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

78 ELECTROSTATICS. [76.

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

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

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

This cone will cut off a small element ${\displaystyle d\omega }$ from the surface of the sphere, and small elements ${\displaystyle dS_{l}}$, ${\displaystyle dS_{2}}$ , &c. from the closed surface at the various places where the line ${\displaystyle OP}$ intersects it.

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

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

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

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

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

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

${\displaystyle \sum R\,\cos \epsilon \;dS=0\,\!}$,
 and therefore ${\displaystyle \iint R\,\cos \epsilon \,dS=0}$,

the integration being extended over the whole closed surface.

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

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

.

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

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

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

Since in air the displacement is equal to the induction divided