132 



Prof. W. M Thornton 



the 



To express this in terms of practical units we have that 

 the electrical displacement q is proportional to x. Let q 

 (on unit area) = mx, then when there is no absorption 



The work done on unit volume =^Fnu' 0> 





4iirm ' t 



k v< 



When there is 



— 5- •3-. and on thickness t, w=-r— , 



absorption m x = - . . and the work done on thickness 



V L 4:777)1 t ' 



t is as before (equation (3)) 



k 9 A: 



w = - — . tr-f - — . va. 



tint \nrt 



To represent this graphically let OA (fig. 3) be the dis- 

 placement x caused by the external field and AB that of 



Fig. 3. 



A X. B 



interattraction. Let AC be the force F<? and DE = 2 /e? 2 . 

 The area OAC is then the work done in polarization when 

 there is no absorption, and the area ODE that done through 

 interaction but supplied by the external circuit. The curve 

 CE is that of fig. 2, displaced vertically to pass through C. 

 The shaded areas together represent the total work of 

 polarization. 



Let F be the breakdown gradient, the areas are then 

 respectively kv 2 /$7rt and kav/^iri, and their sum has been 

 found to be constant. 



F is, however, not constant but is less in thick specimens. 

 Let F change by an amount SF, the displacement x 1 must 

 be such that F8^ =.t'iSF. Since x /~F is a constant quantity 

 the condition is that #]=# > provided that the position of the 

 arc OE is not appreciably changed, which is so since OC is 

 nearly tangential to the curve. We have, therefore, that in- 

 order that the total work of breakdown should be independent 

 of thickness, the extra displacement caused by interattraction 

 should be the same as that initially produced by the applied 

 field. The total displacement is therefore as a consequence 

 proportional to the applied field at breakdown. 



