No epsilon or mu factor in the equations

[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?

 

[Meir Achuz]

"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.

 

[Born2bwire]

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.

 

[jtbell]

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.

 

[PJC01]

It is true that in Gaussian units, there is no epsilon or mu factor in the equations. However, this does not mean that the permittivity (epsilon) and permeability (mu) of a medium are not important. In fact, they play a crucial role in determining the behavior of electromagnetic fields in a given medium.

The equation for electric field in Gaussian units is given by E = Q/r^2, where Q is the charge and r is the distance. This equation does not explicitly include epsilon, but it is implicitly present in the value of Q. Similarly, the equation for magnetic field is B = mu*I/r, where I is the current and r is the distance. Again, while mu is not explicitly present, it is implicitly present in the value of I.

The mistake here is assuming that the absence of epsilon and mu in the equations means that they do not affect the behavior of charges and fields in different media. In reality, the permittivity and permeability of a medium determine how charges and fields interact with each other, and switching between media can indeed produce or make charges "vanish" due to the different values of epsilon and mu.

In conclusion, while the equations in Gaussian units may not explicitly include epsilon and mu, these factors are still crucial in determining the behavior of electromagnetic fields and should not be overlooked.