OCCURRING IN SPHERICAL HARMONIC ANALYSIS. 
413 
COS if/ 
k cos ip 
gk COS \f/ 
k 2 COS 2 1p 
1 \ 1 [2* f27T 
jj-- Jdxf/=- j cos xf/e * cos ' p d\jj= I sin' 2 1 fje /c cos ^difi 
Combining these results, we find for the potential at (f, g, h) clue to a circular plate 
of unit density the expression 
cdpsin 2 xjjeJ a cos ^I/rdxfj .(b) 
If we move the origin to a point at distance v on the negative side of the axis of z, 
we find 
2crf s ++v)^dxb 
Jo r 
in which form the expansion of —— readily gives Clerk Maxwell’s expansion for 
the potential of unit current in the circle (‘Electricity and Magnetism,’ vol. ii., p. 301 
(6'»- 
The result expressed by (b) may be written 
V= 
a- 
T sin 2 ip 
Jf +g 2 + (/c + aj cos ip) 
Hence the potential due to unit current in the circle is 
. sfV 2 f d 1 
a j ^ ,.m xjj^ + ^_j _ a j cog ^ 
I# 
= aj\ sin xb 
J J 0 C 
cl 
Ijr \Zf 3 +g 2 + (h + aj COS ip) 
aj cos ip 
%dxfj 
o Vf+f +O+«/ cos f) 
t p aj cos ip 
)\/f i +g~ + (k + aj cos ip) 
%d\jj 
jfty 
(c) 
This result shows that the potential of a circular current is the same as that of an 
imaginary bar in the axis of z joining the points whose distances from the origin are 
aj and — aj, the density at a point xji being —2 cot xjj. Now the integral (c) is true at 
all points whose distances from the origin are greater than a. We can, however, 
determine the corresponding result for points within the radius a by the ordinary 
theory of inversion. We have seen that for points outside the radius a the potential 
is the same as that of a bar joining two imaginary points. If z be the distance of any 
point of this bar from the origin and z the corresponding inverted distance, we have 
