Let
On the assumption that the electric field between the plates is uniform, and that the velocity of the carrier is proportional to the electric field, the velocity of the positive carrier towards B is
and, in the course of the next half alternation,
towards the plate A.
If x1 is less than d, the greatest distances x1, x2 passed over by the positive carrier during two succeeding half alternations is thus given by
Suppose that the positive carriers are produced at a uniform rate of q per second for unit distance between the plates. The number of positive carriers which reach B during a half alternation consists of two parts:
(1) One half of those carriers which are produced within the distance x1 of the plate B. This number is equal to
(2) All the carriers which are left within the distance x1 from B at the end of the previous half alternation. The number of these can readily be shown to be
The remainder of the carriers, produced between A and B during a complete alternation, will reach the other plate A in the course of succeeding alternations, provided no appreciable recombination takes place. This must obviously be the case, since the positive carriers travel further in a half alternation towards A than they return towards B during the next half alternation. The carriers thus move backwards and forwards in the changing electric field, but on the whole move towards the plate A.
The total number of positive carriers produced between the plates during a complete alternation is 2dqT. The ratio ρ of the number which reach B to the total number produced is thus given by
Substituting the values of x1 and x2, we find that
In the experiments, the values of E₀, E1, d, and T were varied, and the results obtained were in general agreement with the above equation.
The following were the results for thorium: