Variation of Atmospheric Electrical Quantities. 283 



the plate. The reading of the gold leaf is taken (g^), and 

 also after again earthing (g ). During this operation the 

 compensator must be fully out. The charge on the plate is 

 represented by (#1 — <7o). 



{!>) The charge lost in five minutes from the test-plate is 

 obtained by removing the cap and adjusting the compensator 

 until the reading of the gold leaf is again </ , and remains 

 at <7 for the five-minute period. The cap is then replaced 

 and the compensator fully drawn out. The gold leaf is 

 again read (g 2 ) and also after earthing (g '). The leakage 

 in the test-plate due to the air-earth current is represented 



°y {92-90')' 



(c) The first operation (a) is repeated to find the charge 

 on the plate {93-90")- 



The conductivity is given by 



X=^S9{(g 2 -g , )/(g 1 +g 3 -g - 9o ")} xl0~ 25 e.m.u., 



where 5"89 is a universal constant, and no instrumental 

 calibration is involved *. The formula may be proved as 

 follows. The surface density of the charge on the test- 

 plate is given by 



a = electric force i^ir = -. 9nn e.s.u. 



for a potential gradient of 100 volts per metre (1 volt 

 per cm.). Supposing the loss of charge on the test-plate 

 is p per cent, per minute, then the current is equal to 



1 1 



x ^-e.s.u. per second = 1*473/? x 10~ 17 amp./sec. 



For a potential of v volts per cm., the current 



= l-!73py X 10" 17 amp./cm. 2 

 The percentage loss per minute (for five-minute intervals) 



=p=l00(g 2 -g ')/[5xi(g 1 +g3-9o~9o")l 

 = ±0(99-9o)/ (91 +93-9o-9o"). 



* This assumes the constancy in scale value for the divisions in the 

 eyepiece. 



U2 



