What is the change in field in a sphere if Coulomb's force changes with 1/r^3

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 If some charge is given to a solid metallic sphere, the field inside remains zero and by Gauss’s law all the charge resides on the surface. Suppose now that Coulomb’s force between two charges varies as 1/r 3 . Then, for a charged solid metallic sphere, what is the change in field inside the sphere?

Prem Kumar

This is an interesting question.

Non-Coulombic electrostatic potential and Gauss' Law

The first point is,  what would happen if Coulomb's force law went like 1/r^3?

It would completely change the nature of what we mean by electrostatics (and magnetism).

The famous Gauss' Law of electrostatics would not apply. Gauss's law says that if you take the electric field and calculate its ``flux" through a closed surface (like a sphere enclosing a point charge), the result is proportional to the total charge enclosed by the surface, no matter what the shape or size of the surface.

This  is only possible because we live in 3 spatial dimensions and the Coulomb force law is an inverse square law. (If we lived in 4 spatial dimensions, then Coulomb's law would have to be 1/r^3 in order for Gauss's Law to apply).

You can check that with a 1/r^3 Coulomb law, the flux from a point charge through a sphere of radius  R falls off as 1/R (in reality the flux is independent of R).

Field inside Metal

Having said that, let us still try to think of the academic question you've posed. Suppose the force law has this unphysical form and you put some positive charge on a solid metallic sphere. Now, here comes a key point -- what do you mean by a metal ?  A metal or conductor is one with free charges (electrons) that can move around to cancel all electric  field in the metal. If the field were not zero, the electrons would move around and current would flow until the fields cancelled out. So the answer to your question, namely, what is the field inside a solid metallic sphere once charges are put on it -- it  should still be zero.

In fact,  the charge on the sphere would not reside only on the surface. If it did, you could use your new Coulomb law and linear superposition to calculate the electric field inside the sphere and the answer is non-zero (this is easy to see).You should, in principle, be able to use the vanishing of the electric field inside the solid metallic sphere, and assuming linear superposition principle, to find what the charge distribution would have to be. This is not an easy exercise for the 1/r^3 potential, but would be interesting to attempt. The exercise has been done in the following article which appears not to be an open access paper:


f I find the result quoted in it I will post it here.

Prem Kumar

You can download a pdf version of a paper from the link:

If Coulomb's law were not inverse square: The charge distribution ...

This paper does the calculation described in the answer above i.e. it determines the charge distribution inside a conductor when the force law is not Coulomb.

Mahesh Godavarti
If the sphere was a hollow metallic sphere, how would that affect the distribution? Would the field still be zero inside the sphere?
Prem Kumar
If the sphere were hollow and of infinitesimal thickness, and we placed a uniform charge distribution on the surface, the field inside would not be zero. It can be determined by first calculating the potential due to the charge distribution inside the hollow cavity assuming the superposition principle. Up to overall constants of proportionality the result is (it is a simple integral to do), the potential is f(r) = 1/r Log[(R+r)/(R-r)] where R > r and R = radius of sphere and 'r' is radial distance of a point from centre. The radial electric field is the derivative of this with respect to 'r'. It vanishes at the origin as it must by symmetry, but is otherwise non-zero. If the sphere was hollow and had finite thickness, similar considerations would apply although the problem would become technically more complex.