_Charge on Ion._--By this method we can determine the number of ions per unit volume of an ionized gas. Knowing this number we can proceed to determine the charge on an ion. To do this let us apply an electric force so as to send a current of electricity through the gas, taking care that the current is only a small fraction of the saturating current. Then if u is the sum of the velocities of the positive and negative ions produced in the electric field applied to the gas, the current through unit area of the gas is neu, where n is the number of positive or negative ions per cubic centimetre, and e the charge on an ion. We can easily measure the current through the gas and thus determine neu; we can determine n by the method just described, and u, the velocity of the ions under the given electric field, is known from the experiments of Zeleny and others. Thus since the product neu, and two of the factors n, u are known, we can determine the other factor e, the charge on the ion. This method was used by J. J. Thomson, and details of the method will be found in _Phil. Mag._ [5], 46, p. 528; [5], 48, p. 547; [6], 5, p. 346. The result of these measurements shows that the charge on the ion is the same whether the ionization is by Röntgen rays or by the influence of ultra-violet light on a metal plate. It is the same whether the gas ionized is hydrogen, air or carbonic acid, and thus is presumably independent of the nature of the gas. The value of e formed by this method was 3.4 × 10^-10 electrostatic units. H. A. Wilson (_Phil. Mag._ [6], 5, p. 429) used another method. Drops of water, as we have seen, condense more easily on negative than on positive ions. It is possible, therefore, to adjust the expansion so that a cloud is formed on the negative but not on the positive ions. Wilson arranged the experiments so that such a cloud was formed between two horizontal plates which could be maintained at different potentials. The charged drops between the plates were acted upon by a uniform vertical force which affected their rate of fall. Let X be the vertical electric force, e the charge on the drop, v1 the rate of fall of the drop when this force acts, and v the rate of fall due to gravity alone. Then since the rate of fall is proportionate to the force on the drop, if a is the radius of the drop, and [rho] its density, then Xe + (4/3)[pi][rho]ga³ v1 ---------------------- = ---, 4/3[pi][rho]ga³ v or Xe = (4/3)[pi][rho]ga³(v1 - v)/v. But v = 2/9ga²[rho]/[mu], so that / / [mu]³ v^(3/2)(v1 - v) Xe = \/ 2.9[pi] - / ------ · ---------------. \/ g[rho] v Thus if X, v, v1 are known e can be determined. Wilson by this method found that e was 3.1 × 10^-10 electrostatic units. A few of the ions carried charges 2e or 3e.