The other case we shall consider is the very important one in which the velocity of the negative ion is exceedingly large compared with the positive; this is the case in flames where, as Gold (_Proc. Roy. Soc._ 97, p. 43) has shown, the velocity of the negative ion is many thousand times the velocity of the positive; it is also very probably the case in all gases when the pressure is low. We may get the solution of this case either by putting k1/k2 = 0 in equation (8), or independently as follows:--Using the same notation as before, we have i = n1k1Xe + n2k2Xe, d --(n2k2X) = q - [alpha]n1n2, dx dX -- = 4[pi](n1 - n2)e. dx In this case practically all the current is carried by the negative ions so that i = n2k2Xe, and therefore q = [alpha]n1n2. Thus n2 = i/k2Xe, n1 = qk2Xe/[alpha]i. Thus dX 4[pi]e²k2qX 4[pi]i -- = ----------- - ------, dx [alpha]i k2X or dX² 8[pi]e²k2qX² 8[pi]i -- - ------------ = - ------. dx [alpha]i k2 The solution of this equation is [alpha] i² X² = ------- ----- + C[epsilon]^(8[pi]e²k2qx/[alpha]i) q k2²e² Here x is measured from the positive electrode; it is more convenient in this case, however, to measure it from the negative electrode. If x be the distance from the negative electrode at which the electric force is X, we have from equation (7) [alpha] i² X² = ------- ----- + C¹[epsilon]^(8[pi]e²k2qx/[alpha]i) q k2²e² To find the value of C¹ we see by equation (7) that d²X² k1k2 1 --- ------- ------ = q - [alpha]n1n2; dX² k1 + k2 8[pi]e hence _ _ _ | dX² k1k2 1 |^x1 / x1 | --- ------- ------ | = | (q - [alpha]n1n2)dx. |_dX k1 + k2 8[pi]e_| _/0 The right hand side of this equation is the excess of ionization over recombination in the region extending from the cathode to x1; it must therefore, when things are in a steady state, equal the excess of the number of negative ions which leave this region over those which enter it. The number which leave is i/e and the number which enter is i0/e, if it is the current of negative ions coming from unit area of the cathode, as hot metal cathodes emit large quantities of negative electricity i0 may in some cases be considerable, thus the right hand side of equation is (i - i0)/e. When x1 is large dX²/dx = 0; hence we have from equation [alpha]i(i - i0) k1 + k2 C¹ = ---------------- -------, qk1k2e² k2 and since k1 is small compared with k2, we have [alpha]i² / k2 i - i0 \ X² = --------- (1 + -- ------ [epsilon]^{-8[pi]e²k2·qx/[alpha]·i} ). qk2²e² \ k1 i /