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Dielectric Displacement Currents

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DIELECTRIC DISPLACEMENT CURRENTS In electrostatics, conductors are regarded as substances in which electricity can move more or less freely, and insulators as substances through which electricity cannot pass. An electric field produces a current in a conductor, but not in an insulator. If a condenser, consisting of two parallel metal plates with air or a vacuum between them, is connected to a battery, a current flows through the battery and wires leading to the condenser for a frac tion of a second while the condenser is being charged. If the space between the plates is filled with a substance of specific inductive capacityK, the current is increased K times. The charging current flows into one plate and out of the other, just as though there were a flow of electricity right through the insulator between the plates. Such phenomena led Faraday to suggest that there is something analogous to a current in the insulator, or dielectric, between the plates during the charging process. He supposed that the electric field produces a separation of the positive and negative elec tricities in the atoms of the dielectric, so that the positive elec tricity is slightly displaced towards the negatively charged plate, and the negative electricity towards the positively charged plate. The separation is proportional to the field strength, so that, while the field is changing, there is a current. Maxwell adopted Faraday's ideas and pointed out that such currents in the dielectric should produce a magnetic field, just as currents in a conductor do. He supposed the dielectric current to be equal to the current in the wires leading from the battery to the con denser. According to this assumption, there is a dielectric cur rent even when the insulator between the plates is a vacuum.

The Deduction of the Electromagnetic Equations.

If the electric intensity between the plates is F, and the charge per square centimetre on the positive plates is S, in electrostatic units, then we have F = 47rS/K, so that the dielectric current density is equal to K aF , since, according to the assumption 47r at just mentioned, it is equal to ds/dt. If the substance between the plates is not a perfect insulator, there will be a conduction current in addition to the dielectric current. If v denotes the conductivity of the substance, the conduction current density is rF, so that the total current is QF+ K where we may suppose K, F 47r at and a to be all expressed in electrostatic units, so that K= i for a vacuum, and the unit of conductivity is that of a conductor one centimetre long and one square centimetre cross section through which a potential difference of one electrostatic unit produces a current of one electrostatic unit of charge per second. The work required to take a magnetic pole of strength m once round a current of strength i, in electromagnetic units of current, is equal to 47rmi ergs. Hence, if we imagine a small plane closed curve, drawn anywhere in a medium in which there are currents flowing, the work required to take a unit pole once round the closed curve will be equal to 47r times the total current flowing through the curve. This result may be expressed analytically as follows:—Take axes x, y, z, (fig. 24) and consider a small rectangular area ABCD with its sides parallel to the y and z axes. Let BA=CD=by and DA=CB= az. Let the components of the magnetic field in the direction of the axes be noted by and the components of the current density by The current through the rectangle ABCD is equal to so that the work required to take a unit pole round the rectangle is The work to take a unit pole from B to A along BA, and D to C along DC, will be equal to by multiplied by the dif ference between at CD and at AB, or to —4 317 bz. In the az same way, the work to take a unit pole along AD and along CD is equal to bz 19 so that the work to take the unit pole round aHZ — the rectangle along ADCBA is equal to l aybz.

y aHZ Hence we have = ay az By considering the currents through small rectangles parallel to the xz and xy planes, we get in the same way Introducing this notation we see that the six differential tions we have obtained are equivalent to the vector equations aF usually very large compared with the dielectric current, K — 47r at which can then be neglected, and in good insulators the con duction current is usually negligible. These equations express a relation between the electric and magnetic fields in a medium of conductivity ar and specific inductive capacity K. They in volve the assumption that dielectric currents produce the same magnetic field as ordinary conduction currents, and also that the theory of the magnetic field due to currents in thin wire circuits is applicable to currents in a medium in three dimensions. The latter assumption is easily justifiable, since it is clear that any distribution of currents in a medium can be regarded as made up of currents flowing along thin filaments coinciding in direction with the lines of flow.

Another similar relation between the electric and magnetic fields in a medium can be obtained from the result that the in duced electromotive force round any circuit is equal to the rate of diminution of the number of unit tubes of magnetic force passing through the circuit. We assume that this result is true for any small plane curve drawn in the medium, and that the electromotive force is equal to the work required to take a unit charge of electricity round the curve against the forces exerted on it by the electrostatic field. The work to take a unit charge round the rectangle ABCD (fig. 24) along ADCBA is (i — az ay az BySz, as in the case of the unit pole; and the number of unit tubes passing through the rectangle is since along a unit tube µHa = where is the magnetic permeability of the medium and a the cross section of the tube, so that a= I/ µH. Hence we have If V denotes any vector, and V its components along the directions of the axes x, y, z, then we may derive another vector, denoted by curl V from it. the components of which are defined by the equations:— Thus, the component of the curl of a vector along the normal to a small plane area is equal to the line integral of the vector round the boundary of the area divided by the area. The equations (7) and (8) therefore are merely the mathematical expression of the results that the work to take a unit pole once round a closed curve is equal to 47r times the current through the curve, and that the work to take a unit charge once round a closed curve is equal to the rate of diminution of the number of unit tubes of magnetic force passing through the curve.

The equations (I) to (6), or (7) and (8), were first obtained by Clerk Maxwell, and they form the basis of his electromagnetic theory of light and electric waves. They are of fundamental im portance for modern electrical theory. In the case of a perfect in sulator, for which or = o, we have from (I) K = — at av az If p= o, and K has the same value everywhere, this gives x v 32 Fx a2 Fx 32 F„ a2 F Z aF + aF + F= = o, so µK 2 = 2 2 .,2 ax ay az at In the case of an electromagnetic field in which F and H constant over all planes perpendicular to the x axis, this equation reduces to µK = 2 Fx • A solution of this equation isate = f (x — vt), where f (x — vt) denotes any function of x — vt and v= I/ V (µK) . According to this, such a field travels along x with the velocity I/-V (µK). For, if x is increased from to and t also increased from to and if — — then is equal to so that f — is equal to f (x2— • The equation v= 1/ (µK) gives the velocity of electric waves in a medium of magnetic permeability and specific inductive capacity K. µ and K must, of course, both be expressed in units of the same system either the electrostatic or the electro magnetic. The force in dynes between two charges e and e' in electrostatic units is equal to where r is the distance between the charges in centimetres. For a vacuum, K= in electrostatic units of specific inductive capacity. If one electro magnetic unit of charge is equal to c electrostatic units, then, if E and E' denote the values of e and e' in electromagnetic units, we have E = e/c and E' =e' / c. The force between the charges is therefore equal to and this must be equal to where K' is the specific inductive capacity of the medium in electromagnetic units. Hence K' = so that for a vacuum K' = For a vacuum µ = I in electromagnetic units. so that the velocity, for a vacuum, is equal to r/ 102K) = i/ I(c = c cm. per sec.

Thus it appears that the velocity of electric waves in a vacuum in centimetres per second should be equal to the number of electrostatic units of electricity in one electromagnetic unit.

Ratio of Electrostatic to Electromagnetic Units.—The number c can be found by making a measurement of some elec trical quantity in electrostatic units and also in electromagnetic units. For example, the capacity of a condenser may be calculated in electrostatic units from its dimensions and then determined in electromagnetic units, by comparison with standards of re sistance of known value in electromagnetic units. If a condenser of capacity C in electromagnetic units is charged with charges +E and —E electromagnetic units, then its electrical energy is 2 ergs. If the charges and capacity when expressed in electrostatic units are denoted by E' and C', then 2E"/C. But E = E'/c so that = which gives C'/C = In this and other ways it has been shown that c is nearly equal to 3 X It follows that the velocity of electromagnetic waves in a vacuum should be equal to 3 X I cm. per sec. The velocity of light in vacuo is equal to 3 X cm. per sec., so that, following Clerk Maxwell, we may conclude that light consists of electro magnetic waves. It will be noted that the argument consists in throwing Faraday's ideas into mathematical form, which shows that an electromagnetic disturbance will be propagated as a wave; then finding from the equations the velocity of such a wave, and showing that it equals the velocity of light, as ordi narily determined.

equal, current, units, unit, electromagnetic, electrostatic and magnetic