188 Prof. D. N. Mallik on Lines of 



21. As another example, we shall take the case o£ two 

 equal and opposite charges in a uniform field. 

 The equation of a line of force is obviously 



o 



q (cos + cos 0') — V— = 'la— r + s + t, 



7 ap x 



being the intersection of the curves 



q cos 6*— — = Q — r-\-s, 



ap 1 



and the radii of the circle of radius q, from the negative 

 charge as centre, for which we have q cos 0' — q — t. 



In figure VI. we have the first set of lines as in fig. IV. with 

 their index numbers, and the intersections of these and the 

 radii drawn from the negative charge at B, such that the 

 final sum of the index numbers is constant, supply the 

 points through which the lines of force are drawn. 



We have taken the relative positions of the charges, to suit 

 the case of an ideal magnet in a uniform magnetic field with 

 the north pole directed to the north. 



In order to suit this case ? 6 and 0' are measured from BA 

 and AB produced. 



22. Put 6 / = tt. y = 0. 



We have cos 6 = 3- r+s + t . 



q 



.'. is imaginary, if r + s + t < 2q or > 4q. 



Similarly for 6'. 



The only lines of force issuing from either charge are 

 therefore those which lie between 2q and Aq, and all these 

 lines of force pass through both A and B. 



23. Let r + s + t = 2q. 

 The equation is 



COS0 + COS0'— -^- = 0. 



qap 



This is obviously satisfied by 



6 = 0, & = ir, y = 0, 



and 6 = 7T, 6' = 0, y = 0. 



That is, the portions of the axis lying beyond A, B (on 

 either side) are the two branches of the line of force in 



this case. 



