# If the electrostatic Coulomb force followed inverse cube law, what's implication on Gauss' law?

If the electrostatic force between two charged particles varied according to the inverse of the cube of distance between them, how would it affect the Gauss Law?

Well we know if a metallic sphere is given some charge, the charge resides on the surface and the electric field inside is zero. I was told that the fact that the charge is on the surface can also be explained by Gauss Law but I cannot understand how.

Moreover, as said before, if the force varied as the inverse of the cube of the distance in between, how will the field inside the sphere and charge distribution be affected?

I can see that as electric flux through a surface is defined as the closed integral of electric field over the surface, that would surely imply the electric flux varies with distance, which would otherwise be independent of the distance, if the inverse square law is followed.

I thought a variation in flux implies variation in field. Am I correct?

An electrostatic force according to $1/r^3$, where $r$ is the distance between the charges, is inconsistent with Gauss' law. The electric flux through the surface of a sphere around a charge $q$ can be calculated by applying the divergence theorem to Gauss' law: $\int_V \nabla \cdot \vec E dV=\int_S \vec Ed\vec S =4 \pi r^2 E_r= \frac{q}{\epsilon_0}$, where $E_r$ is the radial component of the electric field, $V$ is the sphere's volume and $S$ the sphere's surface.
Hence, you get a quadratic dependence on $r$:
$E_r=\frac{q}{4\pi r^2 \epsilon_0}$.
• The question "if the force varied as the inverse of the cube of the distance in between, how will the field inside the sphere and charge distribution be affected?" is not really meaningful, since you need Gauss' law to calculate the electric field from the charge density. However, a $1/r^3$ dependence contradicts Gauss' law. – ggg Dec 15 '17 at 14:29
• Actually, if the electrostatic force varied according to $1/r^3$ we would not live in 3D (+time) world, but in a 4D (+time). In a 3D world as ours, the force will never vary $\sim 1/r^3$, because this dependency on $r$ would violate charge conversation. – Frederic Thomas Dec 15 '17 at 14:33