Page:Radio-activity.djvu/87

 between two parallel plates at a distance d apart. Suppose that the ionization is confined to a thin layer near the surface of the plate A (see Fig. 1) which is charged positively. When the electric field is acting, there is a distribution of positive ions between the plates A and B.

Let n_{1} = number of positive ions per unit volume at a distance x from the plate A,

K_{1} = mobility of the positive ions,

e = charge on an ion.

The current i_{1} per square centimetre through the gas is constant for all values of x, and is given by

i_{1} = K_{1}n_{1}e(dV/dx).

By Poisson's equation

d^2V/dx^2 = 4πn_{1}e.

Then     i_{1} = (K_{1}/(4π))(dV/dx)(d^2V/dx^2).

Integrating     (dV/dx)^2 = 8πi_{1}x/K_{1} + A,

where A is a constant. Now A is equal to the value of dV/dx when x = 0. By making the ionization very intense, the value of dV/dx can be made extremely small.

Putting A = 0, we see that

dV/dx = ±[sqrt](8πi_{1}x/K_{1}).

This gives the potential gradient between the plates for different values of x.

Integrating between the limits 0 and d,

V = ±(2/3)[sqrt](8πi_{1}/K_{1})d^{3/2},

or     i_{1} = (9V^2/(32πd^3))K_{1}.