Force due to given Static Cliarcjes. 185 



15. It is, now, obviously possible to generalize the equation 



Thus with three charges, e l} e s , e 3 , the lines of force are 



given by , 



e 1 co 1 + e 2 co 2 + e 3 co s = constant, 



or ni + n 2 + n s = constant. 



In order to trace the curves, we have first to draw the lines 

 corresponding to the charges e 1} e 2 (say), and then draw the 

 tubes corresponding to <? 3 (i. e., lines with index numbers, as 

 before) intersecting the lines of force due to the first two. 



The points of intersection of these two sets of lines, such 

 that the sum of the index numbers of the three charges is 

 constant, will be on a line of force. 



16. We may write the above equation in the form 



01 COS 0! H- #2 COS #2 + ^3 COS # 3 = (qi + ? 2 + Qz ~ * + S + t), 



where r + s + t = constant defines a line of force. 



Moreover, # l5 2 , # 3 are all measured from the initial line, 

 clockwise or counter-clockwise, according as the charges are 

 Ive or positive. 



17. We shall consider the particular case in which e 2 is 

 negative and is situated between ? l5 e$ which are to be both 

 positive. 



In this case, 6 2 is measured clockwise and 6 X and 3 are 

 measured counter-clockwise. In other words, the index 

 numbers corresponding to q 2 on the one hand and those for 

 qi, q s on the other, are enumerated as in case II. 



18. Putting 6 1 = 0, Z = 7f, 

 w 7 e have 



q 2 cos 6 2 = q 2 + 2q s - r + s + 1 



a -, , 2q 3 — r + s + t 

 or cos 09 = 1 + 



92 



where 6 2 is the inclination to the axis of the tangent to a line 

 of force at e 2 . 



Now 6 2 is imaginary, 



)if 



i.e., 



2q 3 — r + s + t >0, 

 if r + s + t<2g z . 



Also, (2) if 



2q 3 - r + s + t 

 92 < "' 



i.e. r + s + 1 > 2q d + 2q 2 . 



