Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/311

balance and suspended disk, sufficient apertures being left to see the fiducial marks.

The guard-ring, case, and suspended disk are all in metallic communication with each other, but are insulated from the other parts of the apparatus.

Now let it be required to measure the difference of potentials of two conductors. The conductors are put in communication with the upper and lower disks respectively by means of wires, the weight is taken off the suspended disk, and the lower disk is moved up by means of the micrometer screw till the electrical attraction brings the suspended disk down to its sighted position. We then know that the attraction between the disks is equal to the weight which brought the disk to its sighted position.

If $W$ be the numerical value of the weight, and $g$ the force of gravity, the force is $Wg$, and if $A$ is the area of the suspended disk, $D$ the distance between the disks, and $V$ the difference of the potentials of the disks,

$Wg={\frac {V^{2}A}{8\pi D^{2}}}$

or

$V=D{\sqrt {\frac {8\pi gW}{A}}}$

If the suspended disk is circular, of radius $R$, and if the radius of the aperture of the guard-ring is $R'$, then

$A={\frac {1}{2}}\pi \left(R^{2}+R'^{2}\right)$, and $V=4D{\sqrt {\frac {gW}{R^{2}+R'^{2}}}}$.

218.] Since there is always some uncertainty in determining the micrometer reading corresponding to $D=0$, and since any error

^[1]

1. Let us denote the radius of the suspended disk by $R$, and that of the aperture of the guard-ring by $R'$, then the breadth of the annular interval between the disk and the ring will be $B=R'-R$. If the distance between the suspended disk and the large fixed disk is $D$, and the difference of potentials between these disks is $V$, then, by the investigation in Art. 201, the quantity of electricity on the suspended disk will be

   $Q=V\left\{{\frac {R^{2}+R'^{2}}{8D}}-{\frac {R'{}^{2}-R{}^{2}}{8D}}{\frac {\alpha }{D+\alpha }}\right\}$

   where

   $\alpha =B{\frac {\log _{e}2}{\pi }}$, or $\alpha =0.220635(R'-R)$

   If the surface of the guard-ring is not exactly in the plane of the surface of the suspended disk, let us suppose that the distance between the fixed disk and the guard-ring is not $D$ but $D+z=D'$, then it appears from the investigation in Art. 225 that there will be an additional charge of electricity near the edge of the disk on account of its height $z$ above the general surface of the guard-ring. The whole charge in this case is therefore

   $Q=V\left\{{\frac {R^{2}+R'^{2}}{8D}}-{\frac {R'{}^{2}-R{}^{2}}{8D}}{\frac {\alpha }{D+\alpha }}+{\frac {R+R'}{D}}\log _{e}{\frac {4\pi (R+R')}{D'-D}}\right\}$