147-149] Molecular Theory 131 



distance e apart and are charged to surface density up. There is no 

 intensity except between the plates, and here the intensity of the field is 

 //.. 



Thus if R is the total intensity outside the slab, that inside will be 

 4}Trnp. If K is the inductive capacity of the material of the slab, and 

 that of the free ether outside the slab, we have 



so that ^ = p ........................... (74). 



It remains to determine the ratio p/R. The potential of a doublet is 

 ~ while that of the field R may be taken to be Rx + C. Thus the total 

 potential of a single doublet and the external field is 



and this makes the surface r = a an equipotential if 3 = R. Thus the 



surfaces of the molecules will be equipotentials if we imagine the molecules 

 to be spheres of radius a, and the centres of the doublets to coincide with 

 the centres of the spheres, the strength of each doublet being Ra 3 . 



Putting IJL = Ra 3 , equation (74) becomes 



K =4-7ryia 3 (75)*. 



Now in unit volume of dielectric, the space occupied by the n molecules 



j- na 3 . Calling this quantity 6, we have -~ ^^ 

 o .ti 



lations only hold on the hypothesis that 6 is small, 



is -j- na 3 . Calling this quantity 6, we have -~ ^^ = 30, or, since our calcu- 



* Clausius (Mech. Warmetheorie, 2, p. 94) has obtained the relation 



K-K 47r 



by considering the field inside a sphere of dielectric. The value of K must of course be inde- 

 pendent of the shape of the piece of the dielectric considered. The apparent discrepancy in the 

 two values of K obtained, is removed as soon as we reflect that both proceed on the assumption 

 that K-K is small, for the results agree as far as first powers of K-K Q . Pagliani (Accad. dei 

 Lincei, 2, p. 48) finds that in point of fact the equation 



K K 



agrees better with experiment than the formula of Clausius. 



92 



