How Does the Inverse Square Law Apply to Force of Attraction Between Charges?

[esvion]

The force of attraction formula between two charges is

$$\frac{(k)(e1)(e2)}{r^2}$$

How does the inverse of r² fit into the equation? I understand the concept of how distance would need to be the inverse in the function, but why is the distance (r) in the inverse squared? Is this the same principle of why s^-2 is the acceleration formula and time is square in the inverse?

Thanks.

 

[Doc Al]

Coulomb's law is an example of an inverse square law, something quite common in physics. Read about it here: http://hyperphysics.phy-astr.gsu.edu/Hbase/Forces/isq.html"

 

[esvion]

I see! thanks!

 

[rcgldr]

http://hyperphysics.phy-astr.gsu.edu/Hbase/Forces/isq.html"

Note that inverse square law is applies to point or spherical sources. For an infinitely (or very large) long line or cylinder, the ratio of force versus perpendicular distance to the line is 1/r. For an infinitely (or very large) plane, the force is constant (independent of distance).

Found the link for the other cases at the same site: electrical field

For the infinite line case, the field strenth is a function of charge "density" over the perpendicular distance "z" to the line ( ... / z).

For the infinite disc (plane) case, the limit as "R" approaches infinity, the [1 - z/sqrt(z^2 + R^2) ] term approaches [1 - 0], and the field strength is constant, independent of distance

 

[biggiekjac]

Things get really cool when you start checking out far field proportionalities in systems of multipoles!