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?
Best Answer
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 $r$:
$E_r=\frac{q}{4\pi r^2 \epsilon_0}$.