ELECTROMAGNETIC UNIT OF ELECTRICITY TO THE ELECTROSTATIC UNIT. 589 
j so that y = Air cot a, and x ranges from oo to | Att, so if 
h = ^ Ktt Q,ot a. . (4) 
c = iA7r.(5) 
this will give the portion GH of the diagram. When sin 6 is greater than unity we 
may put 
d = ^ 77 + id, 
the right hand side of (3) now equals 
I 
A 
y + id — ^ cot a 
log- 
cos a (e^ + e ■*) -f i. sin a (e® — e 
cos « (e^ + e~ — i sin a (e-^ — 
+ 
■g ITT cot a 
calling the quantity under the logarithm P + iQ, we see since 
P d" tQ 
that 
P — iQ 
cos a {e^ + 6 + i sin a (€•’ — 6 •*) 
cos a (e^ + — i sin « (e''^ — 
cos a + e~^') — i sin « (e-^ — e”'*) 
cos « (e^ + e“ ) + I sin « (e^ — ’ 
multiplying these together we see that P^ + Q'^ = 1. So that log (P + iQ) is wholly 
imaginary. Thus we see that as d ranges from 0 to 00 , and t therefore from I to 00 , 
X = gAvr and y ranges from gA-n- cot a to 00 , thus this range of values of t gives 
the portion GF of the figure. Since the real part of the right hand side of the 
equation (3) changes sign with 6 or t, we see that the other portions of the figure are 
given by the negative values of t. 
Since 
rf)+ Lxjj =B log 
we see that as long as t is between — a and a, that is for the portion AB of the 
figure, 
(f) njj = (.ttB + real quantities, 
so that xfj = ttB, and, therefore, the potential is constant over AB; when t is not 
between these values, that is for the other portion of the figure B log {(t — + cc)} 
is real, and therefore xjj = 0. 
Thus these equations give us the solution of the problem when AB is maintained 
at the potential ttB and ODE, FGH are at zero potential. 
