# Dimension of the electric charge in CGS and in SI?

The Bohr magneton is (see e.g. Wikipedia) in SI units:
$$\mu_B=\frac{e\hbar}{2m_e}$$
and in CGS units:
$$\mu_B=\frac{e\hbar}{2m_ec}$$
So the dimension of the electric charge in SI, ##[q_{SI}]##, is related to the dimension of the electric charge in CGS, ##[q_{CGS}]##, by:
$$[q_{CGS}]=[q_{SI}].velocity$$
Now, the electrostatic force between two charges ##q_1## and ##q_2## separated by a distance ##r## is in SI:
$$F=\frac{q_1q_2}{4\pi\epsilon_0r^2}$$
and in CGS:
$$F=\frac{q_1q_2}{r^2}$$
So the dimension of the permitivity, ##[\epsilon_0]##, is given by:
$$[\epsilon_0]=\frac{[q_{SI}]^2}{[q_{CGS}]^2}=velocity^{-2}$$
which is not true.
So I guess I make a mistake somewhere, and I would be grateful for any help.

Dale
Mentor
2020 Award
The dimension of charge in electrostatic cgs units is ##M^{1/2} L^{3/2} T^{-1}##. The dimension of charge in SI is ##Q##. They do not have compatible dimensions so you have to be careful in conversions.

So the dimension of the permitivity, [ϵ0][ϵ0][\epsilon_0], is given by:
##\epsilon_0## doesn’t even exist in cgs, so this approach is fundamentally mistaken. The vacuum permittivity is a defined constant in SI. You obtain its dimensionality in SI through its definition, not through comparison with other unit systems where it doesn’t exist.

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Thanks Dale.

Actually, I think I can answer my own post:
The Bohr magneton does not have the same dimension in SI and in CGS, although this is, of course, the same physical quantity in both system.

One has the same "paradox" with the magnetic field: ##B_{CGS}=\sqrt{\frac{4\pi}{\mu_0}}B_{SI}## with ##\mu_0## in ##N/A^2##. So when one says 1 tesla=10,000 gauss, the 10,000 is not a "pure" number, it has a physical dimension.

BvU