The Mathematical Theory of Electricity and Magnetism (5th ed, 1927) — part 12 of 39

(positive) term of expression (121) is checked by a tendency to contraction (to increase t, and therefore K) represented by the second (now negative)

term of expression (121). If — is not only positive, but is numerically

large, expression (121) may be negative and the dielectric will contract. In this case the decrease in energy resulting on the increase of K produced by contraction will more than outweigh the gain resulting from the diminution of the volume occupied by dielectric.

202, 203] Stresses in Dielectric Media 181

These considerations enable us to see the physical significance of all the

X 2 terms in expression (120), except the first term -^- {Kx — 1). To interpret

this term we must examine the conditions near the edge of the dielectric slab, for it is only here that X1 has a value different from zero. We see at once that this term represents a pull at and near the edge of the dielectric, tending to suck the dielectric further between the plates — in fact this force alone gives rise to the tendency to motion of the slab as a whole, which was discovered in § 139.

Keturning to the general systems of forces of § 199, we may say that the first system (which as we have seen always tends to drag the surface of the dielectric into the region in which K has the greater value) represents the tendency for the system to decrease its energy by increasing the volume occupied by dielectrics of large inductive capacity, whilst the second system (which tends to compress or expand the dielectric in such a way as to increase its inductive capacity) represents the tendency of the system to decrease its energy by increasing the inductive capacity of its dielectrics. That any increase in the inductive capacity is invariably accompanied by a decrease of energy has already been proved in § 191.

Electrostriction.

203. It will now be clear that the action of the various tractions on the surface of a dielectric must always be accompanied not only by a tendency for the dielectric to move as a whole, but also by a slight change in shape and dimensions of the dielectric as this yields to the forces acting on it. This latter phenomenon is known as electrostriction. It has been observed experimentally by Quincke and others. A convenient way of shewing its existence is to fill the bulb of a thermometer-tube with liquid, and place the whole in an electric field. The pulls on the surface of the glass result in an increase in the volume of the bulb, and the liquid is observed to fall in the tube. From what has already been said it will be clear that a dielectric may either expand or contract under the influence of electric forces.

The stresses in the interior of a dielectric, as given in § 199, may also be accompanied by mechanical deformation. Thus it has been observed by Kerr and others, that a piece of non-crystalline glass acquires crystalline properties when placed in an electric field. Such a piece of glass reflects light like a uniaxal crystal of which the optic axis is in the direction of the lines of force.

182 General Analytical Theorems [ch. vii

Green's Equivalent Stratum.

204. Let S be any closed surface enclosing a number of electric charges, and let P be any point outside it. The potential at P due to the charges inside S is

Vp = 1 1 1 -dxdydz,

.P

Fio. 56.

where r is the distance from P to the element dxdydz, and the integration extends throughout S. By Green's Theorem (equation (101))

IJJ(UWV-VWU)dxdydz=IJ(ud^-Vd£)dS, where the normal is now drawn outwards from the surface S.

In this equation, put U = -, then, since V2F = — 4<7rp, we have as the value of the first term,

fjfuWdccdydz = - 4>ttVp.

And since V2Z7 = 0, the second term vanishes. The equation accordingly becomes

-**- Bf£)-r*®}«: (m)-

205. Suppose,  first,  that  the  surface  S  is  an  equipotential.     Then 
     

= vfffv*fy dxdydz

= 0,

so that equation (122) becomes

VP=. U —±L±ldS (123).

204-207] Green's Equivalent Stratum 183

Thus the potential of any system of charges is the same at every point outside any selected equipotential which surrounds all the charges, as that of a charge of electricity spread over this equipotential, and having surface

1 97 density — j— ~— . Obviously, in fact, if the equipotential is replaced by a

conductor, this will be the density on its outer surface.

206. If the surface is not an equipotential, the term / / V =- (-) dS

will not vanish. Since, however, jj,^- (-] is the potential of a doublet of

strength /u, and direction that of the outward normal, it follows that

1 1 7^- (-} dS is the potential of a system of doublets arranged over the

surface S, the direction at every point being that of the outward normal, and the total strength of doublets per unit area at any point being V.

Thus the potential Vp may be regarded as due to the presence on the surface S of

1 dV (i) a surface density of electricity — j— -~— ;

V

(ii) a distribution of electric doublets, of strength - — per unit area,

and direction that of the outward normal.

207. Equation (122) expresses the potential at any point in the space

dV outside S in terms of the values of V and -~- over the boundary of this space.

We have seen, however, that the value of the potential is uniquely determined

dV by the values either of V or oi — over the boundary of the space. In actual

electrostatic problems, the boundaries are generally conductors, and therefore

equipotentials. In this case equation (123) expresses the values of the

dV potential in terms of -~- only, amounting in fact simply to

rP=jfUs.

What is generally required is a knowledge of the value of VP in terras of the values of V over the boundaries, and this the present method is unable to give. For special shapes of boundary, solutions have been obtained by various special methods, and these it is proposed to discuss in the next chapter.

184

General Analytical Theorems

[ch. VII

EXAMPLES.

1. If the electricity in the field is confined to a given system of conductors at given potentials, and the inductive capacity of the dielectric is slightly altered according to any law such that at no point is it diminished, and such that the differential coefficients of the increment are also small at all points, prove that the energy of the field is increased.

2. A slab of dielectric of inductive capacity K and of thickness x is placed inside a parallel plate condenser so as to be parallel to the plates. Shew that the surface of the slab experiences a tension

3. For a gas K=-l + 8p, where p is the density and 6 is small. A conductor is immersed in the gas : shew that if 62 is neglected the mechanical force on the conductor is 2jr(72 per unit area. Give a physical interpretation of this result.

CHAPTER VIII

METHODS FOR THE SOLUTION OF SPECIAL PROBLEMS

The Method of Images. Charge induced on an infinite uninsulated plane.

208. The  potential  at  P  of  charges  e  at  a  point  A  and  -  e  at  another 
     

point A' is

V = — —

AP A'P

.(124),

and this vanishes if P is on the plane which bisects A A' at right angles. Call this plane the plane S. Then the above value of V gives V=0 over the plane S, V= 0 at infinity, and satisfies Laplace's equation in the region to the right of 8, except at the point A, at which it gives a point charge e.

i *

V x ' / '

V \ \ • '/.»'

"X » X X \ ' I ' '

S < ! i\ V

' / \ ' Sx

' I V Vx

Fig. 57.

These conditions, however, are exactly those which would have to be satisfied by the potential on the right of S if S were a conducting plane at zero potential under the influence of a charge e at A. These conditions amount to a knowledge of the value of the potential at every point on the boundary of a certain region — namely, that to the right of the plane S — and of the charges inside this region. There is, as we know, only one value of the

186 Methods for the Solution of Special Problems [oh. viii

potential inside this region which satisfies these conditions (cf. § 186), so that this value must be that given by equation (124).

To the right of 8 the potential is the same, whether we have the charge — e at A' or the charge on the conducting plane 8. To the left of S in the latter case there is no electric field. Hence the lines of force, when the plane S is a conductor, are entirely to the right of S, and are the same as in the original field in which the two point-charges were present. The lines end on the plane S, terminating of course on the charge induced on S.

We can find the amount of this induced charge at any part of the plane by Coulomb's Law. Taking the plane to be the plane of yz, and the point A to be the point (a, 0, 0) on the axis of x, we have

47TO- = R = — —- ex

—li

!•

dx (V(ic - a)2 + y" + z" /(x + af + y- + z2

where the last line has to be calculated at the point on the plane S at which we require the density. We must therefore put x = 0 after differentiation, and so obtain for the density at the point 0, y, z on the plane S,

2ae

4>7rcr = —

(a2 + f + z*f '

or, if a2 + y1 + z- = r2, so that r is the distance of the point on the plane S from the point A,

ae

2irr

•125

•099 •044 •021

•012 •007

Thus the surface density falls off inversely as the cube of the distance from the point A. The distribution of electricity on the plane is represented graphically in fig. 58, in which the thickness of the shaded part is proportional to the surface density of electricity. The negative electricity is, so to speak, heaped up near the point A under the influence of the attraction of the charge at A. The field produced by this distribution of electricity on the plane S at any point to the right of 8 is, as we know, exactly the same as would be produced by the point charge — e at A'.

209. This problem affords the simplest illustration of a general method for the solution of electrostatic problems, which is known as the " method of images." The principle underlying this method is that of finding a system of electric charges such that a certain surface, ultimately to be made into a conductor, is caused to coincide with the equipotential V = 0. We then replace the charges inside this equipotential by the Green's equivalent

Fio. 5S.

208-210] Images 187

stratum on its surface (cf. § 204). As this surface is an equipotential, we can imagine it to be replaced by a conductor and the charges on it will be in equilibrium. These charges now become charges induced on a conductor at potential zero by charges outside this conductor.

From the analogy with optical images in a mirror, the system of point charges which have to be combined with the original charges to produce zero potential over a conductor are spoken of as the " electrical images " of the original charges. For instance, in the example already discussed, the field is produced partly by the charge at A, partly by the charge induced on the infinite plane : the method of images enables us to replace the whole charge induced on the plane by a single point charge at A'. So also, if A were a candle placed in front of an infinite plane mirror, the illumination in front of the mirror would be produced partly by the candle at A, partly by the light reflected from the infinite mirror ; the method of optical images enables us to replace the whole of this reflected light by the light from a single source at A'.

210. In an electrostatic field produced by any number of point charges, we can, as we have seen, select any equipotential and replace it by a con- ductor. The charges on either side of this equipotential are then the "images" of those on the other side.

Thus if we can write the equation of any surface in the form

• • ^ + C + .-.=0 (125),

where r is the distance from a point outside the surface, and r', r", . . . are the distances from points inside the surface, then we may say that charges e', e", ... at these latter points are the images of a charge e at the former point.

The method of images may be applied in a similar way to two-dimensional problems. Suppose that the equation of a cylindrical surface can be expressed in the form

o - 2e log r - 2e' log r' - 2e" log r" — . .. - 0,

where r is the perpendicular distance from a fixed line on one side of the surface, and r', r", . . . are perpendicular distances from fixed lines on the other side. Then line-charges of line-densities e', e", ... at these latter lines may be taken to be the image of a line-charge of line-density e at the former line.

Illustrations of the use of images in three dimensions are given in §§ 211 — 219. An illustration of the use of a two-dimensional image will be found in § 220.

188 Methods for the Solution of Special Problems [ch. viii

Charges induced on Intersecting planes,

211. It  will  be  found  that  charges 
     

e at x, y, 0,

— e at — x, y, 0,

— e at x, — y, 0,

-er

__^

e at — x, — y, 0

give zero potential over the planes x = 0, y = 0. The potential of these charges is therefore the same, in the quadrant in which x, y are both positive, as if the boundary of this quadrant were a conductor put to earth under the in- fluence of a charge e at the point x, y, 0.

It will be found that a conductor consisting of three planes intersecting at right angles can be treated in the same way.

212. The  method  of  images  also  supplies  a  solution  when  the  conductor 
     

Fig. 59.

7T

consists of two planes intersecting at any angle of the form — , where n is

any positive integer. If we take polar coordinates, so that the two planes

7T

are 6 = 0, 6 = - , and suppose the charge to be a charge e at the point r, 6, we shall find that charges

e at (r, 6), (r, B + ^), (r. 0 + ^), ...,

e at (,.,*), (r,-(« + ^)). (l -(« + £)).....

give zero potential over the planes

0 = 0, $=-.

211-213]

Images

189

Charge induced on a sphere.

213. The  most  obvious  case,  other  than  the  infinite  plane,  of  a  surface 
     

whose equation can be expressed in the form (125), is a sphere.

Fig. 61.

If R, Q are any two inverse points in the sphere, and P any point on the surface, we have

RP:PQ = 00:OQ,

so that

OQ

PQ PR

°Ji = 0.

00

Thus the image of a charge e at Q is a charge — e ^ at R, or the

image of any point at a distance / from the centre of a sphere of radius a

ect is a charge - j at the inverse point, i.e. at a point on the same radius

a2 distant -j from the centre.

Let us take polar coordinates, having the centre of the sphere for origin and the line OQ as 0 = 0. Our result is that at any point 8 outside the sphere, the potential due to a charge e at Q and the charge induced on the surface of the sphere, supposed put to earth, is

ea

~ QS RS e ea

Vr2+/2-2/rcos0 / „ a< a2 a

/<V +7^ 7rc

where r, 6 are the coordinates of S.

190 Methods for the Solution of Special Problems [ch. viii

214. We  can  now  find  the  surface-density  of  the  induced  charge.     For 
     

at any point on the sphere

= B_ l_dV

4nr 47r dr *

in which we have to put r=a after differentiation. Clearly

dv

dr

ea\r — j cos 0 1 (r2 +/2 - 2/r cos fffi , ( , a4 _ a- n\ $ '

e(r— /cos #)

Putting r — a we obtain

— /cos #

a8/8 -a8/ cos 0 (a2 +/2 - 2/a cos 6>)t (a2/2 + a4 - 2a3/ cos 0)$

!

4?r Va2 + f* - i

e { a —f2/a \

4tt ((a2 +y2 _ 2/a cos 6>)ti

e (f2-a2)

4tt a.£Q3 '

Thus the surface-density varies inversely as SQ*, so that it is greatest at C and falls off continually as we recede from the radius OC. The total

on

charge on the sphere is — j , as can be seen at once by considering that the total strength of the tubes of force which end on it is just the same as would

Fig. 62.

214-216] Images 191

be the total strength of the tubes ending on the image at R if the conductor were not present.

Figure 62 shews the lines of force when the strength of the image is a quarter of that of the original charge, so that f= 4>a. It is obtained from fig. 19 by replacing the spherical equipotential by a conductor, and annihi- lating the field inside.

Superposition of Fields.

215. We have seen that by adding the potentials of two separate fields at every point, we obtain the potential produced by charges equal to the total charges in the two fields. In this way we can arrive at the field produced by any number of point charges and uninsulated conductors of the kind we have described. The potential of each conductor is zero in the final solution because it is zero for each separate field.

There is also another type of field which may be added to that obtained by the method of images, namely the field produced by raising the conductor or conductors to given potentials, without other charges being present. By superposing a field of this kind we can find the effect of point charges wThen the conductors are at any potential.

216. For instance, suppose that, as in fig. 62, we have a point charge e and the conductor at potential 0. Let us superpose on to the field of force already found, the field which is obtained by raising the conductor to potential V when the point charge is absent. The charge on the sphere in the second field is aV, so that the total charge is

it ea aV-j.

By giving different values to V, we can obtain the total field, when the sphere has any given charge or potential.

If the sphere is to be uncharged, we must have V=-^, so that a point charge placed at a distance / from the centre of an uncharged sphere raises it to potential -, , a result which is also obvious from the theorem of § 104.

192 Methods for the Solution of Special Problems [ch. viii

Sphere in a uniform field of force.

2YJ. A uniform field of force of which the lines are parallel to the axis of x may be regarded as due to an infinite charge E at x = R, and a charge — E at X — — R, when in the limit E and R both become infinite. The intensity at any point is

2E R*

parallel to the axis of x, so that to produce a uniform field in which the intensity is F parallel to the axis of x, we must suppose E and R to become infinite in such a way that

-R>=-R

dV

Since, in this case, F = — ~— , the potential of such a field will clearly

be -Fx + G.

"Suppose that a sphere is placed in a uniform field of force of this kind, its centre being at the origin. We can suppose the charge E at x = R to have an image of strength

Ea _ a?

"X at X~R>

while the other charge has an image

Ea aa

These two images may be regarded as a doublet (cf. § 64) of strength -p- x -p , and of direction parallel to the negative axis of x. The strength

it -it

-TP— W-

Thus we may say that the image of a uniform field of force of strength F is a doublet of strength Fa3 and of direction parallel to that of the intensity of the uniform field.

The potential of this doublet is

Fa3 cos 6 r» '

and that of the field of original field of force is

• Fx + C,

or, in polar coordinates, — Fr cos Q + G,

217] Images

so that the potential of the whole field

= — F cos 0 ( r

193

a)

=)+o

.(126).

a

Fig. 63.

As it ought, this gives a constant potential G over the surface of the sphere.

Fig. U.

The lines of force of the uniform field F disturbed by the presence of a doublet of strength Fa3 are shewn in fig. 63. On obliterating all the lines of force inside a sphere of radius a, we obtain fig. 64, which accordingly shews the lines of force when a sphere of radius a is placed in a field of intensity F. These figures are taken from Thomson's Reprint of Papers on Electrostatics and Magnetism (pp. 488, 489)*.

• I am indebted to Lord Kelvin for permission to use these figures.

13

194 Methods for the Solution of Special Problems [ch. viii

218. Line of no electrification. The theory of lines of no electrification has already been briefly given in § 98. We have seen that on any conductor on which the total charge is zero, and which is not entirely screened from an electric field, there must be some points at which the surface-density a- is positive, and some points at which it is negative. The regions in which a is positive and those in which c- is negative must be separated by a line or system of lines on the conductor, at every point of which a = 0. These lines are known as lines of no electrification.

If R is the resultant intensity, we have at any point on a line of no electrification,

R = 4tto- = 0,

so that every point of a line of no electrification is a point of equilibrium. At such a point the equipotential intersects itself, and there are two or more lines of force.

If the conductor possesses a single tangent plane at a point on a line of no electrification, then one sheet of the equipotential through this point will be the conductor itself: by the theorem of § 69, the second sheet must intersect the conductor at right angles.

These results are illustrated in the field of fig. 64. Clearly the line of no electrification on the sphere is the great circle in a plane perpendicular to the direction of the field. The equipotential which intersects itself along the line of no electrification ( V = G) consists of the sphere itself and the plane containing the line of no electrification. Indeed, from formula (126),

it is obvious that the potential is equal to C, either when 6 = — , or

when r = a.

The intersection of the lines of force along the line of no electrification is shewn clearly in fig. 64.

Plane face with hemispherical boss.

219. If we regard the whole equipotential V= C as a conductor, we obtain the distribution of electricity on a plane conductor on which there is a hemispherical boss of radius a. If we take the plane to be a; = 0, we have, by formula (126),

V-C = -Fcos0(r-^)=-Fx(l-^).

At a point on the plane,

1 (dV\ F f, as

' 4tt \dx J,=0 4tt J1 r3j ' and on the hemisphere

47r \dr Jr=a 4-7T '

218-220] Images 195

The whole charge on the hemisphere is found on integration to be

I

f f- 3 cos 6 ) lira? sin 0 d6 = f Fa2,

0 = 0 V47T

while, if the hemisphere were not present, the charge on the part of the plane now covered by the base of the hemisphere would be

(s)m'=ift'

Thus the presence of the boss results in there being three times as much electricity on this part of the plane as there would otherwise be : this is compensated by the diminution of surface-density on those parts of the plane which immediately surround the boss.

Capacity of a telegraph-wire.

220. An important practical application of the method of images is the determination of the capacity of a long straight wire placed parallel to an infinite plane at potential zero, at a distance h from the plane. This may be supposed to represent a telegraph-wire at height h above the surface of the earth.

Let us suppose that the wire has a charge e per unit length. To find the field of force we imagine an image charged with a charge — e per unit length at a distance h below the earth's surface. The potential at a point at distances r, r' from the wire and image respectively is, by §§ 75 and 100,

C — 2e log r + 2e log r',

and for this to vanish at the earth's surface we must take C= 0. Thus the potential is

2e log - . ° r

At a small distance a from the line-charge which represents the telegraph- wire, we may put r' = 2/i, so that the potential is

2e log — ,

° a

from which it appears that a cylinder of small radius a surrounding the wire is an equipotential. We may now suppose the wire to have a finite radius a, and to coincide with this equipotential. Thus the capacity of the wire per unit length is

«£?

13—2

196 Methods for the Solution of Special Problems [ch. vm

Infinite series of Images.

221. Suppose  we  have  two  spheres,  centres  A,  B  and  radii  a,  b,  of  which 
     

the centres are at distance c apart, and that we require to find the field when

Fig. 65.

both are charged. We can obtain this field by superposing an infinite series of separate fields (cf. § 116).

Suppose first that A is at potential V while B is at potential zero. As a first field we can take that of a charge Va at A. This gives a uniform potential V over A, but does not give zero potential over B. We can reduce the potential over B to zero by superposing a second field arising from

the image of the original charge in sphere B, namely a charge at B',

c

b2 where BB' = — . This new field has, however, disturbed the potential over

A. To reduce this to its original value we superpose a new field arising

from the image of the charge at B' in A, namely a charge . ?j at A',

C 0

c

c

where A A' = ^ . This field in turn disturbs the potential over B, and so

c

c

we superpose another field, and so on indefinitely. The strengths of the

various fields, however, continually diminish, so that although we get an

infinite series to express the potential, this series is convergent. As we shall

see, this series can be summed as a definite integral, or it may be that a good

approximation will be obtained by taking only a finite number of terms.

The total charge on A is clearly the sum of the original charge Va plus the strengths of the images A', A", ... etc., for this sum measures the aggregate strength of the tubes of force which end on A. Similarly the charge on B is the sum of the strengths of the images at B', B",

To obtain the field corresponding to given potentials of both A and B we superpose on to the field already found, the similar field obtained by raising B to the required potential while that of A remains zero.

221, 222] Images 197

If 9ii, 922, Qu are the coefficients of capacity and induction, the total charge on A when B is to earth and V— 1 is qn ; similarly that on B is qu. In this way we can find the coefficients qu> q12 from the series of images already obtained. The result is found to be

o?b asb*

qn - a + c,_ fta + (c, _ &2)2 _ ^ + ...,

ab a"-b*

?12~ c c(c2-62-a2) + -' and from symmetry

v*-b + ^z^-2 + (c2 _ a,f _ b,ci + • -.,

As far as — , these results clearly agree with those of § 116.

222. The series for qn, q12, q22 have been put in a more manageable form by Poisson and Kirchhoff.

Let A, denote the position of the sth of the series of points A', A", ..., and Bg the sth of the series £', B", ... ; then Ag is the image of Bt in the sphere of radius a, and similarly B, is the image of At_\ in the sphere of radius b. Let ag = AA,y bg=BBg, and let the charges at A„ Ba be ea, e'g respectively.

Then ag (c — b,) = a2 since A, is the image of Bg,

68(c-a,_!) = 62 „ B3 „ „ At_v

Further, by comparing the strengths of a charge and its image,

(127),

ea=

a , , b

L ^ 81 ** 8 — ^S — 1}

c-bg c-as.x

ab

C> (c-bJic-a,^)0'-*

and similarly

d - ah c'

' (c-oil_1)(o-6-_1),r-1'

We have therefore

eg-i

ab a3 bg asc — <za

(c — b,)(c — a,_i) a 6 a&

and

(c-&, + i)(c-aa)_ c{c-ag) b ab ab a'

By addition we eliminate

««»

and obtain

or, if we put - = ua,

e3 ( eg c2-a2- b2

^2 ^2 _ A2

(128),

at

and from symmetry it is obvious that the same difference equation must be satisfied by a

quantity u't—~r. e *

The solution of the difference equation (128) may be taken to be

ug = Aa> + Bp», where a, /3 are the roots of

;2_c a b ; + l=0.

198 Methods for the Solution of Special Problems [oh. vin

The product of these roots is unity, so that if a is the root which is less than unity, we can suppose

so that

and similarly

We now have

e,=

AiP + B'

*- S A>

8 A'cP+B"

08 a8 g„ = a + e1 + e2 + ... = ct + 2 Aa-,s + g,

To determine A, B, we have

^+5

a

=a,

a26

^^+.8 c2-62' A B 1

so that where Thus

and 7ll=a<l-a{I^ + TZ^a+1-^4 + ...}.

To determine J', 5', we have

-e i «(i-£2)'

aa'(l-gg)

e»~ l_^2a28 »

el=^

a&

^'a2 + 5'

^ = -77

a«6»

4'a4 + .B' c(c2-a2~62)'

from which, in the same way,

?12=__(l-a2)j— +_+i__6 + ...J.

The value of q™ can of course be written down by symmetry from that of qn.

as The coefficients each depend on a sum of the type 2 - — pr~^a • This series has been

J. g U

expressed in terms of definite integrals by Poisson. From the known formula

r sin pt , feP + l] 1 Jo e2,T'-l* tep-lj 2/> we obtain, on putting jd = log £2 a2",

"8 _i. °' ft/" a' sin (log £'«»)*,,,

l_^2a28-fa log|2a««~ Jo ^^

From this follows

s a* _ 1 y f° 2 a' sin (2 log g + 2s log a)*

l-£2a28 2(l-a) 21og£ + 2slog« WJ0 e2rt-l

= _i r ?* dt « r sin(2<logg)-asin(2<log£/«) ,,

2(l-a) jol-a2' + 1 Jo (e27rJ-l)[l-2acos(2doga) + a2J^

i

222, 223]

Images

199

The series has also been expressed in finite terms by E. W. Barnes (Quart. Journ. Math. 138 (1903), p. 155) in terms of Double Gamma Functions, but neither of these forms is convenient for numerical computation.

A. Russell (Proc. Phys. Soc. 23 (1911), p. 352) has shewn how the original series can be rearranged in a rapidly convergent form. If n is an integer, to be chosen subsequently,

ar

S=co

2„ 1 _ £2„23 *=0 1 — 5 a

3=71-1 gS

s=n-l = 2

a

2 a«( 2 £2Pa2P*)

1=71 ^P=0 '

P=co (gan)2p p=0 1 a

3=0 W2a28

The larger n is chosen to be the more rapidly the second series converges, although of course large values for n require the computation of a large number (n) of terms in the original series. As an example, given by Russell, suppose that a=7r, b=r, c=10r; it is sufficient to take n — \ and the series are found to be

gr11 = 7r+fr{l+0-0O89509 + O'0OOO929 + 0-0000009 + ...} = 7-5765970r, -£i2=0-7r + T%r{l +0-0003580 + 0-0000001 + ...} = 08143266r, ?22=ri601124r.

As a second example Russell takes a = 98r, 6=10-8r, and c=a + b + 0'2r, so that the spheres are almost in contact ; the values of the coefficients are obtained to seven figures on taking n = 4 and computing seven terms of the second series.

223. Having calculated the coefficients, we can obtain the relations between the charges and potentials, and can find also the mechanical force between the spheres. If this force is a force of repulsion F, we have

dWE_ Ldpn dpi2 F F , dp2, „

or again

dc dc dc oc

The following table, applicable to two spheres of equal radius, taken to be unity, is compiled from materials given by Lord Kelvin*.

c

Pn ( =P2s)

Pl2

g u (=322)

212

2 dc \ dc J

dpi2 dc

1.9911

2 dc

3? 12

dc

Eatio of charges for equilibrium

2-0

•721

•721

00

— CO

CD

co

co

co

1

2-1

•915

•509

1-584

-•884

•154

•453

1-138

2349

•391

2-2

•939

•475

1-431

-•724

•0826

•305

•529

1-127

•294

2-5

•968

•406

1-253

-•525

•0300

•181

•174

•413

•169

30

•986

•335

1-146

-•389

•0101

•115

•066

•186

•089

35

•993

•286

1-099

-•317

•00437

•0825

•0344

•114

•053

40

•996

•250

1-072

-•269

•00216

•0628

•0207

•079

•034

5 0

•998

•200

1-044

-•209

•00065

•0401

•0096

•048

•016

6-0

•999

•167

1-030

-•172

•00026

•0278

•0053

•031

•009

CO

1-0

0

1-0

0

0

0

0

0

0

• Papers on Electrostatics and Magnetism, p. 96, § 142.

200 Methods for the Solution of Special Problems [oh. vni

Images in dielectrics.

224. The method of images can also be applied to find the field produced by point charges when half of the field is occupied by dielectric, the boundary of the dielectric being an infinite plane.

We begin by considering the field produced by a single charge e at P, it being possible to obtain the most general field by the superposition of simple fields of this kind.

We shall shew that the field in air is the same as that due to a charge e at P and a certain charge e' at P', the image of P, while the field in the dielectric is the same as that due to a certain charge e" at P, if the whole field were occupied by air.

Fig. 66.

Let PP' be taken for axis of x, the origin 0 being in the boundary of the dielectric, and let OP = a. Then we have to shew that the potential YA in air is

V.= e + e'

*/(x + af + y2 + z2 VO - a)2 + y2 + z% '

while that in the dielectric is

V(a? + a)2 + f + z* '

These potentials, we notice, satisfy Laplace's equation in each medium, everywhere except at the point P, and they arise from a distribution of charges which consists of a single point charge e at P. The potential in air at the point 0, y, z on the boundary is

VA =

e + e

Va2 + y2 + z* '

224, 225] Images 201

while that in the dielectric at the same point is

7„ =

V a2 + y2 + z2

Thus the condition that the potential shall be continuous at each point of the boundary can be satisfied by taking

/

e" = e + e (129).

The remaining condition to be satisfied is that at every point of the

dV . . dV

boundary, ^— in air shall be equal to K -^- in the dielectric ; i.e. that

K-— = -^t when a; = 0. ox ox

Now, when x = 0,

RdVD^ Ke"a

dx (a2+y2 + z2f dVA ea e'a

dos (a2 + y1 + z*f (a3 + y2 + z2f ' so that this last condition is satisfied by taking

Ke" = e-e' (130).

Thus the conditions of the problem are completely satisfied by giving e, e" values such as will satisfy relations (129) and (130); i.e. by taking

2 ^

"= TTK

K-l

.(131).

e =-TTKe )