STRESSES IN THE DIELECTRIC 



135 



any point in the medium there is a stress along the lines of force, a 

 tension 2-7rD 2 /K, where D is the electric strain. Since E = 4-TrD/K, 

 we may put the tension equal to KE 2 /8?r. 



Pressure transverse to the lines of strain. A portion 

 of the medium, say the portion between two cross-sections a a 2 of 

 a tube of strain, could not be in equilibrium under these forces 

 unless the lines of force were parallel and the field uniform. 

 Equilibrium under the end pulls is obviously impossible unless ai is 

 parallel to a z . To show that the field must also be uniform, let 

 D! and D 2 be the strains at a t and o 2 ; the difference of the pulls 

 on the two areas is 



and since D^ D 2 a 2 , this may be put 



which only vanishes if D 1 = D 2 or a l = u 2 , and this is the condition 

 for a uniform field. When the field is not uniform there must be 

 forces across the sides of the tube to make equilibrium possible. 



Value of the pressure in a simple case. A value for 

 the side forces is suggested by considering a special case. Let a 

 particle charged with Q be placed at O, Fig. 90, and let the 



DA 



o 

 FIG. 90. 



C F 



opposite charge be so far away that the lines of force radiate 

 straight away from O. Consider the equilibrium of the shell 

 between two hemispheres, ABC radius r and DEF radius r+dr, 

 drawn with O as centre. We have a tension normal to the surface 



O 2 O 



inwards across ABC equal to 27rD 2 /K = , since D = 



Now a closed surface is in equilibrium under a uniform tension 

 or pressure, so that if we put an equal tension, Q 2 /87rK?* 4 , on the 

 diametral plane AC we have equilibrium for the hemisphere 

 OABCO. Or the tensions over the curved surface ABC have a 



02^7-2 Q2 



resultant inwards equal to the total tension 5 ^ ,= srr-o across 

 the diametral plane. 



