754 Sir J. J. Thomson on the Application of the 



Hence the force on P along P" P due to the other electrons 



IS 2 



3-04 - 



We now proceed to calculate the force on P due to the 

 positive charges if the force between an electron and a 

 positive charge E at a distance r is represented by 



r l \ rJ' 



The force on P along PP" due to the positive charges 

 will be 



8E«/1 1 1 1 \ 



a 2 \3 3 ' 2 ll 3 ' 2 27 3 / 2 ~ l "{2+(2«+l) 2 } 3,2 ~ t " "V 



16B«/1 1 . 1 \ 



a? V3 2 + ll 2 27 2 ^'-'/ 



= ^/ 1 . 89 _^_ 1 . 95 \ 

 o? \ a J 



In order that the positive charges on the atoms should be 

 numerically equal to the charges on the electrons, E must 

 equal 4c ; so that the force due to the positive atoms is 







- 2 (7-56-- 

 a 1 \ a 



tsY 





H. 



suce for 



equilibrium 













e 2 

 3*04 % = 



a 2 



= h-56- 



a J a z 



so 



that 



c 



= •58. 







Though this value of a would ensure the equilibrium of the 

 system, it can be shown that the equilibrium is unstable. We 

 might do this by obtaining the frequencies of the vibrations 

 of the electrons about their position of equilibrium, but as the 

 equilibrium proves to be unstable it is unnecessary to take 

 this trouble, and we shall content ourselves with showing that 

 the equilibrium is unstable for a particular displacement of 

 the electrons. 



The displacement we shall consider when P is displaced 

 vertically upwards through a distance p, Q vertically down- 

 wards through the same distance, P vertically upwards. 

 Thus the displacements of the electrons on the line P Q R..S 

 are all of equal magnitude, but alternate displacements have 

 opposite signs. The same is true for the electrons on the three 



