DISCHARGE OF NEGATIVE ELECTRICITY FROM HOT PLATINUM. 
269 
from the platinum surface. Let p denote the gas pressure due to the electrons and 
let p denote the electric volume density, so that p = ne. Then, when there is 
equilibrium, we have —clp/dx + Fp = 0, where F denotes the electric force inside the 
layer. Outside the layer, in the same way, —dpfdx + F'p — 0, where F r denotes the 
force outside. Then at the layer x = t and F t + 47rcr = F' t ; also, since p must be 
continuous and dF/dx = Arrp, we have dF/dx t = dF'/dx t . 
At a great distance from the platinum we have for equilibrium F' = 0 and 
dF'fdx = 0, and at x — 0, dF/dx = 4frp 0 , where p„ = n 0 e and n 0 denotes the number of 
electrons per cubic centimetre in the platinum. These relations enable V, the 
difference of potential between the layer and the platinum, to be calculated. 
Let p = — ftp, where /3 is a constant at constant temperature, so that 
dp _ pdp _ /3 d 2 F 
dx dx 47r dx 2 ' 
Hence 
^ dx 2 +F dx °’ 
which gives 
p dr 2 “ 
In the same way 
/3 f + i F " = c '= 0 - 
Hence 
13 oc 
— 22- =-h constant, 
F' 2 
which gives 
P (f' — v) = 
When x = 0, we have 477-p 0 /3 + |-F 0 2 = c, and when x = t, since 
dF _ dF' _ F? 
dx t dx t 2(3 
we get F f 2 —F' t 2 = 2c, or F 2 — (F ( +47tct) 2 = 2c, which gives F* = — c/^ncr — ‘Zttct. Now 
in the case of hot platinum we know that only a very small fraction of the electrons 
which collide with the surface escape, so that p outside the layer is very small; hence, 
when t is small and cr large, it is clear that F' ( will be small compared with F t . 
Consequently F ( 2 = 2c very nearly, so that 2c = ( — c/47tct — 2ncr) 2 , which gives c = 8n 2 a- 2 , 
so that Fj = — 47rcr. Hence 
/3~ + ±F 2 = 8ttV. 
CutJC 
