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Force to a Point Force |
There are two marvelous theorems concerning forces such as gravitation, electrostatic or radiation pressure which are inversely proportional to the square of distance. These theorems say that if the charge which produces the force is distributed uniformly over a sphere then
Although it is seldom articulated there is a third theorem that says
This is an attempt to consider the general problem. Let r be the radius of the sphere and ρ the charge density. Let ρf(y) be the force from an infinitesimal element of area r²dφdθ at a point y distance from it.
For the ring between φ-dφ/2 and φ+dφ/2 the force is
where he radius of the ring, z, is equal to rsin(φ). The distance y is given by
The component of the force in the x-direction is
The variable of integration can be changed from φ to y. Thus
Therefore, since y=x-r when φ=0 and y=x+r when φ=π and cos(φ)=(x²+r²−y²)/(2xr)
For convenience let ρf(y)y be denoted as g(y). Then
For an inverse distance squared force F(x,r) reduces to the form K/x² where K is proportional to the amount of charge on the sphere. In particular, for an inverse distance squared force, F is independent of r.
Now consider ∂F/∂r which is
Let the indefinite integral ∫g(y)dy be denoted as G(y). Thus the above equation can represented as
If ∂F/∂r is identically equal to zero then
This equation can be solved as
The solution to this equation is
where k is an arbitrary constant.
Since g(x)=xρf(x) this means the solution for f(x) is
where h is an arbitrary constant.
Thus it is found that the only force function that allows a force distributed over a sphere to be reduced to a point force at the center of the sphere is the inverse distance squared form.
Concider the case in which f(y)=Ke^{-y}/y², the form that applies when the particle which carries the force decays.
(To be continued.)
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