On the Drop of Potential in Vacuum-tube Discharge. 491 



where Y is the fall o£ potential from a to 0, and s the dis- 

 tance. The force accelerating the particle is 



V 



— e. 

 s 



and the acceleration produced 



\e f 



sin 



Fig. 1. 



£ 



The kinetic energy of the mass when it reaches is given by 

 the expression 



where u is the velocity of the particle at a, and Ye the work 

 done by the field on the charge e in moving from a to 0. 

 The energy with which the particle rebounds is 



{smtf + Ye)*?, 



where k is the coefficient of restitution*, 0<k<l. Under 

 the energy of rebound it moves against the field with a 

 decreasing velocity, coming to rest say at a } from which 

 position it is again driven by the field to the electrode, 

 reaching it with the same energy with which it left it. The 

 energy with which it rebounds in the second case must be 



(imu?+Ye)kS 



likewise for the third 



(imu 2 + Ye)k«, 



and so on. 



We wish to find the time required for the particle to come 

 to rest at the electrode — it being conceived as the time of 

 discharge. This may be obtained by summing the intervals 

 between successive impacts, considering the time of impact 

 as relatively negligible. 



* See Tait and Steele, ; Dynamics of a Particle/ pi 345. * 

 •1 K2 



