Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/288

The charge on the disk due to unit potential of the large disk, supposing the density uniform, would be ${\frac {R^{2}}{4A}}$.

The charge on one side of a straight groove of breadth $B$ and length $L=2\pi R$, and of infinite depth, would be

${\frac {1}{4}}{\frac {RB}{A+\alpha '}}$

But since the groove is not straight, but has a radius of curvature $R$, this must be multiplied by the factor $\left(1+{\frac {1}{2}}{\frac {B}{R}}\right)$.

The whole charge on the disk is therefore

${\frac {R^{2}}{4A}}+{\frac {1}{4}}{\frac {RB}{A+\alpha '}}\left(1+{\frac {B}{2R}}\right)$

(39)

$={\frac {R^{2}+R'^{2}}{8A}}-{\frac {R'^{2}-R^{2}}{8A}}\cdot {\frac {\alpha '}{A+\alpha '}}$

(40)

The value of a cannot be greater than

$\alpha '={\frac {B\log 2}{\pi }},\ =0.22B$ nearly.

If $B$ is small compared with either $A$ or $R$ this expression will give a sufficiently good approximation to the charge on the disk due to unity of difference of potential. The ratio of $A$ to $R$ may have any value, but the radii of the large disk and of the guard-ring must exceed $R$ by several multiples of $A$.

EXAMPLE VII. – Fig. XII.

202.] Helmholtz, in his memoir on discontinuous fluid motion^[1], has pointed out the application of several formulae in which the coordinates are expressed as functions of the potential and its conjugate function.

One of these may be applied to the case of an electrified plate of finite size placed parallel to an infinite plane surface connected with the earth.

Since

$x_{1}=A\phi$ and $y_{1}=A\psi$,

and also

$x_{2}=Ae^{\phi }\cos \psi$ and $y_{2}=Ae^{\phi }\sin \psi$,

are conjugate functions of $\phi$ and $\psi$, the functions formed by adding $x_{1}$ to $x_{2}$ and $y_{1}$ to $y_{2}$ will be also conjugate. Hence, if

${\begin{array}{l}x=A\phi +Ae^{\phi }\cos \psi \\y=A\psi +Ae^{\phi }\sin \psi \end{array}}$

1. Königl. Akad. der Wissenschaften, zu Berlin, April 23, 1868.