Method of treating Electrostatic Theorems. 25 



4:77"} 



to E#. Since E answers to -^ we evidently have here the 



correspondence we seek. And the force at any point r' 

 corresponds to the force of restitution per unit volume of 

 dielectric. 



9. The case of two parallel plates may be very simply 

 treated by this method. 



Let S = the surface-area of either, d the distance between 

 them, x the displacement. 



The plates are supposed to be so large compared with the 

 distance between them that all the lines of displacement run- 

 ning from one plate to the other may be considered straight, 

 and x may be considered as the displacement of every particle 

 of matter between the plates. Then the energy of the strain 



= i.S.cZ.E.^ 2 . 



The difference of pressures (V) is evidently the springback of 

 the matter in a tube of unit area reaching from one plate to 

 the other, and is therefore E . d . x. 



From the value of the energy, it is evident that there is a 

 force urging the plates together equal to J . S . E . x 2 . 



Hence this force (F, say) 



Y 2 .S 



^*E.<F 



or 



v=V¥ ; 



4tt 



which, if ^ be written for E, is the usual formula for the 



electrometer. 



If Q be the charge on the positive plate, Q=S . x 

 Hence 



Y = B.d.x 



_E^a\Q 



~ S 



s 



The capacity of the condenser is therefore 



E.d 



10. Using our present hypothesis, the " method of images " 

 may be employed as in the ordinary hypothesis to solve the 

 problems of the charged sphere near an infinite plate, two 

 charged spheres whose radii are not small compared with the 

 distance between their centres, and similar problems. 



As an example consider the first problem. 



