S26 Prof. 0. W. Richardson : 



B then coincide and we have 

 , ZV 



X — 4~2 2' 



Uq — — 2 > 



L — 2 



The thermionic current starting from any region A of a 

 plane which flows to another region B of the same plane in 

 a uniform electric field perpendicular to the plane is therefore 

 given by 



= n j 1 dx Q dy \ ^* w e kmw ° 2 dw I I 



B 



km z 2 e 2 



4 iii 2 Wq 2 



km 2 2 e2 r .. 



When Ze is negative this solution corresponds to the 

 physical case, but when Ze is positive we see that if u is 

 positive u — n is negative. In this case the points of inter- 

 section correspond to negative values of the time and have 

 no physical existence. They are, however, identical with 

 the points of intersection of the ions starting from the back 

 of the strip which corresponds to the case when the limits 

 for w are from to — <x> . 



§ 7. Inclined Planes. 



If the plane B is inclined to the plane A, the electric force 

 being still uniform and perpendicular to the plane A, the 



trajectories will be unaltered and -=~ &c. will still be given 



by equations (13). Let the equation to B be 



z— z — a(x— osq) = 0, 



the line of intersection of the two planes being thus parallel 

 to the axis of y. Then 



