256 ENERGY OF CURRENTS. 



Let us assume that the permanent state has been attained, 

 and consider two successive layers. The flow of electricity which 

 traverses the first is equal to the sum of the flow I' which traverses 

 the second, and of the flow / which escapes by the lateral surface, 

 that is 



or 



/=!_!'= -dl. 



As we have 



._ i _Vdx 



?L p> 



dx 

 it follows from equation (i) that 



Making /3 2 = , this equation becomes 



the same as for the permanent state of a wire when there is an escape 

 at the surface (220). To determine the constants of the integral 



let us suppose that the lengths are calculated from the middle of the 

 battery, and that, the whole being symmetrical, the potential is zero 

 when x = ; it follows that 



V- A (<*-*-*). 



If Vi is the potential at the ends of the battery and / its 

 length, we have 



