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

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 $\epsilon$,

$dS=r^{2}\,\sec \epsilon \,d\omega$;

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

$R\,\cos \epsilon \,dS=\pm e\,d\omega$;

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 $r$,

$\sum R\,\cos \epsilon \;dS=0\,\!$,

and therefore

$\iint R\,\cos \epsilon \,dS=0$,

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 $ed\omega$, and then has an equal number of entrances and issues, so that in this case

$\sum R\cos \epsilon \,dS=e\,d\omega$

.

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

$\iint R\,cos\epsilon \,dS=e\,\iint d\omega =4\pi \,e$.

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