Force due to given Static Charges. 181 



For, if 0, 0\ as before (3) define P, 

 qa cos — (q — r)a 

 — qa cos 0'= (q — s)a. 



cos# — cos #' = 2 . 



v ~\~ s 

 Putting l = 0, we have cos = 3 . 



q 



Or, is imaginary, if r + s<2q or > 4^, i.e. r + s, equal 

 to or greater than 2q define the lines which alone issue from 

 one of the charges and the rest from the other. 



For r + s = 2q, we have cos — cos 0' = 0. 



This is the line of force, passing through the point of 

 equilibrium, bisecting the line joining the two charges. 



7. Case IV. — Let the charges be unequal and of the 

 same sign. 



The construction is the same as in case II. ; only the index 

 numbers corresponding to the centre B have to be marked as 

 in case III,, i. e. both counter-clockwise. 



We have qa cos = (q—r)a. 



— q'a cos r = (q — s)a. 



qcos0 — q'cos0 ! = (q -\-q' — r + s), 



i. e., e cos 0—a' cos 0' = constant. 



Put 0' = 0. 



Then cos<9= l + ^'~ r + '\ 



y 



and is imaginary if 2q' > r + s. 



That is, r + s = or < 2q', define the lines which do not 

 issue from A. 



If r + s = 2q, q cos — q'cos0' = q — q'. 



As in case II., the bipolar equation of the line is 



(a-p'-p) \(a-p' +p) t - («_ p +/ >')f| = 0, 



L p P J 



i. e., a—p'—p = 0, 



or the portion of the axis between A and B and 



/ i , \9 a—p + p' , 

 (a-p +p)J~, £-^-q' = 0. 



r P 



