IDEA OF POTENTIAL. 175 



where Q is the coulombs per centimeter on the respective 

 wires; e' and e" are both towards the right as shown in Fig. 

 115 and therefore both to be considered as positive. 



The total volts per centimerer at A# is e' + e", and the volts 

 along A* is (e' + e")-&x. Therefore: 





and by integrating this expression from x = R to x = D R 

 we get the voltage or potential difference between the wires, 

 namely : 



(4) 



Now the charge on one mile (161,000 centimeters) of the line 

 (positive on one wire and negative on the other) is 161,000 Q. 

 Therefore using the value of Q from equation (4) we get: 



161,000 7T 



Charge on one mile of line 



(5) 



and the expression in the brackets is the capacity in farads of 

 one mile of the double line. To get equation (i) the Naperian 

 logarithm is to be reduced to the common logarithm. 



105. Maximum electric stress between the wires of a trans- 

 mission line. The electrical stress at the surface of the wire to 

 the left (or right) in Fig. 115 may be thought of as being made up 

 of two parts, one part being due to the charge on the wire to 

 the left and the other part being due to the charge on the wire 

 to the right. These two parts are given by equations (i) and 

 (2) of Art. 104 if R be written for x. Therefore the total 

 electrical stress at the surface of one of the wires in Fig. 115 is: 



(i) 



- R 



