Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/118

78 issue from the surface, and then it may enter and issue any number of times alternately, ending- by issuing from it.

Let $$\epsilon$$ be the angle between $$OP$$ and the normal to the surface drawn outwards where $$OP$$ cuts it, then where the line issues from the surface $$cos \epsilon$$ will be positive, and where it enters $$cos \epsilon$$ will be negative.

Now let a sphere be described with centre $$O$$ and radius unity, and let the line $$OP$$ describe a conical surface of small angular aperture about $$O$$ as vertex.

This cone will cut off a small element $$d \omega$$ from the surface of the sphere, and small elements $$dS_l$$, $$dS_2$$, &c. from the closed surface at the various places where the line $$OP$$ intersects it.

Then, since any one of these elements $$dS$$ intersects the cone at a distance $$r$$ from the vertex and at an obliquity $$\epsilon$$,

and, since $$R = er^{-2}$$, we shall have

the positive sign being taken when $$r$$ issues from the surface, and the negative where it enters it.

If the point $$O$$ is without the closed surface, the positive values are equal in number to the negative ones, so that for any direction of $$r$$,

the integration being extended over the whole closed surface.

If the point $$O$$ is within the closed surface the radius vector $$OP$$ first issues from the closed surface, giving a positive value of $$e d\omega$$, and then has an equal number of entrances and issues, so that in this case.

Extending the integration over the whole closed surface, we shall include the whole of the spherical surface, the area of which is $$4\pi$$, so that

Hence we conclude that the total induction outwards through a closed surface due to a centre of force $$e$$ placed at a point is zero when is without the surface, and $$4\pi e$$ when $$O$$ is within the surface.

Since in air the displacement is equal to the induction divided