104 STATIC ELECTRICITY 



QV QV 



still be J charge X potential, or -g- in A and -- in B. For 



the work done in raising element q through potential v will still be 

 qv whatever the medium between the conductors. 



Energy per cubic centimetre. Let us consider a tube 

 of unit cross-section going from plate to plate and let d be the 

 distance between the plates. The energy to be assigned to this 



. 

 tube in B is |<r V -- j^ ? since V = 



Since there are d cubic centimetres in the tube the energy 



. 27ro- 2 ^TrD* KE'* KK 



per cubic centimetre is ^ ^ u , snice 1 = -^ - 



The pull per unit area of charged surface. Thi- can 



only be measured directly in liquid and gaseous dielectric-. Let us 

 suppose that, keeping the charges the same, the distance bet w cm 

 the plates in B is increased by 1 cm. The energy stored in each 



cubic centimetre added is ^ , and as this energy is imparted by 

 the work done in separating the plates we require n pull per square 

 centimetre of TT to give the required energy. Hence the pull 



, 

 on the charged surface is ^ = ^ 



We shall assume that there is the same pull when the dielectric 

 is solid. 



Force between charged bodies. To account for the pull 



9 a 



rc on the surface by action at a distance according to tin- 

 inverse square law, we must suppose that each element of charge 

 acts with force ^ ,. - a in the medium with dielectric constant 

 K, on the element of charged surface having cr on it. Hence 

 charge q acts on charge q at distance d with force j? *i-. 



The energy in the field. In Chapter VI we showed 



E* 



that if we assign energy at the rate 77 = S-TrD 1 per cubic centi- 



metre to each element of volume in a field where K = 1, we just 

 account for the total energy of the system. 



Let us take two electrical systems identical as regards con- 

 ductors and the charges upon them, but one having air as the 

 dielectric with specific inductive capacity 1, and the other having 

 a dielectric with specific inductive capacity K. The work done in 

 charging the two systems respectively will be \ QV and QV/K. 

 The number of unit cells will be respectively QV and QV/K, so 

 that we may assign half a unit of energy to each unit cell in 

 each case. But the cells will be larger in the second case in the ratio 



