256 Prof. J. J. Thomson on the Theory of Conduction of 



In a steady state the number of positive ions in unit volume 

 at a given place remains constant, hence 

 j 

 -^(k 1 n 1 X)=q—an 1 n 2 , (5) 



and 



— — (k 2 n 2 X) =9. — *n x n 2 . 



Substituting in either of these equations the values of n Y n 2 

 previously found we get, since di/d,v = 0, 



lire k^k.dA lx)~ 9 XVft + ifc^V 4tt rfar/V ** dx t 

 If we put X 2 = 2y, 



this equation becomes 



1 ^1^2 ^P * i h \/ ___h \ 



Aire^ + k, P dy ~ q 2ye' 2 (k l + k 2 ) 2 \ ±ir P )\ ^ P )' 



I have not been able to integrate this equation in the 



feneral case when q is finite and k x not equal to k 2 . We can, 

 owever, integrate it when q is constant and k x =:k 2 = k. In 

 this case the equation may be written 



77 



d I k * 2 2 \ « 1/ ^ 2 gV J. 



the solution of which is 



where C is a constant of integration. 



If the current through the gas passes between two parallel 

 plates maintained at a constant potential-difference, d~K/dx = 

 midway between the plates ; at the positive plate n x — 0, while 

 w 2 = at the negative plate; hence if X , X x be respectively 

 the values of X midway between the plates and at either plate, 

 we have, putting p = 0, to get X 



qek 

 $7rek 



