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Differences between SI and Gaussian units
(3) In Gaussian units, the k in Coulomb’s law is chosen to be dimensionless, whereas in SI units the k (which involves 0 ) has units.3 The result is that the esu can be expressed in terms of other Gaussian units, whereas the analogous statement is not true for the coulomb. This is the most important difference between the two systems.
A.4 Three units versus four Let us now discuss in more detail the issue of the number of units in each system. The Gaussian system has one fewer because the esu can be expressed in terms of other units via Eq. (A.4). This has implications with regard to checking units at the end of a calculation. In short, less information is gained when checking units in the Gaussian system, because the charge information is lost when the esu is written in terms of the other units. Consider the following example. In SI units the electric field due to a sheet of charge is given in Eq. (1.40) as Ďƒ/2 0 . In Gaussian units the field is 2Ď€ Ďƒ . Recalling the units of 0 in Eq. (1.3), the units of the SI field are kg m C−1 s−2 (or kg m A−1 s−3 if you want to write it in terms of amperes, but we use coulombs here to show analogies with the esu). This correctly has dimensions of (force)/(charge). The units of the Gaussian 2Ď€ Ďƒ field are simply esu/cm2 , but since the esu is given by Eq. (A.4), the units are g1/2 cm−1/2 s−1 . These are the true Gaussian units of the electric field when written in terms of fundamental units. Now let’s say that two students working in the Gaussian system are given a problem where the task is to find the electric field due to a thin sheet with charge density Ďƒ , mass m, volume V, moving with a nonrelativistic speed v. The first student realizes that most of this information is irrelevant and solves the problem correctly, obtaining the answer of 2Ď€ Ďƒ (ignoring relativistic corrections). The second student royally messes things up and obtains an answer of Ďƒ 3 Vm−1 v−2 . Since the fundamental Gaussian units of Ďƒ are g1/2 cm−1/2 s−1 , the units of this answer are 1/2 −1/2 −1 3 g cm s (cm)3 Ďƒ 3V g1/2 , (A.5) −→ = 2 2 mv (g)(cm/s) cm1/2 s which are the correct Gaussian units of electric field that we found above. More generally, in view of any answer with the Eq. (A.4) we see that n units of (g1/2 cm−1/2 s−1 ) esu g−1/2 cm−3/2 s has the correct units for the field. The present example has n = 3. There are, of course, also many ways to obtain incorrect answers in the SI system that just happen by luck to have the correct units. Correctness of the units doesn’t guarantee correctness of the answer. But the 3 To draw a more accurate analogy: in SI units the defining equation for the ampere
(from which the coulomb is derived) contains the dimensionful constant Îź0 in the force between two wires.