Appendices - 1. The electric field in vacuum

The contribution to the flux of the vector from the lateral surface is equal to zero, since on the lateral surface .

Therefore, the total flux through the entire closed cylindrical surface is equal to

Inside the cylindrical surface under consideration there is a charge

where — is the charge density on the plane. By Gauss's theorem

,

consequently, the magnitude of the field strength of a charged plane is equal to

Let us emphasize that the result obviously does not depend on the distance from the plane at which the bases of the considered cylinder are located. It follows that on each side of the plane, the electric field it creates is uniform.

Using the previously introduced axis perpendicular to the charged plane, the field on both sides of the plane can be described by a single formula, valid for any sign of the charge on the plane

Here — is the unit vector of the axis .

Integrating, taking into account

,

for the dependence on of the potential of the plane's field it is easy to obtain:

The potential is normalized by the condition . Here, as in the example with an infinitely long charged cylindrical surface, the potential grows without bound as the distance goes to infinity, so normalization to zero at infinity is meaningless.

The field lines of a charged plane are shown in Fig. 1.52 and 1.53.

Fig. 1.52. Field of a positively charged plane

Fig. 1.53. Field of a negatively charged plane

Field of a parallel-plate capacitor

Let us determine the field strength created by two infinite parallel planes, charged uniformly and with opposite signs. The charge densities on the planes are equal in absolute value and are, respectively: and (an ideal parallel-plate capacitor). With the help of Fig. 1.54 it is easy to see that in the gap between the planes, the fields created by them are directed the same way, so inside, the total field is twice the field from each of the planes. Outside the planes, the fields created by them are directed in opposite directions, so the total field from both planes is equal to zero (Fig. 1.55).

Fig. 1.54. Electric field of a parallel-plate capacitor

Fig. 1.55. Electric field of oppositely charged planes

Fig. 1.56. Electric field strength of oppositely charged planes

In Appendix 6, an example is worked out involving the motion of a charged particle in a constant electric field.

Potential of the field of a charged disk

As has already been noted more than once, knowing the potential of the field of a point charge and using the superposition principle, it is in principle always possible to calculate the potential of the field created by any distribution of charges.

Let us find, as an example, the potential of the electric field created on the axis of a thin disk of radius R, uniformly charged with a surface charge density (Fig. 1.57). By virtue of axial symmetry, at points on the axis the two components of the field strength perpendicular to the axis are equal to zero: , it remains to find — the component of the field directed along the axis.

Fig. 1.57. Calculation of the potential on the axis of a charged disk

Let us select on the disk a ring of radius s and width ds (shown hatched in Fig. 1.57). The area of the ring is equal to and therefore the charge concentrated on it is

Since all elements of the ring are at the same distance

from the observation point A, the potential , created by the ring at point A, is given by the same formula with replaced by :

The total potential of the field, created by the entire disk at point A, is equal to the sum of the potentials from all possible rings with radii s, where 0 < s < R

At large distances from the center of the disk the square root can be expanded in a series, limiting to the first two terms of the expansion

then the formula simplifies and, as it should, turns into the formula for the potential of a point charge

where — is the total charge of the disk

Using the relation between the field strength and the potential , one can find the field strength on the axis of the disk

Coulomb's law and the dimensionality of space

The space in which we live has three dimensions. In other words, three coordinates are needed (for example, in Cartesian or in spherical systems) to specify the position of a point A (Fig. 1.58). It turns out that the number 3 is closely related to the form of Coulomb's law. We have seen that the Ostrogradsky—Gauss theorem follows from Coulomb's law. The converse is also true: Coulomb's law can be derived from the Ostrogradsky—Gauss theorem. But this theorem is more general in character than Coulomb's law. In particular, it applies to spaces of dimensionality , where need not necessarily be equal to three.

Fig. 1.58. Cartesian and spherical coordinate systems

Indeed, the theorem essentially asserts that field lines begin and end on charges or go off to infinity. The dimensionality of space plays no role here. So let us assume that we live in a space of some dimensionality and see what physics would look like in this strange world. Let us take a point charge and mentally surround it with a sphere of radius Before continuing our acquaintance with -dimensional physics, let us agree on terminology.

Appendix 1. Force of interaction of systems of point charges

Let us begin with the simple case where one of the two interacting systems is a single charge . The charges of the other system will be denoted , where is the number of the charge in the system. Then the force of their interaction, namely: the force acting on the charge , according to the superposition principle, can be written in the form

Here and — are the position vectors of the points at which the corresponding charges are located. The second relation is simply a statement of Newton's third law for this case. It must be emphasized that we are talking, throughout, about the force of interaction.

If two systems of charges, the first: , and the second: , interact with each other, then the force of this interaction can be written in the form of the following double sum

Here, as before, and — are the position vectors of the points at which the corresponding charges are located.

The formula, at least in principle, solves the problem of calculating the Coulomb (electrostatic) interaction forces of arbitrary systems of stationary point charges.

Appendix 2. Force of interaction of systems of continuously distributed charges

By a physically infinitesimal element is meant an element which, on the one hand, is so small that under the conditions of the given problem it can be considered a point mass, and, on the other hand, is so large that the discreteness of the charge (see relation) of this element can be neglected.

Let us make a necessary, in our view, additional clarification regarding the words «physically infinitesimal» volume. Such a volume should not be regarded as the result of a formal, purely mathematical limiting transition to zero. A physically infinitesimal volume is a volume whose size , on the one hand, is small compared with any characteristic length in the problem under consideration (recall that in different problems these are lengths of quite different magnitude) and it can be considered a point. On the other hand, it is macroscopically large, that is, its size is large compared with the average distance between the particles (atoms, molecules, ions, electrons, etc.) making up the substance, so that the number of these particles inside the physically infinitesimal volume is macroscopically large. Correspondingly, the fact that this number changes discretely can be neglected. Thus, the linear size of a physically infinitesimal volume must satisfy the following inequality

Let us also note that quantities of the form can of course be regarded as derivatives, but it is simplest, most convenient, and most productive to regard them as fractions: this is the ratio of the charge in the volume to the magnitude of this volume.

Two bodies and with volumes and are charged with densities and . In the first body, in the vicinity of a point with position vector , let us select a physically infinitesimal volume , inside which there is a charge . Quite similarly, in the second body, in the vicinity of a point with position vector , let us select a physically infinitesimal volume , inside which there is a charge . Both volumes can be considered point charges; in accordance with Coulomb's law, let us write the expression for the force of their interaction:

Note that the first relation differs from the corresponding one only in notation. In the latter expression the values of the charges and are simply substituted. Now, according to the superposition principle, it is necessary to sum over all pairs of point charges and of which the first and second bodies consist. The difference is only that infinitesimal interaction forces of pairs of infinitesimal charges are summed, and there are infinitely many of them. The sum of an infinitely large number of infinitesimal quantities is an integral. Therefore, in the expression that is the result of such a summation, instead of the sums

we must write integrals over the volumes of both bodies As a result we obtain

The reader should be reassured: we will not have to calculate sixfold integrals here, any more than the threefold integrals describing the force of interaction of a single point charge and an extended body

with a charge uniformly distributed over it with density .

Let us consider examples of the application of these formulas.

Appendix 3.

Example 7. A conducting disk of radius rotates with angular velocity Given that the current in the conductor is carried by electrons, determine the potential difference between the axis of the disk and its periphery.

Solution. Let us first obtain an estimate of the result using dimensional analysis. At our disposal are — the charge of the electron and its mass , the angular velocity and the radius of the disk .

The desired formula must have the form:

Substituting the dimensions, we obtain:

from which follow the equations (which, in this case, are also the solution to the problem):

Let us now solve the problem exactly. In order for an electron located inside the disk at a distance from the axis to move in a circle, a centripetal force must act on it:

This force is provided by a redistribution of the electron concentration in the disk, creating a radial electric field E. The equilibrium condition for the electron:

where — is the potential at the center (at the periphery) of the disk.

Here is the linear velocity of points at the periphery of the disk. In principle, the formula obtained can be used to determine the ratio of the electron's charge to its mass. In practice, however, this is extremely difficult. Let us give an estimate. The charge of the electron is , its mass . Let us take the velocity at the periphery of the disk to be equal to 300 m/s. The potential difference arising between the axis and the periphery of the disk will then turn out to be equal to . Such a voltage is very difficult to measure in a moving system.

Appendix 4.

Example 8. A spherical drop of water, carrying an electric charge has potential on its surface. What is the radius of the drop? What will be the value of the potential on the surface of the new spherical drop formed by the merging of two former ones? What is the dependence of the potential on the surface of the new drop, formed by the merging of several old ones, on their number n?

Solution. The potential on the surface of a charged sphere (or ball — in this case it gives the same result) is equal to

from which we find the radius of the drop:

When n drops of volumes merge, a new drop is formed with radius and volume increased by a factor of :

The new drop will also carry an increased charge: From this we find for the potential on its surface:

When two drops merge, we obtain for the potential

Appendix 5.

Let us consider the general derivation of Gauss's theorem. It is based on a direct calculation of the flux of the field-strength vector of a point charge through a closed surface of arbitrary shape. As before, let us place the origin of coordinates at the point in space where, for now, the single charge is located (see Fig. 1.59).

Fig. 1.59. Charge inside the surface

With the solid angle let us «cut out» on opposite sides of the surface two small areas and , with outward normals to the areas and , and field strengths and , respectively. The contributions to the flux from these two areas

The scalar products entering the expressions for the fluxes are, evidently, equal to: and . Both areas under consideration do not lie on spherical surfaces, so their areas are larger by a factor of «|cos(α)|» than those of areas cut out by the same solid angle but on spherical surfaces of the corresponding radii ( and ), so the areas of the surface elements we are considering are equal to and . In this case it is necessary to write precisely the absolute value of the cosine, since the area cannot be negative, while the angles can be obtuse and, correspondingly, the cosines — negative. Substituting the scalar products and the areas of the surface elements into the expressions for and , we obtain

In the formulas written above it is taken into account that, in the case when the charge is located inside the surface both angles ( and ) are acute, both cosines are positive, and in both cases

The expressions for and are identical, so for any surface element we can write ; integrating over the entire surface, as before, we obtain

Let us note that the equality of the fluxes and is explained quite simply: 1) the areas of the surface elements grow with increasing distance to the point where the charge is located, proportionally to the square of this distance, while the magnitude of the field strength decreases inversely proportional to the square of the same distance; 2) by virtue of the central nature of the field of a point charge, the scalar product (in the numerator) and the ratio of the areas (in the denominator) both contain the cosine (in the first case) and the absolute value of the cosine (in the second case) of one and the same angle. Thus, the equality of the fluxes and is a consequence of the properties of the field of a point charge: its centrality and the law of decrease , that is, ultimately — Coulomb's law.

Fig. 1.60. Charge outside the surface

If the charge is located outside the surface , then one of the two angles (see Fig. 1.60) is obtuse, and, for example, as in Fig. 1.59:

,

then the sum

.

In this case, integrating over the entire surface, we have

In the last expression, — is the «far» part of the surface from the charge , on which , and — is the «near» part relative to the charge , on which . Combining both results, and for a surface of arbitrary shape, we obtain:

The derivation of Gauss's theorem in general form can also be found, for example, in the textbook I.E. Tamm, «Fundamentals of the Theory of Electricity», Moscow, Nauka, 1989, p. 18.

Appendix 6.

Formula for calculating the gradients of a central and axially symmetric scalar field

Let us obtain a formula, very useful in practice, for calculating the gradient of a scalar function possessing spherical

or cylindrical