Electric field of a uniformly charged plane. Application of the Gauss theorem to calculate the field of an infinite uniformly charged plane

Earlier we showed that electric field, created by an infinite uniformly charged plate is homogeneous, that is, the field strength is the same at all points, and the strength vector is directed perpendicular to the plane, and its modulus is equal to E o = σ/(2ε o). The family of lines of force of such a field is a set of parallel lines perpendicular to the plate. On fig. 275, 276 also shows a plot of the projection of the field strength vector Ez per axle Z perpendicular to the plate (we place the origin of this axis on the plate). It is clear that the potential of this field depends only on the coordinate z, that is, the equipotential surfaces in this case are planes parallel to the charged plate.



rice. 275



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With the traditional choice of the zero potential level φ(z → ∞), the potential of an arbitrary point is equal to the work of moving a unit positive charge from a given point to infinity. Since the modulus of tension is constant, such work (and, consequently, the potential) turns out to be equal to infinity! Therefore, the indicated choice of the zero potential level is unsuitable in this case.
Therefore, one should use the arbitrariness of choosing the zero level. It is enough to choose an arbitrary point with coordinate z = z o, and assign to it an arbitrary value of the potential φ(z o) = φ o(Fig. 277).

rice. 277
Now, to calculate the value of the potential at an arbitrary point φ(z), we can use the relation between the strength and field potential

Considering that in this case the field strength is constant (at z > 0) this expression is written as

from which follows the desired dependence of the potential on the coordinate (for z > 0)

In particular, one can set an arbitrary value of the potential of the plate itself, that is, put at z = z o = 0, φ = φ o. Then the value of the potential at an arbitrary point is determined by the function

the graph of which is shown in Figure 278.

rice. 278
The fact that the potential with respect to infinity turned out to be infinitely large is quite obvious - after all, an infinite plate also has an infinitely large charge. As we have already emphasized, such a system is an idealization - there are no infinite plates. In reality, all bodies have finite dimensions, so for them the traditional choice of zero potential is possible, although in this case the field distribution can be very complicated. Within the framework of the considered idealization, it is more convenient to use the choice of the zero level used by us.

Let us find the distribution of the potential of the field created by two identical uniformly charged parallel plates, the charges of which are equal in absolute value and opposite in sign 1 (Fig. 279).

Rice. 279
Let us denote the surface charge density on one plate +σ , and on the other −σ . Distance between plates h we will consider it to be much smaller than the dimensions of the plates. We introduce a coordinate system, the axis z which is perpendicular to the plates, we place the origin of coordinates in the middle between the plates. Obviously, for infinitely large plates, all field characteristics (strength and potential) depend only on the coordinate z. To calculate the field strength at various points in space, we use the obtained expression for the field strength created by an infinite uniformly charged plate and the superposition principle.
Each uniformly charged plate creates a uniform field, the intensity modulus of which is equal to E o = σ/(2ε o), and the directions are shown in Figure 280, 281.



rice. 280

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Adding the field strengths according to the principle of superposition, we obtain that in the space between the plates the field strength E = 2E o = σ/ε o twice the field strength of one plate (here the fields of individual plates are parallel), and outside the plates there is no field (here the fields of individual plates are opposite).
Strictly speaking, for plates of finite dimensions, the field is not uniform, the field lines of the field of plates of finite dimensions are shown in Figure 282.



rice. 282
The strongest deviations from uniformity are observed near the edges of the plates (often these deviations are called edge effects). However, in the region adjacent to the middle of the plates, the field can be considered uniform with a high degree of accuracy, that is, edge effects can be neglected in this region. Note that the errors in this approximation are the smaller, the smaller the ratio of the distance between the plates to their dimensions.
To unambiguously determine the distribution of the field potential, it is necessary to choose the level of the zero potential. We will assume that the potential is equal to zero in a plane located in the middle between the plates, that is, we set φ = 0 at z = 0.
Despite the arbitrariness in choosing the zero level of the potential, our choice can be logically justified based on the symmetry of the system. Indeed, the system of charges under consideration mirrors itself upon mirror reflection with respect to the plane z = 0 and a simultaneous change in the signs of the charges. Therefore, it is desirable that the potential distribution also has the same symmetry: it is restored by mirror reflection with a simultaneous change in the sign of all field functions. The method we have chosen for choosing the zero potential satisfies this symmetry.



rice. 283
Denote the potential of a positively charged plate +φo, then the potential of the negatively charged plate will be equal to φo. These potentials are easy to determine using the found value of the field strength between the plates and the relationship between the strength and potential difference electric field. The equation of this connection in this case has the form +φo − φo = Eh. From this relation, we determine the values ​​of the potentials of the plates φ o = σh/(2ε o). Taking into account that the field is uniform between the plates (therefore, the potential changes linearly), and there is no field outside the plates (therefore, the potential is constant here), the dependence of the potential on the coordinate z has the form (Fig. 284)

rice. 284

 Tasks for independent work.
 1. In all the examples considered, perform the inverse operation: according to the found distribution of the potential using the formula E x = −∆φ/∆x calculate the strengths of the considered fields.
 2. Strictly derive formula (6).
 3. Qualitatively explain the following "paradox". In the field of a flat capacitor, the “infinity” potential is ambiguously defined: when moving in the positive direction of the axis Z the potential of "infinity" turned out to be equal −φ o; when moving in the negative direction of the axis Z − +φo, when moving along the axes X or Y− equals zero. So what is the potential of "infinity" in a real system of two plates of finite sizes?

1 Such a system is called flat capacitor, we will study these devices in more detail later.

We will assume that the charge is positive. The plane is charged with a constant surface density. It follows from symmetry that the intensity at any point of the field has a direction perpendicular to the plane (Fig. 2.10). Obviously, at points that are symmetric with respect to the plane, the field strength is the same in magnitude and opposite in direction.

Let us single out an area on the charged plane. Surround this area with a closed surface. As a closed surface, we imagine a cylindrical surface with generators perpendicular to the plane and bases of magnitude located symmetrically relative to the plane. Apply to this surface the Gauss theorem . There will be no flow through the lateral part of the surface, since it is equal to zero at each of its points. For bases, it is the same as . Therefore, the total flow through the surface will be equal to . There is a charge inside the surface. According to the Gauss theorem, the following condition must be satisfied: , where . (3)

The result obtained does not depend on the length of the cylinder, i.e. at any distance from the plane, the field strength is the same in magnitude. The pattern of tension lines looks as shown in Fig. 2.11. For a negatively charged plane, the directions of the vectors will be reversed. If the plane is of finite dimensions, then the result obtained will be valid only for points whose distance from the edge of the plate significantly exceeds the distance from the plate itself (Fig. 2.12).