Gauss's law in cgs unit system

1. Nov 19, 2014

Karol

1. The problem statement, all variables and given/known data
Is there a constant parrallel to the ε0 permittivity in the Gauss law in c.g.s?

2. Relevant equations
Coulomb force in m.k.s: $F=\frac{1}{4\pi\varepsilon_0}\frac{qq'}{r^2}$
Coulomb force in c.g.s: $F=\frac{qq'}{r^2}$
Gauss's law in m.k.s: $\frac{N}{A}=\varepsilon_0 E$

3. The attempt at a solution
The Coulomb's constant k was transformed to $\frac{1}{4\pi\varepsilon_0}$ in order that the number of field lines N extruding from a surpace round a point charge will equal the net chrage inside the surface:
$$\frac{N}{A}=\varepsilon_0 E\rightarrow \frac{N}{4\pi r^2}=\varepsilon_0 \frac{q}{4\pi\varepsilon_0 r^2}\rightarrow N=q$$
So the $\frac{1}{4\pi}$ term was chosen for that in order to cancel with the area of a sphere.
I think, in order that N=q will be also in c.g.s we also need $\frac{1}{4\pi}$ term:
Coulomb's law in c.g.s: $F=\frac{1}{4\pi B}\frac{qq'}{r^2}\rightarrow \frac{1}{4\pi B}=1$

2. Nov 19, 2014

ehild

In the cgs system, the Coulomb force is $F=\frac{q_1 q_2}{r^2}$,
ε0=1, D=εE in materials, D=E in vacuum.
Gauss' Law for a single point charge in vacuum $\oint{E_n dA} = 4\pi q_{enclosed}$

3. Nov 19, 2014

Karol

I don't know yet this integral.
So i am right, right? Coulomb's law in c.g.s: $F=\frac{1}{4\pi}\frac{qq'}{r^2}$

4. Nov 19, 2014

ehild

No. $F=\frac{qq'}{r^2}$
The electric field around a point charge q is radial, and the magnitude is E=q/r2. Integrating is to a sphere of radius r, with the charge in the centre is (4πr2)E = 4πq