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}} $|undefined

or

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

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}}} $.|undefined

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