If the number of positive ions per unit volume is m and v is the velocity, we have nue+mve = i, where i is the current through unit area of the gas. Since nue = i0[epsilon]^nx and i = i0[epsilon]^nl, when l is the distance between the plates, we see that nu / mv = [epsilon]^(nx) / ([epsilon]^(nl) - [epsilon]^(nx)), n v [epsilon]^(nx) -- = -- · -------------------------------. m u [epsilon]^(ne) - [epsilon]^(nx) Since v/u is a very small quantity we see that n will be less than m except when [epsilon]^nl - [epsilon]^nx is small, i.e. except close to the anode. Thus there will be an excess of positive electricity from the cathode almost up to the anode, while close to the anode there will be an excess of negative. This distribution of electricity will make the electric force diminish from the cathode to the place where there is as much positive as negative electricity, where it will have its minimum value, and then increase up to the anode. The expression i = i0[epsilon]^[alpha]l applies to the case when there is no source of ionization in the gas other than the collisions; if in addition to this there is a source of uniform ionization producing q ions per cubic centimetre, we can easily show that qe i = i0[epsilon]^{[alpha]l} + -------(e^{[alpha]l} - 1). [alpha] With regard to the minimum energy which must be possessed by a corpuscle to enable it to produce ions by collision, Townsend (loc. cit.) came to the conclusion that to ionize air the corpuscle must possess an amount of energy equal to that acquired by the fall of its charge through a potential difference of about 2 volts. This is also the value arrived at by H. A. Wilson by entirely different considerations. Stark, however, gives 17 volts as the minimum for ionization. The energy depends upon the nature of the gas; recent experiments by Dawes and Gill and Pedduck (_Phil. Mag._, Aug. 1908) have shown that it is smaller for helium than for air, hydrogen, or carbonic acid gas.