Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/322

 Let $$\sigma $$ be the surface-density of a plane, which we shall suppose to be that of $$xy$$.

The potential due to this electrification will be

$V=-4\pi\sigma z $

Now let two disks of radius $$a$$ be rigidly electrified with surface-densities $$-\sigma' $$ and $$+\sigma' $$. Let the first of these be placed on the plane of $$xy$$ with its centre at the origin, and the second parallel to it at the very small distance $$c$$.

Then it may be shewn, as we shall see in the theory of magnetism, that the potential of the two disks at any point is $$\omega\sigma'c $$, where $$\omega $$ is the solid angle subtended by the edge of either disk at the point. Hence the potential of the whole system will be

$V=-4\pi\sigma z+\omega\sigma'c $

The forms of the equipotential surfaces and lines of induction are given on the left-hand side of Fig. XX, at the end of Vol. II.



Let us trace the form of the surface for which $$V = 0$$. This surface is indicated by the dotted line.

Putting the distance of any point from the axis of $$z = r$$, then, when $$r$$ is much less than $$a$$, and $$z$$ is small,

$\omega=2\pi-2\pi\frac{z}{a}+\mathrm{etc.} $

Hence, for values of $$r$$ considerably less than $$\alpha $$, the equation of the zero equipotential surface is

$0=-4\pi\sigma z+2\pi\sigma'c-2\pi\sigma'\frac{zc}{a}+\mathrm{etc.} $

or

$z_{0}=\frac{\sigma'c}{2\sigma+\sigma'\frac{c}{a}} $|undefined

Hence this equipotential surface near the axis is nearly flat.

Outside the disk, where $$r$$ is greater than $$a$$, $$\omega $$ is zero when $$z$$ is zero, so that the plane of $$xy$$ is part of the equipotential surface.

To find where these two parts of the surface meet, let us find at what point of this plane $$\tfrac{dV}{dz}=0 $$.

When $$r$$ is very nearly equal to $$a$$

$\frac{dV}{dz}=-4\pi\sigma+\frac{2\sigma'c}{r-a} $

Hence, when

$\frac{dV}{dz}=0,\ r_{0}=a+\frac{\sigma'c}{2\pi\sigma} $

The equipotential surface $$V =0$$ is therefore composed of a disk-