Nice calculations Horace. They don't take into account the forces exerted by the probably much more numerous -but more distant- charges at the surface of the electrodes, which at some point when the current tends to zero will dominate, but it seems likely that this will not happen before the current gets down to pA levels.
Now are we sure that the current of those airthreads is really several nA? Michel ----- Original Message ----- From: "Horace Heffner" <[EMAIL PROTECTED]> To: <[email protected]> Sent: Wednesday, June 27, 2007 4:18 AM Subject: Re: [Vo]:Air threads > If we assume 10^-9 amp, that's 6.24x10^9 electrons per second. If we > assume a 10 cm path length and 100 kph ion speed we have a transit > time of (10 cm)/(100 kph) = 0.0036 sec., thus (6.24x10^9 q/s)(0.0036 > s) = 2.25x10^7 electrons in the path. That gives a separation of (10 > cm)/(2.25x10^7 q) = 4.44x10^-9 m between charges. That means a force > of (8.98755x10^9 m/F) (q^2)/(4.44x10^-9 m)^2 = 1.17x10^-11 N = > 1.19x10^-12 kgf. > > > That's (1.19x10^12 kgf)/(Me) = 1.28x10^19 m/s2 acceleration, or > 1.3x10^18 g's on a bare electron. Using the mass of nitrogen > molecule as about 2*14 times the mass of a proton we get(1.19x10^12 > kgf)/(4.68x10^-26 kg) = 2.5x10^14 m/s^2 = 2.55x10^13 g's, which is > still impressive. > > If things spread out a cm or so, we have force = (8.98755x10^9 m/F) > (q^2)/(0.01 m)^2 = 2.3x10^-24 N = 2.35x10^-25 kgf. That gives > (2.35x10^-25 kgf)/(4.68x10^-26 kg) = 49 m/s^2 = 5 g's. It still has > some lateral expansion pressure on the jet. > > > Regards, > > Horace Heffner > > > >
