224 Prof. J. J. Thomson on the Electrical 



work done by the electric field on the corpuscles during their 

 free path, we can easily find an expression for c. 



If the electric force X is represented by X cos (qt + e), then 

 the equation of motion of a corpuscle during its free path is 



d 2 v 



and hence 



m- 



dx eX . / - , N . <?X . 



-y = — - sin (at + e) + mu - sm e, 



at q q 



if u is the velocity of projection of the corpuscle at the 

 beginning of the free path when £ = 0. The work done by 

 the electric field on the corpuscle during the free path 



liXdx=* f \xj . dt=fj\X cos (qt + e) J . dt. 



Substituting the value of -j-, we see that the integral 



e 2 X 2 

 = '■ |-{J sin 2 (qt 2 + e) — J sin 2 e — sin e (sin (qt 2 + e) —sin e)} 



+ u— - (sin (pt 2 + €) — sin e) 



1 6 2 X 2 uX e 



= s - (sin (qU + e) — sin e) 2 + — { sin ( pt 2 + e) — sin € } . 



2 m q ■ K * ' q ' 



The last term on the right-hand side will vanish when we 

 take the average, and the expression will reduce to 



2 f!^ sin2 f C o^(f +A 



m q 2 2 \ 2 / 



Since the phase e at the beginning of the free path may 

 have any value, the mean value of the last factor is -J : 

 hence the average rate of absorption of energy per free path is 



X 2 





m ° q 2 



Since there are N free paths completed in unit volume per 

 unit time, the rate of absorption of energy per unit volume 

 is equal to 



,in 2 ^- 



^x/ 2 



m ^° q 2 



This expression must be equal to JcX 2 , for the rate at 



