by a Charged Condenser moving through Space* 15o 



charged bodies. Draw two parallel planes across the circuit 

 at a distance apart dy. Let these planes cut the circuit in 

 the points acdb, and let mnpq be a rectangular element of 

 acdh. Let yfr be the angle made at any point by v with the 

 plane of mnpq, and let P be the electrical polarization 

 resolved parallel to the long axis of acdb. On writing dx 

 for mq the magnetic flux through mnpq is 



±7rvP COS ijr . mn . dx. 



Remembering that mn cos ^fr=dy and writing F for the 

 electric intensity corresponding to P, we have 



Magnetic flux through mnpq = l\¥ 2 dy . dx. 

 Hence the total flux through acdb 



c being the velocity of light. Now F, which is the electric 

 intensity corresponding to P, is composed of two parts, 

 and may be written in the form X — vy, where X is the part 

 which is derivable from a potential, and which can produce 

 mechanical force on a charge e moving with the system, 

 while — vy is a part whose mechanical force — vye on a charge 

 e moving with the system is just counteracted by the mecha- 

 nical force + rye exerted on the charge owing to its motion 

 in the magnetic field, 7 being the magnetic field resolved 

 perpendicular to P. The field 7 is of course also perpendicular 

 to v. Since both ends of the strip AB are at the same 



potential I Xdx = 0, so that 



J A 



Kv c b Kv 2 r B 



Flux through strip = — $ <%! (^ — vy)dx = — ^- dy I —7 



'-" J A C J A 



Kv 2 



— -j- (Flux through strip). 



Hence the flux through the strip is zero, and since the whole 

 curve S may be divided up into strips in this way, the total 

 flux through it is zero *. 



* The absence of any magnetic field in the interior of a conductor, 

 e. g. the conducting compass-needle in Kbntgens experiment, due to the 

 motion of charged bodies along with the conductor, is a particular case 

 of this more general theorem. 



