of Electricity in a uniform plane conducting Surface. 461 



and for the point R we have 



20 + RrX=(rc + l)a + O-lV. 



Consequently for each of these points, which we already know 

 are situated on the same flow-line of the resultant system, we 

 have 



1 + 2 + ... + jfc + 0j=(rc + m)a. 



In the same way, we should get for the points P' and Q' situated 

 on the line of flow of the resultant system which lies next to the 

 line through P, Q, and R on the left, the respective values which 

 follow : 



2,6 + P'i?X= n* + (m + 1) «, 



Z0 + Q!qX=(n + l)u + mct, 

 and therefore for each of them 



X + . .. + k + l ={n + m + l)u. 



The same reasoning gives for the points Q L and R x on the line 

 of flow next on the right to that through P, Q, and R, 



1 + ... + jfc +0/=(rc + 0i — l)a. 

 30. Next suppose the additional pole at L to be a sink; then, 

 since the resultant flow must be everywhere concurrent with both 

 the components, the new flow-lines will pass through the other 

 pair of angles of each quadrilateral ; for instance, three conse- 

 cutive lines will pass through the points P' and Qj, through P", 

 Q, and R 2 , and through Q' and R t ; and by applying the same 

 considerations as above, it will be seen that these lines are cha- 

 racterized by the following values of the expression 



namely : — for the line through P' and Q^ 



(n— m + l)ct; 

 for that through P", Q, and R 2 , 



(n— m)a ; 

 and for that through Q! and R p 



(n — m — l)a. 

 Hence it follows that sources and sinks affect the value of the 

 sum of angles denoted by 20 in opposite ways, and therefore that, 

 as assumed provisionally above (§ 28), sources and sinks must be 

 distinguished by a difference in the intrinsic sign of the angles 

 which the radii vectores drawn from them make with the fixed 

 line. When this is done, the equation 



2<9 = ncc 



