Dimensions of Physical Quantities. Ill 



whence, substituting for e in turn in terms of K and fi, 



[K]=:[L-^TK-i] = [LT-V]. 

 Hence 



A'[LT-V] = 20[L-iTK-i]. 



a;=20[L-2T2K-V-']. 

 And since 



[K-V-']=^''[L'T-2], 



which is correct. 



In these cases the fundamental units of length, mass, and 

 time, and the medium with respect to which the unit charge 

 and the unit pole are defined, are the same in both systems of 

 units ; but the method can be extended to cases in which 

 these conditions do not hold, and which are far more complex 

 than those which are likely to occur in practice. 



If the unit charge were defined to be such that two 

 unit charges at a distance of 1 metre in a medium of which 

 the inductive capacity was twice that of air, repelled each 

 other with a force which would, in one second, produce in a 

 centigram a velocity of 1 metre per second, we may find the 

 number of ordinary electrostatic units to which it is equivalent. 

 We have the general relation 



x[l.{- Mi^ Tr' Ki^] = 1 X [L/ M/ T^-' K/] ; 

 _ /uietreN^/centigramN^/KgN^ 

 'V cm. J \ gram / \]LJ 

 = 10^x10-1 XV 2 = 100^2. 

 Thus if it were desired to find the number of O.G.S. elec- 

 tromagnetic units of charge in the unit electrostatic charge 

 above defined, we should proceed as follows : — 



^[Li* Mi^ /.r^] = 1 X [L2^ M} %- 1 K2*] . 

 But 



^7^ = 171 Ki^LiT, ' ; 



.-. xv,\l.;^ Mi^ Tr' Ki*] = 1 X [L^^ ^} Tr^ K/] ; 



where, as before, quantities with 2 subscript refer to the new 

 unit. Of course Vi must be expressed in terms of centi- 

 metres and seconds, as these are the magnitudes of Lj and Ti 

 respectively. 



.-. A-x3xl0^o=100v/2, .-. ^=^xlO-«. 



As a final example we may take the following problem, in 

 which the data are chosen merely for the purpose of illus- 

 tration: — Find the number of C.G.S. electrostatic units of 



