Force due to given Static Charges. 189 



24. Again, let r + s + t = ±q. 



v 2 

 We have cos0 + cos0'- ^— = — 2. 



Put = 6' and tan = — -^ . 



c 



The values of will define the points o£ intersectionjof the 

 curve with the line perpendicular to AB, through its 

 middle point. 



We have 2 cos — -^— tan 2 6 = — 2. 

 4tqap 



Or 2(cos0 + l)-j^-(sec 2 6>-l) = 0. 



Or cos 6 + 1 = 0, L e. 6 — it. 



This corresponds to the portion o£ the axis lying between 

 A and B. We have also 



2- -^- . ( 1 ~ C0S ^) ~Q 



4zqap ' cos 2 6 



c 2 c 2 



i. 6., 2 cos 2 6 + -, cos 6 — = 0. 



±qap 4>qap 



Since cos 6 is to be negative, this gives one admissible 

 value of cos 6, showing there is another branch o£ the curve, 



r + s + t = ±q. 



25. It can also easily be shown that y is minimum, corre- 

 sponding to this value of cos 6. The curve is therefore 

 convex to the axis (AB) and obviously goes to infinity. 



2Q. We conclude, therefore, that the line of force defined 

 by r + s + t = ±q, consists also of two branches, viz., the 

 portion of the axis between A and B and a branch going off 

 to infinity. This branch, of which the minimum distance 

 from the axis is finite, is convex to the axis. 



27. Without further analysis but simply on the principle 

 of continuity, we conclude : — 



That the lines of force which converge to both A and B 

 divide into two groups : — 



(a) Starting with the portions of the axis lying beyond 

 A and B, we have one group which consists of lines, each of 



