of Electric Oscillations along Parallel Wires. 609 



§ 5. Application to the Ordinary Two-wire Case. 



Let us now take the case of two similar parallel wires each 

 of radius a, at a distance apart b which is large compared 

 with a. Suppose we try, as a first attempt at a solution, to 

 superpose two single-wire solutions of the type given in the 

 last paragraph. The compounded vectors will, of course, 

 satisfy the connecting equations for the dielectric. The 

 difficulty arises when we come to the surface-conditions at 

 the wires. But here we notice that of the three components 

 specified in (4), there is one, the longitudinal electric force, 

 which varies with log r ; the other two contain the factor 



The falling off in the values of the two I - J components 

 is much more rapid than for the (log r) component. Take for 



example ,— — "ttttt- The quantity log — is, in actual cases, 

 1 b . 100 . a J ^jca ' 



roughly of order 10. At the surface of the second wire this 



is reduced by log e 100 = 4*6. So that at the second wire the 



longitudinal electric force due to the field of the first wire is 



about half as great as the force due to the field of the second 



wire itself; whereas the radial electric force and the magnetic 



force only amount to one per cent, of the corresponding values. 



In fact, to neglect at the surface of the second wire the radial 



electric force due to the first, involves the same degree of 



error as if we had the two wires statically charged with equal 



and opposite charges, and neglected the small variations from 



uniform distribution on the wires, due to their mutual 



influence. 



If now we agree to retain at the surface of the second 



wire only the lengthwise component of the field of the first, 



and further, if we neglect the variation of this component 



irom one point to another over the surface of the wire, we 



can arrange the ratio of the internal and external constants 



(d, D above) so as to satisfy the surface condition. We have 



then, instead of (4), 



Inside. Outside. 



Lengthwise electric force... dJ (k 2 a) D[K (ra) — K (c6)]. 



Magnetic forces as before. 



. d = K (ca)-K (ch) kf K^ca) 

 " D J (M hcJ^a)' ' l } 



The smallness of ca enables us to use the small argument 



value for K (cJ) also, in all cases except where — is so great 

 that the problem ceases to resemble experimental conditions. 



