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Gauss's law in cgs unit system

  1. Nov 19, 2014 #1
    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:
    [tex]\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[/tex]
    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. jcsd
  3. Nov 19, 2014 #2

    ehild

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    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}##
     
  4. Nov 19, 2014 #3
    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}##
     
  5. Nov 19, 2014 #4

    ehild

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    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
     
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