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Force of attraction forumla

by esvion
Tags: attraction, force, forumla
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esvion
#1
Nov24-08, 05:09 PM
P: 19
The force of attraction formula between two charges is

[tex]\frac{(k)(e1)(e2)}{r^2}[/tex]

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.
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Doc Al
#2
Nov24-08, 06:01 PM
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Coulomb's law is an example of an inverse square law, something quite common in physics. Read about it here: Inverse Square Law
esvion
#3
Nov24-08, 07:06 PM
P: 19
I see! thanks!

rcgldr
#4
Nov24-08, 09:33 PM
HW Helper
P: 7,032
Force of attraction forumla

Quote Quote by Doc Al View Post
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
#5
Nov26-08, 03:11 AM
P: 6
Things get really cool when you start checking out far field proportionalities in systems of multipoles!


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