In Coulomb's law if the relation was as if electric field intensity was to vary inversely $1/r$ with distance rather than the inverse $1/r^2$ of square of distance, would the Gauss's law still be valid? This was asked in our university tests and I'm clueless about it.

  • $\begingroup$ Do you mean electric rather than magnetic field? $\endgroup$ – BMS Dec 24 '13 at 6:22
  • $\begingroup$ Yes.sorry it was my mistake $\endgroup$ – Devgeet Patel Dec 24 '13 at 7:53

No, this is not possible. It is only possible for $\mathbf{E}$ to be proportional to $\mathrm{r}^{-2}$.

  • $\begingroup$ It'd be better if you explained why it has to be that way. $\endgroup$ – Gonenc Sep 30 '15 at 16:37

Gauss's law would not be valid.

You can imagine the electric field "flowing out" from positive charges and "draining" into negative charges. The "amount" of electric field decreases at the same rate as it spreads (since area of a surface increases by the square of its scale). This means that if no matter how we expand our Gaussian surface, if we don't cross any sources or sinks, the total electric field flowing through it must be conserved.

If the electric field were to vary inversely with distance, the total electric field passing through a Gaussian surface would not depend only on the charge enclosed but also on how you draw it. If we take for instance a spherical surface of radius r around a charge Q, the total electric field passing through that surface would be:

$$ E_{tot}=(4\pi r^2)(k\frac{Q}{r})\propto Qr $$

Note that the total electric field passing through the surface now depends on the size of the Gaussian surface and would not make for a nice physical law.


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