Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/264

 Distribution of Electricity on a Disk.

By making two of the axes of the ellipsoid equal, and making the third vanish, we arrive at the case of a circular disk, and at an expression for the surface-density at any point $$P$$ of such a disk when electrified to the potential $$V$$ and left undisturbed by external influence. If $$\sigma$$ be the surface-density on one side of the disk, and if $$KPL$$ be a chord drawn through the point $$P$$, then

$\sigma=\frac{V}{2\pi^{2}\sqrt{KP\cdot PL}}.$|undefined

Application of the Principle of Electric Inversion.

178.] Take any point $$Q$$ as the centre of inversion, and let $$R$$ be the radius of the sphere of inversion. Then the plane of the disk becomes a spherical surface passing through $$Q$$$$$$, and the disk itself becomes a portion of the spherical surface bounded by a circle. We shall call this portion of the surface the bowl.

If $$S'$$ is the disk electrified to potential $$V'$$ and free from external influence, then its electrical image $$S$$ will be a spherical segment at potential zero, and electrified by the influence of a quantity $$V'R$$ of electricity placed at $$Q$$.

We have therefore by the process of inversion obtained the solution of the problem of the distribution of electricity on a bowl or a plane disk when under the influence of an electrified point in the surface of the sphere or plane produced.

Influence of an Electrified Point placed on the unoccupied part of the Spherical Surface.

The form of the solution, as deduced by the principles already given and by the geometry of inversion, is as follows:

If $$C$$ is the central point or pole of the spherical bowl $$S$$, and if $$a$$ is the distance from $$C$$ to any point in the edge of the segment, then, if a quantity $$q$$ of electricity is placed at a point $$Q$$ in the surface of the sphere produced, and if the bowl $$S$$ is maintained at potential zero, the density $$\sigma$$ at any point $$P$$ of the bowl will be

$\sigma=\frac{1}{2\pi^{2}}\frac{q}{QP^{2}}\sqrt{\frac{CQ^{2}-a^{2}}{a^{2}-CP^{2}}},$|undefined

$$CQ, CP$$, and $$QP$$ being the straight lines joining the points, $$C, Q$$, and $$P$$.

It is remarkable that this expression is independent of the radius of the spherical surface of which the bowl is a part. It is therefore applicable without alteration to the case of a plane disk.