Let q be the number of ions (positive or negative) produced in one cubic centimetre of the gas per second by the ionizing agent, n1, n2, the number of free positive and negative ions respectively per cubic centimetre of the gas. The number of collisions between positive and negative ions per second in one cubic centimetre of the gas is proportional to n1n2. If a certain fraction of the collisions between the positive and negative ions result in the formation of an electrically neutral system, the number of ions which disappear per second on a cubic centimetre will be equal to [alpha]n1 n2, where [alpha] is a quantity which is independent of n1, n2; hence if t is the time since the ionizing agent was applied to the gas, we have dn1/dt = q - [alpha]n1 n2, dn2/dt = q - [alpha]n1 n2. Thus n1 - n2 is constant, so if the gas is uncharged to begin with, n1 will always equal n2. Putting n1 = n2 = n we have dn/dt = q - [alpha]n² (1), the solution of which is, since n = 0 when t = 0, k([epsilon]^{2k[alpha]t} - 1) n = ---------------------------- (2) [epsilon]^{2k[alpha]t} + 1 if k² = q/[alpha]. Now the number of ions when the gas has reached a steady state is got by putting t equal to infinity in the preceding equation, and is therefore given by the equation n0 = k = [root](q/[alpha]). We see from equation (1) that the gas will not approximate to its steady state until 2k[alpha]t is large, that is until t is large compared with 1/2k[alpha] or with 1/2[root](q[alpha]). We may thus take 1/2[root](q[alpha]) as a measure of the time taken by the gas to reach a steady state when exposed to an ionizing agent; as this time varies inversely as [root]q we see that when the ionization is feeble it may take a very considerable time for the gas to reach a steady state. Thus in the case of our atmosphere where the production of ions is only at the rate of about 30 per cubic centimetre per second, and where, as we shall see, [alpha] is about 10^-6, it would take some minutes for the ionization in the air to get into a steady state if the ionizing agent were suddenly applied. We may use equation (1) to determine the rate at which the ions disappear when the ionizing agent is removed. Putting q=0 in that equation we get dn/[alpha]t = -[alpha]n². Hence n = n0/(1 + n0[alpha]t) (3),