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Q & A: why is Coulomb's law inverse square?

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Most recent answer: 11/12/2013
Q:
why in coloumbs law inverse sqaure radius instead of only radius
- bilal mushtaq (age 19)
islamabad pakistan
A:

That's a tough question. Here are some partial answers, not as deep or as clear as they should be.

Often we picture the electrical field by sets of field lines originating at electrical charges and spreading out in space, smoothly without breaks. The strength of the field is represented by the density of the lines. Starting with a point charge , the lines spread out over bigger areas the farther away (r) they go. The area goes as r2, so the field strength goes as 1/r2.

That answer doesn't really get to the reason, because it doesn't say why the whole field-line picture works. The mathematical equivalent of it is to say ∇⋅E=ρ (in some simple units), meaning that the divergence of the electric field is proportional to the electric charge density. That's like saying that the number of field lines originating in some volume is proportional to the electric charge there. Why should this be true?

It may be easier to think of the electrical potential V rather than the field E, because V is just a number, not a vector. The equivalent equation for V is 2V=-ρ, which means that the curvature of V, a measure of how much it differs at some point from the average of nearby points, is proportional to that charge density. Away from charges, that means that at each point V is the average of V in its neighborhood. If you think of that potential V as coming from some sort of stuff that flows around, that makes sense. This gives right away a solution that for a point charge V(r) falls off as 1/r. Since E is the slope of -V, it falls off as 1/r2

Perhaps somebody else can come up with an answer that is easier to see, or better at capturing the underlying theory (QED), or perhaps even both.

Mike W.

 

p.s. Here's another try at the same question. 


(published on 11/12/2013)

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