General formulation: The flux of the electric field strength vector through any arbitrarily chosen closed surface is proportional to the electric charge contained within this surface.

In the CGSE system:

In SI:

- flux of the electric field strength vector through a closed surface.

- the total charge contained in the volume that bounds the surface.

- electrical constant.

This expression is the Gauss theorem in integral form.

In differential form, Gauss's theorem corresponds to one of Maxwell's equations and is expressed as follows

in SI system:

,

in the CGSE system:

Here is the volumetric charge density (in the presence of a medium, the total density of free and bound charges), and is the nabla operator.

For the Gauss theorem, the principle of superposition is valid, that is, the flux of the intensity vector through the surface does not depend on the distribution of the charge inside the surface.

The physical basis of the Gauss theorem is the Coulomb law, or, in other words, the Gauss theorem is an integral formulation of the Coulomb law.

Gauss's theorem for electrical induction (electrical displacement).

For a field in a substance, Gauss's electrostatic theorem can be written differently - through the flux of the vector of electrical displacement (electrical induction). In this case, the formulation of the theorem is as follows: the flux of the electric displacement vector through a closed surface is proportional to the free electric charge contained within this surface:

If we consider the theorem for the field strength in a substance, then as the charge Q it is necessary to take the sum of the free charge inside the surface and the polarization (induced, bound) charge of the dielectric:

,

Where ,
Is the polarization vector of the dielectric.

Gauss's theorem for magnetic induction

The flux of the magnetic induction vector through any closed surface is zero:

.

This is equivalent to the fact that in nature there are no "magnetic charges" (monopoles) that would create a magnetic field, as electric charges create an electric field. In other words, the Gauss theorem for magnetic induction shows that the magnetic field is vortex.

Application of Gauss's theorem

The following quantities are used to calculate electromagnetic fields:

Bulk charge density (see above).

Surface charge density

where dS is an infinitesimal area of ​​the surface.

Linear charge density

where dl is the length of an infinitesimal segment.

Consider the field created by an infinite homogeneous charged plane. Let the surface charge density of the plane be the same and equal to σ. Imagine mentally a cylinder with generatrices perpendicular to the plane, and the base ΔS located symmetrically relative to the plane. By virtue of symmetry. The intensity vector flux is equal to. Applying the Gauss theorem, we get:

,

from which

in the CGSE system

It is important to note that, despite its universality and generality, Gauss's theorem in integral form has relatively limited application due to the inconvenience of calculating the integral. However, in the case of a symmetric problem, its solution becomes much simpler than using the superposition principle.

For a full description of the electrostatic field of a given system of charges in a vacuum, the experimentally confirmed Coulomb's law and the principle of superposition are sufficient. But at the same time, it is possible to characterize the properties of the electrostatic field in a different generalized form, without relying on statements regarding the Coulomb field of a point charge.

Let us set a new physical quantity that describes the electric field - the flux Φ of the vector of the electric field strength. Suppose that in the space containing a given electric field, there is some rather small area Δ S.

Definition 1

Elementary flux of the vector of tension (through the site S) Is a physical quantity equal to the product of the modulus of the vector E →, the area Δ S and the cosine of the angle α between the vector and the normal to the site:

Δ Φ = E Δ S cos α = E n Δ S.

In this formula, E n is the modulus of the normal component of the field E →.

Picture 1 . 3. one . Illustration of the elementary flow Δ Φ.

Example 1

Now we take for consideration some arbitrary closed surface S... We split the given surface into small areas Δ S i, calculate the elementary fluxes Δ Φ i of the field through these small areas, and then find their sum, which ultimately gives us the flux Φ of the vector through the closed surface S(fig. 1. 3. 2):

Φ = ∑ ∆ Φ i = ∑ E m ∆ S i

When it comes to a closed surface, the outer normal is always used.

Picture 1 . 3. 2. Calculation of the flow Ф through an arbitrary closed surface S.

The Gauss theorem or law for an electrostatic field in a vacuum is one of the basic electrodynamic laws.

Theorem 1

The flux of the vector of strength of the electrostatic field E → through an arbitrary closed surface is equal to the algebraic sum of the charges located inside this surface, divided by the electric constant ε 0.

The Gauss equation has the form:

Φ = 1 ε 0 ∑ q in n at r

Proof 1

Let us prove the indicated theory: for this we investigate the spherical surface (or the surface of the ball) S. A point charge q is located at the center of a given surface. Any point of the sphere has an electric field perpendicular to the surface of the sphere and equal in magnitude:

E = E n = 1 4 π ε 0 q R 2,

where R is the radius of the sphere.

The flow Φ through the surface of the ball is written as the product E and the area of ​​the sphere is 4 π R 2. Then: Φ = 1 ε 0 q.

Our next step is to surround the point charge with an arbitrary closed-type surface S; we also define an auxiliary sphere R 0 (Fig. 1. 3. 3).

Picture 1 . 3. 3. The flow of an electric field of a point charge through an arbitrary surface S surrounding the charge.

Take for consideration a cone with a small solid angle Δ Ω at the top. The cone under consideration defines a small area Δ S 0 on the sphere, and on the surface S- platform Δ S. Elementary flows Δ Φ 0 and Δ Φ through these areas are the same. Indeed:

Δ Φ 0 = E 0 Δ S 0, Δ Φ = E Δ S cos α = E Δ S ",

where the expression Δ S "= Δ S cos α defines the area, which is defined by a cone with a solid angle Δ Ω on the surface of a sphere of radius n.

Since ∆ S 0 ∆ S "= R 0 2 r 2, then ∆ Φ 0 = ∆ Φ. From the obtained it follows that the total flux of the electric field of a point charge through an arbitrary surface covering the charge is equal to the flux Φ 0 through the surface of the auxiliary spheres:

Φ = Φ 0 = q ε 0.

We can also demonstrate that when a closed surface S does not cover point charge q, the flux Φ is equal to zero. This case is illustrated in Fig. one . 3. 2. All lines of force of the electric field of a point charge penetrate a closed surface S through. Inside surface S there are no charges, i.e. no breakage or nucleation of field lines is observed in this region.

The generalization of Gauss's theorem to the case of an arbitrary distribution of charges is a consequence of the principle of superposition. The field of any distribution of charges can be written as a vector sum of electric fields of point charges. The flow Φ of the system of charges through an arbitrary closed surface S will consist of flows Φ i of electric fields of individual charges. When the charge q i located inside the surface S, it contributes to the flow equal to q i ε 0. If the charge is located outside the surface, its contribution to the flux is zero.

So, we have proved the Gauss theorem.

Remark 1

Gauss's theorem is, in fact, a consequence of Coulomb's law and the principle of superposition. However, taking the statements of the theorem as the initial axiom, the consequence will be Coulomb's law, in connection with which the Gauss theorem is sometimes called alternative formulation of Coulomb's law.

Based on the Gauss theorem, in certain cases it is easy to determine the strength of the electric field around a charged body (in the presence of the previously guessed symmetry of the given distribution of charges and the general structure of the field).

Example 2

As an example, we can consider a problem in which it is necessary to calculate the field of a thin-walled hollow uniformly charged long cylinder with a radius R... This problem has axial symmetry, and for reasons of symmetry, the electric field must have a direction along the radius. Thus, in order to be able to apply the Gauss theorem, it is optimal to choose a closed-type surface S in the form of a coaxial cylinder of some radius r and length l, closed at both ends (Fig. 1. 3. 4).

Picture 1 . 3. four . Illustration of the field of a uniformly charged cylinder. O O "- axis of symmetry.

If r ≥ R , then the entire flux of the intensity vector will pass through the lateral surface of the cylinder, since the flux through both bases is zero. The formula for the lateral surface area of ​​a cylinder will be written as: 2 π r l . We apply Gauss's law and get:

Φ = E 2 π r l = τ l ε 0.

In the above expression, τ is the charge of the length of the cylinder. Then you can write:

E = τ 2 π ε 0 r.

This expression is independent of the radius R charged cylinder, which means that it is also applicable to the field of a long uniformly charged filament.

To find the field strength inside a charged cylinder, it is necessary to create a closed surface for the case r< R . В соответствии с симметрией задачи поток вектора напряженности через боковую поверхность цилиндра должен быть, и в этом случае он равен Φ = E 2 π r l . Исходя из гауссовской теоремы, этот поток находится в пропорции к заряду, расположенному внутри замкнутой поверхности. Заряд этот равен нулю, откуда вытекает, что электрическое поле внутри однородно заряженного длинного полого цилиндра тоже есть нуль.

In the same way, Gauss's theorem and formula are applicable to determine the electric field in other cases when the charge distribution is characterized by some kind of symmetry, for example, symmetry about the center, plane or axis. In all these cases, it is necessary to choose a closed Gaussian surface of a suitable shape.

Example 3

For example, in the case of central symmetry, it is optimal to choose a surface in the form of a sphere whose center is located at the point of symmetry. When we have symmetry about the axis, a suitable type of closed surface would be a coaxial cylinder, closed at both ends (similar to the example discussed above).

In the absence of symmetry and the impossibility of guessing the general structure of the field, the Gauss theorem cannot be applied to simplify the solution of the problem of determining the field strength.

Example 4

Let us consider another example of the distribution of charges in the presence of symmetry: finding the field of a uniformly charged plane (Fig. 1. 3. 5).

Picture 1 . 3. five . Field of a uniformly charged plane. σ is the surface charge density. S is a closed Gaussian surface.

Here the Gaussian surface S it is optimal to define it as a cylinder of a certain length, closed at both ends. The axis of the cylinder is perpendicular to the charged plane; in turn, the ends of the cylinder are at the same distance from it. In accordance with the symmetry, the field of a uniformly charged plane must everywhere have a normal direction. We apply Gauss's theorem and get:

2 E ∆ S = σ ∆ S ε 0 or E = σ 2 ε 0.

Here σ is the surface charge density or charge per unit area.

The expression that we obtained for the electric field of a uniformly charged plane can also be used for flat charged areas of a finite size: here the distance from the point at which we determine the field strength to the charged area should be significantly less than the size of the area.

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1. The intensity of the electrostatic field created by a uniformly charged spherical surface.

Let a spherical surface of radius R (Fig. 13.7) bears a uniformly distributed charge q, i.e. the surface charge density at any point of the sphere will be the same.

2. The electrostatic field of the sphere.

Let we have a ball of radius R, uniformly charged with bulk density.

At any point A, lying outside the ball at a distance r from its center (r> R), its field is similar to the field of a point charge located in the center of the ball. Then outside the ball

(13.10)

and on its surface (r = R)

(13.11)

At point B, which lies inside the ball at a distance r from its center (r> R), the field is determined only by the charge contained within the sphere of radius r. The flux of the tension vector through this sphere is

on the other hand, according to the Gauss theorem

From the comparison of the last expressions it follows

(13.12)

where is the dielectric constant inside the ball. The dependence of the field strength created by a charged sphere on the distance to the center of the sphere is shown in (Figure 13.10)

3. Field strength of a uniformly charged infinite rectilinear thread (or cylinder).

Suppose that a hollow cylindrical surface of radius R is charged with constant linear density.

Let us draw a coaxial cylindrical surface of radius The flux of the intensity vector through this surface

By Gauss's theorem

From the last two expressions, we determine the field strength created by a uniformly charged thread:

(13.13)

Let the plane have infinite length and the charge per unit area is equal to σ. From the laws of symmetry it follows that the field is directed everywhere perpendicular to the plane, and if there are no other external charges, then the fields on both sides of the plane must be the same. Let us restrict part of the charged plane to an imaginary cylindrical box, so that the box is cut in half and its generators are perpendicular, and the two bases, each with area S, are parallel to the charged plane (Figure 1.10).

The total flow of the vector; tension is equal to the vector times the area S of the first base plus the flow of the vector through the opposite base. The stress flux through the lateral surface of the cylinder is zero, because the lines of tension do not cross them. In this way, On the other hand, by Gauss's theorem

Hence

but then the field strength of an infinite uniformly charged plane will be equal to

This theorem is only a consequence of Coulomb's law and the principle of superposition of electric fields. Here is its wording:

The flux of the electric field strength vector through a closed surface in vacuum is equal to the algebraic sum of the electric charges contained inside this surface, divided by the electric constant  0 .

We begin the proof of the theorem with the simplest case: we calculate the flux of the vector of the field strength of a point charge Q.

The strength of this field is well known (see 1.3)

Taking into account the spherical symmetry of the field, we first choose a sphere of radius r, centered at the point where the charge is Q(fig. 2.5., 1). The flux of the stress vector through this surface is easy to calculate

Here we took into account that:

Fig. 2.5.

Taking into account the last remark, we write the flow (2.7) in the following form:

(2.8)

Thus, for the first simplest case, Gauss's theorem turned out to be true. What follows from this?

The result obtained allows us to conclude that the found flux does not depend on the radius of the Gaussian surface. This is easy to understand: after all, with an increase in the distance from the charge Q surface area is growingproportionately the square of the radius, and the field strength decreasesinversely the square of the radius.

Let us recall, in addition, that the flux of the intensity vector is equal to the number of lines of force penetrating the Gaussian surface. The independence of the flux from the radius of the surface means that the lines of force of the field of a point charge, starting at a positive charge, extend further to infinity without interruption. Hence - further conclusions.

The flux of the vector of the field strength of a point charge through any a closed surface (Fig. 2.5, 2), covering a point chargeQ, is equal to the ratio

This conclusion is beyond doubt, since the flux is equal to the previous unchanged number of lines of force penetrating the closed surface.

The flow of the intensity vector, through an arbitrary closed surface that does not cover the electric charge, is equal to zero (Fig. 2.5, 3).

This conclusion is also easy to understand, since the number of lines of force flowing into a Gaussian surface is equal to the number of lines leaving it. Therefore, the total flux through this surface is zero.

Now we can turn to the consideration of the general case: let an arbitrary closed surface S covers N point charges (Fig. 2.6.). Let us calculate the flux of the total field strength vector through this surface S, taking into account that, in accordance with the principle of superposition, the resulting field is equal to the vector sum of individual fields

Fig. 2.6.

So, using the definition of a flow, we calculate it through an arbitrary closed surface S.

(2.9)

The result obtained is a proof of the validity of the Gauss theorem: the flux of the vector of the strength of the electrostatic field in vacuum through any closed surface is proportional to the algebraic sum of the charges contained inside this surface.

The calculation of the field strength of a large system of electric charges using the principle of superposition of electrostatic fields can be significantly simplified using the Gauss theorem. This theorem defines tension vector flux electric field through an arbitrary closed surface.

For an arbitrary closed surface S the flux of the stress vector through this surface is determined by the expression

(1.23)

where is the projection of the vector onto the normal to the area dS(fig. 1.10); vector whose modulus is dS, and the direction coincides with the direction of the normal to the site ().

Consider a spherical surface of radius r covering a point charge q located in its center (fig. 1.11). In accordance with formula (1.23), the flux of the stress vector through this surface will be equal to:

This result is valid for a closed surface of any shape: if you surround the considered sphere with an arbitrary closed surface, then each line of tension penetrating the sphere will pass through this surface as well.

Let us now consider the general case of an arbitrary closed surface surrounding n charges. In accordance with the principle of superposition, the field strength created by all charges is equal to the vector sum of the field strengths caused by each charge separately; therefore, the flux of the intensity vector of the resulting field will be equal to:

According to (1.24), each of the integrals under the sum sign is equal to. Hence,

(1.25)

those. the flux of the vector of the strength of the electrostatic field in vacuum through an arbitrary closed surface is equal to the algebraic sum of the charges contained inside this surface, divided by the electric constant.

We apply Gauss's theorem to determine the field strength of a uniformly charged infinite plane. In this case, its surface charge density

is the same anywhere in the plane. This means that the lines of tension are perpendicular to the plane at any point, i.e. the field of the charged plane is uniform (Fig. 1.12).

Let us mentally select a cylinder in space, the axis of which is perpendicular to the plane and one of the bases passes through the point of interest to us. According to Gauss's theorem,

On the other hand, since the lines of tension intersect only the bases of the cylinder, the vector flux can be expressed in terms of the electric field strength at both bases of the cylinder, i.e.

Let us give (without derivation) expressions for calculating the strength of the electrostatic field formed by some other charged bodies.