Measures for Electric and Magnetic Quantities. 389 

 Static Measure. Dynamic Measure. 



[«,]=[M*L*T- 1 ], 

 [mJ = [M*Li-T- 2]? 



W=[M*LfT- 2 ], 



[E S ] = [M-L*T _1 ], 

 [EJ^CL-T], 

 [0J = [L]. 



[*] = [M>L*], 

 [m,] = [M*LiT" 1 ] J 

 M=[M*L*T~ 1 ] J 

 [EJ = [M*L-T" 2 ], 

 [R,] = [LT- 1 ], 

 [CJ=[L~ 1 T 2 ]. 



- § 5. Comparison of the Units of the two Systems. 



In the foregoing, in determining the static units the mea- 

 surement based on the electrostatic force, and in determining 

 the dynamic units the measurement based on the electrodynamic 

 force, were employed. The formulas thus formed can there- 

 fore only serve to express the relation in which the units of 

 each system stand to one another, but not to compare, as to 

 their quantity, a unit of one of the systems with the corre- 

 sponding unit of the other. For this latter purpose the ratio 

 between the electrodynamic and the electrostatic force must 

 also be taken into account. 



In the static system the force between two quantities of 

 electricity is expressed simply by the product of the quantities 

 of electricity divided by the square of the distance ; while the 

 force between two quantities of magnetism has for its ex- 

 pression in the static system the product of the quantities of 

 magnetism divided by the square of the distance and then 

 multiplied by a constant factor k, which determines the ratio 

 between the electrodynamic and the electrostatic force. In 

 the case of the quantities of electricity and magnetism con- 

 sidered being assumed as units, the expressions of the two 

 forces are [e 2 L -2 ] and A[m 2 L -2 ]. 



Now, in order, first, to learn more precisely the nature of 

 the factor k, we will in the last expression, in accordance with 

 (3 a), put for [m s ] the product [LT _1 ] . [e g ], by which it is 

 changed into &[L 2 T -2 ] . [e 2 L~ 2 ]. Now, as in this expression 

 the last factor [<? 2 L -2 J represents a force (viz. the unit of 

 force), and the whole expression is also to represent a force, 

 the product &[L 2 T -2 ] must be a pure numerical value, whence 

 it follows that k must be the reciprocal value of the square of 

 a velocity. Consequently, if we choose the symbol K for the 



latter, we can write k=. ^, by which, if at the same time we 

 put the formula [MLT -2 ], representing the force-unit, for 



