The Mathematical Theory of Electricity and Magnetism (5th ed, 1927) — part 21 of 39

the condition to be satisfied is that there shall be no flow of current, and this

dV is expressed mathematically by the condition that -~- shall vanish.

Thus the problem of determining the current-flow in a conductor amounts

mathematically to determining a function V such that equation (312) is satis-

dV fied throughout the volume of the conductor, while either — = 0, or else V has

a specified value, at each point on the boundary. By the method used in § 188, it is easily shewn that the solution of this problem is unique.

It is only in a very few simple cases that an exact solution of the problem can be obtained. There are, however, various artifices by which approxima- tions can be reached, and various ways of regarding the problem from which it may be possible to form some ideas of the physical processes which determine the nature of the flow in a conductor. Some of these will be discussed later (§§ 386—394). At present we consider general characteristics of the flow of currents through conductors.

346 Steady Currents in continuous Media [ch. x

Conditions to be satisfied at the Boundary of two

Conducting Media.

378. The conditions to be satisfied at a boundary at which the current flows from one conductor to another are as follows:

(i) Since there must be no accumulation of electricity at the boundary, the normal flow across the boundary must be the same whether calculated in the first medium or the second. In other words

• ^— must be continuous, r dn

where 5- denotes differentiation along the normal to the boundary. on

(ii) The tangential force must be continuous, or else the potential would not be continuous. Thus

-7— must be continuous,

OS

where =- denotes differentiation along any line in the boundary.

These boundary conditions are just the same as would be satisfied in an electrostatical problem at the boundary between two dielectrics of inductive

capacities equal to the two values of -. Thus the equipotentials in this

electrostatic problem coincide with the equipotentials in the actual current problem, and the lines of force in the electrostatic problem correspond with the lines of flow in the current problem.

Clearly these results could be deduced at once from the differential equation (312) on passing to the limit and making r become discontinuous on crossing a boundary.

Refraction of Lines of Flow.

379. Let any line of flow cross the boundary between two different conducting media of specific resistances rlf t2, making angles e1} e2 with the normal at the point at which it meets the boundary in the two media respectively. The lines of flow satisfy the same conditions as would be satisfied by electrostatic lines of force crossing the boundary between two

dielectrics of inductive capacities — , — , so that we must have (cf. equa- tion (71))

— cot 6j = — cot e2.

Ti T2

Hence rt tan e1 — r2 tan e2,

expressing the law of refraction of lines of flow.

378-381] Boundary Conditions 347

380. As an example of refraction of lines of current flow, we may consider the case of a steady uniform current in a conductor being dis- turbed by the presence of a sphere of different metal inside the conductor. The lines shewn in fig. 78 will represent the lines of flow if the specific resistance of the sphere is less than that of the main conductor. The lines of flow tend to crowd into the sphere, this being the better conductor — in the language of popular science, the current tends to take the path of least resistance.

Charge on a Surface of Discontinuity.

381. If u is the normal component of current flowing across the boundary between two different conductors, we have by Ohm's Law,

tx dn t2 dn '

where =- denotes differentiation along the normal which is drawn in the dn

direction in which u is measured (say from (1) to (2)), and Vlt V2 are the potentials in the two conductors.

If there is no charge on the boundary between the two conductors we must, from equation (70), have the relation

on on

where Klt K2 are the inductive capacities of the two conductors. This condition will, however, in general be inconsistent with the condition which, as we have just seen, is made necessary by the continuity of u. Thus there will in general be a surface charge on the boundary between two conductors of different materials.

The amount of this charge is given at once by equation (72), p. 125. If a denotes the surface density at any point, we have

on on = -(Klr1-R2r2)u (313).

This surface charge is very small compared with the charges which occur in statical

electricity. For instance, if we have current of 100 amperes per sq. cm. passing from one

metallic conductor to another, we take in formula (313),

u = !00 arnperes = 3x 10u electrostatic units,

10~6 r= 10-6 ohms =___ >}

K=l,

the last two being true as regards order of magnitude only. The value of Aircr is of the order of magnitude of Ktu, or Jx 10-6 in electrostatic units. As has been said, the value of 4n-cr at the surface of a conductor charged as highly as possible in air is of the order of 100.

348 Steady Currents in continuous Media [oh. x

382. As an example of the distribution of a surface charge, we may

notice that the surface-density of the charge on the surface of the sphere

dV considered in § 380 will be proportional to either value of — , and therefore

to cos 6, where 6 is the angle between the radius through the point and the direction of flow of the undisturbed current.

Generation of Heat.

383. Consider any small element of a tube of flow, length ds, cross-

1 dV section &>. The current per unit area is, by equations (310), — , so

1 dV that the current flowing through the tube is — — co. The resistance of

tcIs the element of the tube under consideration is — . Hence, as in 5 355, the

w > a >

amount of heat generated per unit time in this element is

<ldV Vrds 1 fdVV

(ldV Vrds 1/dVy

• a- o> — or - —

\T OS j ft) T\OS )

ft)

ds.

. . . . l /dVy

Thus the heat generated per unit time per unit volume is — f -^ — J , and the total generation of heat per unit time will be

mm+(%hm^ ™

Thus the heat generated per unit time is 87r times the energy of the whole field in the analogous electrostatic problem (§ 169).

Rate of generation of heat a minimum.

384. It can be shewn that for a given current flowing through a con- ductor, the rate of heat generation is a minimum when the current distributes itself as directed by Ohm's Law. To do this we have to compare the rate of heat generation just obtained with the rate of heat generation when the current distributes itself in some other way.

Let us suppose that the components of current at any point have no longer the values

1 dV 1 dV_ ldV

t dx ' t dy ' t dz

assigned to them by Ohm's Law, but that they have different values

ldV ldV ldV

t ox t dy t dz

382-385] Generation of Heat 349

In order that there may be no accumulation at any point under this new distribution, the components of current must satisfy the equation of con- tinuity, so that we must have

die , dv dw _ ,„,.~

5- +^r + ;T = 0 (315).

ox oy oz v '

By the same reasoning as in § 383, we find for the rate at which heat is generated under the new system of currents,

///T ((- VTx + •)' + Hw + Vf+("rW+ <*)] *** which, on expanding, is equal to

-~2!IKud^+v%+wd^)dxdydz

• [ffr(wi+v2 + w2)dxdydz (316).

On transforming by Green's Theorem, the second term

= 2fjfvP£ + ^+d^)dxdydz-2ffv (lu + mv + nw) dS.

The volume integral vanishes by equation (315), the integrand of the surface integral vanishes over each electrode from the condition that the total flow of current across the electrode is to remain unaltered, and at every point of the insulating boundary from the condition that there is to be no flow across this boundary. Thus the new rate of generation of heat is represented by the first and third terms of expression (316). The first term represents the old rate of generation of heat, the third term is an essentially positive quantity. Thus the rate of heat generation is increased by any deviation from the natural distribution of currents, proving the result.

385. An immediate result of this is that any increase or decrease in the specific resistance of any part of a conductor is accompanied by an increase or decrease of the resistance of the conductor as a whole. For on decreasing the value of t at any point and keeping the distribution of currents unaltered, the rate of heat production will obviously decrease. On allow- ing the currents to assume their natural distribution, the rate of heat production will further decrease. Thus the rate of heat production with a natural distribution of currents is lessened by any decrease of specific resistance. But if / is the total current transmitted by the conductor, and R the resistance of the conductor, this rate of heat production is RI-. Thus R decreases when t is decreased at any point, and obviously the converse must be true (cf. § 359).

350 Steady Currents in continuous Media [ch. x

The Solution of Special Problems.

Current-flow in an Infinite Conductor.

386. A good approximation to the conditions of electric flow can occasionally be obtained by neglecting the restrictive influence of the boundaries of a conductor, and regarding the problem as one of flow between two electrodes in an infinite conductor. For simplicity, we shall consider only the case in which the conductor is homogeneous.

The conditions to be satisfied by the potential V are as follows. We

must have V = VX over one electrode, and V=V2 over the second electrode,

dV 1

while — must vanish at infinity to a higher order than — and throughout

the conductor we must have V2F = 0 (§ 376). We can easily see (cf. §§ 186, 187) that these conditions determine V uniquely.

Consider now an analogous electrostatic problem. Let the conducting medium be replaced by air, while the electrodes remain conductors. Let the electrodes receive equal and opposite charges of electricity until their difference of potential is Vx—V2. At this stage let -}r denote the electro- static potential at any point in the field. Let yfr1} ^2 be the values of \jr over the two electrodes, so that ^ — ty2 = Vi — V2. Then there will be a constant C (namely K — ^i)> sucn that yjr + C assumes the values V1} K respectively over the two electrodes. Moreover V2-^ = 0 throughout the field, so that V2(-v^ + (7) = 0 throughout the field, and |r=0 at infinity except for terms

-I o

in — (cf. § 67), so that ~- (^ + C) vanishes at infinity to a higher order

than — .

Hence -fy + C satisfies the conditions which, as we have seen, must be satisfied by the potential V in the current problem, and these are known to suffice to determine V uniquely. It follows that the value of V must be Tjr+C.

Thus the lines of flow in the current problem are identical with the lines of force when the two electrodes are charged to different potentials in air.

The normal current-flow at any point on the surface of an electrode is

ldV

t dn'

so that the total flow of current outwards from this electrode

iff|ZdS=_If(|fcdS.

on tJJ on

386, 387] Special Problems 351

If E is the charge on this electrode in the analogous electrostatic problem we have, by Gauss' Theorem,

-11% -*>

4<7rE so that the total flow of current is seen to be .

T

If pn> Pn> P-a are the coefficients of potential in the electrostatic problem

f^PnE-puE,

^=puE-p22E, so that

Yi- V*= fi~ ^2 = (Pn - %2 +P22) E.

If / is the total current, and R the equivalent resistance between the electrodes, we have just seen that

T '

so that

B-^^-j^CPu-apta+J?.) (317).

If we regard the two electrodes in air as forming a condenser, and denote its capacity by 0, we have

so that

B=E^=ss <318>

387. As instances of the applications of formulae (317) and (318) to special problems, we have the following:

I. The resistance per unit length between two concentric cylinders of radii a, b (as, for instance, the resistance between the core of a submarine cable and the sea), is, by formula (318),

II. The resistance per unit length between two straight parallel cylindrical wires of radii a, b, placed with their centres at a great distance r apart, in an infinite conducting medium, is, by formula (317),

rr

— y- (log a — 2 log r + log b)

r , r2 = 2il0S56-

352 Steady Currents in continuous Media [ch. x

III. The resistance between two spherical electrodes, radii a, b, at a great distance r apart, in an infinite conducting medium, is, by formula (317),

4>tt \a b

388. If two electrodes of any shape are placed in an infinite medium at a distance r apart, which is great compared with their linear distances, we

may take p12 in formula (317) equal, to a first approximation, to - . This is

small compared with pn and p&, so that, to a first approximation, we may replace formula (317) by

It accordingly appears that the resistance of the infinite medium may be regarded as the sum of two resistances — a resistance -~ at the crossing of

TT)

the current from the first electrode to the medium, and a resistance -f-= at

the return of the current from the medium to the second electrode. Thus we may legitimately speak of the resistance of a single junction between an electrode and the conducting medium surrounding it.

For instance, suppose a circular plate of radius a is buried deep in the earth, and acts as electrode to distribute a current through the earth. The value of pn for a disc of

radius a is „- , so that the resistance of the junction is — . So also if a disc of radius a

■7-

is placed on the earth's surface, the resistance at the junction is — , and clearly this

also is the resistance if the electrode is a semicircle of radius a buried vertically in the earth with its diameter in the surface.

Flow in a Plane Sheet of Metal.

389. When the flow takes place in a sheet of metal of uniform thickness and structure, so that the current at every point may be regarded as flowing in a plane parallel to the surface of the sheet, the whole problem becomes two-dimensional. If x, y are rectangular coordinates, the problem reduces to that of finding a solution of

da? + df

dV which shall be such that either V has a given value, or else -^— = 0, at every

point of the boundary. The methods already given in Chap, vin for obtain- ing two-dimensional solutions of Laplace's equation are therefore available for the present problem. The method of greatest value is that of Conjugate Functions.

387-390] Special Problems 353

If the conducting medium extends to infinity, or is bounded entirely by

the two electrodes, the transformations will be identical with those already

discussed for two conductors at different potentials (§ 386). If the medium

dV has also boundaries at which — = 0, the procedure must be slightly different.

We must try to transform the two electrodes into lines V= constant, and the other boundaries into lines U= constant, so that the whole of the medium becomes transformed into the interior of a rectangle in the U, V plane.

Let U + iV=f(x + iy)

be a transformation which gives the required value for V over both electrodes,

O XT

and gives ■=— = 0 over the boundary of a conductor. Then V will be the

potential at any point, the lines V = constant will be the equipotentials, and the lines U = constant, being the orthogonal trajectories of the equipotentials, will be the lines of flow.

At any point the direction of the current is normal to the equipotential through the point, and the amount of the current is given by

t on

r\ IT O TT O

But — is equal to -=- , where ^- denotes differentiation in the equipotential. on ^ os os

Thus the current flowing across any piece PQ of an equipotential

[Q = 1 Gds

■QldU, 1

i:v**-<«-<*

If P, Q are any two points in the conductor, a path from P to Q can be regarded as made up of a piece of an equipotential PN, and a piece of a line of flow JSTQ. The* flow across JSfQ is zero, that across PiY is

-(TJN-UP).

T

This is accordingly the total flow across PQ, and since UN= UQ, it may be written as

±(UQ-UP).

390. As an illustration, let us suppose that the conducting plate is a polygon, two or more edges being the electrodes. We can transform this into the real axis in the f-plane by a transformation of the type

%=(s-<hY~\s-<hy~1 (3i9),

23

354

Steady Currents in continuous Media

[CH. X

and this real axis has to be transformed into a rectangle formed (say) by the lines V=V0, V ' =Vi, £7=0, U = G in the TT-plane. The transformation for this will be

dW

ar

= [a-<K£-aP)(?-<>a:-<>r*

•(320),

where a0, ap and aq, ar are the points on the real axis of £ which determine the ends of the electrodes. By elimination of £ from the integrals of equa- tions (319) and (320) we obtain the transformation required.

391. The  following  example  of  this  method  is  taken  from  a  paper  by 
     

H. F. Moulton (Proc. Lond. Math. Soc. in. p. 104).

» Q

2 -plane. Fig. 101.

B

a

TT-plano. Fig. 102.

In fig. 101, let A BCD be a rectangular plate, the piece PQ of one or more sides being one electrode, and the piece RS of one or more other sides being the other electrode. Let the rectangle PQRS in fig. 102 be its transforma- tion in the Tf-plane. In the intermediate £-plane, let the points A, B, G, B transform to £= a, b, c, d respectively, and let the points P, Q, R, S transform to £=£>, q, r, s respectively. Then the transformations are

dz

;«[(?-a)(£-&)(r-c)(r-d)]-*,

If we write

dt;

(b — c) (a — d)

K,

(q-r)(p-s)

(a -c)(b- d) 2m = V(a - c) (6 - d), the integrals are

= \

(p-r)(q-s)

2m' = ^(p-r)(q-s),

y _a (b — d) - b(a — d)sni mz (mod k)

' h-d—(n- d\ sn2 «j.* frnnrl „\ (321),

b — d — (a — d) sn2 mz (mod k)

t>_P (q — s) — q(p — s)sn2m'W (mod X) q — s — ( p — s) sn2 m' W (mod X)

.(322).

The sides AB, AD of the first rectangle are the periods — . of

mm

390-392] Special Problems 355

sn mz (mod k) ; the sides PQ, PS of the second rectangle are the periods in

T ' T '

W, say — , , — 7 , of sn m'W (mod X).

jj

In the TT-plane, the potential difference of the two electrodes is PS, or — , ,

1 L'

while the current is - PQ, or —j- . The equivalent resistance of the plate

t niT r

is accordingly tL'/L, so that the quantity we are trying to determine is L'JL.

Let the coordinates of P, Q, R, S in the 2-plane be zx, z2, z3> z4. In the £-plane the coordinates of these points are p, q, r, s. Hence from equations (321), we have

_ a (b — d) — b (a — d) sn2 mz1 (mod k) ^ (b — d) — {a — d) sn2 mzx (mod k) '

and similar equations for q, r, s. The ratio L'jL of which we are in search is now given by

L' (q —r)(p — s) (sn2 mz2 — sn2 mz3) (sn2 mzl — sn2 W24) L ( p — r) (q — s) (sn2 mz1 — sn2 mz3) (sn2 mz2 — sn2 m^4) '

the whole being to modulus k. The values of sn mz can be obtained from Legendre's Tables.

Moulton has calculated the resistance of a square sheet with electrodes, each of length equal to one-fifth of a side, in the following four cases :

(1) Electrodes at middle of two opposite sides, Resistance = 1/745.R,

(2) Electrodes at ends of two opposite sides and facing one another,

Resistance = 2-408.K,

(3) Electrodes at ends of two opposite sides and not facing one

another, Resistance = 2'589.R,

(4) Electrodes bent equally round two opposite corners of square,

Resistance = 3027 R,

where R is the resistance of the square when the whole of two opposite sides form the electrodes. A comparison of the results in cases (2) and (3) shews how large a part of the resistance is due to the crowding in of the lines of force near the electrode, and how small a part arises from the uncrowded part of the path.

Limits to the Resistance of a Conductor.

392. The result obtained in § 386 enables us to assign an upper and a lower limit to the resistance of a conductor, when this resistance cannot be calculated accurately. For if any parts of the conductor are made into perfect conductors, the resistance of the whole will be lessened, and it may be possible to change parts of the conductor into perfect conductors in such

23—2

356

Steady Currents in continuous Media [ch. x

a way that the resistance of the new conductor can be calculated. This resistance will then be a lower limit to the resistance of the original con- ductor.

As an illustration, we may examine the case of a straight wire of variable

cross-section S. Let us imagine that at small distances along its length we

take cross-sections of infinitely small thickness, and make these into perfect

conductors. The resistance between two such sections at distance ds apart,

t ds will be -77- , where 8 is the cross-section of either. Thus a lower limit to

the resistance is supplied by the formula

'ds

[ds IS'

393. Again, if we replace parts of the conductor by insulators, so causing the current to flow in given channels, the resistance of the whole is increased, and in this way we may be able to assign an upper limit to the resistance of a conductor.

394. As an instance of a conductor to the resistance of which both upper and lower limits can be assigned, let us consider the case of a cylindrical conductor AB terminating in an infinite

conductor G of the same material. This example is

of practical importance in connection with mercury

resistance standards. The appropriate analysis was

first given by Lord Rayleigh, discussing a parallel

problem in the theory of sound.

Let I be the length and a the radius of the tube.

To obtain a lower limit to the resistance, we imagine

a perfectly conducting plane inserted at B. The resistance then consists of

the resistance to this new electrode at B, plus the resistance from this with

It the infinite conductor G. The former resistance is , the latter, bv 5 388

TT/12 'JO '

TTCL*

nr

is t- . so that a lower limit to the whole resistance is 4a

It t

ira'

4a'

ira

which is the resistance of a length I + — - of the tube.

To obtain an upper limit to the resistance, we imagine non-conducting tubes placed inside the main tube AB, so that the current is constrained to flow in a uniform stream parallel to the axis of the main tube until the end B is reached. After this the current flows through the semi-infinite conductor G as directed by Ohm's Law.

392-394] Special Problems 357

The resistance of the tube AB is, as before, — -. To obtain the resist-

7TCL-

ance of the conductor G, we must examine the corresponding electrostatic problem. If / is the total current, the flow of current per unit area over the circular mouth at B is IJTra2. In order that the potentials in the electrostatic problem may be the same, we must have a uniform surface density of electricity

r \ / I \ rl

on the surface of the disc.

The heat generated is I2R, where R is the resistance of the conductor C. It is also

mm+Q+m^ <->■

taken through the conductor G. Now if W is the electrostatic energy of

rl

a disc of radius a, having a uniform surface density a = , „ „ on each side,

we have

where the integral is taken through all space, or again,

where the integral is taken through the semi-infinite space on one side of the disc, i.e. through the space G, if the disc is made to coincide with the mouth B. On substituting for the volume integral in expression (323), we

find that

4ttT7 PR = HLlL (324).

Following Maxwell, we shall find it convenient to calculate W directly from the potential. If a disc of radius r has a uniform surface density a on each side, the potential at a point P on its edge will be

where the integral is taken over one side of the disc, and r is the distance from P to the element dxdy. Taking polar coordinates, with P as origin, the equation of the circle will be r = 2b cos 0 ; we may replace dxdy by rdrdd, and obtain

rr=25cos 9 re=-

Vp = 2a\ \ \drdd = 8b<r.

358 Steady Currents in continuous Media [ch. x

On increasing the radius of the disc to b + db, we bring up a charge kirbadb from infinity to potential 8bcr, so that the work done is

dW = SMWdb,

and integrating from 6 = 0 to b = a, we find for the potential energy of the complete disc of radius a,

Thus, from equation (324),

4ttTT 1287r2a3<7J

R =

Ft 3Pr

rl

or, smce a- =

R =

47r2a2' 8t

3-7r2a Thus an upper limit to the whole resistance is

It 8

T

Tra2 37r2a'

g

which is the resistance of a length I + 5— a of the tube.

Thus we may say that the resistance of the whole is that of a length

1 + act of the tube, where a is intermediate between T and ^— , i.e. between

4 Sir

•785 and -849. Lord Rayleigh*, by more elaborate analysis, has shewn that

the upper limit for a must be less than '8242, and believes that the true

value of a must be pretty close to "82.

The passage of Electricity through Dielectrics.

395. Since even the best insulators are not wholly devoid of conducting power, it is of importance to consider the flow of electricity in dielectrics.

Using the previous notation, we shall denote the potential at any point in the dielectric by V, the specific resistance by t, and the inductive capacity by K. We shall consider steady flow first.

If the flow is to be steady, the equation of continuity, namely

1 (I d-Z^ + 1 (I d-L\ + 1 (I ?L\ = 0 (3^5)

dx \t dx ) dy\r oy ) dz\r dz ) ^ "

must be satisfied. Also if there is a volume density of electrification p, the potential must satisfy equation (62), namely

!(£K(©-£($"» <32e>-

• Theory of Sound, Vol. 11. Appendix A.

394-396] Passage of Electricity through Dielectrics 359

From a comparison of equations (325) and (326), it is clear that steady- flow will not generally be consistent with having p = 0. Hence if currents are started flowing through an uncharged dielectric, the dielectric will acquire volume charges before the currents become steady. When the currents have become steady, the value of V will be determined by equation (325) and the boundary conditions, and the value of p is then given by equation (326).

From equations (325) and (326), we obtain

p = -^TT\d-Xd-x (Kt) +*ydy {Kt) + -dldz {Kr)\ -(327)-

The condition that p shall vanish, whatever the value of V, is that Kr shall be constant throughout the dielectric : if this condition is satisfied the value of p necessarily vanishes at every point for all systems of steady currents. The most important case of this condition being satisfied occurs when the dielectric is homogeneous throughout. If Kr is not constant throughout the dielectric, equation (327) shews that we can have p = 0 at every point provided the surfaces F=cons. and Kr = cons, cut one another at right angles at every point, i.e. provided Kr is constant along every line of flow.

We have already had an illustration (§ 381) of the accumulation of charge which occurs when the value of Kr varies in passing along a line of flow

Time of Relaxation in a Homogeneous Dielectric.

396. Let a homogeneous dielectric be charged so that the volume density at any point is p.

If any closed surface is taken inside the dielectric, the total charge inside this surface must be

\ \pdxdydz, while the rate at which electricity flows into the surface will, as in § 375, be

1 1 (lu + mv + nw) dS,

where u, v, w are the components of current and I, m, n are the direction cosines of the normal drawn into the surface. Since this rate of flow into the surface must be equal to the rate at which the charge inside the surface increases, we must have

ll(lu+ mv + nw) dS = -r \ \pdxdydz

=SSiddtdxdydz-

360 Steady Currents in continuous Media [ch. x

The integral on the left may, by Green's Theorem, be transformed into

-///(fs+S+lf)***'

and this again is equal, by equations (310), to

■d2V d2V d2V\ . . ,

m

Thus we have

ff[[i fd2V dv dv\ dp) , , , A

and since this is true whatever surface is taken, each integrand must vanish separately, and we must have, at every point of the dielectric,

d2V d2V d2V = dp dx2 + By2 + dz2 ~T dt'

We have also, as in equation (326),

d2V d2V d2V= 4tt/j dx2 + dy2 + dz2~ K '

so that Tt=-KTp-

dp _ 47T

The integral of this equation is

"IT*

where p0 is the value of p at time £ = 0.

Thus the charge at every point in the dielectric falls off exponentially

4>7T Kt

with the time, the modulus of decay being -^^ . The time -r— , in which

At 47r

all the charges in the dielectric are reduced to 1/e times their original

value, is called the "time of relaxation," being analogous to the corresponding

quantity in the Dynamical Theory of Gases*.

The relaxation-time admits of experimental determination, and as t is easily determined, this gives us a means of determining K experimentally for conductors. In the case of good conductors, the relaxation-time is too small to be observed with any accuracy, but the method has been employed by Cohn and Arons-f- to determine the inductive capacity of water. The value obtained, A~=73-6, is in good agreement with the values obtained in other ways (cf. § 84).

• Cf. Maxwell, Collected Works, n. p. 681, or Jeans, Dynamical Theory of Gases, p. 294. t Wied. Ann. xxviii. p. 454.

396, 397] Passage of Electricity through Dielectrics 361

Discharge of a Condenser.

m

397. Let us suppose that a condenser is charged up to a certain potential, and that a certain amount of leakage takes place through the dielectric between the two plates. Then, as we have just seen, the dielectric will, except in very special cases, become charged with electricity.

Now suppose that the two plates are connected by a wire, so that, in ordinary language, the condenser is discharged. Conduction through the wire is a very much quicker process than conduction through the dielectric, so that we may suppose that the plates of the condenser are reduced to the same potential before the charges imprisoned in the dielectric have begun to move. For simplicity, let us suppose that the plates of the condenser are both reduced to potential zero. Then the surface of the dielectric may, with fair accuracy, be regarded as an equipotential surface, the potential being zero all over it. It follows that there can be no lines of force outside this equipotential : all lines of force which originate on the charges im- prisoned in the dielectric, and which do not terminate on similar charges, must terminate on the surface of the dielectric. Thus we shall have a system of charges on the surface of the dielectric, these charges being equal in magnitude but opposite in sign to those of the Green's "equivalent stratum " corresponding to the system of charges imprisoned in the dielectric. This system of charges on the surface of the dielectric is of the kind which Faraday would call a " bound " charge (cf. § 141).

Suppose the plates of the condenser to be again insulated. The system of charges inside the dielectric and at its surface is not an equilibrium dis- tribution, so that currents will be set up in the dielectric, and a general rearrangement of electricity will take pkce. The potentials throughout the dielectric will change, and in particular the potentials of the condenser-plates at the surface of the dielectric will change. In other words, the charge on these plates is no longer a " bound " charge, but becomes, at least partially, a "free" charge. On joining the two plates by a wire, a new discharge will take place.

This is Maxwell's explanation of the phenomenon of "residual discharge." It is found that, some time after a condenser has been discharged and insulated, a second and smaller discharge can be obtained on joining the plates, after this a third, and so on, almost indefinitely. It should be noticed that, on the explanation which has been given, no residual discharge ought to take place if the dielectric is perfectly homogeneous. It thus becomes possible to test the theory by experiments on homogeneous dielectrics.

Rowland and Nichols* tested calcspar, which is a perfectly homogeneous crystal, and found no trace of residual discharge. Hertz -f- found traces of a

• Phil. Mag. [5] vol. n. p. 414 (1881). t Wied. Ann. xx. (18S3), p. 279.

362 Steady Currents in continuous Media [ch. x

residual discharge in a homogeneous fluid, benzene, but found that these dis- appeared as impurities were removed from the fluid; Arons* obtained the same result with paraffin. Finally Muraokaf experimented with various oils, paraffin, resin, turpentine and xylol. Residual discharges were not found in the oils singly, but appeared as soon as two or more were mixed together. These facts are in agreement with Maxwell's theory of residual discharge and afford strong confirmation of the theory. On the other hand there are a large number of experimental facts which are difficult to explain in terms of Maxwell's theory alone, and which seem to suggest that the theory is incomplete.

EXAMPLES.

1. The ends of a rectangular conducting lamina of breadth c, length a, and uniform thickness r, are maintained at different potentials. If f(x, y) be the specific resistance p at a point whose distances from an end and a side are x, y, prove that the resistance of

the lamina cannot be less than

■ , or greater than ■

T

1° <fy'

Jo p

dy

/;

a p dx 0J0r

2. Two large vessels filled with mercury are connected by a capillary tube of uniform bore. Find superior and inferior limits to the conductivity.

3. A cylindrical cable consists of a conducting core of copper surrounded by a thin insulating sheath of material of given specific resistance. Shew that if the sectional areas of the core and sheath are given, the resistance to lateral leakage is greatest when the surfaces of the two materials are coaxal right circular cylinders.