Gaussian Unit of Charge and Force

Question

I just read that in the Gaussian Units of charge The Final equation in Coulomb's law is as simple as $$\boldsymbol{F}=\frac{q_1q_2}{r^2}$$

No $\epsilon_0$ no $4\pi$ like you have in the $\mbox{SI}$ units of measurement .

The permittivity constant was the factor in the $\mbox{SI}$ system of Coulombic Force that determined the intensity of force in a medium.

In the Gaussian system i see no such constant . So that would mean that some other factor would govern the Quantity of charge on a body.

What is that factor ?

(I vaguely remember it as being related to the speed of light $c_0$)

You can do the same thing in SI by redefining the charge unit so that a charge Q has a normalized charge $Q\over\sqrt{4\pi\epsilon_0}$, and always use normalized Coulombs. The best convention leaves the $4\pi$, but absorbs the $\epsilon_0$, and this is often quickly said by using the charge-unit convention that $\epsilon_0=1$, that is, the normalized charge is $Q\over \epsilon_0$. The analogous convention in gravity sets $G={1\over 8\pi}$.

user758556 · Accepted Answer · 2012-06-05 05:45:29Z

I believe you're thinking of the statcoulomb, which is the CGS/Gaussian system unit of charge.

The article linked above details some methods of conversion, which you can read for yourself, but probably the most important ones to know that:

1) $1\ Statcoulomb = \frac{{(gram)^{1/2}}(cm)^{3/2}}{sec}$

2) $1\ Coulomb \leftrightarrow \sqrt{4\pi \epsilon_0}\frac{c}{(1m/s)} Statcoulomb$

1 Answer