﻿Lines of Force and Equipotential Surfaces. 833 



and from (3), by the aid of these relations, 



-^r = Ar w cos (n0 — cc) + terms of higher degree in r. 

 The directions of the lines of force are therefore given by 



2o + 7T 2tt + 37T 20L + (2n-l)7T 



U ~ 2n ' 2n ' ' 2n 



It appears that at a point where n equipotential lines 

 intersect, n lines of force will also intersect. Further, these 

 lines of force make eqnal angles ir\n with each other and 

 bisect the angles between the equipotential lines. 



We pass on to the case of a three-dimensional field having 

 an axis of symmetry. Let P be a point of equilibrium at a 

 distance a from the axis of symmetry. Take the foot of the 

 perpendicular from P to the axis of symmetry as the origin 

 of cylindrical coordinates «r> z. <£ and yjr no longer satisfy 

 the same differential equation, but 



p 2 + ^ + p=0, .... (4) 

 and 



Take polar coordinates through P in the meridian plane, 

 the initial line being parallel to the axis of symmetry. 

 Then 



z = rcos0, vr = a + rsin6. ... (6) 



Transforming (4) and (5) to the new coordinates, they 

 become 



(a + ,si„^)(g + i| + i^) ± ( S in^ + cos^^) = 0, (7) 



where the upper sign is taken for the potential and the lower 

 sign for the force-function. 



In the neighbourhood of P let <p be expressed in the form 



£ = r w 3„ + r»+ 1 Vi + r M+2 a»+2 + ■ • ■ ; 



where \, S»+i, are functions of 6 only. Substituting 



in (7) the coefficients of the various powers of r in the 

 expression so obtained must vanish separately. Equating 

 the coefficient of r n ~ 2 to zero, we have 



Phil. Mag. S. 6. Vol. 29. No. 174. June 1915. 3 H 



