In most measurements of radio-activity the material is spread over one plate only. In such a case the ionization is to a large extent confined to the volume of the air close to the active plate. The potential gradient in such a case is shown in Fig. 9. The dotted line shows the variation of potential at any point between the plates when no ionization is produced between the plates; curve _A_ for weak ionization, such as is produced by uranium, curve _B_ for the intense ionization produced by a very active substance. In both cases the potential gradient is least near the active plate, and greatest near the opposite plate. For very intense ionization it is very small near the active surface. The potential gradient varies slightly according as the active plate is charged positively or negatively. [Illustration: Fig. 9.] =47. Variation of current with voltage for surface ionization.= Some very interesting results, giving the variation of the current with voltage, are observed when the ionization is intense, and confined to the space near the surface of one of two parallel plates between which the current is measured. The theory of this subject has been worked out independently by Child[80] and Rutherford[81]. Let _V_ be the potential difference 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 _n₁_ = number of positive ions per unit volume at a distance _x_ from the plate _A_, _K₁_ = mobility of the positive ions, _e_ = charge on an ion. The current _i₁_ per square centimetre through the gas is constant for all values of _x_, and is given by $$ i_1 = K_1n_1e \frac{dV}{dx} $$ By Poisson’s equation $$ \frac {d^2 V} {da^2} = 4\pi n_1 e $$ Then $$ i_1 = \frac {K_1} {4\pi} \frac {dV} {dx} \frac {d^2 V} {dx^2} $$ Integrating $$ (\frac {dV} {dx})^2 = \frac {8\pi i_1x} {K_1} + A $$ 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 $$ \frac {dV} {dx} = \pm \sqrt {\frac {8\pi i_1 x} {K_1}} $$ This gives the potential gradient between the plates for different values of _x_. Integrating between the limits 0 and _d_, $$ V = \pm \frac{2}{3} \sqrt {\frac {8\pi i_1} {K_1} d^{ rac{3}{2}}} $$ or $$ i_1 = \frac {9V^2} {32\pi d^3} K_1 $$ If _i₂_ is the value of the current when the electric field is reversed, and _K₂_ the velocity of the negative ion, $$ i_2 = \frac {9V^2} {32 \pi d^3} K_2 $$ and