764 Sir J. J. Thomson on 



Let £ = — 1 — £, let b be distance from the end of the tube 

 to the edge of the plate, then 



& = C(-l-?-logf), 



an approximate solution is 



%=e [ oh 



b + V 



log(l + ^o)=~( 6 -^> 



If d is the distance of the screen from the edge of the 

 plate, £ A is given by the equation 



-<*=C(* A -log(l+* A )), 



and when d is large compared with C, we can easily get a 

 solution of this equation by successive approximation. 



Two sets of plates were used in the course of the experi- 

 ments. For one set 2*5 cm. long and *2 cm. apart, 



Ctt = -1, I = 8-7, 



h = 25, 



d = 6-2, 

 for these we find 



t A =- 189-4G log (1 + 1 ) = - 79-5, 



this oives 



y=-^X19-7, 



u mv 2. 



where X is the potential difference divided by the distance 

 between the plates. 



For the second pair of plates, which were 5'0 cm. Jong 

 and 3 cm. apart, 



CV = '15, 



b = 5*0, 



d = 3'7, 



I = 8-7, 

 for them 



t A = - 73-42 log (l + * )= -105-7, 

 hence 



</=— 3 X32-2, 



We shall now proceed to consider the values of ejm for 

 the different types of rays. First, with regard to the 

 secondary rays. The values of e\m were measured when 



