Method of treating Electrostatic Theorems. 31 



to its original state, i. e. by pushing the sphere back to its old 

 place, and then, considering the strain of the medium within 

 the sphere, find the difference between the energy of a sphere 

 of elasticity E x displaced a distance b, and a sphere of 

 elasticity E 2 displaced a distance a + b. This difference added 

 to the first result will give us what we want. 



First, then, consider the external medium only. If the 

 sphere have a displacement b + x, the external medium having 

 a displacement b, the difference of pressures at P and P' is 

 (by the last section) 



Ei& . 2r cos $ — E^ cos (f>. 



Let SA be the small area of section of a tube running from P 

 to P'. If we sum the expression E,r cos cj)(2b—x) . SA over 

 the central section of the sphere, we get the total resultant 

 pressure which the external medium exerts on the sphere. 

 (Of course when things are in equilibrium x has such a value 

 that this pressure is balanced by the spring -back of the medium 

 within the sphere itself.) 



Since S2^cos0.<iA=V (the volume of the sphere) this 

 sum becomes 



The energy required is therefore 



E t . \{2b-x)lx. 



o - 1 



=E 1 V(«A_0 



Next consider the medium within the sphere. When the 

 sphere is in the field we have a sphere of elasticity E 2 dis- 

 placed a distance (a + b) ; when it is absent a sphere of elasticity 

 E x displaced a distance b. The difference of the energies is 



yb£ - VE 2 1^ 



i 



=VE X ^- ^.(a + 6)E, (6-|) (by last section) 



Total loss of energy caused by the presence of the sphere 

 =3E,.V4 



