Since e = 4×10^{−10}, α = 2×10^{−6}, and *k*_{1} 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 X_{1}/X_{0} 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 X_{1} to X_{0}
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 *k*_{1} is
not equal to *k*_{2}, but we can get an approximation to the solution
when *q* is constant. The equations (1), (2), (3), (4) are satisfied
by the values—

*n*

_{1}=

*n*

_{2}= (

*q*/α)

^{1/2}

k_{1}n_{1}Xe = |
k_{1} | I, |

k_{1} + k_{2} |

k_{2}n_{2}Xe = |
k_{2} | I, |

k_{1} + k_{2} |

X =( | α | )12 | I | . |

q | e(k_{1} + k_{2}) |

These solutions cannot, however, hold right up to the surface of
the plates, for across each unit of area, at a point P, *k*_{1}I/(*k*_{1} + *k*_{2})*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 *q*λ positive
ions, and only this number if there are no recombinations. Hence
the solution cannot hold when *q*λ is less than *k*_{1}I/(*k*_{1} + *k*_{2})*e*, or where
λ is less than *k*_{1}I/(*k*_{1} + *k*_{2})*qe*.

Similarly the solution cannot hold nearer to the negative plate
than the distance *k*_{2}I/(*k*_{1} + *k*_{2})*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
X_{0} = (α/*q*)^{1/2} · I/*e*(*k*_{1} + *k*_{2})
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 X_{1} and X_{2} are the values of X at the positive and negative
plates respectively, and X_{0} the value of X outside the layer,

X_{1} = X_{0}(1 + |
k_{1} | 1 | )12, | X_{2} = X_{0}(1 + |
k_{2} | 1 | )12, | ||

k_{2} | ε | k_{1} | ε |

where ε = α/4π*e*(*k*_{1} + *k*_{2}). 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

X_{1} = X_{0}( |
k_{1} | )12 | 1 | , | X_{2} = X_{0}( |
k_{2} | )12 | 1 | . |

k_{2} | √ε | k_{1} | √ε |

Therefore

_{1}/X

_{2}=

*k*

_{1}/

*k*

_{2},

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
V_{1}, V_{2} at the two layers when
1/ε is large can be shown to be
given by the equations

V_{1} = 8π²( | ε | )32k_{1}( |
k_{1} | )12i², |

qα | k_{2} |

V_{2} = 8π²( | ε | )32k_{2}( |
k_{2} | )12i², |

qα | k_{1} |

hence

_{1}/V

_{2}=

*k*

_{1}²/

*k*

_{2}²,

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

*i*+ B

*i*².

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 | (n_{1}k_{1}X) = q, |

dx |

d | (n_{2}k_{2}X) = −q. |

dx |

The solutions of which if *q* is constant are

n_{1}k_{1}X = qx, |

n_{2}k_{2}X = I/e − qx = q(l − x), |

if l is the distance between the plates, and *x* = 0 at the positive
electrode. Since

*d*X/

*dx*= 4π(

*n*

_{1}−

*n*

_{2})

*e*,

we get

1 | dX² |
= qx{ | 1 | + | 1 | }− q | l | , | |

8π | d²x | k_{1} |
k_{2} | k_{2} |

or

X² | = q | x² | ( | 1 | + | 1 | )− q |
lx | + C, |

8π | 2 | k_{1} |
k_{2} | k_{2} |

where C is a quantity to be determined by the condition that
∫*l*0 X*dx* = V,
where V is the given potential difference between the
plates. When the force is a minimum *d*X/*dx* = 0, hence at this point

x = | lk_{1} |
, l − x = | lk_{2} | . |

k_{1} + k_{2} | k_{1} + k_{2} |

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 *k*_{1}/*k*_{2} = 0 in equation (8),
or independently as follows:—Using the same notation as before,
we have

*i*=

*n*

_{1}

*k*

_{1}X

*e*+

*n*

_{2}

*k*

_{2}X

*e*,

d | (n_{2}k_{2}X) = q − αn_{1}n_{2}, |

dx |

dX | = 4π (n_{1} − n_{2})e. |

dx |

In this case practically all the current is carried by the negative
ions so that *i* = *n*_{2}*k*_{2}X*e*, and therefore *q* = α*n*_{1}*n*_{2}.

Thus

*n*

_{2}=

*i*/

*k*

_{2}X

*e*,

*n*

_{1}=

*qk*

_{2}X

*e*/α

*i*.

Thus

dX | = | 4πe²k_{2}qX | − | 4πi | , |

dx | αi | k_{2}X |

or

dX² | − | 8πe²k_{2}qX² | = − | 8πi | . |

dx | αi | k_{2} |

The solution of this equation is

X² = | α | i² |
+ Cε ^{8πe² k2qx/αi}. | |

q | k²_{2}e² |

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 |
+ C¹ε^{8πe² k2qx/αi}. | |

q | k_{22}e² |

To find the value of C¹ we see by equation (7) that

d²X² | k_{1}k_{2} |
1 | = q − αn_{1}n_{2}; | ||

dX² | k_{1} + k_{2} | 8πe |

hence

[ | dX² | k_{1}k_{2} |
1 | ]x_{1} = ∫x_{1}0
(q − αn_{1}n_{2})dx. | ||

dX | k_{1} + k_{2} |
8πe |

The right hand side of this equation is the excess of ionization
over recombination in the region extending from the cathode to *x*_{1};
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 *i*_{0}/*e*, if *i*_{0} is the current of negative ions coming from unit area