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Q & A: Why inverse square distance in Coulomb law?

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Most recent answer: 08/16/2012
Q:
why there is inverse square distance in coulumb`s law? why not only the distance or anything else?
- mumtaz ahmad (age 21)
srinagar j&k india
A:
That's a very deep question, probably too deep for me. I'll give an initial answer and then if some colleagues help out maybe we can deepen or simplify it. What I'll try to do is show that the inverse square law follows from some more generic properties.

For starters, notice that the classical gravity force also falls off as the square of the distance. So that suggests that the reason isn't too specific to electrical forces.

 Let's start out with some basics. The source of the electrical force is a scalar (simple number, not a vector) property: charge. (For gravity it's a different scalar, mass.) The force is a vector property, meaning it has not only a strength but a direction. The electrical force only depends on positions, not velocities. (The related force that depends on velocities is given another name, magnetism.) So long as the force doesn't violate energy conservation, we can then write it as the spatial derivative (gradient) of a potential energy which depends only on the distance between the point charges. We'll assume that the energy depends linearly on each of the charges involved. So let's say that the energy of a point particle at position r  depends on some potential V(r) that is itself a linear function of the distribution of other charges, ρ(r).

What sort of spatial dependencies then follow? You could imagine that the energy was zero unless the two point charges were at exactly the same place. I believe that would be logically consistent, but it would not give any sort of noticeable force for classical point particles, since they are essentially never at the same exact place.

How else could V(r)  depend on ρ(r)? The first derivative of V with respect to r is a vector, so that can't match up with ρ. The simplest scalar derivative is a second derivative, the divergence of the gradient, called del-squared or the Laplacian. So the simplest expression we could imagine by which an interesting force's potential energy depends on the source distribution is of the form:

 del-squared of V(r) = constant*ρ(r).


My dad remembered his best teacher writing that expression on the board, then saying with a smile "Sometimes this is called the inverse square law." The reason is that it's a straightforward mathematical fact that the solution for a point-like ρ(r) at the origin is a V(r) that falls of as 1/r. Taking its gradient gives force falling of as 1/r2.

Now this doesn't prove that the force law has to be just like that, only that it's a simple natural form. There are other forces which have a different and slightly more complicated form, cutting off more sharply at longer distances. It's not surprising that those short-range nuclear forces were discovered later, since they're not noticeable on the distance scale of the sorts of experiments that could be done before say 1900.

Mike W.

(published on 08/16/2012)

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