Transporting and Accessing Charge at EHV, Part I

I'll now revert to the dual topics of i) transporting charge against a voltage gradient, without necessarily paying a full energy penalty for doing so; and ii) accessing that charge at very high voltage, thus gaining net energy.

Electrostatic vs electromagnetic charge transport

There is a very simple method of transporting charge "shielded" from an external voltage gradient. It is used daily in billions of electrical devices around the world (although the shielding effect is usually incidental). It is simply the familiar current-carrying metallic wire conductor. We recall that electric current is simply the rate at which electric charge flows; i = dq/dt.

Is that all that needs to be said? These days, probably not — a lot more words are probably required. But if so, rather than use my own words, I'll again quote the foremost expert, Professor Noël Felici, on the matter of electrostatic vs electromagnetic current [Ref 1]:— 

Figure 1. Comparison of electrostatic and electromagnetic generators.
(a) Electrostatic air-blast generator. Ash particles move from the left to the right. They pick up a negative charge from the ionizing wire (A), travel in the insulating pipe (B), the walls of which carry stationary charges of opposite sign, and are discharged by the blades of the collector (C). For a significant current to be developed, the electric field near the walls must be strong, with the attendant risk of breakdown.
(b) Electromagnetic generator. The copper bar is displaced in a magnetic field (not shown) giving rise to a Lorentz force (arrow) driving electrons to the right. Their electric field is caught by interspersed positive ions, and no macroscopic electric field related to current will appear.

The caption of the above figure is exactly as Felici wrote it, except for my added emphasis on the last phrase. Again with my emphasis, he comments further:—

"To settle the matter, let us consider an "electrostatic" and an "electromagnetic" generator. In the first one (Pauthenier's air-blast machine, figure 1(a)), we have electrified ash particles moving in an insulating pipe against the pull of the field due to the voltage between terminals. In the second (figure 1(b)), mobile electrons are driven in a copper bar against a similar pull by Lorentz forces, which replace the air blast. What significant differences do we find? In both cases static charges are driven against a true static field due to the voltage on terminals by an extraneous force, and this gives rise, in modern parlance, to an e.m.f. In the "electrostatic" generator, the electric field of the charge carriers extends down to the walls of the insulating pipe and should be caught there by static charges of opposite sign carried by the surface of the walls or, better, by electrodes at cascaded potentials in belt machines. At any rate, the field of the charge carriers is macroscopic; if too strong somewhere, it entails gas breakdown and thus limits the possible current output. Besides, the ash particles are not spontaneously electrified; they need a charging (and discharging) arrangement to accomplish their turn of duty. This is not so with the "electromagnetic" (or "electrodynamic") generator. The field of the electrons drifting in the bar is not caught at its surface but at the atomic level by stationary copper ions. Although incredibly strong, it cannot cause breakdown anyway, since breakdown (in an insulator) would require the field to run over many electronic mean free paths; besides, electrons are charged particles with an enormous charge-to-mass ratio.

Thus, the difference between the two machines reduces to the fact that the field of the charge carriers does not extend so far in the second case as in the first."

Transporting charge without paying a full energy penalty

So, the field of the charge carriers (electrons) in a wire conductor does not extend far enough for those electrons to be strongly influenced by an external voltage gradient, such as necessarily exists between a dome at high voltage and Earth. This means we could indeed transport charge against the gradient without paying a full energy penalty. There is still a problem to be solved in accessing the charge, once transported. I'll look at that next time.

Ref 1: "Electrostatics and electrostatic engineering," N. J. Felici, 1967 Static Electrification Conference.