From the values which have been found for k2 and [alpha], we know that 8[pi]ek2/[alpha] is a large quantity, hence the second term inside the bracket will be very small when eqx is equal to or greater than i; thus this term will be very small outside a layer of gas next the cathode of such thickness that the number of ions produced on it would be sufficient, if they were all utilized for the purpose, to carry the current; in the case of flames this layer is exceedingly thin unless the current is very large. The value of the electric force in the uniform part of the field is equal to i/k2e·[root]([alpha]/q), while when i0 = 0, the force at the cathode itself bears to the uniform force the ratio of (k1 + k2)^½ to k1^½. As k1 is many thousand times k2 the force increases with great rapidity as we approach the cathode; this is a very characteristic feature of the passage of electricity through flames and hot gases. Thus in an experiment made by H. A. Wilson with a flame 18 cm. long, the drop of potential within 1 centimetre of the cathode was about five times the drop in the other 17 cm. of the tube. The relation between the current and the potential difference when the velocity of the negative ion is much greater than the positive is very easily obtained. Since the force is uniform and equal to i/k2e·[root]([alpha]/q), until we get close to the cathode the fall of potential in this part of the discharge will be very approximately equal to i/k2e·[root]([alpha]l/q), where l is the distance between the electrodes. Close to the cathode, the electric force when i0 is not nearly equal to i is approximately given by the equation i /[alpha]\^½ X = --------- (---------) [epsilon]^{-4[pi]e²k2qx/[alpha]i}, e(k1k2)^½ \ q / , and the fall of potential at the cathode is equal approximately to _[oo] / | X dx, _/0 that is to i /[alpha]\^½ [alpha] i --------- (---------) ----------. e(k1k2)^½ \ q / 4[pi]e²k2q The potential difference between the plates is the sum of the fall of potential in the uniform part of the discharge plus the fall at the cathode, hence /[alpha]\^½ i / i[alpha]² 1 \ V = (---------) --- ( il + --------- ------------ ). \ q / ek2 \ 4[pi]e²q [root](k1k2)/ The fall of potential at the cathode is proportional to the square of the current, while the fall in the rest of the circuit is directly proportional to the current. In the case of flames or hot gases, the fall of potential at the cathode is much greater than that in the rest of the circuit, so that in such cases the current through the gas varies nearly as the square root of the potential difference. The equation we have just obtained is of the form V = Ai + Bi²,