900 Dr. A. C. Crehore on 



Hence the sum of (1) to (5) inclusive gives the whole 

 electrostatic force between these two groups of charges, 

 namely 



F =| 2 (-I20)6 4 r- 6 (8) 



This is still a very large repulsive force, and attempts 

 were made in former papers to balance this against the 

 force due to the revolutions of the electrons in orbits. With 

 the present model, however, we are deprived even of this 

 possibility because the electrons cannot revolve in an orbit 

 completely around the centre of the atom according to 

 electromagnetic theory. To estimate the magnitude of this 

 force, let b be made as small as possible and taker=10 _8 cm. 

 The smallest value of b admissible is the minor radius of the 

 electron, say 10 ~ 13 cm., so that 



F= j (-0-012) dynes. .... (9) 



This is equal to the repulsion between two electrons at a 

 distance of about 9 cm. No forces are available due to the 

 rotation of the electrons on their axes to balance this great 

 force. For some time this difficulty has been a stumbling- 

 block to further progress. 



It was not until the idea was used that the negative 

 electrons and the positive charge, as discussed in the pre- 

 ceding paper, do not have a spherical shape that it was 

 perceived the electrostatic force just calculated must bo 

 revised, because the assumed form of an oblate spheroid 

 cannot be treated as a sphere for external points. Let us, 

 therefore, recompute the electrostatic force for the com- 

 bination of charges represented in fig. 4, the hydrogen 

 molecule. 



To undertake this brings us at once into unexplored 

 territory, for the solution depends in part upon the electro- 

 static force between two solid oblate spheroids of charge — 

 a problem that has not been solved in its generality. We 

 shall, therefore, limit ourselves in this preliminary investi- 

 gation to the case of two spheroids having a common axis of 

 revolution, and besides this approximate the spheroid by 

 dividing it into two parts — first the sphere inscribed within 

 it, and second the rest of the spheroid. Let the charge of 

 the inscribed sphere be denoted by E x and of the rest by E 2 , 

 so that 



Et + E^ = e for the negative electron. . . (10) 



The reason for making this division is because the sphere 



