> Zubrin has taken a familiar approximation -- that the field near a flat charged plate is uniform -- and forgotten that it's only an approximation, and that th

> Zubrin has taken a familiar approximation -- that the field near a flat charged plate is uniform -- and forgotten that it's only an approximation, and that the keyword is "near". Near is all that matters in a plasma, since charged particles leaving the device will get screened and decouple from it within a few Debye lengths. Zubrin of course knows this. Read the paper. > As particles leave the vicinity of the drive, they will be attracted back toward it, reversing the momentum transfer. As particles leave the vicinity of the drive, they will get screened off and decouple from the drive. On top of that, there is a more general problem with the "correct explanation", even in a vacuum: it only considers the effect of the capacitor on the ion, but not of the ion on the capacitor. When the ion "climbs a big hill to get through the first plate", it loses kinetic energy. When the ion is accelerated between the plates, it gains kinetic energy. When the ion climbs back out of the potential on the other side of the capacitor, it again loses kinetic energy. Where is all that energy being transferred to? Right, to the capacitor. When the capacitor is absorbing energy, mutually repelling charges on each plate are being forced a little closer together. It's like loading a spring. When the capacitor is releasing energy, those charges are moving further away from each other again, like releasing the spring. Of course, this means that the capacitor's electric field is changing as the ion goes by. If there is no power source hooked up to the capacitor, the moving back and forth of charges on its plates will inevitably cause some losses; energy will be dissipated as heat, and in the end, you will actually have reduced the passing ion's speed. But of course, the capacitor in a drive is hooked up to a power source. It does work to keep the charges on the plates locked in place as the ion goes by. If it is perfectly successful (it isn't, of course, but it can get pretty close), the charges on the plates will not move at all, and the capacitor's field will be constant throughout the ion's passage. So the field is constant because you have a power source doing work, spending energy, to keep it constant while the ion is passing through. Where does all that energy end up in the end? Not in the field (hey, it's constant). Some of it is dissipated as heat, but the rest goes to the ion, as kinetic energy.