Page:Elementary Text-book of Physics (Anthony, 1897).djvu/80

66 drawn outward from the surface by $$a$$ and $$a'$$ respectively. The components of the forces $$\frac{m}{r^2}$$ and $$\frac{m}{r'^2}$$ drawn outward normal to the surfaces $$s$$ and $$s'$$ are $$\frac{m}{r^2} \cos{\alpha}$$ and $$\frac{m}{r'^2} \cos{\alpha '}$$ respectively. Hence the flux of force through these elements is $$\frac{m}{r^2}s \cos{\alpha} + \frac{m}{r'^2}s' \cos{\alpha '}$$. But $$s \cos{\alpha}$$ and $$s' \cos{\alpha '}$$ are equal to the normal cross-sections of the tube of force at the distances $$r$$ and $$r'$$ from $$O$$, the minus sign being inserted because one of the two cosines is negative; and since the tube of force is a cone, Hence the flux of force through these two elements, due to the mass at the point $$O$$, is equal to zero. Since similar tubes of force may be drawn from the point so as to include all the elements of the surface $$ABC$$, and since to each pair of elements thus determined the same proposition applies, it follows that the total flux of force due to the mass $$m$$ through the surface is equal to zero. The same proposition will hold for the flux of force due to any other particle situated outside the surface, and therefore holds true for any mass whatever situated outside the surface.

The flux of force through a closed surface containing any number of particles is equal to $$4 \pi M$$, where $$M$$ is the mass of all the particles. To prove this, let us consider a single particle $$m$$ situated at the point $$O$$. About this point describe a sphere of radius $$r$$. The force at each point of the sphere is $$\frac{m}{r^2}$$, and the total flux of force through the sphere is equal to this force multiplied by the area of the sphere, or to $$\frac{m}{r^2}\cdot 4 \pi m = 4 \pi m.$$ Now to prove a similar proposition for any closed surface enclosing the mass $$m$$, we describe about the point $$O$$ a sphere which is entirely enclosed by the surface. Since the region enclosed between this sphere and