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

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

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-