by a Charged Condenser moving through Space, 167 



Returning to the argument on page 154 we are now to 

 realize that in taking the flux through the strip we are to 

 take account of the true polarization at each point in the 

 dielectric, and our expression for the flux through the strip 

 AB becomes 



Flux = 4th; dyf P dx, 



where P is the resolved portion of the true polarization in 

 the direction dx. Now, as we have seen, P is made up of 



KF 



two parts - — £ and — N, so that 



throSh X strip = > [ J W-*«ffl)* + £ { K,F-^)dx 



+ f (K 1 F-47rc 2 N)^r] 

 and since as we have seen 



over any path, the above equation becomes 



Flux vdy f * , vdy { B , _ vdy C B ,^ . , 



,, ! , . = 2 I Fdx=^?\ fdx=-f\ (X-vy)d.v, 



through strip c 2 J A c* J A « r J A 



where X and <y have the significance accorded them on 

 page 153, and / is the true electric intensity corresponding 

 to F * at a point in the dielectric. 



Since 



J A 



we ha\e 



Flux 

 throu 



\ Xdx=0, 



M 



lux. v [ v 



i , . = Tydy \ycLv = ^c/y (Flux), 



gh strip c 2 J c 2 J v 



from which it follows that the flux through the strip is zero. 



* / is the x component of the actual force at a point ; it is a quantity 

 varying rapidly from point to point, being in fact in opposite directions 

 on the two sides of an element of a doublet. F is the ordinarily defined, 

 smoothed out value of this electric force. It can onlv be realized as 



y- where $V is the alteration of potential in the length $s, (js being a 



length small compared with ordinary dimensions, but large compared 

 with the distance between two doublets. The line integrals of f and F 

 over any finite path are of course the same. 



