Units - MKSA vs Gaussian

Coulomb's law has the form:

$$F = K_C \frac{q_1 q_2}{r^2}.$$ (1)

Ampere's law has the form:

$$F = K_A \frac{i_1 i_2}{r_{12}^2} \; d{\bf{s}_1} \times d{\bf{s}_2} \times {\bf{\hat{r}}_{12}},$$ (2)

where the current and charge are related by i₁ = dq₁/dt for all unit systems. The two constants K_C and K_A are related so that their ratio K_C/K_A has the units of (m/s)² and it is experimentally known that in both the MKSA and CGS (Gaussian) unit systems, it the value K_C/K_A = c², where c is the speed of light.

The choices for these constants in the MKSA and Gaussian units are given below:

CGS (Gaussian) MKSA

$$\frac {1}{4 \pi \epsilon_0}$$ K_C 1

$$\frac{1}{c^2} \frac {\mu_0}{4 \pi}$$ K_A

Here, $\frac {\mu_0}{4 \pi} \equiv 10^{-7} N/A^2$ and $\frac {1}{4 \pi \epsilon_0} = c^2 \cdot 10^{-7} N/A^2 = 8.98755\times10^9 N \cdot m^2/C^2$.

Below is a table comparing MKSA and Gaussian unit systems. The fundamental units for each system are labeled ``fixed" and are used to define the derived units.

Variable 2c||MKSA 2c||Gaussian MKSA/Gaussian

Unit Relation Unit Relation

length m fundamental cm fundamental 100

mass kg fundamental gm fundamental 1000

time s fundamental s fundamental 1

$$kg \cdot m^2/s gm \cdot cm^2/s$$ force N dyne 10⁵

$$\frac{1}{10 c}$$ current A fundamental statampere statcoulomb/s

$$A \cdot s \sqrt{dyne \cdot cm^2} \frac{1}{10 c}$$ charge C statcoulomb

One advantage of the Gaussian system is that all of the field vectors: ${\bf{E,D,B,H,P,M}}$ have the same dimensions, and in vacuum, ${\bf{B=H}}$ and ${\bf{E=D}}$ and the dielectric and permittivity constants $\epsilon$ and $\mu$ are unitless.

CGS (Gaussian) MKSA

$$\nabla \cdot {\bf{D}} = 4 \pi \rho \nabla \cdot {\bf{D}} = \rho$$

$$\nabla \cdot {\bf{B}} = 0 \nabla \cdot {\bf{B}} = 0$$

$$\nabla \times {\bf{E}} = - \frac{1}{c} \frac{\partial{\bf{B}}}{\partial t} \nabla \times {\bf{E}} = - \frac{\partial{\bf{B}}}{\partial t}$$

$$\nabla \times {\bf{H}} = \frac{4 \pi}{c} {\bf{J}} + \frac{1}{c} \frac{\partial{\bf{D}}}{\partial t} \nabla \times {\bf{H}} = {\bf{J}} + \frac{\partial{\bf{D}}}{\partial t}$$

$${\bf{F}} = q ({\bf{E}} + \frac{{\bf{v}}}{c} \times {\bf{B}} {\bf{F}} = q ({\bf{E}} + {\bf{v}} \times {\bf{B}}$$

$$u = \frac{1}{8 \pi} ({\bf{E}}\cdot{\bf{D}}+{\bf{B}}\cdot{\bf{H}}) u = \frac{1}{2} ({\bf{E}}\cdot{\bf{D}}+{\bf{B}}\cdot{\bf{H}})$$

$${\bf{S}} = \frac{c}{4 \pi} ({\bf{E}} \times {\bf{H}}) {\bf{S}} = ({\bf{E}} \times {\bf{H}})$$