda understood that, but I'd like a more detailed explanation, especially of the "power loss" $\endgroup$ Commented Feb 26, 2012 at 14:33

1 Answer 1


The electric field of a transmission line decreases with the distance from the line, so there is a potential difference in the radial direction. This potential difference can cause the glow in fluorescent tubes. I believe such tubes can glow even if it is a DC transmission line, as the charge can drift in the air. You may wish to consider a comparison with a corona discharge, which often takes place near transmission lines (both DC and AC), even if there are no fluorescent tubes around, because the electric field strength can be so high near the conductor that causes air breakdown. Corona discharge causes extra power losses, so do glowing fluorescent tubes. While there are losses even in open AC circuits due to reactive loads, extra load leads to extra losses. So those fluorescent bulbs do cause extra losses, they don't use just "wasted energy". Furthermore, I am not sure it is safe to fool around transmission lines.

  • $\begingroup$ Could you explain how the E field is formed? That's part of my confusion; I see no cause for a field. No charges &c, the wire is neutral. Unless the time varying B field creates an E field or something. But that wouldn't happen in the case of DC current. $\endgroup$ Commented Feb 27, 2012 at 13:42
  • $\begingroup$ I don't quite see how the wire can be neutral everywhere. The wire has finite resistivity, so the wire potential varies (approximately linearly) along the wire, otherwise there is no direct current. Therefore, the wire cannot be neutral everywhere, so typically there is a potential difference between the wire and the ground, therefore, there is electric field. $\endgroup$
    – akhmeteli
    Commented Feb 27, 2012 at 19:45
  • $\begingroup$ I understood that as well. From a basics point of view, where does the E come from? E should still obey Gauss' law. It looks as if it's emenating from the wire and latching onto the sheet, which implies that the wire has net charge density.. Where exactly is the cjarge source of it. Field causes potential, not the other way around. The only source of E I cam see is a time varying B. But that doesnt explain where theE comes from in the DC case. $\endgroup$ Commented Feb 28, 2012 at 0:48
  • $\begingroup$ @Manishearth: My understanding is as follows. If there is direct current, that means there is a closed circuit. As there is no superconductivity (I don't consider any exotic systems here), that means there is a DC generator in the circuit, which basically separates negative and positive charges. These charges get distributed over the surface of the wire and create electric field. $\endgroup$
    – akhmeteli
    Commented Feb 28, 2012 at 21:14
  • $\begingroup$ @akhmateli Really? But then the same mechanism won't work for AC. And the magnetic-field-thing is just something I wanted to throw out there, it won't really be large enough to give a large enough field. Ideally, AC and DC should have the same source of field, but charge distribution doesn't seem to work for AC. $\endgroup$ Commented Feb 29, 2012 at 0:28

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