Do extremely high-voltage power lines emit positrons? Stretching across China, from the Xinjiang region in the west to the province of Anhui in the east, there's a ±1,100 kV high-voltage direct-current transmission system. I'm not 100% sure what "±1,100 kV" means, but I think it means that the system uses two conductors, one with a voltage of +1,100 kV relative to ground and one with a voltage of -1,100 kV relative to ground.
Now, the rest mass of an electron is about $510\ \mathrm{keV}/c^2$. That got me thinking. Imagine if we had an electron and a positron somewhere near the positive conductor of the transmission line. If the electron were sucked into the line, and the positron were flung away, the total amount of kinetic energy the two particles would gain is 1,100 keV, right? However, my understanding of what rest mass means is that 1,020 keV is enough energy to create an electron and a positron. If I understand particle physics correctly (which I certainly don't), doesn't this mean that electron–positron pairs should be created near the transmission line and flung apart just like this, with the two particles containing about 80 keV of kinetic energy in total?
The way that Hawking radiation is sometimes described to laypeople like me is that a particle–antiparticle pair spontaneously appears near the event horizon of the black hole, and the kinetic energy produced when one of the two particles falls in is enough to "pay off the debt" and make the particles "real" (or something like that!). If that explanation is in any way vaguely accurate, then it seems like maybe the same thing should happen in the presence of a very strong electric field.
So, do high-voltage transmission lines like these really emit positrons, or am I totally off base here?
 A: 
If the electron were sucked into the line, and the positron were flung away, the total amount of kinetic energy the two particles would gain is 1,100 keV, right?

Correct so far.

doesn't this mean that electron–positron pairs should be created near the transmission line and flung apart just like this[?]

No, because the hypothetical electron and positron don't have the energy until they've been acted on by the field for some time. The possible electron-positron pair that hasn't been created yet doesn't have any energy, because it hasn't been acted on by the field.
Now, if your wires were surrounded by vacuum, and you dropped an electron in near the negative wire, it would be be accelerated toward the positive wire. By the time it got there, it would have enough kinetic energy that if it collided (speaking very crudely) with some other appropriate particle, it might conceivably create some third particle with ~2000 keV of energy (based on a quick Wikipedia exploration it seems the most likely particle you could make this way would be an x-ray photon, but don't ask me for any more detail than that --- I'm not by any means a particle physicist).
But real-world power lines aren't surrounded by vacuum, so none of this can happen because any electron being moved around by the field between the two wires will be constantly interacting with the molecules of the air and losing kinetic energy in small steps as it moves, rather than all at once in a particle creating event.
A: As pointed out by another answer, what matters is the field strength in volts per metre. The energy would have to be imparted within the incredibly short lifetime and travelling distance of the virtual pair.
Bringing the power lines closer together increases the field strength. But many orders of magnitude before any particle-pair interactions, the field will begin to ionise the air around the lines. Initially this has two main consequences; current begins to leak across the gap and at night the air around the negative line glows gently as electrons from the line recombine with the air ions. The effect looks very pretty. Bring them closer still and you eventually get an avalanche discharge, an electric arc, between them.
Even in a vacuum they will arc across, the voltage collapse and the lines break or coalesce long, long before any particle pairs can be created.
A: Your analysis doesn't make sense because the units don't match up. $1100 \, \text{kV}$ is not more than twice $510 \, \text{keV}/c^2$, because the two quantities can't be compared at all. It's like saying $4$ meters is twice as big as $2$ minutes.
It's indeed possible to create electron-positron pairs, but you need a tremendously large electric field, given by the Schwinger limit,
$$E = \frac{m_e^2 c^3}{e \hbar} \sim 10^{18} \, \text{V}/\text{m}.$$
Power lines don't have electric fields anywhere near this big, and it's good that they don’t, because this is eight orders of magnitude higher than the field needed to rip the electrons off atoms.
A: If an elementary charge would fly from earth to the power line, without encountering elastic interaction in its path, it would indeed have enough energy to create an electron positron pair upon arrival. This does not happen because the electric field of the cable does not reach the electric breakdown limit of air, which is about 3 kV per mm. https://en.wikipedia.org/wiki/Electrical_breakdown
A: 
doesn't this mean that electron–positron pairs should be created near the transmission line and flung apart

To understand this from a semiclassical view, the key point is that pair creation from vacuum is limited by the uncertainty principle $\Delta H\, \Delta t \lesssim \hbar$ (using $H$ for energy to avoid confusion with electric field $E$). You have noted that if energy to create the pair is "borrowed", it seems that enough energy can be gained from the electric field to "pay it back" and make the pair "real".
However, the energy has to be paid back within a specific amount of time, of order $\hbar/m_e c^2$, not just "eventually". The longest distance an electron and positron can separate in this time is $\hbar/m_e c$, the Compton wavelength, which is the inherent quantum "fuzziness" in an electron's position. Thus, the energy that can be gained from the electric field $E$ before the clock runs out is $e\hbar E/m_e c$. This must cover the borrowed energy of order $m_e c^2$. Thus, creating real pairs from vacuum requires
$$E \sim \frac{m_e^2 c^3}{e \hbar},$$
as noted by knzhou.
A different potential mechanism that has been mentioned is acceleration of a preexisting free electron in the field, leading to a collision with sufficient energy to create a real positron. But this is unlikely given the medium present (air).
A: Extremely unlikely, but possible.
First, you have to find a free electron somewhere in the air. There are always few of them.
Next, it needs some luck to accelerate all the way from one electrode to the other without hitting some air molecule and wasting its energy. Still possible.
If you accelerate a single electron over 1100 kV and smash it on some matter, you have quite low, but still measurable chance of hitting a nucleus directly. Hitting an electron is not going to work - the other electron will recoil with the ~half of the energy.
Next it has to kick a virtual positron out of virtuality and into existence. It has barely the needed energy and almost no energy budget left to spend on different non-idealities. But possible.
I am too lazy to do a "fermi approximation", but I have a gut feeling that having a good detector and great deal of time, you may get a positron or two.
Lightnings (as @John Doty commented) produce positrons roughly the same way. Then again, they try harder, use more resources and were still just recently caught producing positrons.
