492 Prof. Q. Majorana : Theoretical and 



the PO axis ; the AQ segment will trace the 2tt . TQ . QA 

 surface. It is possible in equation (1) to substitute for day 

 this surface, reduced to unit distance from P, that is to say,. 

 dividing it by x 2 . We have 



. y tfmTQ.QA _hx 



= * 2x2 e - 



QD is to be traced parallel to OP ; B projected normally to 



QD on c ; we have Qc=di/. I call PQO = « ; PQQ^tf ; it 

 appears from the figure that 



dy=QB sin 6 ; QA= QB cos a ; 



therefore QA= . ^ cos a. 

 sin 



Moreover, 



TQ = R sin ; x= \/W + r*-2rij. 



Differentiating 



, _ xdx 



From the triangle OPQ we have 



r 2 = x 2 + R 2 — 2xK cos a, 



,, , tf 2 + R 2 -r 2 



therelore cosa = 



2a?R 



Hence we have 



dm A . R 2 — r 2 \ _ H:c 



e dx. 



*~<^^ : ) 



Let us call dF the total flux of action that is put forth by 

 the dm particle, and that succeeds in emanating itself f rom 

 the sphere ; this flux will be given by the integral of <fi, taken 

 between R -f r and R — r limits, 



dF=-k—-\ (1+ — )e ax dx; 



4rJ R+r x or J 



and carrying out the integration 



E-r 



H* / D5 „ 3 ^ /» -Hr 



^ F= ^j-f_! + &l!),-H, +H(E; _,. 2) p 



E+r 



