We have now to multiply the charge of an element of the upper disk into the potential due to the lower disk, and integrate for the whole surface of the upper disk,

Between the limits of integration we may write with a sufficient degree of approximation,

so that when b is very small compared with a, the value of γ cannot differ greatly from that given by equation (11). Hence we may write the expression (10)

The corresponding quantity for the action of the upper disk on itself is got by putting A, = A, and b = 0, and is

=

1=

In the actual case 4,-4,-1E, where E is the whole charge, and the capacity is

or, since in our approximation we have neglected (-)

be expressed with sufficient accuracy in the form

(14)

showing that the capacity of two disks very near together is equal to that of an infinitely thin disk of somewhat larger radius.

If the space between the two disks is filled up, so as to form a disk of sensible thickness, there will be a certain charge on the curved surface, but at the same time the charge on the inner sides of the disks will disappear, and that on the outer sides near the edges will be diminished, so that the capacity of a disk of sensible thickness is very little greater than that given by (15).

We may apply this result to estimate the correction for the thickness of the square plates used by Cavendish. The factor by which we must multiply the thickness in order to obtain the correction for the diameter of an infinitely thin plate of equal capacity is

1

α

log 7.

2π

The correction is in every case much smaller than Cavendish supposed.

NOTE 21, ARTS. 277, 452, 473, 681.

Calculation of the Capacity of the Two Circles in Experiment VI.

9.3

The diameter of one of the circles was 9.3 inches, so that its capacity when no other conductor is in the field is = 2.960. The distance between their centres was 36, 24, and 18 inches, which we may call C1, Cal and

π

The height of the centres of the circles above the floor was about 45 inches, so that the distance of the image of the circle would be about 90 inches and that of the image of the other circle would be about

r = (90° + c2)1.

Hence, if P is the potential of the circles when the charge of each

where the first term is due to the circle itself, the second and third to the other circle, as in Note 11, and the two last to the images of the two circles. We thus find for the three distances

P1 = 0·3438,

P1 = 0·3567,

P1 = 0·3689.

3

The capacity is 2P-', and the number of "inches of electricity," according to the definition of Cavendish, is 4P-',

or

for the three cases.

11.636,

11.212,

10.844,

The large circle was 18.5 inches in diameter and its centre was 41 inches from the floor, so that its charge would be 12.69 inches of electricity.

Hence the relative charges are as follows:

I am not aware of any method by which the capacity of a square can be found exactly. I have therefore endeavoured to find an approximate value by dividing the square into 36 equal squares and calculating the charge of each so as to make the potential at the middle of each square equal to unity.

The potential at the middle of a square whose side is 1 and whose charge is 1, distributed with uniform density, is

4 log (1+√√2) = 3·52549.

In calculating the potential at the middle of any of the small squares which do not touch the sides of the great square I have used this formula, but for those which touch a side I have supposed the value to be 3.1583, and for a corner square 2.9247.

and the capacity of a square whose side is 1 will be 0·3607.

The ratio of the capacity of a square to that of a globe whose diameter is equal to a side of the square is therefore 0·7214.

In Art. 654 Cavendish deduces this ratio from the measures in Art. 478 and finds it 0.73, which is very near to our result. If, however, we take the numbers given in Art. 478, we find the ratio 0.79. From Art. 281 we obtain the ratio 0.747.

The ratio of the charge of a square to that of a circle whose diameter is equal to a side of the square is by our calculation 1.133.

In Art. 648 Cavendish says that the ratio is that of 9 to 8 or 1.128, which is very close to our result, but in Arts. 283* and 682 he makes it 1.153.

The numbers in Art. 281 from which Cavendish deduces this would make it 1.1514.

The numbers given in Art. 478 would make it 1.176.

Cavendish supposes that the capacity of a rectangle is the same as that of a square of equal area, and he deduces this from a comparison of the square 15.5 with the rectangle 17·9 × 13·4.

It is not easy to calculate the capacity of a rectangle in terms of its sides, but we may be certain that it is greater than that of a square of equal area.

For if we suppose the electricity on the square rendered immoveable, and if we cut off portions from two sides of the square and place them on the other two sides so as to form a rectangle, we are carrying electricity from a place of higher to a place of lower potential, and are therefore diminishing the energy of the system.

If we now make the electricity moveable, it will re-arrange itself on the rectangle and thereby still further diminish the energy. Hence the energy of a given charge on the rectangle is less than that of the same charge on the square, and therefore the capacity of the rectangle is greater than that of the square.

NOTE 23, ARTS. 288 and 542.

On the Charge of the Middle Plate of Three Parallel Plates.

The plates used by Cavendish were square, but for the purpose of a rough estimate of the distribution of electricity between the three plates we may suppose them to be three circular disks.

First consider two equal disks on the same axis, at a distance small compared with the radius of either.

If the disks were in contact, the distribution on each would be the same as on each of the two surfaces of a single disk, and it would be entirely on the outer surface.

* In Art. 283 of this book the number is printed 1.53. It should be 1.153.

If the distance between the disks is very small compared with their radii, the force exerted by one of the disks at any point of the other will be nearly but not quite normal to its surface. The component in the plane of the disk will be directed outwards from the centre, so that the density will be greater near the edge than in a single disk having the same charge, but as a first approximation we may assume that the sum of the surface-densities on both sides of any element of the disk is the same as if the other disk were away.

But the density on the outer surface of the disk will be increased, and the density on the inner surface diminished, by a quantity numerically equal to the normal component of the repulsion of the other disk divided by 47, and the whole charge of the outer surface will be increased, and the whole charge of the inner surface diminished, by a quantity equal to the charge of that part of the other disk, the lines of force from which cut the disk under consideration.

Hence the charges of the inner and outer surfaces of the disk are

respectively, where the value of the elliptic co-ordinate w is that corresponding to the edge of the other disk.

If a is the radius of either disk, and c the distance between them,

If we now place another equal disk on the same axis at a distance c from one of them, the potential being the same for all three, the new disk will greatly diminish the charge of the surface of the disk which is next to it, but it will not have much effect on the charges of the other surfaces.

The result will therefore be that the charges of the two outer disks will together be greater, but not much greater, than that of a single disk at the same potential, but the charge of each of the surfaces of the middle disk will be the same as that of one of the inner surfaces of a pair of disks at distance c. Hence the charge of the middle disk will be to that of the two outer disks together as w to a.

If we substitute for the square plates of twelve inches in the side disks of 13.8 inches diameter which would have nearly the same capacity, then if the distance between the outer disks is 1.15 inches, c = .575 and @= 1.936 and a = 35 w, or the charge of the middle disk would be 3-5 times greater if the other disks had been removed.

If the distance between the outer disks is 1.65 inches, c = 875 and 2.293, whence a = 2.2 w, or the charge of the middle disk would have been 2.2 times greater if the outer disks had been removed.

It is evident, however, that in the assumed distribution the potential is less at the edges of the outer disks than at their centres. The elec