180 Prof. D. N. Mallik on Lines of 



Now, 6' is imaginary, i£ r+ s>2g'. 



That is, the lines of force issuing from g at A will not 

 converge to e f at B, when r + s>2<j'. 



11 r + s = 2q f , 



we have 



q cos 6 + q 1 cos 6' = q — q' '. 



Transferring to bipolars, p and /o', we have 



smce a 2 + p' 2 -2a /5 'cos6' = p 2 



and a 2 + /o 2 -2ap cos 0' = p' 2 , 



i. £., the line of force corresponding to r + s = 2q', consists of 



(1) a + p-p'=0, 



which is the portion of the axis beyond A and B, and 



(2) the curve , 



(a + p + p') q - + (a-p'-p)'± = 0. 



r H 



This curve intersects the axis, a + p—p' = 0, at the point 

 defined by , 



p*-p'*> 



i. e., it passes through the point of equilibrium. 



The line r + s = 8 in the fig. II. (where £ = 12, 9' = 4) 

 obviously separates the lines which converge to e (at B) and 

 those which do not so converge. 



(6 is imaginary if r + s>2q, but the maximum number of 

 lines is only 2q, therefore all the lines issue from A.) 



6. Case III. — (Two equal charges of the same sign.) 



Here, e l = e 2 = e. 



The equation is 



cos 6 — cos 6' = constant. 



The same construction being made, as in the first case, the 

 index numbers corresponding to the centre B have to bs 

 marked 1, 2, 3, &c. beginning on the same side as B, that is,, 

 both counter-clockwise. 



