ELECTRIC OSCILLATIONS AND ELECTRIC WAVES. 203 



The horizontal lines in Fig. 139 represent the wires of a trans- 

 mission line, and the heavy portions represent an element of the 

 line. Let e be the voltage across the transmission line at the 



w i re i > &x + i+Ai wire 



toire \d ir c\ wire 



x *-**- AX *r-i+Ai 



*_V \'.x m \j 



i - _..' x|x "" > 



Fig. 139. 



point ad and let i be "the current in the line at the same point" 

 (meaning outflowing current in one wire and returning current 

 in the other wire) as shown in Fig. 139. The voltage across the 

 line at the point be is e + Ae, and the current in the line at the 

 same point is i + A^. 



The capacity of the element abed is C-A#, where C is the 

 capacity of unit length of the line. Therefore the charge q 

 "on the element" (positive charge on ab and negative charge on 

 cd) is g = C-Ax X e* and the rate of decrease of g, namely, 



de 



-77 C-A#, is equal to Ai. Therefore we have : 



r-- m 



C dt~ ~dx 



The net electromotive force around the elementary circuit 

 abed is (e + Ae) minus e, and this electromotive force causes the 

 current in the circuit to decreasef at a definite rate such that 



Ae = L-Ax X -r according to Art. 33 of Chapter III, where 



* The charge is greater than C. Ax X e and less than C. Ax X (e + Ae) and when 

 Ax approaches zero, the expression for q approaches C.Ax X e as a limit. 



t The electromotive force e + Ae is associated with an electric field from wire 

 to wire and the arrow shows the direction of this field or the direction of e + Ae 

 as it would be indicated by a voltmeter. Evidently, however, an excessive charge 

 on the wires at be and a large electromotive force from b to c would tend to 

 create a current opposite to *. 



