Unraveling the Mystery of 4π in Coulomb's Law

[Rats_N_Cats]

Why the 4π in Coulomb's Law, SI version? The CGS version does well without it...

<br /> \mbox{thanks in advance!}<br />

 

[Pythagorean]

http://en.wikipedia.org/wiki/Statcoulomb

They're not dimensionally equivalent. The 4pi is a result of the spherical geometry of the field created by a point charge.

 

[Rats_N_Cats]

http://en.wikipedia.org/wiki/Statcoulomb

They're not dimensionally equivalent. The 4pi is a result of the spherical geometry of the field created by a point charge.

I've read that link you provided...I understand the difference between statcoulomb and coulomb, that they're not dimensionally equivalent. however this comes from the ε_o, which has a dimension. It doesn't say why the 4π enters the picture

 

[Rats_N_Cats]

And what's the "spherical geometry of the field created by a point charge"? Could anyone elaborate on that?

 

[Born2bwire]

You should read up on rationalized units. The difference between CGS and MKS in terms of where the 4\pi is located depends on how you rationalize Maxwell's equations. So it is not that CGS does not have the 4\pi, it is that CGS moves the 4\pi to the Maxwell equations.

http://en.wikipedia.org/wiki/Gaussian_units#.22Rationalized.22_unit_systems

 

[Rats_N_Cats]

So both CGS and SI have the 4π, only in different places? But what is its necessity? Wouldn't it be possible to define the electromagnetic units such that the factor of 4π is eliminated?

 

[Born2bwire]

It is necessary because of the spherical geometry inherent in the physics. For example, if I have a point source antenna that creates spherical waves, the energy across any spherical surface centered about our source must remain constant in a lossless medium. That is, if we have a lossless medium, then the energy emitted must remain constant. If we emit spherical waves, then the entire energy spread across a given wavefront remains constat as it propagates out in space. If we were to look at the energy at a single point on the wavefront as the wave expanded/propagated, then we would see that the fields would drop off as 1/(4 \pi r^2) since the surface of the wavefront is expanding as a spherical surface.

This is where we get the 4\pi from. In terms of statics, we can look at Gauss' Law. If I place a single point charge at the center of a spherical Gaussian surface, then the total flux through the Gaussian surface of the electric field is proportional to the charge. Through the use of spherical symmetry we can actually derive the actual electric field from this relationship. The result is of course Coulomb's law and once again due to the spherical geometry we acquire the 4\pi factor. But since Coulomb's law is incorporated into Maxwell's Equations, we can move the 4\pi off of Coulomb's law to Gauss' law and not change the resulting physics.

 

[UgOOgU]

To see how the unit systems of electromagnetic quantities are constructed and understand the relations between its equations I recommend:
Jackson, J.D. Appendix on Units and Dimensions on Classical Electrodynamics.

 

[Rats_N_Cats]

<br /> \mbox{hmmm...got that}<br />
Thanks, born2bwire! Your explanation was good.