ELECTRIC FIELD. 



glass, and, since the flux density is the same in both, we have: 



ko6o = kgCg (l) 



Now e a being the electric field intensity (volts per centimeter) 

 in the glass it is evident that e g x is the total voltage across the 

 glass. Similarly, e y is the total voltage across the oil. There- 

 fore the total voltage from plate to plate is e g x + e y. That is: 



E = egX + e y (2) 



When E, k a , k , x and y are given, these two equations enable 

 e g and e to be calculated. 



; oif.V.'.j 



B 



2cm I 2cm_J 



70000 volt9 



Fig. 93. 



Example. A layer of glass 2 centimeters thick (k = 6) and 

 a layer of air 2 centimeters thick (k = i) are subjected to a total 

 electromotive force of 70,000 volts between parallel metal plates, 

 as shown in Fig. 93, and from equations (i) and (2) we find the 

 following: Electrical stress in the glass 5,000 volts per centimeter; 

 Electrical stress in the air 30,000 volts per centimeter. 



The presence of a layer of dielectric of large inductivity throws 



an excess of stress on a layer of a dielectric of small inductivity. 



Thus with air alone between the metal plates in Fig. 93 (layer of 



air 4 centimeters thick), the electrical stress in the air would be 



II 



