Force due to given Static Charges. 179 



In the figure I., p = 2, 7 = 6. 



(a) Putting 0' = 0, we have 



. »• + $ 



cos 6 = 1 , 



and is obviously the inclination of the tangent to a line of 

 force at A. 



Now, is imaginary if r + s>2q. 



(b) If 0' = 7T, 



r-\-s 



cos = 3- 



9 



r + s 



and in this case is imaginary if < 2, 



or > 4. 



Hence we conclude that lines of force for which r-\-s"jp2q 

 are those situated on one side of AB and those for which r + 5 

 lies between 27 and 47 are situated on the other side of AB. 



4. Case II. — Two unequal opposite charges. 

 Here, e 3 = — e\ e x = e. 



The equation is 



e cos + e' cos 0' = const. 



We have only now to take AO=qa and BD = 7'a (fig. II.), 

 such that 



£_! 



q'-e" 



For obviously if 0, 0' define any point P, satisfying the 

 condition r + s = constant, as in case I., we have, as before, 



7 a cos = (7 — r)a, 



q'a cos 0'= (q' — s)a. 



7 cos + 7' cos 0' = (7 -f 7' — ?' + 5), 



or £?. cos + <?'. cos 0'= const. 



5. Suppose 7 > 7'. 

 Putting = 



we have q' cos 0' = q'—r + s. 



Or, CO s0'=l-^ti; 



(0' being the inclination to the axis of the tangent to a line 

 of force at B). 



N2 



