With the same notation as before, the breadth of either of the outer layers will in time dt increase by X(k1 + k2)dt, and the number of ions in it by X(k1 + k2)ndt; these ions will reach the plate, the outer layers will receive fresh ions until the middle one disappears, which it will do after a time l/X(k1 + k2), where l is the thickness of the slab ab of ionized gas; hence N, the number of ions reaching either plate, is given by the equation _ / l/X(k1+k2) n0X(k1 + k2) X(k1 + k2) / n0[alpha]l \ N = | --------------dt = ---------- log( 1 + ---------- ). _/ 0 1 + n0[alpha]t [alpha] \ X(k1 + k2) / If Q is the charge received by the plate, X / Q0[epsilon]\ Q = Ne = -------------- log ( 1 + ----------- ), 4[pi][epsilon] \ 4[pi]X / where Q0 = n0le is the charge received by the plate when the electric force is large enough to prevent recombination, and [epsilon] = [alpha]4[pi]e(R1 + R2). We can from this result deduce the value of [epsilon] and hence the value of [alpha] when R1+R2 is known. _Distribution of Electric Force when a Current is passing through an Ionized Gas._--Let the two plates be at right angles to the axis of x; then we may suppose that between the plates the electric intensity X is everywhere parallel to the axis of x. The velocities of both the positive and negative ions are assumed to be proportional to X. Let k1X, k2X represent these velocities respectively; let n1, n2 be respectively the number of positive and negative ions per unit volume at a point fixed by the co-ordinate x; let q be the number of positive or negative ions produced in unit time per unit volume at this point; and let the number of ions which recombine in unit volume in unit time be [alpha]n1n2; then if e is the charge on the ion, the volume density of the electrification is (n1 - n2)e, hence dX -- = 4[pi](n1 - n2)e (1). dx If I is the current through unit area of the gas and if we neglect any diffusion except that caused by the electric field, n1ek1X + n2ek2X = I (2). From equations (1) and (2) we have 1 / I k2 dX \ n1e = ------- ( - + ----- -- ) (3), k1 + k2 \ X 4[pi] dx / 1 / I k1 dX \ n2e = ------- ( - - ----- -- ) (4), k1 + k2 \ X 4[pi] dx / and from these equations we can, if we know the distribution of electric intensity between the plates, calculate the number of positive and negative ions. In a steady state the number of positive and negative ions in unit volume at a given place remains constant, hence neglecting the loss by diffusion, we have d --(k1n1X) = q - [alpha]n1n2 (5). dx d - --(k2n2X) = q - [alpha]n1n2 (6). dx