| 1 | dX² | = qx{ | 1 | + | 1 | }- q | l | , | |
| 8π | d²x | k1 | k2 | k2 |
or
| X² | = q | x² | ( | 1 | + | 1 | )- q | lx | + C, |
| 8π | 2 | k1 | k2 | k2 |
where C is a quantity to be determined by the condition that ∫l0 Xdx = V, where V is the given potential difference between the plates. When the force is a minimum dX/dx = 0, hence at this point
| x = | l k1 | , l − x = | l k2 | . |
| k1 + k2 | k1 + k2 |
Hence the ratio of the distances of this point from the positive and negative plates respectively is equal to the ratio of the velocities of the positive and negative ions.
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 − αn1n2, |
| dx |
| dX | = 4π (n1 − n2)e. |
| dx |