Page:Elementary Text-book of Physics (Anthony, 1897).djvu/79

 If the force at $$P$$ and at the other points on the path be directed from $$O$$, the work done in the successive elements of the path is numerically equal to the expressions already obtained, but is opposite in sign; so that the work done by such forces, as the test unit moves from $$P$$ to $$X$$, is equal to $$m \left(\frac{1}{OP} - \frac{1}{OX} \right)\cdot$$ When the point $$X$$ is at an infinite distance from $$O$$, the work done in moving the test unit from $$P$$ to $$X$$ equals $$\frac{m}{OP}\cdot$$ This is the potential at the point $$P$$, due to a repulsive force with its centre at $$O$$. In this case the test unit at an infinite distance has no potential energy, so that $$\frac{m}{OP}$$ expresses its potential energy at $$P$$.

'''56. Flux of Force. Tubes of Force.'''—Still retaining the convention that the forces of the field are due to mass attraction and follow the law of inverse squares, we will now prove certain propositions which are of great importance in the theories of gravitation, electricity, and magnetism.

If in a field an area $$s$$ be described so small that the force is the same for all points of it, the product of the area and the normal component of the force is called the elementary flux of force over or through that area. We will show that the total flux of force, that is, the sum of all the elementary fluxes, taken over a closed surface in the field which does not contain any masses is equal to zero.

We consider first the flux of force arising from a mass $$m$$ situated at the point $$O$$. Let $$ABC$$ (Fig. 24) represent a closed surface not containing the mass $$m$$; draw a tube of force cutting this surface in the elements $$s$$ and $$s'$$. The forces due to the mass $$m$$ at points in these areas will be $$\frac{m}{r^2}$$ and $$\frac{m}{r'^2}$$ respectively. We represent the angles between the common direction of these forces and the normals to the elements $$s$$ and $$s'$$