of positive electricity, so that its potential is I. Now the energy of the charged condenser is QV2 and the total number of tubes of force in the field is Q, so that the quotient of the energy of the charged condenser by the number of tubes of force is V/2. Next suppose that the distance between the plates is decreased. The result will be that the capacity of the condenser will be increased, and so, if the charge Q on the plate A remains the same, its potential will decrease. Let us, however, increase the potential of A, that of B being still kept zero, till the potential of A has its previous value V, and let the new charge on A be Q. The energy of the condenser is now QV2, while Q tubes of force occupy the field. Hence the quotient of the energy by the number of tubes is V/2, that is, has the same value as before. If we suppose that each tube of force contributes an equal amount to the energy of the field, then the contribution by each tube is the same in the two cases, namely 2. The question, however, arises: Are we justified in supposing that each tube contributes an equal amount to the energy of the field? for some of the tubes will be short and stretch almost straight from one plate to the other, while others may be quite long and sweep round in a great curve from one plate to the other. Now we have seen above that as long as the difference of potential of the plates, that is, the difference of potential between the two ends of the tubes of force, is kept the same, the quotient obtained by dividing the energy by the number of tubes is the same whatever the relative positions of the plates. For instance we get the same result whether the plates are placed near together and parallel to one another, so that almost all the tubes stretch straight from one plate to the other, or the plates are turned so that one is at right angles to the other, and hence a large proportion of the tubes have to curve round from one plate to the other. Considerations such as these lead us to consider that each tube of force in an electrical field contributes an equal amount to the energy of the field.

We have next to see how the energy stored up in a tube of force is distributed along its length. Consider a single tube; this will start from a small area of the plate A, on which there will be a unit of positive electricity. Now if it were possible to move this portion of the surface of A along the tube of force to the plate B, this tube of force would be annihilated. In the first place let us suppose that the removal of this portion of the charge of A does not affect the potential of the plate. Under these circumstances the work done in carrying the unit from one end of the tube to the other would be V. Thus the destruction of the tube of force has been accompanied by the performance of V units of work, while the energy contained within the tube we have seen is only half this quantity. The reason for this difference is that the supposition we have made as to the potential of A remaining the same after the removal of the unit is erroneous. As the portion of the plate A carrying the unit charge is moved away from the plate A the potential will gradu

ally fall. The result of this fall of the potential of A is that the quantity of energy stored up in each of the tubes which are left stretching from A to B is reduced. Now it can be shown that the loss on this account is exactly equal to the loss on account of the destruction of the tube, and, further, that the rates at which the losses occur as the tube is gradually destroyed by the motion of the portion of the plate A carrying the end is the same in the two cases. Hence the work done during the movement of the unit charge from a point P of the tube to a neighbouring point is equal to the sum of the energy contained within the portion of the tube included between P and Q, and the loss of energy of the condenser owing to the decrease of the difference of potential between its plates. Since these two losses of energy are equal, the energy contained in the part of the tube between P and Q is half the work which is done when the unit is carried from P to Q. Now the electrical force, F, varies along the tube, but if we consider the two points P and Q sufficiently near together, we may consider that F remains constant over this distance. Hence the work done in carrying the unit from P to Q will be F.PQ, for the force F acts alone the tube, that is, along the direction of the path PQ. Thus the energy included in the tube of force between P and Q is F.PQ 2. Hence the energy stored in unit length of the tube is F2, so that the energy stored up in unit length of a tube of force is numerically equal to half the electrical force at the part of the tube considered.

The cross-section of a tube of force at a point where the force is F being 47F, the volume of unit length of the tube in this part of the field is 4 F. But the energy stored up in unit length of the tube is F2. Hence 4/Fc.c. of the field contains F2 units of energy, i.e. ergs. Therefore the energy contained in unit volume, that is, one c.c., of the field, at a part where the electrical force is F, is F2/8′′.

If the difference of potential between the ends of a tube of force is F', and we draw the equipotential surfaces so that the difference of potential between consecutive surfaces is one unit, these surfaces will divide each tube into small portions or cells. The whole energy stored up in the tube being V/2, the energy stored in each cell will be half an erg.

As we shall see later, the method of looking upon the energy possessed by a charged conductor as stored up in the field leads to many important generalisations, and the calculations we have made above will help to give the reader some mental grasp of what is implied when a certain region is said to be a field of electrical force, and to appreciate how completely this field is mapped out by the tubes of force.

458. Strength of the Field near a Charged Conductor.—If the density of the charge at a given point of the surface of a charged conductor is σ, then a tubes of force will start from the unit of area of the surface of the conductor, and since the tubes of force intersect a conducting surface at right angles, the cross-section of a tube in the immediate neighbourhood of the surface will be 1o. If the surface density is not

uniform, the same will still hold good in virtue of the manner in which we have defined the measure of the surface density when it is variable, for if dq is the charge on a small element of surface, da, surrounding the given point, the surface density is dq/da; but since dq tubes of force leave the small area da, the cross-section of a single tube is da/dq. Now we have seen in § 455 that the strength of a field is equal to 4π/s, where s is the cross-section of the tube of force of the field at the given point. Hence the strength of the field in the immediate neighbourhood of a charged conductor, at a point where the surface density of the charge is σ, is 4′′σ. We have in the above argument assumed that the tubes of force leave the surface of the conductor at the point considered on one side only, as would be the case if the portion of the surface considered forms part of the outside of a closed surface, for under these conditions there is no force within the surface, and so all the tubes of force must leave the portion of the conductor on the one side. In the case of an unclosed conductor, such as a plane, there are two ways of regarding the problem. If, as is usual, we take o' as the charge on both sides of unit area of the plane, then the lines of force will start out equally from each side of the plane, so that the number of tubes of force leaving each square centimetre on either side will be σ'/2, and the cross-section of a tube of force in the immediate neighbourhood of the surface will be 2/o'. Hence in this case the strength of the field will be given by F=2πσ'.

B

459. Mechanical Force Exerted on each Unit of Area of a Charged Surface.--Let ABC (Fig. 447) represent a section of a closed conductor which is charged so that the surface density over a small area of the surface at AC is σ. Let us consider the electrical force at two points, P, and P, one just outside and the other just inside the surface of the conductor. We may consider that the force at these two points is made up of two parts, namely, the force due to the portion AC of the charged surface and that due to the part CBA. Let the force at the point P1, due to the portion CBA, be F1, then, since the two points P1 and P2 are by supposition very near together, the force due to the part CBA of the conductor, which is at a comparatively great distance from both points, will be the same for both, namely F. Also, since the portion AC of the surface is small, it is practically plain, and so the force exerted at the two points P1 and P2, which are similarly situated, will be equal in magnitude but opposite in direction. Thus if the force due to AC at P1 is F2 acting along the outward drawn normal to the surface, the force at P2 will be - F. Hence, adding together the two component forces for each point, we get that the force at P1 is F1+F, while the force at P is F1-F. But the point P, being within

2

FIG. 447.

a closed conductor, the force there is zero, while we have just seen the the force at a point, such as P1, just outside the surface of a conductor charged to a surface density σ, is 4′′σ. We therefore get that

Now F is the force at P1 or P, due to the portion CBA of the charged conductor, and therefore if we imagine that the portion AC of the surface were disconnected from the rest, without in any way altering the distribation of the charge, the electrical force acting at the point where this portion of the surface is situated is F1 or 2′′σ. Now if a is the area of the portion AC, the charge on this portion is σa, and so the mechanical force in the direction of the normal, which this portion of the conductor would experience owing to the action of the rest of the conductor, is σa. F1 or

2πσα.

Although for clearness we have supposed the portion considered to be separated from the rest of the surface, the same force is exerted, although this division does not occur. We thus see that in the case of a charged conductor, on every small element of area, a, there is exerted an outward force 202, where σ is the density of the charge on the element in question. This force acts everywhere normally to the surface, for the force at the point P, is normal to the surface, since the lines of force, as has been shown, everywhere cut a conducting surface at right angles. The outward normal force exerted on the unit of area is 2σ2. This force is of the nature of an hydrostatic pressure, and if the conductor is expansible, as for instance is the case with a soap-bubble, it will cause the conductor to expand when it is electrified.

The electrical force, F, at P1, that is, the force due to the whole charged conductor, being 4σ, we get that the mechanical force experienced by unit area of the surface of the conductor is Fo/2 or F2/8π.

460. Tension along the Tubes of Force.--We have seen in the last section that each unit of area of a charged conductor experiences an outward mechanical force, due to the charge, which amounts per unit area to Fo/2, where F is the electrical force just outside the portion of the charged surface considered, and σ is the density of the charge on this portion of the surface. Now the number of tubes of force which

start from unit area of the surface is σ, and hence, if we suppose that each of these tubes exerts a tension on the surface of the conductor equal to F/2, the total tension exerted on unit area will be Fo/2, that is, is of the actual amount which occurs. Hence we are able to account for the mechanical forces which act on bodies when placed in an electric field, if we suppose that each of the tubes of force is in a state of tension, the magnitude of the tension being at each point equal to half the electrical force at that point.

1

If we take a unit of area on an equipotential plane, passing through a point P, that is, at right angles to the tubes of force, the number of tubes which cross this unit of area will be F 4, where F is the force at P. Hence, as the tension along each tube of force is F 2, the tension in the air across the unit of area is F 8.

As we have already mentioned, this tension is not alone sufficient to account for the distribution of the tubes of force, but it can be shown that if in addition we imagine that there exists a pressure at right angles to the lines of force, of which the magnitude is F8 per unit area, then this distribution can be accounted for.

461. Dielectrics other than Air.-We have hitherto confined our discussion of the state of the electric field to the case where the only dielectric present was air, and we have now to proceed to consider what alterations will have to be made in the expressions we have deduced, when the whole or part of the field is occupied by other dielectrics.

Suppose that we have two infinite planes, A and B, placed parallel to one another at a distance d apart, and that the plane A is given a positive charge, such that the surface density is o. Let the plane Å be kept at zero potential, and the potential of A be la when the dielectric separating the planes is air, and when the dielectric separating the planes has a specific inductive capacity K.

Since the planes are infinite, the field of force between the planes must be uniform, so that the tubes of force are all parallel, stretching straight across from one plane to the other, and have everywhere the same cross-section. Since the density of the charge on the plane A is σ, the number of tubes of force which leave unit area of the surface is σ, and hence the cross-section of each tube is 10. The cross-section of the tubes will be the same whatever the dielectric, for we suppose that the density of the charge on the plane A is kept the same in all cases.

Now the capacity of unit area of the plane A will bear the same ratio to the capacity of the whole plane as does unit area to the total area of the plane, so that we may, if we like, confine our attention to unit area taken on each of the planes. The capacity being the ratio of the charge on one plate of a condenser to the difference of potential between the plates, the capacity Ca of unit area of the planes, when the dielectric is air, is given by

Ca=σ Va,

for the charge on unit area of either plane is σ, and the difference of potential between the planes is Va.

In the same way the capacity of unit area, when the dielectric is not air, is given by

Ck=σ Vk.

Now the specific inductive capacity of the dielectric is defined as the ratio of the capacity of a condenser having the given dielectric separating