532 THE ELECTROMAGNETIC FIELD. [PT. III. CH. XIII. 



wires are small enough, or the tubes thin enough, or in any case 

 if the conductivity is great enough. Ignoring the manner of 

 distribution of the current, then, we consider only the total cur- 

 rent / in the wire at any point. This is variable from point to 

 point, and is, like V, a function of x and t. We shall suppose 

 that the phenomena are exactly symmetrical in the two conductors 

 as far as the currents go, so that / has equal values with opposite 

 signs at corresponding points in the two conductors. 



Let us consider the charge that exists at any instant on the 

 portion of one conductor between the points x and x -4- dx. Since 

 the capacity per unit length is K, the difference of potential V. 

 we have for the charge dq, 



dq=VKdx. 



If the current flowing in the positive direction at x is /, that at 

 x + dx is 



so that the total gain of charge of the element in unit time is 

 9/, d 



We accordingly obtain the differential equation connecting the 

 current and difference of potential, 



--*? 



Considering now the flow of the current, we have in the pair of 

 corresponding elements of the two wires the electrostatic electro- 



motive force 5 dx and the electromotive force of induction 



dx 



-Ldx^-, the resistance being Rdx. Accordingly the current 

 ot 



equation is 



w * 



The equations (l) and (2) are the equations of the problem. 

 Differentiating (i) by t and (2) by a; we may eliminate /, obtaining 



