Force due to given Static Charges. 187 



19. The above investigation may be summarized as 

 follows : — 



Starting with the line r-\-s + t = 2q 3 , we have one series 

 of lines, each of which consists of a branch issuing from q% 

 and going off to infinity and a closed branch joining q 1 and 

 q 2 and another series, each consisting of a branch to infinity 

 issuing from q x and a closed branch joining q 2 , q$. 



20. The line which separates the two sets will obviously be 

 the line passing through a point of equilibrium. 



Now the point of equilibrium is defined by distances 



n, r 2 , r z from e l9 e 2 , e 3 , 

 where ffifa -f c 2 ) _ <h_ _ gsfa + gg) 



and ci, c 2 are the distances of e\ and e 3 from e 2 . 

 Also 



(r 2 2 + Ctfs) (ci + c 2 ) = i\ 2 c 2 + r 2 \ . 



In figure V., 



9i = 15, ? 2 = 10, q 3 = 10 ; 



c, = 4, c 2 = 2. 

 This gives r + s + 1 = 28*95. 



The corresponding line of force is not therefore shown in 

 the figure. And we conclude, further, that the lines 

 r + $ + j = 20 to 28 belong to the first set and 28 to 50 belong 

 to the second set (art. 19). 



In the figure IV. given at the end of Maxwell's Treatise, 

 vol. i., we have 



q\ = 15, 





q 2 = 12, q 3 = 20. 



ci = 9, 





c 2 = 16. 



r, = 15, 





r % = 12, r 3 = 20. 



COS 0! = |, 



cos 



;0 2 = O, COS0 3 =-f. 



r+s+t = 



?l + ?2 + ?3-(2lCOS0 1 + . . 



= 



47- 



-9 + 16 



= 



54: 



= a whole number. 



If we therefore draw a diagram with these values of q v q 2 , 

 one of the lines so drawn will pass through the point of 

 equilibrium, which will obviously be the point at which the 

 two branches of this line of force will intersect. 



