- #1

- 21

- 2

$$\mu_B=\frac{e\hbar}{2m_e}$$

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.