# No epsilon or mu factor in the equations

1. Apr 25, 2010

### vin300

I seem to have a silly problem.In Gaussian units, there's no epsilon or mu factor in the equations, so esu must be medium dependant.Now if this is true, then I can "produce" or make "vanish" charges just by switching between media which obviously isn't true.So the mistake?

2. Apr 25, 2010

### clem

Re: Gaussian

"so esu must be medium dependant" Why?
In Gaussian units, D and E in a medium are related by D=\epsilon E.
This epsilon is called the permittivity of the material.
It is dimensionless, and is equal to the "dielectric constant" in SI.
There just is no 4piepsilonzero or muzero/4pi, since free space has unit permittivity and permeability.

3. Apr 25, 2010

### Born2bwire

Re: Gaussian

To clarify on clem's statements, we still retain the relative permittivity and permeability factors and drop the vacuum constants in favor of c (which sometimes can be also set to unity in some units). I find it a bit annoying myself because we often have the habit of assuming vacuum in our problems and thus drop out the appearance of \epsilon and \mu all together since they are now unity. However, this is annoying when you want to look at the problem in an arbtrary homogeneous medium because now you have removed the relationship with the permeability and permittivity from the final solution. So sometimes converting back from Gaussian to MKS can be difficult.

4. Apr 25, 2010

### Staff: Mentor

Re: Gaussian

As an aside, you might consider making your thread titles a wee bit more descriptive. When I saw "Gaussian", I though you might be asking about Gaussian probability distributions or Gaussian wave packets or some such thing.