Inverse cube law

[physica]

I have been reading about how dipoles produce an inverse cube law. The sad thing is that I can't find any maths behind it. Am I reading a crackpot theory? If not could someone post the maths or give a link to the maths?

[Carrock]

  On 10/22/2014 at 11:38 AM, physica said:

I have been reading about how dipoles produce an inverse cube law. The sad thing is that I can't find any maths behind it. Am I reading a crackpot theory? If not could someone post the maths or give a link to the maths?

There is quite a lot of useful information here:

https://www.physicsforums.com/threads/magnetic-force-inverse-cubed-law.587204/

[swansont]

Write down the field of two monopoles of opposite charge, separated by an infinitesimal amount. Expand the denominator

[Enthalpy]

The cube is correct for a dipole, I confirm.

 

Remember this is for a static field or slowly varying. If the charges, currents... vary significantly within one propagation time between the poles, then you obtain a part as a propagating field which decreases less quickly.

[physica]

Thanks guys this has helped a lot.

[physica]

Coming back to this topic, why would Gauss's law still be valid if Coulomb's was replaced by an inverse cube law?

[Enthalpy]

It's still valid for each charge of the dipole. Only the net effect, as the difference between two forces in 1/R², changes as 1/R³.

[swansont]

  On 11/6/2014 at 4:37 PM, physica said:

Coming back to this topic, why would Gauss's law still be valid if Coulomb's was replaced by an inverse cube law?

 

It wouldn't be. Consider a spherically symmetric scenario — the law works because the flux drops off as r^2 and so does the surface area of a sphere. That's not the case for an inverse-cube relation. IOW, E.dA is not a constant — it varies with r.

[Enthalpy]

If the dipole consists of two finite charges with finite separation, each feels a force as 1/R² from a distant one. In that sense, Coulomb's law still applies.