Page 30 - Maxwell House

10 Chapter 1

Treating dimensions as algebraic quantities, one can see their relationship as

−2
kg ∙ A −1 ∙ s ~ k ∙ m −1 ∙ A
In the above expression, the symbol “~” represents a proportionality, not an approximation.
Solving this relationship as an ordinary equation we have

−2
−2 ∙ A
k ~ m ∙ kg ∙ s
According to Table 1.5, line 35 the units for the factor k and the magnetic permeability
0
coincides and we can expect that
=
0
That is the correct scalar equity for magnetic fields in a vacuum.
In the same manner, the affiliation between electric field strength E and electric displacement
strength D can be established in a vacuum as

= [C∙ m ]
−2
0
Here is the permittivity of free space.
0
Example #2. Now check the unit dimensions of factor k in Lorentz’s equation we will present
later

= ( + x )
Here is force exerted by electric and magnetic fields and is the speed of some charged
−2
particle. According to Table 1.5, lines 5, 21, 3, and 25 the units for force is [kg ∙ m ∙ s ],
−1
−1
for electric field strength E is [m ∙ kg ∙ −3 ∙ A ], for speed v is [m ∙ s ], and for magnetic
−2
inductance strength B is [kg ∙ A −1 ∙ s ]. Treating dimensions as algebraic quantities, one can
see their relationship as
−2
−1
kg ∙ m ∙ −2 ~ ∙ {m ∙ kg ∙ −3 ∙ A −1 + (m ∙ s ) ∙ (kg ∙ A −1 ∙ s )}
Pay attention that both terms in the curly brackets must be and are of the same unit dimension.
Solving this relationship as an ordinary equation we have
−2
kg∙m∙
~ = A ∙ s = C
m∙kg∙ −3 ∙ A −1
Examining line 9 of Table 1.5 one can come to the conclusion that the factor k must be the
electrical charge . Therefore, we can expect that

= ( + x )

Example #3. The last example is the unit dimension of factor k in the electric charge
conservation law called Gauss’s law (presented later)

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