The Mathematical Theory of Electricity and Magnetism (5th ed, 1927) — part 4 of 39

If the inner and outer spheres are in electrical contact, their potentials are the same ; and if, as experiment shews to be the case, there is no charge on the inner sphere, then the whole potential must be that just found. This expression must, accordingly, have the same value whether c represents the radius of the outer sphere or that of the inner. Since this is true whatever the radius of the inner sphere may be, the expression must be the same for all values of c. We must accordingly have

2acV ., v /./ v — =r- =7 (« + c) -J (a - c),

where V is the same for all values of c. Differentiating this equation twice with respect to c, we obtain

0=/"(a + c)-/"(a-c).

Since by definition, /(r) depends only on the law of force, and not on a or c it follows from the relation

/" (a + c) =/" (a - o), that f" (r) must be a constant, say G.

40 Electrostatics — Field of Force [ch. n

Hence f(r) = A + Br + \ Gr\

and b\ definition f(r) =1(1 <f>(r)dr) rdr,

so tha ■•, on equating the two values of/" (r),

B + Cr = r I <f> (r) dr.

r°° B

There fore (f>(r)dr=C + - ,

J r r

so that the law of force is that of the inverse square.

48. Maxwell has examined what charge would be produced on the inner sphere if, instead of the law of force being accurately B/?°2, it were of the form B/r2+v, where q is some small quantity. In this way he found that if q were even so great as Yi6M> ^ne cnarge on the inner sphere would have been too great to escape observation. As we have seen, the limit which Cavendish was able to assign to q was ■£$.

It may be urged that the form Bjr2+<* is not a sufficiently general law of force to assume. To this Maxwell has replied that it is the most general law under which conductors which are of different sizes but geometri- cally similar can be electrified similarly, while experiment shews that in point of fact geometrically similar conductors are electrified similarly. We may say then with confidence that the error in the law of the inverse square, if any, is extremely small. It should, however, be clearly understood that experiment has only proved the law B/r2 for values of r which are great enough to admit of observation. The law of force between two electric charges which are at very small distances from one another still remains entirely unknown to us. ^— __^

III. The Equations of Poisson and Laplace.

49. There is still a third way of expressing the law of the inverse

square, and this can be deduced most readily from Gauss' Theorem.

Let us examine the small rectangular parallel- epiped, of volume dxdydz, which is bounded by the six plane faces

x = £ ± \dx, y = v± \dy, z=K± &z- We shall suppose that this element does not con- tain any point charges of electricity, or part of Fig. 11. any charged surface, but for the sake of generality

we shall suppose that the whole space is charged

A

/

47-49] Equations of Laplace and Poisson 41

with a continuous distribution of electricity, the volume-density of electrifi- cation in the neighbourhood of the small element under consideration being p. The whole charge contained by the element of volume is accordingly pdxdydz, so that Gauss' Theorem assumes the form

NdS = ^irpdxdydz (16).

The surface integral is the sum of six contributions, one from each face of the parallelepiped. The contribution from that face which lies in the plane x = i~ — \dx is equal to dydz, the area of the face, multiplied by the mean value of N over this face. To a sufficient approximation, this may be supposed to be the value of JV" at the centre of the face, i.e. at the point £ — \dx, 7), £, and this again may be written

ftl)

\dx J ' ,, '

so that the contribution to I INdS from this face is

<dV\

dydz\dx)

Similarly the contribution from the opposite face is

" dydz (a?)

the sign being different because the outward normal is now the positive axis of x, whereas formerly it was the negative axis. The sum of the contributions from the two faces perpendicular to the axis of x is therefore

-*M«U„, -£),.„„„! <■'►

dV The expression inside curled brackets is the increment in the function —

when x undergoes a small increment dx. This we know is dx ^- ( -^— J , so that expression (17) can be put in the form

• fa? dxdydz.

The whole value of j JNdS is accordingly

/d'V d*v d2V\ . , ,

U? + w + 1*) dxdydz'

dy and equation (16) now assumes the form

dV dV d*V . ,1QN

^+w+^=~^p ( }'

42 Electrostatics — Field of Force [ch. n

This is known as Poisson's Equation; clearly if we know the value of the potential at every point, it enables us to find the charges by which this potenti il is produced.

50. In   free  space,  where  there  are  no  electric  charges,  the  equation 
    

assumes the form

927 opy &y

and ths is known as Laplace's Equation. We shall denote the operator

d*_ &_ d*_ da? By2 dz1

by V2, m that Laplace's equation may be written in the abbreviated form

V2F=0 (20).

Equations (18) and (20) express the same fact as Gauss' Theorem, but express it in the form of a differential equation. Equation (20) shews that in a region in which no charges exist, the potential satisfies a differential equation which is independent of the charges outside this region by which the potential is produced. It will easily be verified by direct differentiation that the value of V given in equation (10) is a solution of equation (20).

We can obtain an idea of the physical meaning of this differential equation as follows.

Let us take any point O and construct a sphere of radius r about this point. The mean value of V averaged over the surface of the sphere is

^ 1

4>7T)

VdS

= -^ ffv sin 0ddd<f>,

where r, 0, (f> are polar coordinates, having O as origin. If we change the radius of this sphere from r to r + dr, the rate of change of V is

BV 1 ffdV .

dVdS

4nrr2 JJdr

= 0, by Gauss' Theorem,

shewing that V is independent of the radius r of the sphere. Taking r = 0, the value of V is seen to be equal to the potential at the origin 0.

This gives the following interpretation of the differential equation :

V varies from point to point in such a way that the average value of V taken over any sphere surrounding any point 0 is equal to the value of V at 0.

49-54] Maxima and Minima of Potential 43

Deductions from Law of Inverse Square.

51. Theorem. The potential cannot have a maximum or a minimum value at any point in space xuhich is not occupied by an electric charge.

For if the potential is to be a maximum at any point 0, the potential at every point on a sphere of small radius r surrounding 0 must be less than that at 0. Hence the average value of the potential on a small sphere surrounding 0 must be less than the value at 0, a result in opposition to that of the last section.

A similar proof shews that the value of V cannot be a minimum.

52. A second proof of this theorem is obtained at once from Laplace's

equation. Regarding V simply as a function of x, y, z, a necessary condition

. . d-V d2V d2V

for V to have a maximum value at any point is that -~-j , -^-j and -=-y shall

each be negative at the point in question, a condition which is inconsistent with Laplace's equation

dx* ay2 dz2 So also for V to be a minimum, the three differential coefficients would have to be all positive, and this again would be inconsistent with Laplace's equation.

53. If V is a maximum at any point 0, which as we have just seen

dV must be occupied by an electric charge, then the value of -^- must be

negative as we cross a sphere of small radius r. Thus 1 1 -~-dS is negative

where the integration is taken over a small sphere surrounding 0, and by Gauss' Theorem the value of the surface integral is — kire, where e is the total charge inside the sphere. Thus e must be positive, and similarly if V is a minimum, e must be negative. Thus :

If V is a maximum at any point, the point must be occupied by a positive charge, and if V is a minimum at any point, the point must be occupied by a negative charge.

54. We have seen (§ 36) that in moving along a line of force we are moving, at every point, from higher to lower potential, so that the potential continually decreases as we move along a line of force. Hence a line of force can end only at a point at which the potential is a minimum, and similarly by tracing a line of force backwards, we see that it can begin only at a point of which the potential is a maximum. Combining this result with that of the previous theorem, it follows that :

Lines of force can begin only on -positive charges, and can end only on negative charges.

44 Electrostatics — Field of Force [oh. ii

It is of course possible for a line of force to begin on a positive charge, and go to infinity, the potential decreasing all the way, in which case the line of force has, strictly speaking, no end at all. So also, a line of force may come from infinity, and end on a negative charge.

Obviously a line of force cannot begin and end on the same conductor, for if it did so, the potential at its two ends would be the same. Hence there can be no lines of force in the interior of a hollow conductor which contains no charges ; consequently there can be no charges on its inner surface.

Tubes of Force.

55, Let us select any small area dS in the field, and let us draw the lines of force through every point of the boundary of this small area. If dS is taken sufficiently small, we can suppose the electric intensity to be the same in magnitude and direction at every point of dS, so that the directions of the lines of force at all the points on the boundary will be approximately all parallel. By drawing the lines of force, then, we shall obtain a " tubular " surface — i.e., a surface such that in the neighbourhood of any point the surface may be regarded as cylindrical. The surface obtained in this way is called a " tube of force." A normal cross-section of a " tube of force " is a section which cuts all the lines of force through its boundary at right angles. It therefore forms part of an equipotential surface.

56. Theorem. If talt w2 be the areas of two normal cross-sections of the same tube of force, and Rlt R2 the intensities at these sections, then

R1(01 = R2(02.

Consider the closed surface formed by the two cross-sections of areas

o)1, w2, and of the part of the tube of force joining them. There is no charge inside this

surface, so that by Gauss' theorem, ljNdS = 0.

If the direction of the lines of force is from «! to o)2, then the outward normal intensity Fig. 12. over &>2 is Rit so that the contribution from this

area to the surface integral is R2(o2. So also over Wj the outward normal intensity is —Ru so that a^ gives a contribution — RiCO!. Over the rest of the surface, the outward normal is perpendicular to the electric intensity, so that JVr = 0, and this part of the surface contributes

nothing to IjNdS. The whole value of this integral, then, is

R2(i)i — RiCOx, and since this, as we have seen, must vanish, the theorem is proved.

54-58]

Tubes of Force

45

57. Coulomb's Law. If R is the outward intensity at a point just outside a conductor, then R = 4>7ra, where a is the surface density of electri- fication on the conductor.

We have already seen that the whole electrification of a conductor must reside on the surface. Therefore we no longer deal with a volume density of electrification p, such that the charge in the element of volume dxdydz is p dxdydz, but with a surface-density of electrification a such that the charge on an element dS of the surface of the conductor is adS.

The surface of the conductor, as we have seen, is an equipotential, so that by the theorem of p. 29, the intensity is in a direction normal to the surface. Let us draw perpendiculars to the surface at every point on the boundary of a small element of area dS, these per- pendiculars each extending a small distance into the conductor in one direction and a small distance away from the conductor in the other direction. We can close the cylindrical surface so formed, by two small plane areas, each equal and parallel to the original element of area dS. Let us now apply Gauss' Theorem to this closed surface. The normal intensity is zero over every part of this surface except over the cap of area dS which is outside the conductor. Over this cap the outward normal in- tensity is R, so that the value of the surface integral of normal intensity taken over the closed surface, consists of the single term RdS. The total charge inside the surface is adS, so that by Gauss' Theorem,

RdS=4,7r*dS (21),

and Coulomb's Law follows on dividing by dS.

58. Let us draw the complete tube of force which is formed by the lines of force starting from points on the boundary of the element dS of the surface of the conductor. Let us suppose that the surface density on this element is positive, so that the area dS forms the normal cross-section at

Fig. 13.

Fig. 14.

the positive end, or beginning, of the tube of force. Let us suppose that at the negative end of the tube of force, the normal cross-section is dS', that

46 Electrostatics — Field of Force [oh. ii

the surface density of electrification is a-', a' being of course negative, and that the intensity in the direction of the lines of force is R'. ■ Then, as in

equati )n (21),

R'dS' = - 4>ir*'dS',

since ,he outward intensity is now - R'.

Since R, R' are the intensities at two points in the same tube of force at which the normal cross-sections are dS, dS', it follows from the theorem

of 5 53, chat

RdS = R'dS'

and hence, on comparing the values just found for RdS and R'dS', that

crdS = — cr'dS'.

Since crdS and a-'dS' are respectively the charges of electricity from which the tube begins and on which it terminates, we see that :

The negative charge of electricity on which a tube of force terminates is numerically equal to the positive charge from which it starts.

If we close the ends of the tube of force by two small caps inside the conductors, as in fig. 14, we have a closed surface such that the normal intensity vanishes at every point. Thus, by Gauss' Theorem, the total charge inside must vanish, giving the result at once.

59. The numerical value of either of the charges at the ends of a tube of force may conveniently be spoken of as the strength of the tube. A tube of unit strength is spoken of by many writers as a unit tube of force.

The strength of a tube of force is <rdS in the notation already used, and this, by Coulomb's Law, is equal to -j— RdS where R is the intensity at the

47T

end dS of the tube. By the theorem of § 56, RdS is equal to Rlw1 where _Rj, «! are the intensity and cross-section at any point of the tube. Hence jRjO)! = 47T times the strength of the tube. It follows that :

The intensity at any point is equal to 4>tt times the aggregate strength per unit area of the tubes which cross a plane drawn at right angles to the direction of the intensity.

In terms of unit tubes of force, we may say that the intensity is 4nr times the number of unit tubes per unit area which cross a plane drawn at right angles to the intensity.

The conception of tubes of force is due to Faraday: indeed it formed almost his only instrument for picturing to himself the phenomena of the Electric Field. It will be found that a number of theorems connected with the electric field become almost obvious when interpreted with the help of the conception of tubes of force. For instance we proved on p. 37 that

..(22),

58-62] Tubes of Force 47

when a number of charged bodies are placed inside a hollow conductor, they induce on its inner surface a charge equal and opposite to the sum of all their charges. This may now be regarded as a special case of the obvious theorem that the total charge associated with the beginnings and termi- nations of any number of tubes of force, none of which pass to infinity, must be nil.

Examples of Fields of Force. *

60. It will be of advantage to study a few particular fields of electric force by means of drawing their lines of force and equipotential surfaces.

I. Two Equal Point Charges.

61. Let A, B be two equal point charges, say at the points x = — a, + a. The equations of the lines of force which are in the plane of x, y are easily found to be

a#=F = y

dx ~ X ~ (PB3 - PA*

x + a{TWTPA3

where P is the point x, y.

This equation admits of integration in the form

x + a x — a /nn.

-pj- + -p]f = cons (23).

From this equation the lines of force can be drawn, and will be found to lie as in fig. 15.

62. There are, however, only a few cases in which the differential equations of the lines of force can be integrated, and it is frequently simplest to obtain the properties of the lines of force directly from the differential equation. The following treatment illustrates the method of treating lines of force without integrating the differential equation.

From equation (22) we see that obvious lines of force are

dy (i) y = 0, ^- = 0, giving the axis AB;

(ii) x = 0, PA=PB, ^ = oo, giving the line which bisects AB at

right angles. These lines intersect at G, the middle point of AB. At this point, then,

^- has two values, and since Jf- = ^ > it follows that we must have X = 0, dx ox A

F=0. In other words, the point C is a point of equilibrium, as is otherwise

obvious.

48

Electrostatics — Field of Force

[ch. n

The same result can be seen in another way. If we start from A and draw a small tube surrounding the line AB, it is clear that the cross-section of the tube, no matter how small it was initially, will have become infinite by the time it reaches the plane which bisects AB at right angles — in fact the cress-section is identical with the infinite plane. Since the product of the cross-section and the normal intensity is constant throughout a tube, it follows that at the point G, the intensity must vanish.

Fig. 15.

At a great distance R from the points A and B, the fraction

PB* - PA* TB* + PA*

vanishes to the order of 1/R, so that

dx x '

except for terms of the order of 1/ifr Thus at infinity the lines of force become asymptotic to straight lines passing through the origin.

Let us suppose that a line of force starts from A making an angle 6 with BA produced, and is asymptotic at infinity to a line through C which makes an angle <f> with BA produced. By rotating this line of force about the axis AB we obtain a surface which may be regarded as the boundary of a bundle of tubes of force. This surface cuts off an area

2tt (1 - cos 6) r*

62]

Charges +e, +e

49

from a small sphere of radius r drawn about A, and at every point of

this sphere the intensity is e/r2 normal to the sphere. The surface again

cuts off an area

2tt (1 - cos <f>) R2

from a sphere of very great radius R drawn about G, and at every point of this sphere the intensity is 2e/R2. Hence, applying Gauss' Theorem to the part of the field enclosed by the two spheres of radii r and R, and the surface formed by the revolution of the line of force about AB, we obtain

2tt (1 - cos 6) r2 x -2- 2?r (1 - cos <j>) R* x j| = 0,

from which follows the relation

sin \Q = /2 sin \ </>.

In particular, the line of force which leaves J. in a direction perpendicular to AB is bent through an angle of 30° before it reaches its asymptote at infinity.

The sections of the equipotentials made by the plane of xy for this case are shewn in fig. 16 which is drawn on the same scale as fig. 15. The equa- tions of these curves are of course

= cons.,

PA ' PB

curves of the sixth degree. The equipotential which passes through G is of interest, as it intersects itself at the point G. This is a necessary conse-

Fig. 16.

Indeed the conditions

quence of the fact that G is a point of equilibrium, for a point of equilibrium, namely

dZ=o ?I=o, 8-?=o,

dx dy dz

may be interpreted as the condition that the equipotential (V= constant) through the point should have a double tangent plane or a tangent eone at the point.

j. 4

50

Electrostatics — Field of Force

[ch. u

II. Point charges + e, — e.

6 J. Let charges ± e be at the points x = ± a (A, B) respectively. The diffeiential equations of the lines of force are found to be

dy _ Y _ y

cte ~ X == /PB3 + PA3\ '

and i he integral of this is

x + a x — a

PA PB The lines of force are shewn in fig. 17

= cons.

Fig. 17.

III. Electric Douhlet.

64. An important case occurs when we have two large charges -f e, — e, equal and opposite in sign, at a small distance apart. Takiag Cartesian coordinates, let us suppose we have the charge + e at a, 0, 0 and the charge — e at — a, 0, 0, so that the distance of the charges is 2a.

The potential is

e e

V(a; - ay + y2 + z2 V(# + a)2 + y% + z% '

and when a is very small, so that squares and higher powers of a may be neglected, this becomes

2eax

(x2 + y2 + z2)*

If a is made to vanish, while e becomes infinite, in such a way that 2ea retains the finite value /x, the system is described as an electric

i

63, 64]

Charges +e, — e

51

doublet of strength //, having for its direction the positive axis of x. Its potential is

fix

(a? + y- + z2)$'

Fig. 18.

or, if we turn to polar coordinates and write x — r cos 9, ig

fi cos 9

.(24).

The lines of force are shewn in fig. 18. Obviously the lines at the centre of this figure become identical with those shewn in fig. 17, if the latter are shrunk indefinitely in size.

4-2

52

Electrostatics — Field of Force

[ch. II

IV. Point charges + 4e, — e. Fig. 19 represents the distribution of the lines of force when the

electric field is produced by two point charges, + 4e at A and — e at B.

At infinity the resultant force will be 3e/r2, where r is the distance from a point near to A and B. The direction of this force is outwards. Thus no lines of force can arrive at B from infinity, so that all the lines of force which enter B must come from A. The remaining lines of force from A go to infinity. The tubes of force from A to B form a bundle of aggregate

Fig. 19.

strength e, while those from A to infinity have aggregate strength Se. The two bundles of tubes of force are separated by the lines of force through G. At G the direction of the resultant force is clearly indeterminate, so that G is a point of equilibrium. As the condition that G is a point of equilibrium we have

AC BG*

So that AB = BG. At G the two. lines of force from A coalesce and then separate out into two distinct lines of force, one from G to B, and the other from C to infinity in the direction opposite to GB.

The equipotentials in this field, the system of curves

4 J_ PA PB " cons-'

are represented in fig. 20, which is drawn on the same scale as fig. 19.

65]

Charges +4e, — e

53

Since G is a point of equilibrium the equipotential through the point G must of course cut itself at G. At G the potential

4e GA

e GB

AB'

since GA=2GB. From the loop of this equipotential which surrounds B, the potential must fall continuously to — oo as we approach B, since, by the theorem of § 51, there can be no maxima or minima of potential between this loop and the point B. Also no equipotential can intersect itself since there are obviously no points of equilibrium except G. One of the inter-

Fm. 20.

mediate equipotentials is of special interest, namely that over which the potential is zero. This is the locus of the point P given by

A 1

= 0,

PA PB

and is therefore a sphere. This is represented by the outer of the two closed curves which surround B in the figure.

In the same way we see that the other loop of the equipotential through G must be occupied by equipotentials for which the potential rises steadily to the value + oo at A. So also outside the equipotential through G, the potential falls steadily to the value zero at infinity. Thus the zero equi- potential consists of two spheres — the sphere at infinity and the sphere surrounding B which has already been mentioned.

54

Electrostatics — Field of Force

[ch. II

V. Three equal charges at the corners of an equilateral triangle.

66. As a further example we may examine the disposition of equi- pote ibials when the field is produced by three point charges at the corners of an equilateral triangle. The intersection of these by the plane in which the charges lie is represented in fig. 21, in which A, B, G are the points at which the charges are placed, and I) is the centre of the triangle ABG.

] t will be found that there are three points of equilibrium, one on each of the lines AD, BD, CD. Taking AD = a, the distance of each point of equilibrium from D is just less than £ a. The same equipotential passes through all three points of equilibrium. If the charge at each of the points

Fig. 21.

A, B, G is taken to be unity, this equipotential has a potential

304 a

The

equipotential has three loops surrounding the points A, B, G. In each of

these loops the equipotentials are closed curves, which finally reduce to

small circles surrounding the points A, B, G. Those drawn correspond to

325 35 375 , 4

, — , , and - .

a a a a

the potentials

304

Outside the equipotential , the equipotentials are closed curves

a

66]

Charges +e, +e, +e

55

surrounding the former equipotential, and finally reducing to circles at in-

2 225 25 ^"75

finity. The curves drawn correspond to potentials - , , — , and - — .

r r a a a a

There remains the region between the point B and the equipotential l At B the potential is , so that the potential falls as we recede from the

a

a

304

equipotential and reaches its minimum value at B. The potential at

a

B is of course not a minimum for all directions in space : for the potential increases as we move away from B in directions which are in the plane ABG, but obviously decreases as we move away from B in a direction per-

Fm. 22.

pendicular to this plane. Taking B as origin, and the plane ABG as plane of xy, it will be found that near D the potential is

a 4a3 °

Thus the equipotential through D is shaped like a right circular cone in the immediate neighbourhood of the point B. From the equation just found, it is obvious that near B the sections of the equipotentials by the plane ABG will be circles surrounding B.

56 Electrostatics — Field of Force [ch. n

From a study of the section of the equipotentials as shewn in fig. 21, it is easy to construct the complete surfaces. We see that each equipotential for which V has a very high value consists of three small spheres surrounding the point.0, A, B, 0. For smaller values of V, which must, however, be greater

than , each equipotential still consists of three closed surfaces surround- ing A, B, C, but these surfaces are no longer spherical, each one bulging out towards the point D. As V decreases, the surfaces continue to swell out,

304 until, when V = , the surfaces touch one another simultaneously, in a

(Ju

way which will readily be understood on examining the section of this equi- potential as shewn in fig. 21. It will be seen that this equipotential is

shaped like a flower of three petals from which the centre has been cut away.

3

As V decreases further the surfaces continue to swell, and when V = -, the

a

space at the centre becomes filled up. For still smaller values of V the

equipotentials are closed singly-connected surfaces, which finally become

spheres at infinity corresponding to the potential V = 0.

The sections of the equipotentials by a plane through DA perpendicular to the plane ABO are shewn in fig. 22.

Special Properties of Equipotentials and Lines of Force.

The Equipotentials and Lines of Force at infinity. 67. In § 40, we obtained the general equation

7=2 e_l

[(a? - ^)a + (y - yiT + (z - z*yf '

If r denotes the distance of x, y, z from the origin, and rx the distance of #i> 2/i> 2i> fr°m tne origin, we may write this in the form

[r2 - 2 {ccx1 + yVl + zzx) + rfft '

At a great distance from the origin this may be expanded in descending powers of the distance, in the form

t^_^Mi , xasi + yyi + zzi , 3(^1 + y<y1 + -g-g1)2 In3 ) y--r\1+ J3 +2 r< 2>+"7"

The term of order - is — - . r r

The term of order - is - %ex {xxl + yyx + zzx).

Thus by taking the origin at this centroid, the term of order - will

r2

66-68] Equipotentials and Lines of Force 57

If the origin is taken at the centroid of ex at <&,, ylt zu e2 at xit y%> z2, etc., we have

Xex xx = 0, tex yx = 0, Xe1z1 = 0.

by taking disappear.

The term of order — is r3

3 1

^3 2ex (xxx + yyx + zzx)* - ~ texrx\

Let A, B, G, be the moments of inertia about the axes, of ex at x1, yx, zx, etc., and let I be the moment of inertia about the line joining the origin to x, y, z\ then

2W = ^(A+B+C),

2ex (xxx + yyx + tutf = r2 (Xe^ - 1), and the terms of order - become

A+B+C-3I

2r3

Thus taking the centroid of the charges as origin, the potential at a great distance from the origin can be expanded in the form

F_Se A+B + G-SI

Thus except when the total charge 2e vanishes, the field at infinity is the same as if the total charge %e were collected at the centroid of the charges. Thus the equipotentials approximate to spheres having this point as centre, and the asymptotes to the lines of force are radii drawn through the centroid. These results are illustrated in the special fields of force considered in §§ 61 — 66.

The Lines of Force from collinear charges.

68. When the field is produced solely by charges all in the same straight line, the equipotentials are obviously surfaces of revolution about this line, while the lines of force lie entirely in planes through this line. In this important case, the equation of the lines of force admits of direct integration.

Let %, 1%, B, ... be the positions of the charges ex, e2, e3, — Let Q, Q' be any two adjacent points on a line of force. Let iV^ be the foot of the perpendicular from Q to the axis i?^, . . . , and let a circle be drawn perpen- dicular to this axis with centre N and radius QN. This circle subtends at i? a solid angle

2tt (1 - cos ex),

58

Electrostatics — Field of Force

[ch. n

where ^ is the angle Qi?iV. Thus the surface integral of normal force arising from e,, taken over the circle QN, is

273-0! (1 — cos #j)

and the total surface integral of normal force taken over this surface is

2-77-26! (1 — cos #i).

If we draw the similar circle through Q', we obtain a closed surface bounded by these two circles and by the surface formed by the revolution

Fig. 23.

of QQ\ This contains no electric charge, so that the surface integral of normal force taken over it must be nil. Hence the integral of force over the circle QN must be the same as that over the similar circle drawn through Q'. This gives the equations of the lines of force in the form

(integral of normal force through circle such as QN) = constant,

which as we have seen, becomes

2ex cos #, = constant.

Analytically, let the point i? have coordinates a^, 0, 0, let ^ have coordinates a2, 0, 0, etc. and let Q be the point x, y, z. Then

cos 6X =

ijj Wl

V(# - arf + y2 + z*'

and the equation of the surfaces formed by the revolution of the lines of force is

2 x^ - = constant.

V(# - tfj)2 + y2 + z*

It will easily be verified by differentiation that this is an integral of the differential equation

dx X '

68, 69] Equipotentials and Lines of Force 59

Equipotentials which intersect themselves.

69. We have seen that, in general, the equipotential through any point of equilibrium must intersect itself at the point of equilibrium.

Let x, y, z be a point of equilibrium, and let the potential at this point be denoted by V0. Let the potential at an adjacent point x + f, y -f 77, z + £, be denoted by Vtjr,,(. By Taylor's Theorem, if/(#, y, z) is any function of x, y, z, we have

where the differential coefficients of / are evaluated at x, y, z. Taking f(x, y, z) to be the potential at x, y, z, this of course being a function of the variables x, y, z, the foregoing equation becomes

ay + &+H d^ + ^vdxTy

If x, y, z is a point of equilibrium,

Tr ydV dV „dV , . (y?rV _„ 32F \ /orx

aF=ar=8F=0

3# 3^/ 3.z

/ wy &y \

/ 3217- 32 y %

Referred to a;, 7/, z as origin, the coordinates of the point x+%, y+ 77, 0 + £ become £, 77, £, and the equation of the equipotential V=G becomes

d2V . «fc 32F

In the neighbourhood of the point of equilibrium, the values of £, 77, £ are small, so that in general the terms containing powers of £, 77, £ higher than squares may be neglected, and the equation of the equipotential V= C

becomes

fry gay

In particular the equipotential V=V0 becomes identical, in the neighbourhood of the point of equilibrium, with the cone

3a;2 ^ 3#3?/ Let this cone, referred to its principal axes, become

a¥* + b7)'* + c?a=*0 (26),

then, since the sum of the coefficients of the squares of the variables is an

invariant,

32F 327 d2V A

60 Electrostatics — Field of Force [ch. ii

Now a + b + c = 0 is the condition that the cone shall have three per- pendicular generators. Hence we see that at the point at which an equipo ential cuts itself, we can always find three perpendicular tangents to the equipotential. Moreover we can find these perpendicular tangents in an infinite number of ways.