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3. Electric and Magnetic Units: Gaussian and SI

    Michael Fowler, UVa

Introduction

As you are well aware, there are two systems of units in wide use in the physics of electricity and magnetism, and you need to be able to function in both of them. Gradually, the SI (système international) is taking over, but many physicists still use the Gaussian system, or some variant of it, because it’s much more convenient at the microscopic level.

Furthermore, it’s not like miles and kilometers, with a simple numerical ratio between the basic units: in E&M, the basic units are dimensionally different!

Gaussian (cgs) Units:

We’ll begin with the Gaussian system. The units of length, mass and time are the centimeter, gram and second, hence the name cgs. 

The unit of force, the dyne, imparts unit acceleration to unit mass, so will accelerate one gram at one cm/sec².  These are nice units for typical tabletop physics investigations. The unit of charge (the statcoulomb) is defined as the charge that repels an identical charge one cm away with a force of one dyne.

Thus the force between two charges is

$\overrightarrow{F}=\frac{q_1{q}_2}{r^2}\widehat{\overrightarrow{r}}.$ 

Now the dimensions of force are $ML{T}^{-2},$ so evidently the dimensions of charge are

$\left[q\right]=\sqrt{M{L}^3{T}^{-2}}=M^{1/2}{L}^{3/2}{T}^{-1}.$ 

The electric field from a charge is

$\overrightarrow{E}=\frac{q}{r^2}\widehat{\overrightarrow{r}}$ 

from which ${\left[E\right]}_{\text{Gaussian}}=M^{1/2}{L}^{-1/2}{T}^{-1}.$ 

The formula for the Lorentz force on a charge $q$ moving at velocity $\overrightarrow{v}$ (in Gaussian units) is

$\overrightarrow{F}=q\left(\overrightarrow{E}+\left(\frac{\overrightarrow{v}}{c}\right)\times \overrightarrow{B}\right)$ 

So evidently in this system the electric and magnetic fields have the same dimensionality. This is particularly clear in the Gaussian units formula for the energy density in an electromagnetic field:

$U=\frac{1}{8\pi }\left({\overrightarrow{E}}^2+{\overrightarrow{B}}^2\right).$

SI Units

In using SI, the awkward fractional dimensionalities found above are avoided by introducing another dimension:  current, $I.$ 

(Current rather than charge, no doubt, because this is the system used by electrical engineers.) 

The dimension of charge is therefore

$\left[q\right]=IT.$ 

This makes it necessary to introduce a dimensional constant in the electrostatic force law:

$\overrightarrow{F}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}\overrightarrow{r}$ 

where we must have $\left[{\epsilon }_0\right]=M^{-1}{L}^{-3}{T}^4{I}^2.$  (Check this!) 

Recall that in SI units, unlike Gaussian, the unit of charge is not defined by the above equation: it follows from the unit of current, which is such that for two long parallel wires one meter apart, each carrying unit current, the force per meter on one of them is given (in Newtons) by

$F=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}=\frac{{\mu }_0}{2\pi }=2\times {10}^{-7}.$ 

Here evidently

$\left[{\mu }_0\right]=ML{T}^{-2}{I}^{-2}.$

${\mu }_0$ is defined to have the numerical value $4\pi \times {10}^{-7},$ this sets the unit of current$—$the ampère$—$and hence of charge, the coulomb.

Naturally, we still want to define the electric field as the force per unit charge on a test charge at some point, $\overrightarrow{F}=q\overrightarrow{E},$ so we must have

${\left[E\right]}_{\text{SI}}=ML{T}^{-3}{I}^{-1}.$ 

The field from a point charge is  $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\overrightarrow{r},$ this is dimensionally consistent.

Notice now that the electric field energy density must be proportional to $E^2$ in both systems, but since the electric field has different dimensionalities, there must be a dimensional multiplier in SI to agree with the Gaussian $E^2/8\pi$.  In fact, the SI expression is $\frac{1}{2}{\epsilon }_0{E}^2.$ (Check the dimensions!)

In SI units, the magnetic field is defined in terms of a force on a moving charge $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B},$ but in contrast to the Gaussian system, there is no $1/c$ term, so the electric and magnetic fields have dimensions differing by a velocity.  Since $q\overrightarrow{v}$ is an element of current, from the force on part of a parallel wire from another infinite parallel wire is as given above, we have a circling magnetic field of magnitude $B={\mu }_{0}I/2\pi r$ from a current in an infinite straight wire.  (Check how this follows from the force between two wires.)

The dimensionality of the magnetic field ${\left[B\right]}_{SI}=M{T}^{-2}{I}^{-1},$ check it from $\left[{\mu }_0\right]=ML{T}^{-2}{I}^{-2},$ or from the electric field dimensionality.

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