It is said that Coulomb's 'inverse-square' law (and Gauss's Law) are empirical facts. I'm wondering how do we know that Coulomb's law is inverse-square, and what are the possible consequences if it's not an inverse-square law. Suppose I write out the 'generalized' Coulomb's law for a point charge of the form:

$$ \vec{E}=\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{r^{2+\delta}} $$

and the gravitational field of the form:

$$ \vec{g}=\frac{GM}{r^{2+\delta}} $$

where $\delta$ represents a deviation from the inverse square. What are the implications in each case if $\delta\neq0$?

I have an example in Coulomb's case, but I'm not pretty sure how can I explain that:

Imagine two isolated concentric spherical conducting shells charged with total charges Qa and Qb and radii a and b such that a > b. If we connect the two shells with a thin conducting wire, if $\delta\neq0$, there will be some charge left on the inner shell.

Why would that be true? Thanks!

  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/22010/2451 , physics.stackexchange.com/q/93/2451 and links therein. $\endgroup$ – Qmechanic Sep 30 '20 at 20:24
  • 1
    $\begingroup$ @Qmechanic Thanks! $\endgroup$ – ZR- Sep 30 '20 at 20:37
  • 1
    $\begingroup$ I the inner shell is charged with electrons and the outer one is not charged, all electrons will flow from the inner shell to the outer one. This will also happen when $\delta$ is not equal to zero. If $\delta$ is equal to one, this will happen slower because the force reduces faster (linearly with $\frac{1}{r^3}$). In both systems, the electrons move to a state of least potential energy.Where did you find that electrons should be left on the inner shell? Can you provide a link? $\endgroup$ – Deschele Schilder Sep 30 '20 at 22:38
  • 1
    $\begingroup$ @Deschele Schilder Thanks! The equation to find the charge of the inner shell is given by $Q_b = (\frac{4\pi\epsilon_0 aV}{m-1})\delta F(m)$ it's on my note but I forgot where it comes from. $\endgroup$ – ZR- Sep 30 '20 at 23:47
  • $\begingroup$ If $\delta \neq 0$ then the forces are weaker if $\delta > 0$ and stronger if $\delta < 0$. $\endgroup$ – PiKindOfGuy Oct 1 '20 at 3:43

Any inverse-square relation for an interaction comes from this notion of flux and flux density.

Let's say that instead of an electric field, it's just the intensity of a spherically radiating object. Let's say a 100 watt light bulb. Radiation intensity is about how much power of radiation passes through a unit of area (say meter${}^2$). So a sphere of radius $r$ surrounding that light bulb will have a total of 100 watts of power crossing the spherical boundary from inside the sphere to the outside. This means that the 100 watts is distributed equally over the entire surface area of the sphere, which is $4 \pi r^2$.

So the intensity $I$, power per unit surface area, times the amount of surface is equal to the total power $P$.

$$ I \cdot 4 \pi r^2 = P $$

For the light bulb case above, $P$=100 watts. This makes the intensity an inverse-square quantity in our 3-dimensional space:

$$ I = \frac{P}{4 \pi r^2} $$

That is the root to the notion of an inverse-square law. The exponent in the denominator must be exactly 2 in a 3-dimensional space.

Now for E&M or for gravity, it is a different quantity than power that is being conserved and distributed over the surface area of a sphere of radius $r$. It is electric flux or gravitational flux that is spreading out in 3-dimensional space. Then the flux density must be an inverse-square quantity. And field strength is proportional to flux density. If you double the flux density, you also double the field strength.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.