Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/236

 Let $$a$$ be the radius of the sphere.

Let $$f$$ be the distance of the electrified point $$A$$ from the centre $$C$$.

Let $$e$$ be the charge of this point.

Then the image of the point is at $$B$$, on the same radius of the sphere at a distance $$\tfrac{a^{2}}{f}$$, and the charge of the image is $$-e\tfrac{a}{f}$$.

We have shewn that this image will produce the same effect on the opposite side of the surface as the actual electrification of the surface does. We shall next determine the surface-density of this electrification at any point $$P$$ of the spherical surface, and for this purpose we shall make use of the theorem of Coulomb, Art. 80, that if $$R$$ is the resultant force at the surface of a conductor, and $$\sigma$$ the superficial density,

$R=4\pi\sigma,\,$

$$R$$ being measured away from the surface.

We may consider $$R$$ as the resultant of two forces, a repulsion $$\tfrac{e}{AP^{2}}$$ acting along $$AP$$, and an attraction $$e\frac{a}{f}\frac{1}{PB^{2}}$$ acting along $$PB$$.

Resolving these forces in the directions of $$AC$$ and $$CP$$, we find that the components of the repulsion are

$\frac{ef}{AP^{3}}$ along AC, and $\frac{ea}{AP^{3}}$ along CP.|undefined

Those of the attraction are

$-e\frac{a}{f}\frac{1}{BP^{3}}$ along AC, and $-e\frac{a^{2}}{f}\frac{1}{BP^{3}}$ along CP.|undefined

Now $$BP=\tfrac{a}{f}AP$$, and $$BC=\tfrac{a^{2}}{f}$$, so that the components of the attraction may be written

$-ef\frac{1}{AP^{3}}$ along AC, and $-e\frac{f^{2}}{a}\frac{1}{AP^{3}}$ along CP.|undefined

The components of the attraction and the repulsion in the direction of $$AC$$ are equal and opposite, and therefore the resultant force is entirely in the direction of the radius $$CP$$. This only confirms what we have already proved, that the sphere is an equipotential surface, and therefore a surface to which the resultant force is everywhere perpendicular.