Since e = 4×10−10, α = 2×10−6, and k1 for air at atmospheric pressure = 450, β is about 2.3 for air at atmospheric pressure and it becomes much greater at lower pressures.
Thus X1/X0 is always greater than unity, and the value of the ratio increases from unity to infinity as β increases from zero to infinity. As β does not involve either q or I, the ratio of X1 to X0 is independent of the strength of the current and of the intensity of the ionization.
No general solution of equation (8) has been found when k1 is not equal to k2, but we can get an approximation to the solution when q is constant. The equations (1), (2), (3), (4) are satisfied by the values—
| k1n1Xe = | k1 | I, |
| k1 + k2 |
| k2n2Xe = | k2 | I, |
| k1 + k2 |
| X =( | α | )12 | I | . |
| q | e(k1 + k2) |
These solutions cannot, however, hold right up to the surface of the plates, for across each unit of area, at a point P, k1I/(k1 + k2)e positive ions pass in unit time, and these must all come from the region between P and the positive plate. If λ is the distance of P from this plate, this region cannot furnish more than qλ positive ions, and only this number if there are no recombinations. Hence the solution cannot hold when qλ is less than k1I/(k1 + k2)e, or where λ is less than k1I/(k1 + k2)qe.
Similarly the solution cannot hold nearer to the negative plate than the distance k2I/(k1 + k2)qe.
The force in these layers will be greater than that in the middle of the gas, and so the loss of ions by recombination will be smaller in comparison with the loss due to the removal of the ions by the current. If we assume that in these layers the loss of ions by recombination can be neglected, we can by the method of the next article find an expression for the value of the electric force at any point in the layer. This, in conjunction with the value X0 = (α/q)12 · I/e(k1 + k2) for the gas outside the layer, will give the value of X at any point between the plates. It follows from this investigation that if X1 and X2 are the values of X at the positive and negative plates respectively, and X0 the value of X outside the layer,
| X1 = X0(1 + | k1 | 1 | )12, | X2 = X0(1 + | k2 | 1 | )12, | ||
| k2 | ε | k1 | ε |
where ε = α/4πe(k1 + k2). Langevin found that for air at a pressure of 152 mm. ε = 0.01, at 375 mm. ε = 0.06, and at 760 mm. ε = 0.27. Thus at fairly low pressures 1/ε is large, and we have approximately
| X1 = X0( | k1 | )12 | 1 | , | X2 = X0( | k2 | )12 | 1 | . |
| k2 | √ε | k1 | √ε |
Therefore
 |
| Fig. 12. |
or the force at the positive plate is to that at the negative plate as the velocity of the positive ion is to that of the negative ion. Thus the force at the negative plate is greater than that at the positive. The falls of potential V1, V2 at the two layers when 1/ε is large can be shown to be given by the equations
| V1 = 8π2( | ε | )32k1( | k1 | )12i2, |
| qα | k2 |
| V2 = 8π2( | ε | )32k2( | k2 | )12i2, |
| qα | k1 |
hence
so that the potential falls at the electrodes are proportional to the squares of the velocities of the ions. The change in potential across the layers is proportional to the square of the current, while the potential change between the layers is proportional to the current, the total potential difference between the plates is the sum of these changes, hence the relation between V and i will be of the form
Mie (Ann. der. Phys., 1904, 13, P. 857) has by the method of successive approximations obtained solutions of equation (8) (i.) when the current is only a small fraction of the saturation current, (ii.) when the current is nearly saturated. The results of his investigations are represented in fig. 12, which represents the distribution of electric force along the path of the current for various values of the current expressed as fractions of the saturation current. It will be seen that until the current amounts to about one-fifth of the maximum current, the type of solution is the one just indicated, i.e. the electric force is constant except in the neighbourhood of the electrodes when it increases rapidly.
Though we are unable to obtain a general solution of the equation (8), there are some very important special cases in which that equation can be solved without difficulty. We shall consider two of these, the first being that when the current is saturated. In this case there is no loss of ions by recombination, so that using the same notation as before we have
| d | (n1k1X) = q, |
| dx |
| d | (n2k2X) = −q. |
| dx |
The solutions of which if q is constant are
| n1k1X = qx, |
| n2k2X = I/e − qx = q(l − x), |
if l is the distance between the plates, and x = 0 at the positive electrode. Since
we get
| 1 | dX2 | = qx{ | 1 | + | 1 | }− q | l | , | |
| 8π | d 2x | 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 = | lk1 | , l − x = | lk2 | . |
| 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
| d | (n2k2X) = q − αn1n2, |
| dx |
| dX | = 4π (n1 − n2)e. |
| dx |
In this case practically all the current is carried by the negative ions so that i = n2k2Xe, and therefore q = αn1n2.
Thus
Thus
| dX | = | 4πe2k2qX | − | 4πi | , |
| dx | αi | k2X |
or
| dX2 | − | 8πe2k2qX2 | = − | 8πi | . |
| dx | αi | k2 |
The solution of this equation is
| X2 = | α | i2 | + Cε 8πe2 k2qx/αi. | |
| 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)
| X2 = | α | i2 | + C¹ε8πe2 k2qx/αi. | |
| q | k2 2e2 |
To find the value of C¹ we see by equation (7) that
| d 2X2 | k1k2 | 1 | = q − αn1n2; | ||
| dX2 | k1 + k2 | 8πe |
hence
| dX2 | k1k2 | 1 | x1 = (q − αn1n2)dx. | |||
| dX | k1 + k2 | 8πe |
![{\displaystyle {\Bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5009a1a3c1fcd7483273ceb0ed46bba07856fae5)
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 i0 is the current of negative ions coming from unit area
