﻿Determination of the Ratio of the Electrical Units. 107 

 when integrated over the circle r = b, 



Force= ~ { \ drt k 9 A 2 [J (kr)] 2 =i%k 2 b 2 A 2 J Q f \^ 



by a known theorem. The values of kb are 2*404, 5*520, 

 8-654, &c. 



When z=0, Y= SAJ (*r) sinh kl ; 



and the values of A can be found if we know that of V over 

 the whole circle r = b. As usual we have 



VJ (*r)rtfr= A sinh kl * J 2 (kr)rdr 

 Jo 



= P 2 AsinhHJ /2 (H>); 



so that 



Force 



P 2 [f VJ (kr)rdrT 

 = ^~T 2 ~Jnh 2 kl.J ' 2 (kb) 



This would determine the force if we knew the value of V 

 over the whole circle r = b. We know that V = l from ?' = 

 to r=a, and that from r = a to r = b it falls from 1 to 0, but 

 we do not know the precise law of this fall. A pretty good 

 estimate of the integral would be made by taking V = l up 

 to a radius %(a + b) and afterwards V = 0. In this case 



we get* 



f 



Jo 



and 



YJ Q (kr)rdr= -M^lj^L^a + b)] ; 



Fnrrr _v fa + ^) 2 jQ ,2 [P(a + ^)] 



* 01Ce -^ 8b\ sinh 2 kl. J »(kb) * 



If the object be merely to find an upper limit to the force of 

 attraction, we may suppose the value V= 1 to extend up to 

 r = 6, and this in practical cases will not really alter the 

 result very much. Writing a = b in the above formula, we 

 get the simple expression 



Force =:JX [sinh kl]~ 2 . 



This is the force with which the piston is attracted towards C 

 when the potential-difference is unity, and it expresses the 

 excess of the force by which C is drawn in over that which 



* See for example ■ Theory of Sound,' § 204, equation (8). 



