San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 On The Reduction of a Distributed Force to a Point Force

## Background

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

• at a point outside of the sphere the force is the same as if the charge were concentrated at the center of the sphere.
• at a point inside of the sphere the force is zero.

Although it is seldom articulated there is a third theorem that says

• at a point on the sphere the force is the same as if one half of the charge were concentrated at the center of the sphere

## Analysis

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

#### dF = ρf(y)[2πz]rdφ and the component in the x direction dFcos(φ) = ρf(y)[2πz]rcos(&phil)dφ

where he radius of the ring, z, is equal to rsin(φ). The distance y is given by

#### y² = x² + r² - 2xrcos(φ)

The component of the force in the x-direction is

#### F(x,r) = ∫0πρf(y)(2πr²)sin(φ)cos(φ)dφ which reduces to F(x,r) = 2πr²∫0πρf(y)sin(φ)cos(φ)dφ

The variable of integration can be changed from φ to y. Thus

#### 2ydy = 2xrsin(φ)dφand hence dφ = ydy/(xrsin(φ))

Therefore, since y=x-r when φ=0 and y=x+r when φ=π and cos(φ)=(x²+r²−y²)/(2xr)

#### F(x,r) = 2πr²∫x-rx+r[ρf(y)y(x²+r²−y²)/(2(xr)²)]dy which reduces to F(x,r) = (π/x²)∫x-rx+r[ρf(y)y(x²+r²−y²)]dy

For convenience let ρf(y)y be denoted as g(y). Then

## Reduction of a Distributed Force to a Point Force

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

#### ∂F/∂r = (π/x²){[g(x+r)(2xr) − g(x-r)(2xr)] + 2r∫x-rx+rg(y)dy}

Let the indefinite integral ∫g(y)dy be denoted as G(y). Thus the above equation can represented as

#### ∂F/∂r = (π/x²){[G'(x+r)(2xr) − G'(x-r)(2xr)] + 2r[G(x+r)−G(x-r)]}

If ∂F/∂r is identically equal to zero then

#### x[G'(x+r)−G'(x-r)/(2r)] = −[G(x+r)−G(x-r)/(2r)] this holds for all r and hence it holds for all r as r goes to zero Thus limr→0 {x[G'(x+r)−G'(x-r)/(2r)]} = − limr→0 [G(x+r)−G(x-r)/(2r)] which is xG"(x) = −G'(x) or, equivalently xg'(x) = −g(x)

This equation can be solved as

#### dg/g = −dx/x

The solution to this equation is

#### g(x) = k/x

where k is an arbitrary constant.

Since g(x)=xρf(x) this means the solution for f(x) is

#### f(x) = h/x²

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.

## Analysis for Nonreductive Force Forms

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.)