The Mathematical Theory of Electricity and Magnetism (5th ed, 1927) — part 3 of 39

Law II. When one body electrifies another by conduction, the total electrification of the two bodies remains the same ; that is, the one loses as much positive or gains as much negative electrification as the other gains of positive or loses of negative electrification.

Law III. When electrification is produced by friction, or by any other known method, equal quantities of positive and negative electrifi- cation are produced.

Definition. The electrostatic unit of electricity is that quantity of positive electricity which, when placed at unit distance from an equal quantity, repels it with unit of force.

Law IV. The repulsion between two small bodies charged respect- ively with e and e' units of electricity is numerically equal to the product of the charges divided by the square of the distance.

These are the forms in which the laws are given by Maxwell. Law I, it will be seen, includes II and III. As regards the Definition and Law IV, it is necessary to specify the medium in which the small bodies are placed, since, as we shall see later, the force is different when the bodies are in air, or in a vacuum, or surrounded by other non-conducting media. It is usual to assume, for purposes of the Definition and Law IV, that the bodies are in air. For strict scientific exactness, we ought further to specify the density, the temperature, and the exact chemical composition of the air. Also we have seen that when the electricity is not insulated on small bodies, but is free to move on conductors, the forces of Law IV must be regarded as acting on the charges of electricity themselves. When the electricity is not free to move, there is an action and reaction between the electricity and matter, so that the forces which really act on the electricity appear to act on the bodies themselves which carry the charges.

CHAPTER II

THE ELECTROSTATIC FIELD OF FORCE

Conceptions used in the Survey of a Field of Force I. The Intensity at a point.

30. The space in the neighbourhood of charges of electricity, considered with reference to the electric phenomena occurring in this space, is spoken of as the electric field.

A new charge of electricity, placed at any point 0 in an electric field, will experience attractions or repulsions from all the charges in the field. The introduction of a new charge will in general disturb the arrangement of the charges on all the conductors in the field by a process of induction. If, however, the new charge is supposed to be infinitesimal, the effects of induction will be negligible, so that the forces acting on the new charge may be supposed to arise from the charges of the original field.

Let us suppose that we introduce an infinitesimal charge e on an infinitely small conductor. Any charge e^ in the field at a distance r, from the point 0 will repel the charge with a force ee^r?. The charge e will experience a similar repulsion from every charge in the field, so that each repulsion will be proportional to e.

The resultant of these forces, obtained by the usual rules for the com- position of forces, will be a force proportional to e — say a force Re in some direction OP. We define the electric intensity at 0 to be a force of which the magnitude is R, and the direction is OP. Thus

The electric intensity at any point is given, in magnitude and direction, by the force per unit charge which would act on a charged particle placed at this point, the charge on the particle being supposed so small that the distribution of electricity on the conductors in the field is not affected by its presence.

The electric intensity at 0, defined in this way, depends only on the permanent field of force, and has nothing to do with the charge, or the size, or even the existence of the small conductor which has been used to explain

30, 31] Lines of Force 25

the meaning of the electric intensity. There will be a definite intensity at every point of the electric field, quite independently of the presence of small charged bodies.

A small charged body might, however, conveniently be used for exploring the electric field and determining experimentally the direction of the electric intensity at any point in the field. For if we suppose the body carrying a charge e to be held by an insulating thread, both the body and thread being so light that their weights may be neglected, then clearly all the forces acting on the charged body may be reduced to two: —

(i) A force Re in the direction of the electric intensity at the point occupied by e,

(ii) the tension of the thread acting along the thread.

For equilibrium these two forces must be equal and opposite. Hence the direction of the intensity at the point occupied by the small charged body is obtained at once by producing the direction of the thread through the charged body. And if we tie the other end of the thread to a delicate spring balance, we can measure the tension of the spring, and since this is numerically equal to Re, we should be able to determine R if e were known. We might in this way determine the magnitude and direction of the electric intensity at any point in the field.

In a similar way, a float at the end of a fishing-line might be used to determine the strength and direction of the current at any point on a small lake. And, just as with the electric intensity, we should only get the true direction of the current by supposing the float to be of infinitesimal size. We could not imagine the direction of the current obtained by anchoring a battleship in the lake, because the presence of the ship would disturb the whole system of currents.

II. Lines of Force.

31. Let us start at any point 0 in the electric field, and move a short distance OP in the direction of the electric intensity at 0. Starting from P let us move a short distance PQ in the direction of the intensity at P,

Q

Fio. 4.

and so on. In this way we obtain a broken path OPQR..., formed of a number of small rectilinear elements. Let us now pass to the limiting case in which each of the elements OP, PQ, QR, ... is infinitely small. The broken path becomes a continuous curve, and it has the property that at every point on it the electric intensity is in the direction of the tangent

26 Electrostatics — Field of Force [oh. ii

to the curve at that point. Such a curve is called a Line of Force. We may therefore define a line of force as follows : —

A Ane of force is a curve in the electric field, such that the tangent at every point i ? in the direction of the electric intensity at that point.

If we suppose the motion of a charged particle to be so much retarded by frictional resistance that it cannot acquire any appreciable momentum, then a charged particle set free in the electric field would trace out a line of force. In the same way, we should have lines of current on the surface of a lake, such that the tangent to a line of current at any point i oincided with the direction of the current, and a small float set free on the lake would describe a current-line.

32. The resultant of a number of known forces has a definite direction, so that there is a single direction for the electric intensity at every point of the field. It follows that two lines of force can never intersect ; for if they did there would be two directions for the electric intensity at the point of intersection (namely, the two tangents to the lines of force at this point) so that the resultant of a number of known forces would be acting in two directions at once. An exception occurs, as we shall see, when the resultant intensity vanishes at any point.

The intensity R may be regarded as compounded of three components X, Y, Z, parallel to three rectangular axes Ox, Oy, Oz.

The magnitude of the electric intensity is then given by

R* = X2 + F2 + Z\

and the direction cosines of its direction are

X Y £ R' R' R'

These, therefore, are also the direction cosines of the tangent at x, y, z to the line of force through the point. The differential equation of the system of lines of force is accordingly

dx _dy _dz X~T~~Z'

III. The Potential

33. In moving the small test-charge e about in the field, we may either have to do work against electric forces, or we may find that these forces will do work for us. A small charged particle which has been placed at a point 0 in the electric field may be regarded as a store of energy, this energy being equal to the work (positive or negative) which has been done in taking the charge to 0 in opposition to the repulsions and attractions of the field. The energy can be reclaimed by allowing the particle to retrace its path. Assume the charge on the moving particle to be so small that

31-33] The Potential 27

the distribution of electricity on the conductors in the field is not affected by it. Then the work done in bringing the charge e to a point 0 is pro- portional to e, and may be taken to be Ve. The amount of work done will of course depend on the position from which the charged particle started. It is convenient, in measuring Ve, to suppose that the particle started at a point outside the field altogether, i.e. from a point so far removed from all the charges of the field that their effect at this point is inappreciable — for brevity, we may say the point at infinity. We now define V to be the potential at the point 0. Thus

The potential at any point in the field is the work per unit charge which has to be done on a charged particle to bring it to that point, the charge on the particle being supposed so small that the distribution of electricity on the conductors in the field is not affected by its presence.

In moving the small charge e from x, y, z to x + dx, y + dy, z + dz, we shall have to perform an amount of work

• (Xdx + Ydy + Zdz) e,

so that in bringing the charge e into position at x, y, z from outside the field altogether, we do an amount of work

-ej(Xdt

Ix + Ydy + Zdz),

where the integral is taken along the path followed by e.

Denoting the work done on the charge e in bringing it to any point x, y, z in the electric field by Ve, we clearly have

y=-fX S (Xdx + Ydy + Zdz) (6),

" 00

giving a mathematical expression for the potential at the point x, y, z.

The same result can be put in a different form. If ds is any element of the path, and if the intensity R at the extremity of this element makes an angle 6 with ds, then the component of the force acting on e when moving along ds, resolved in the direction of motion of e, is Re cos 6. The work done in moving e along the element ds is accordingly

— Re cos dds,

so that the whole work in bringing e from infinity to x, y, z is

[x, y, z

— el Rcos dds,

J oo

and since this is equal, by definition, to Ve, we must have

V=- fX'y'ZR cos 6ds (7).

28 Electrostatics— Field of Force [ch. ii

We see at once that the two expressions (6) and (7) just obtained for V are identical, on noticing that 0 is the angle between two lines of which the directioi . cosines are respectively

X Y Z , dx dy dz

R' R' R ds' ds' ds'

a Xdx Ydy Zdz We thereiore have ™*6 = rTs+ Rds+Rds>

so that R cos 0ds = Xdx + Ydy + Zdz,

and the identity of the two expressions becomes obvious.

If the Theorem of the Conservation of Energy is true in the Electro- static Field, the work done in bringing a small charge e from infinity to any point P must be the same whatever path to P we choose. For if the amounts of work were different on two different paths, let these amounts be VP e and VP'e, and let the former be the greater. Then by taking the charge from P to infinity by the former path and bringing it back by the latter, we should gain an amount of work {VP — VP') e, which would be contrary to the Conservation of Energy. Thus VP and VP must be equal, and the potential at P is the same, no matter by what path we reach P. The potential at P will accordingly depend only on the coordinates x, y, z of P.

As soon as we introduce the special law of the inverse square, we shall find that the potential must be a single-valued function of x, y, z, as a consequence of this law (§ 39), and hence shall be able to prove that the Theorem of Conservation of Energy is true in an Electrostatic field. For the moment, however, we assume this.

34. Let us denote by W the work done in moving a charge e from P to Q. In bringing the charge from infinity to P, we do an amount of work

Fio. 5.

which by definition is equal to VP e where VP denotes the value of V at the

point P. Hence in taking it from infinity to Q, we do a total amount of

work VP e + W. This, however, is also equal by definition to Vq e. Hence

we have

Vpe+ W =VQe,

or W = (VQ-VJ>)e (8).

33-86] The Potential 29

35. Definition. A surface in the electric field such that at every point on it the potential has the same value, is called an Equipotential Surface.

In discussing the phenomena of the electrostatic field, it is convenient to think of the whole field as mapped out by systems of equipotential surfaces and lines of force, just as in geography we think of the earth's surface as divided up by parallels of latitude and of longitude. A more exact parallel is obtained if we think of the earth's surface as mapped out by "contour-lines" of equal height above sea-level, and by lines of greatest slope. These reproduce all the properties of equipotentials and lines of force, for in point of fact they are actual equipotentials and lines of force for the gravitational field of force.

Theorem. Equipotential surfaces cut lines of force at right angles.

Let P be any point in the electric field, and let Q be an adjacent point on the same equipotential as P. Then, by definition, Vp = Vq, so that by equation (8) W = 0, W being the amount of work done in moving a charge e from P to Q. If R is the intensity at Q, and 6 the angle which its direction makes with QP, the amount of this work must be — Re cos 0 x PQ, so that

Re cos 0 = 0.

Hence cos 0=0, so that the line of force cuts the equipotential at right angles. As in a former theorem, an exception has to be made in favour of the case in which P = 0.

36. Instead of P, Q being on the same equipotential, let them now be on a line parallel to the axis of x, their coordinates being x, y, z and x + dx, y, z respectively. In moving the charge e from P to Q the work done is — Xedx, and by equation (8) it is also (Vq — Vp) e. Hence

-Xdx=VQ-VP.

Since Q and P are adjacent, we have, from the definition of a differential coefficient,

dV_VQ-VP

dx dx

hence we have the relations

--& --?■ --f o

results which are of course obvious on differentiating equation (6) with respect to x, y and z respectively.

Similarly, if we imagine P, Q to be two points on the same line of force we obtain

BV R==-~ds-'

where ^- denotes differentiation along a line of force. Since R is necessarily

... gp- ,

positive, it follows that -j-~ is negative, i.e. V decreases as s increases, or the

OS

30 Electrostatics — Field of Force [oh. ii

intensity is in the direction of V decreasing. Thus the lines of force run from higher to lower values of V, and, as we have already seen, cut all equipot •ntials at right angles.

37. At a point which is occupied by conducting material, the electric charges, as has already been said, must be in equilibrium under the action of the forces from all the other charges in the field. The resultant force from all these charges on any element of charge e is however Re, so that we must have B = 0. Hence X = Y = Z = 0, so that

dec dy dz

In other words, V must be constant throughout a conductor for electro- static equilibrium to be possible. And in particular the surface of a conductor must be an equipotential surface, or part of one. The equi- potential of which the surface of a conductor is part has the peculiarity of being three-dimensional instead of two-dimensional, for it occupies the whole interior as well as the surface of the conductor.

In the same way, in considering the analogous arrangement of contour-lines and lines of greatest slope on a map of the earth's surface, we find that the edge of a lake or sea must be a contour-line, but that in strictness this particular contour must be regarded as two-dimensional rather than one-dimensional, since it coincides with the whole surface of the lake or sea.

If V is not constant in any conductor, the intensity is in the direction of V decreasing. Hence positive electricity tends to flow in the direction of V decreasing, and negative electricity in the direction of V increasing. If two conductors in which the potential has different values are joined by a third conductor, the intensity in the third conductor will be in direction from the conductor at higher potential to that at lower potential. Electricity will flow through this conductor, and will continue to flow until the redistribution of potential caused by the transfer of this electricity is such that the potential is the same at all points of the conductors, which may now be regarded as forming one single conductor.

Thus although the potential has been defined only with reference to single points, it is possible to speak of the potential of a whole conductor. In fact, the mathematical expression of the condition that equilibrium shall be possible for a given system of charges is simply that the potential shall be constant throughout each conductor. And when electric contact is established between two conductors, either by joining them by a wire or by other means, the new condition for equilibrium which is made necessary by the new physical condition introduced, is simply that the potentials of the two conductors shall be equal.

36-38] The Potential 31

The earth is a conductor, and is therefore at the same potential through- out. In all practical applications of electrostatics, it will be legitimate to regard the potential of the earth as zero, a distant point on the earth's surface replacing the imaginary point at infinity, with reference to which potentials have so far been measured. Thus any conductor can be reduced to potential zero by joining it by a metallic wire to the earth.

Mathematical expressions of the Law of the Inverse Square. I. Values of Potential and Intensity.

38. We now discuss the values of the potential and components of electric intensity when the space between the conductors is air, so that the electric forces are determined by Coulomb's Law.

If we have a single point charge ex at a point P, the value of R, the resultant intensity at any point 0, is

PO*' and its direction is that of PO. Hence if 6 is the angle between OP and

Fig. 6.

00', the line joining 0 to an adjacent point 0', the work done in moving a charge e from 0 to 0'

= eR cos 0 . 00'

= eR(OP-0'P)

= — eRdr,

where OP = r, O'P = r + dr. Hence the work done against the repulsion of the charge ex in bringing e from infinity to 0' by any path is

-e Rdr =-e \dr = — \

where rx = O'P.

If there are other charges e2> ^s> ••• the work done against all the repulsions in bringing a charge e to 0' will be the sum of terms such as the above, say

\ri r2 rs J

32 Electrostatics — Field of Force [ch. n

where r2, r3, ... are the distances from 0' to e2, e3, ..., so that by definition

F = -x+^+-3 + (10).

n r2 r3

39. It is now clear that the potential at any point depends only on the coordinates of the point, so that the work done in bringing a small charge from infinity to a point P is always the same, no matter what path we choose, the result assumed in § 33.

It follows that we cannot alter the amount of energy in the field by moving charges about in such a way that the final state of the field is the same as the original state. In other words, the Conservation of Energy is true of the Electrostatic Field.

40. Analytically, let us suppose that the charge e1 is at x[, y1} zx\ e2 at x2, y2, z2 ; and so on. The repulsion on a small charge e at x, y, z resulting from the presence of ex at xlt yXi zx is

exe (x-x.y+iy-y.y + iz-z.r and the direction-cosines of the direction in which this force acts on the charge e, are

^Ifl VjlVl etc

[(x - x,f + (y- 2/x)2 + (z - zrf] i ' [O - xxf + (y - yxy + (z - ttf$ '

Hence the component parallel to the axis of x is

e^e (x — Xi) [(x-xy + iy-yrf + iz-zj-f

By adding all such components, we obtain as the component of the electric intensity at x, y, z,

Z = 2 6l (x ~ Xl) l (11),

[(x-xj + iy-yj + iz-zj]? and there are similar equations for Y and Z.

We have as the value of V at x, y, z, by equation (6), V = - * (Xdx + Ydy + Zdz)

J CO

_ ^ r.r, y, z vgi {(x - Xl) dx + (y — yr) dy + (z — zz) dz]

J 00

= 2

[(x - x,y + (y - y,f + (z- gffl

[{x-x.f + iy-y^+^-z^ giving the same result as equation (10).

38-42] Gauss' Theorem 33

41. If the electric distribution is not confined to points, we can imagine it divided into small elements which may be treated as point charges. For instance if the electricity is spread throughout a volume, let the charge on any element of volume dx'dy'dz be pdx'dy'dz so that p may be spoken of as the " density " of electricity at x, y , z . Then in formula (11) we can replace #i by pdx'dy'dz', and xlt ylt zx, by x , y', z'. Instead of summing the charges 6j, ... we of course integrate pdx'dy'dz' through all those parts of the space which contain electrical charges. In this way we obtain

p (x — x') dx'dy'dz

[(« - O' + (y-2/')= + (*-0sI

-III — f-

3 , etc.,

and V = ((f pdx'dy'dz'

JJJ[(x - x'f +(y- y'f + (z- zjf

These equations are one form of mathematical expression of the law of the inverse square of the distance. An attempt to perform the integration, in even a few simple cases, will speedily convince the student that the form is not one which lends itself to rapid progress. A second form of mathe- matical expression of the law of the inverse square is supplied by a Theorem of Gauss which we shall now prove, and it is this expression of the law which will form the basis of our development of electrostatical theory.

II. Gauss' Theorem.

42. Theorem. If any closed surface is taken in the electric field, and if N denotes the component of the electric intensity at any point of this surface in the direction of the outward normal, then

SI

NdS = 4>ttE,

where the integration extends over the whole of the surface, and E is the total charge enclosed by the surface.

Let us suppose the charges in the field, both inside and outside the closed surface, to be e1 at %, e2 at li, and so on. The intensity at any point is the resultant of the intensities due to the charges separately, so that at any point of the surface, we may write

N = N, + N2+ (12),

where Nlf i\T2, ... are the normal components of intensity due to e1} e2, ... separately.

Instead of attempting to calculate 1 1 NdS directly, we shall calculate

separately the values of llNidS, jJN'2dS, .... The value of JJNdS will,

by equation (12), be the sum of these integrals.

j. 3

34 Electrostatics— Field of Force [ch. n

Let us take any small element dS of the closed surface in the neighbour- hood of a point Q on the surface and join each point of its boundary to the point &, Let the small cone so formed cut off an element of area da from

Fig. 7.

a sphere drawn through Q with i? as centre, and an element of area dco from a sphere of unit radius drawn about Px as centre. Let the normal to the closed surface at Q in the direction away from 7^ make an angle 6 with PXQ.

The intensity at Q due to the charge ex at Px is eJRQ2 in the direction P}Q, so that the component of the intensity along the normal to the surface in the direction away from i? is

cos 6

nor

The contribution to I jJSfjdS from the element of surface is accordingly

± nk» cos 6 dS,

the + or — sign being taken according as the normal at Q in the direction away from Pt is the outward or inward normal to the surface.

Now cos 6 dS is equal to da; the projection of dS on the sphere through Q having /? as centre, for the two normals to dS and d<r are inclined at an angle 6. Also da = P1Q2doy. For do; do are the areas cut off by the same cone on spheres of radii PXQ and unity respectively. Hence

P%™a6d8=e-p§-=e>dl°'

If Px is inside the closed surface, a line from P{ to any point on the unit sphere surrounding i? may either cut the closed surface only once as at Q (fig. 8) — in which case the normal to the surface at Q in the direction away from i? is the outward normal to the surface — or it may cut three times, as at Q', Q", Q'" — in which case two of the normals away from i? (those at Q', Q'" in fig. 8) are outward normals to the surface, while the third normal away from Pt (that at Q" in the figure) is an inward normal — or it may

42]

Gauss' Theorem

35

cut five, seven, or any odd number of times. Thus a cone through a small element of area dw on a unit sphere about Px may cut the closed surface any odd number of times. However many times it cuts, the first small area cut

off will contribute exd(o to \NxdS, the second and third small areas if they

Fig. 8.

occur will contribute — exdco and + exdco respectively, the fourth and fifth if they occur will contribute — exdaj and + exdw respectively, and so on. The total contribution from the cone surrounding dco is, in every case, + exdu>.

Fio. 9.

•Summing over all cones which can be drawn in this way through Tx we obtain

the whole value of I JNxdS, which is thus seen to be simply ex multiplied by

•the total surface area of the unit sphere round i?, and therefore 4nrex.

3—2

36 Electrostatics — Field of Force [ch. n

On the other hand if 7? is outside the closed surface, as in fig. 9, the cone through any element of area dco on the unit sphere may either not cut the clc jed surface at all, or may cut twice, or four, six or any even number of times. If the cone through dco intersects the surface at all, the first pair of elements of surface which are cut off by the cone contribute — e^dco and

■-exdco respectively to I \NydS. The second pair, if they occur, make a similar

contribution and so on. In every case the total contribution from any small cone through i? is nil. By summing over all such cones we shall include the contributions from all parts of the closed surface, so that if Px is outside

the surface I \NxdS is equal to zero.

We have now seen that llN-^dS is equal to krrex when the charge ex is

inside the closed surface, and is equal to zero when the charge ex is outside the closed surface. Hence

(JxfdS = (JN, dS + (JN2dS + ...

= 4tt x (the sum of all the charges inside the surface)

which proves the theorem.

Obviouslv the theorem is true also when there is a continuous distribution of electricity in addition to a number of point charges. For clearly we can divide up the continuous distribution into a number of small elements and treat each as a point charge.

dV

Since N, the normal component of intensity, is equal by § 36 to — -~— ,

where =- denotes differentiation along the outward normal, it appears that

we can also express Gauss' Theorem in the form

"dV

SI

dS=- 4>ttE. on

Gauss' theorem forms the most convenient method at our disposal, of expressing the law of the inverse square.

We can obtain a preliminary conception of the physical meaning under- lying the theorem by noticing that if the surface contains no charge at all, the theorem expresses that the average normal intensity is nil. If there is a negative charge inside the surface, the theorem shews that the average normal intensity is negative, so that a positively charged particle placed at a point on the imaginary surface will be likely to experience an attraction to the interior of the surface rather than a repulsion away from it, and vice versa if the surface contains a positive charge.

42-46] Gauss' Theorem 37

Corollaries to Gauss' Theorem.

43. Theorem. If a closed surface he drawn, such that every point on it is occupied by conducting material, the total charge inside it is nil.

We have seen that at any point occupied by conducting material, the electric intensity must vanish. Hence at every point of the closed surface,

N = 0, so that 1 1 NdS = 0, and therefore, by Gauss' Theorem, the total charge

inside the closed surface must vanish.

The two following special cases of this theorem are of the greatest importance.

44. Theorem. There is no charge at any point which is occupied by con- ducting material, unless this point is on the surface of a conductor.

For if the point is not on the surface, it will be possible to surround the point by a small sphere, such that every point of this sphere is inside the conductor. By the preceding theorem the charge inside this sphere is nil, hence there is no charge at the point in question.

This theorem is often stated by saying : — The charge of a conductor resides on its surface.

45. Theorem. If we have a hollow closed conductor, and place any number of charged bodies inside it, the charge on its inner surface luill be equal in magnitude but opposite in sign, to the total charge on the bodies inside.

For we can draw a closed surface entirely inside the material of the conductor, and by the theorem of § 43, the whole charge inside this surface must be nil. This whole charge is, however, the sum of (i) the charge on the inner surface of the conductor, and (ii) the charges on the bodies inside the conductor. Hence these two must be equal and opposite.

This result explains the property of the electroscope which led us to the conception of a definite quantity of electricity. The vessel placed on the plate of the electroscope formed a hollow closed conductor. The charge on the inner surface of this conductor, we now see, must be equal and opposite to the total charge inside, and since the total charge on this conductor is nil, the charge on its outer surface must be equal and opposite to that on the inner surface, and therefore exactly equal to the sum of the charges placed inside, independently of the position of these charges.

The Cavendish Proof of the Law of the Inverse Square.

46. We have deduced from the law of the inverse square, that the charge inside a closed conductor is zero. We shall now shew that the converse theorem is also true. Hence, in the known fact, revealed by the

38

Electrostatics— Field of Force

[ch. n

observations of Cavendish and Maxwell, that the charge inside a closed conductor is zero, we have experimental proof of the law of the inverse square - /hich admits of much greater accuracy than the experimental proof of Coulomb.

The theorem that if there is no charge inside a spherical conductor the law of force must be that of the inverse square is due to Laplace. We need consider this converse theorem only in its application to a spherical conductor, this be ng the actual form of conductor used by Cavendish. The apparatus illustrated in fig. 10 is not that used by Cavendish, but is an improved form designed by Maxwell, who repeated Cavendish's experiment in a more delicate form.

Two spherical shells are fixed by a ring of ebonite so as to be concentric with one another, and insulated from one another. Electrical contact can be established between the two by letting down the small trap-door B through which a wire passes, the wire being of such a length as just to establish contact when the trap-door is closed. The experiment is conducted by electrifying the outer shell, opening the trap-door by an insulating thread without discharging the conductor, afterwards dis- charging the outer conductor and testing whether any charge is to be found on the inner shell by placing it in electrical contact with a delicate electroscope by means of a conducting wire inserted through the trap- door. It is found that there are no traces of a charge on the inner sphere.

FlG- 10- 47. Suppose we start to find the law of electric

force such that there shall be no charge on the inner

sphere. Let us assume a law of force such that the repulsion between two

charges e, e at distance r apart is ee'<j>(r). The potential, calculated as

explained in § 33, is

Ze f (f>(r)dr (13),

J r

where the summation extends over all the charges in the field.

Let us calculate the potential at a point inside the sphere due to a charge E spread entirely over the surface of the sphere. If the sphere is of radius a, the area of its surface is 47ra2, so that the amount of charge per unit area is EJ^na2, and the expression for the potential becomes

V ' = ll^{~Mr)dr}a*smed0d4> (14),

the summation of expression (13) being now replaced by an integration which

46, 47] Cavendish's Proof of Law of Force 39

extends over the whole sphere. In this expression r is the distance from the point at which the potential is evaluated, to the element a2 sin 0ddd<j> of spherical surface.

If we agree to evaluate the potential at a point situated on the axis 0=0 at a distance c from the centre, we may write

r2 = a2 + c2- 2ac cos 9.

Since c is a constant, we obtain as the relation between dr and dO, by differentiation of this last equation,

rdr = ac sin ddd (15).

If we integrate expression (14) with respect to <£, the limits being of course <f) = 0 and <f> = 2tt, we obtain

F= \E J 6 " ( f <f>(r) dr} sin Odd,

or, on changing the variable from 6 to r, by the help of relation (15)

~r=a+c / f« \ rdr

rr=a+c / /"= \

V=%E <j>(r)dr)

J r=a-c \J r '

ao If we introduce a new function f(r), defined by

/(r)=[y 4>{r)dr}rdr,

we obtain as the value of V,

V = ^c[f{a + c)-f{a-G)]'

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