206 ADVANCED ELECTRICITY AND MAGNETISM. 



of length. This equation may be established as follows: The 

 capacity of the element abed is C-Ax so that C-kx X e f 

 is the charge on ab (or negative charge on cd). But all of this 

 charge must flow past the point b during Ajt/F' of a second, 

 so that the current in the line is found by dividing C-Ax X e' 

 by Aac/F' which gives equation (6). 



Imagine a current distribution and a voltage distribution 

 traveling along together and each sustaining the other according 

 to equations (5) and (6), then i = i', e = e\ and V = V so 

 that equation (6) may be written: 



i = CeV (7) 



Solving equations (5) and (7) for F, we get 



And eliminating F from equations (5) and (7), we get 



| Li 2 = \Ce 2 (9) 



This equation means that the magnetic energy per unit length of 

 line (%Li 2 ) is equal to the electric energy per unit length of line 

 (| Ce 2 ) in a wave of the kind which is represented by equations 

 (5) and (7). 



Equation (8) gives the velocity at which two mutually sustain- 

 ing current and voltage distributions must travel, and two such 

 mutually sustaining distributions of e and i constitute an elec- 

 tromagnetic wave. If L is expressed in henrys per mile (see 

 Art. 39), and if C is expressed in farads per mile (see Art. 104), 

 then equation (8) gives F as 186,000 miles per second. 



The voltage and current distributions in a traveling wave 

 must be alike according to equations (5) and (7) ; that is, where 

 i is large, e is correspondingly large. The relation between e 

 and i is shown best by equation (9) from which we get 



(10) 



