66-68] Equipotentials and Lines of Force 57 



If the origin is taken at the centroid of e at # 1} y lt z ly e^ at # 2 , y 2 , z 2 , etc., 

 we have 



i#i = 0, 2X2/1 = 0, Se^! = 0. 



Thus by taking the origin at this centroid, the term of order - will 

 disappear. 



The term of order is 



3 1 



<p 2$ O#! + yyi + zz^f - 



Let A, B, C, be the moments of inertia about the axes, of e l at # 1} y lf z lt 

 etc., and let / be the moment of inertia about the line joining the origin to 

 x t y,z\ then 



and the terms of order become 



(as*?! + yyi + zz^f = r 2 Qerf - 1), 



me 



A+B + C-3I 



Thus taking the centroid of the charges as origin, the potential at a great 

 distance from the origin can be expanded in the form 



v= Ze A + B + C-ZI 

 r r 3 



Thus except when the total charge 2<e vanishes, the field at infinity is 

 the same as if the total charge 2e were collected at the centroid of the 

 charges. Thus the equipotentials approximate to spheres having this point 

 as centre, and the asymptotes to the lines of force are radii drawn through 

 the centroid. These results are illustrated in the special fields of force 

 considered in 61 66. 



The Lines of Force from collinear charges. 



68. When the field is produced solely by charges all in the same straight 

 line, the equipotentials are obviously surfaces of revolution about this line, 

 while the lines of force lie entirely in planes through this line. In this 

 important case, the equation of the lines of force admits of direct integration. 



Let Pj, P 2 , P 3) ... be the positions of the charges e ly e%, e 3 , .... Let Q, Q r be 

 any two adjacent points on a line of force. Let N be the foot of the 

 perpendicular from Q to the axis PjPa, ..., and let a circle be drawn perpen- 

 dicular to this axis with centre N and radius QN. This circle subtends 

 at P l a solid angle 



2?r a - cos #!>, 



