Ivan Anderson wrote: > Yes, > > We went through this before, but I still have trouble resolving the > issues in my mind. > > The variables as I see them : > voltage > current > current density of electrode > particle density > > Any one variable is dependent on all the others. > Rise in voltage, leads to rise in current, leads to a rise in current > density, leads to a rise in particle density which travels towards the > cathode at a higher speed. > > Limit current = limit in current density = lower voltage = limit > particle density which travels at lower speed. > > Control voltage = control current (by removing wetted depth of > electrode) = higher current density = higher particle density travelling > at higher speed. > > The velocity of an ion in an electric field is the product of the field > strength > E(Vcm^-1) and its electrophoretic mobility U(cm^2 V^-1 s^-1). > The ion is instantly accelerated to the velocity where the resistive > forces (viscosity of medium etc.) equal the attractive forces, where > upon the ion travels at a constant rate. > > v = EU(cm s^-1) > > For silver ions (effective radius of 0.15nm) U = 6.4 x 10^-6. > Larger ions = slower speeds. > Higher charge = higher speeds. > > So at a field strength of 1Vcm^-1 the velocity = 6.4 x 10^-6 cm s^-1 > which is very slow unless you are an ion. > > Obviously it is not the voltage that is the main diffusion factor in CS > production but thermal or mechanically induced currents.
As I have stated before, my information comes from work with HVAC, and may not be applicable to LVDC. Using 10,000 volts in a 2 inch chamber with .1" of electrode exposed, we find that the most of the voltage drop is within a couple of tenth's of an inch of the electrode. I just ran the electrodes down to about an inch long, and the voltage-current went from 10,000 volts and 100 mA to 5000 volts and 150 mA. So the bulk resistivity of the CS between the electrodes is under 33 K ohm. The total resistance when the electrodes are only 1/10 inch long is 10,000/.1 = 100 K ohms. A quick an dirty approximation of the resistance within 1/5 inch of the electrode would be about a minumum 33 K ohm. So we have essentially 100 K, with 33K near the each electrode and 33K bulk between the electrodes. If we assume most of the "electrode" drop is within 1/5 inch of each electrode, then we get 3,333 volts drop across .2 inches, or a gradient of 16,666 volts per inch. 16,666 v/in = 6561 volts per cm. Using your numbers from above, we find that the drift velosity is .04 cm per second. But I believe this is really too low. The cross section .2 inch from the electrodes will be approximately .3 inch in diameter, with an area of .07 inches. The cross section of the electrode surface is about .032^2*3.14*.1 = .00032, so the voltage gradient right at the surface of the electrode would be around 10 times higher, and would give a velocity of .4 cm/sec. I have made a lot of approximations here, but it has been too long for me to figure out the calculous any more. :< I do believe the answer is within an order of magnitude of the correct answer though. I do believe that you are correct, this does not apply to LVDC, where the voltages are too low. Marshall -- The silver-list is a moderated forum for discussion of colloidal silver. To join or quit silver-list or silver-digest send an e-mail message to: [email protected] -or- [email protected] with the word subscribe or unsubscribe in the SUBJECT line. To post, address your message to: [email protected] Silver-list archive: http://escribe.com/health/thesilverlist/index.html List maintainer: Mike Devour <[email protected]>
