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
dX/dx = 4[pi](n1 - n2)e,
we get
1 dX^2 / 1 1 \ l
----- ---- = qx ( -- + -- ) - q --,
8[pi] d^2x \k1 k2 / k2
or
X^2 x^2 / 1 1 \ lx
----- = q --- ( -- + -- ) - q -- + C,
8[pi] 2 \k1 k2 / k2
where C is a quantity to be determined by the condition that
_
/ l
| Xdx = V,
_/0
where V is the given potential difference between the plates. When the
force is a minimum dX/dx = 0, hence at this point
lk1 lk2
x = -------, l - x = -------.
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 - [alpha]n1n2,
dx
dX
-- = 4[pi](n1 - n2)e.
dx
In this case practically all the current is carried by the negative
ions so that i = n2k2Xe, and therefore q = [alpha]n1n2.