# Page:Popular Science Monthly Volume 80.djvu/433

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THE KINETIC THEORY OF MATTER

this case the drop was positive, that a negative ion had been caught from the air. The next time recorded under F, namely, 34.8, indicates that another negative ion has been caught. The next time, 84.5, indicates the capture of still another negative ion. This charge was held for two trips, when the speed changed back again to 34.6, showing that a positive ion had now been caught which carried precisely the same charge as the negative ion which before caused the inverse change in time, i. e., that from 34.8 to 84.5.

 G. F. 13.6 12.5 13.8 12.4 13.4 21.8 13.4 34.8 13.6 84.5 13.6 85.5 13.7 34.6 13.5 34.8 13.5 16.0 13.8 34.8 13.7 34.6 13.8 21.9 13.6 13.5 13.4 13.8 33.4 ——— Mean 13.595

Now all of the successive values of the charge carried by the drop throughout the experiment can be easily computed from the constant speed under gravity and the successive values of the speed in the electric field. To find the absolute values of these charges it is indeed necessary to know the weight of the drop, and the determination of this weight may involve an error of a fraction of a per cent, at most, but since this weight remains constant throughout the experiment the relative values of the successive charges can be found with absolute certainty and with great precision without any knowledge of this weight. They are in fact simply proportional to the successive values assumed by the sum of the two speeds, viz., that under gravity and that in the field.

1. For in the case of bodies moving slowly and uniformly through a resisting medium any two forces produce velocities which are proportional to the forces. The downward force due to gravity is here mg and the upward force due to the field is Fe, in which F denotes the strength of the field and e the charge on the drop. Hence, if v1 is the downward velocity due to gravity and v2 the upward velocity due to the excess of the upward pull of the field over the downward pull of gravity, we have

${\frac {v_{2}}{r_{1}}}={\frac {mg}{Fe-mg}}\ or\ e={\frac {mg}{Fv_{1}}}\ (v_{1}+v_{2})$  