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 the SI system:

- the flux of the electric field strength vector through the 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 stress vector through the surface does not depend on the distribution of the charge inside the surface.

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

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

For a field in matter, 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.

Electric field strength vector flux. Let a small area DS(Figure 1.2) intersect the lines of force of the electric field, the direction of which is with the normal n angle to this site a... Assuming that the vector of tension E does not change within the site DS, define tension vector flux across the site DS how

DFE =E DS cos a.(1.3)

Since the density of the lines of force is equal to the numerical value of the tension E, then the number of lines of force crossing the siteDS, will be numerically equal to the flow valueDFEacross the surfaceDS... We represent the right-hand side of expression (1.3) as the scalar product of vectors E andDS= nDS, where nIs the unit normal vector to the surfaceDS... For an elementary site d S expression (1.3) takes the form

dFE = E d S

Across the entire site S the intensity vector flux is calculated as an integral over the surface

Electric induction vector flux. The flux of the electric induction vector is determined similarly to the flux of the electric field strength vector

dFD = D d S

Some ambiguity is noticeable in the definitions of flows, due to the fact that for each surface two normals of the opposite direction. For a closed surface, the outer normal is considered positive.

Gauss's theorem. Consider point positive electric charge q inside an arbitrary closed surface S(fig. 1.3). The flux of the induction vector through an element of the surface d S is equal to
(1.4)

Component d S D = d S cos asurface element d S in the direction of the induction vectorDconsidered as an element of a spherical surface of radius r, in the center of which is the chargeq.

Considering that d S D/ r 2 is equal elementary bodily corner dw, under which from the point of location of the chargeqsurface element d is visible S, we transform expression (1.4) to the form d FD = q d w / 4 p, whence, after integration over the entire space surrounding the charge, i.e., within the solid angle from 0 to 4p, we get

FD = q.

The flux of the electric induction vector through a closed surface of arbitrary shape is equal to the charge contained inside this surface.

If an arbitrary closed surface S does not cover point charge q(Fig. 1.4), then, having built a conical surface with apex at the point where the charge is located, we divide the surface S into two parts: S 1 and S 2. Vector stream D across the surface S we find as the algebraic sum of flows through surfaces S 1 and S 2:

.

Both surfaces from the point of location of the charge q visible at one solid angle w... Therefore the flows are equal

Since the flow through a closed surface is calculated using outward normal to the surface, it is easy to see that the flux Ф 1D < 0, тогда как поток Ф2D> 0. The total flux Ф D= 0. This means that the flux of the electric induction vector through a closed surface of arbitrary shape does not depend on the charges located outside this surface.

If the electric field is created by a system of point charges q 1 , q 2 ,¼ , q n, which is covered by a closed surface S, then, in accordance with the principle of superposition, the flux of the induction vector through this surface is defined as the sum of fluxes created by each of the charges. The flux of the electric induction vector through a closed surface of arbitrary shape is equal to the algebraic sum of the charges covered by this surface:

It should be noted that the charges q i do not have to be point-like, a necessary condition is that the charged area must be completely covered by the surface. If in the space bounded by a closed surface S, the electric charge is distributed continuously, then it should be assumed that each elementary volume d V has a charge. In this case, on the right-hand side of expression (1.5), the algebraic summation of charges is replaced by integration over the volume enclosed inside the closed surface S:

(1.6)

Expression (1.6) is the most general formulation Gauss theorem: the flux of the electric induction vector through a closed surface of arbitrary shape is equal to the total charge in the volume enclosed by this surface, and does not depend on the charges located outside the considered surface... Gauss's theorem can also be written for the flux of the electric field strength vector:

.

An important property of the electric field follows from the Gauss theorem: lines of force begin or end only on electric charges or go to infinity... We emphasize once again that, despite the fact that the electric field strength E and electrical induction D depend on the location of all charges in space, the fluxes of these vectors through an arbitrary closed surface S determined only those charges that are located inside the surface S.

Differential form of Gauss's theorem. Note that integral form Gauss's theorem characterizes the relationship between the sources of the electric field (charges) and the characteristics of the electric field (strength or induction) in the volume V arbitrary, but sufficient for the formation of integral relations, value. Dividing the volume V for small volumes V i, we get the expression

which is true both as a whole and for each term. Let's transform the resulting expression as follows:

(1.7)

and consider the limit to which the expression on the right-hand side of the equality, enclosed in curly braces, tends for an unlimited division of the volume V... In mathematics, this limit is called divergence vector (in this case, the electric induction vector D):

Vector divergence D in Cartesian coordinates:

Thus, expression (1.7) is transformed to the form:

.

Taking into account that with unbounded division, the sum on the left side of the last expression goes over into the volume integral, we get

The resulting ratio must be satisfied for any arbitrarily chosen volume V... This is possible only if the values ​​of the integrands are the same at each point in space. Therefore, the divergence of the vector D is related to the charge density at the same point by the equality

or for the vector of the strength of the electrostatic field

These equalities express Gauss's theorem in differential form.

Note that in the process of passing to the differential form of Gauss' theorem, a relation is obtained that has a general character:

.

The expression is called the Gauss - Ostrogradsky formula and connects the integral over the volume of the divergence of a vector with the flow of this vector through a closed surface bounding the volume.

Questions

1) What is the physical meaning of the Gauss theorem for an electrostatic field in vacuum

2) There is a point charge in the center of the cube.q... What is the vector flux E:

a) through the full surface of the cube; b) through one of the cube faces.

Will the answers change if:

a) the charge is not in the center of the cube, but inside it ; b) the charge is outside the cube.

3) What is linear, surface, volumetric charge density.

4) Indicate the relationship between the bulk and surface charge density.

5) Can a field outside of oppositely and uniformly charged parallel infinite planes be nonzero?

6) An electric dipole is placed inside a closed surface. What is the flow through this surface

Let us introduce the concept of the flux of the electric induction vector. Consider an infinitely small area. In most cases, it is necessary to know not only the size of the site, but also its orientation in space. Let's introduce the concept of a vector-platform. Let us agree by a vector-site to mean a vector directed perpendicular to the site and numerically equal to the size of the site.

Figure 1 - To the definition of the vector - site

Let's call the vector flow across the site
dot product of vectors and
... In this way,

Vector stream through an arbitrary surface is found by integrating all elementary streams

(4)

If the field is uniform and flat located perpendicular to the field, then:

. (5)

The above expression determines the number of lines of force penetrating the site per unit of time.

Ostrogradsky-Gauss theorem. Divergence of electric field strength

The flux of the electric induction vector through an arbitrary closed surface is equal to the algebraic sum of free electric charges covered by this surface

(6)

Expression (6) is the O-G theorem in integral form. Theorem 0-D operates with an integral (total) effect, i.e. if
it is not known whether this means the absence of charges at all points of the investigated part of the space, or, that the sum of positive and negative charges located at different points of this space are equal to zero.

To find the located charges and their magnitude for a given field, you need a relation connecting the electric induction vector at a given point with a charge at the same point.

Suppose we need to determine the presence of charge at the point a(fig. 2)

Figure 2 - To the calculation of vector divergence

We apply the O-G theorem. The flux of the electric induction vector through an arbitrary surface bounding the volume in which the point is located a, is equal to

The algebraic sum of charges in the volume can be written in the form of the volume integral

(7)

where - charge per unit volume ;

- volume element.

To obtain a connection between the field and the charge at the point a we will decrease the volume, pulling the surface to the point a... In this case, we divide both sides of our equality by the value ... Passing to the limit, we get:

.

The right side of the resulting expression is, by definition, the volumetric charge density at the considered point in space. The left side represents the limit of the ratio of the flux of the electric induction vector through a closed surface to the volume bounded by this surface when the volume tends to zero. This scalar quantity is an important characteristic of the electric field and is called divergence vector .

In this way:

,

hence

, (8)

where - bulk charge density.

Using this ratio, the inverse problem of electrostatics is simply solved, i.e. finding distributed charges over a known field.

If the vector is given, then its projections are known
,
,
on the coordinate axes as a function of coordinates and to calculate the distributed density of charges that created a given field, it turns out to be enough to find the sum of three partial derivatives of these projections with respect to the corresponding variables. At those points for which
no charges. At points where
is positive, there is a positive charge with a bulk density equal to
, and at those points where
will have a negative value, there is a negative charge, the density of which is also determined by the divergence value.

Expression (8) represents Theorem 0-Г in differential form. In this form, the theorem shows that the sources of the electric field are free electric charges; the lines of force of the electric induction vector begin and end, respectively, on positive and negative charges.

The most difficult is the study of electrical phenomena in an inhomogeneous electrical environment. In such a medium, ε has different values, changing abruptly at the boundary of dielectrics. Suppose that we determine the field strength at the interface between two media: ε 1 = 1 (vacuum or air) and ε 2 = 3 (liquid - oil). At the interface at the transition from vacuum to dielectric, the field strength decreases three times, and the flux of the strength vector decreases by the same amount (Fig. 12.25, a). An abrupt change in the vector of the strength of the electrostatic field at the interface between two media creates certain difficulties in calculating the fields. As for the Gauss theorem, under these conditions it completely loses its meaning.

Since the polarizability and strength of dissimilar dielectrics are different, the number of lines of force in each dielectric will also be different. This difficulty can be eliminated by introducing a new physical characteristic of the field - electric induction D (or vector electrical displacement ).

According to the formula

ε 1 Е 1 = ε 2 Е 2 = Е 0 = const

Multiplying all parts of these equalities by the electrical constant ε 0, we obtain

ε 0 ε 1 Е 1 = ε 0 ε 2 Е 2 = ε 0 Е 0 = const

We introduce the notation ε 0 εЕ = D then the penultimate relation takes the form

D 1 = D 2 = D 0 = const

The vector D, equal to the product of the electric field strength in the dielectric and its absolute dielectric constant, is calledvector of electrical displacement

(12.45)

Electrical displacement unit - pendant per square meter(Cl / m 2).

Electrical displacement is a vector quantity, it can also be expressed as

D = εε 0 E = (1 + χ) ε 0 E = ε 0 E + χε 0 E = ε 0 E + P

(12.46)

Unlike the strength E, the electrical displacement D is constant in all dielectrics (Figure 12.25, b). Therefore, the electric field in an inhomogeneous dielectric medium is conveniently characterized not by the strength E, but by the displacement vector D. The vector D describes the electrostatic field created by free charges (i.e. in a vacuum), but with such a distribution in space that exists in the presence of a dielectric, since the bound charges arising in dielectrics can cause a redistribution of free charges that create the field.

Vector field is plotted with electrical displacement lines in the same way as a field depicted by lines of force.

Electrical displacement line - these are lines, tangents to which at each point coincide in direction with the vector of electrical displacement.

The lines of the vector E can begin and end at any charges - free and bound, while the lines of the vectorD- only on free charges. Vector linesDunlike lines of tension, they are continuous.

Since the electric displacement vector does not experience a discontinuity at the interface between the two media, all the lines of induction emanating from charges surrounded by some closed surface will penetrate it. Therefore, for the vector of electrical displacement, the Gauss theorem fully retains its meaning for an inhomogeneous dielectric medium.

Gauss's theorem for the electrostatic field in a dielectric : the flux of the electric displacement vector through an arbitrary closed surface is equal to the algebraic sum of the charges contained within this surface.

(12.47)

The law of interaction of electric charges - Coulomb's law - can be formulated differently, in the form of the so-called Gauss theorem. Gauss's theorem is obtained as a consequence of Coulomb's law and the principle of superposition. The proof is based on the inverse proportionality of the interaction force of two point charges to the square of the distance between them. Therefore, Gauss's theorem is applicable to any physical field where the inverse square law and the principle of superposition apply, for example, to a gravitational field.

Rice. 9. Lines of electric field strength of a point charge crossing the closed surface X

In order to formulate the Gauss theorem, let us return to the picture of the lines of force of the electric field of a stationary point charge. The lines of force of a solitary point charge are symmetrically arranged radial straight lines (Fig. 7). Any number of such lines can be drawn. Let us denote their total number through Then the density of the lines of force at a distance from the charge, i.e., the number of lines crossing the unit surface of the sphere of radius is Comparing this relation with the expression for the field strength of a point charge (4), we see that the density of the lines is proportional to the field strength. We can make these values ​​numerically equal by appropriately choosing the total number of lines of force N:

Thus, the surface of a sphere of any radius enclosing a point charge intersects the same number of lines of force. This means that the lines of force are continuous: in the interval between any two concentric spheres of different radii, none of the lines is cut off and no new ones are added. Since the lines of force are continuous, the same number of lines of force intersect any closed surface (Fig. 9), covering the charge

Lines of force have direction. In the case of a positive charge, they go out from the closed surface surrounding the charge, as shown in Fig. 9. In the case of a negative charge, they enter the interior of the surface. If the number of outgoing lines is considered positive, and the number of incoming lines is negative, then in formula (8) we can omit the sign of the modulus of the charge and write it in the form

Stream of tension. Let us now introduce the concept of the flux of the field strength vector through the surface. An arbitrary field can be mentally divided into small areas, in which the intensity changes in magnitude and direction so little that within this area the field can be considered uniform. In each such area, the lines of force are parallel straight lines and have a constant density.

Rice. 10. To determine the flux of the field strength vector through the site

Let us consider how many lines of force penetrate a small area, the direction of the normal to which forms an angle a with the direction of the lines of tension (Fig. 10). Let be a projection onto a plane perpendicular to the lines of force. Since the number of lines crossing the same, and the line density, according to the accepted condition, is equal to the modulus of the field strength E, then

The quantity a is the projection of the vector E onto the direction of the normal to the site

Therefore, the number of lines of force crossing the site is

The product is called the flux of the field strength through the surface. Formula (10) shows that the flux of the vector E through the surface is equal to the number of lines of force crossing this surface. Note that the flux of the intensity vector, as well as the number of lines of force passing through the surface, is a scalar.

Rice. 11. The flow of the vector of intensity E through the site

The dependence of the flow on the orientation of the site relative to the lines of force is illustrated in Fig.

The field strength flux through an arbitrary surface is the sum of fluxes through the elementary areas into which this surface can be divided. By virtue of relations (9) and (10), it can be argued that the flux of the field strength of a point charge through any closed surface 2 enclosing the charge (see Fig. 9), as the number of lines of force emerging from this surface is equal. In this case, the normal vector to the elementary areas the closed surface should be directed outward. If the charge inside the surface is negative, then the lines of force enter the inside of this surface and the flux of the field strength vector associated with the charge is also negative.

If there are several charges inside a closed surface, then, in accordance with the principle of superposition, the streams of their field strengths will add up. The total flux will be equal to where should be understood as the algebraic sum of all charges inside the surface.

If there are no electric charges inside a closed surface, or their algebraic sum is equal to zero, then the total flux of field strength through this surface is zero: how many lines of force enter the volume bounded by the surface, the same amount goes out.

Now we can finally formulate the Gauss theorem: the flux of the electric field strength vector E in vacuum through any closed surface is proportional to the total charge inside this surface. Mathematically, Gauss's theorem is expressed by the same formula (9), where by the algebraic sum of charges is meant. In absolute electrostatic

system of CGSE units, the coefficient and Gauss's theorem are written in the form

In SI and the flow of tension through a closed surface is expressed by the formula

Gauss's theorem is widely used in electrostatics. In some cases, it can be used to easily calculate the fields created by symmetrically located charges.

Balanced source fields. Let us apply the Gauss theorem to calculate the electric field strength of a sphere with a radius uniformly charged over the surface. For definiteness, we will consider its charge to be positive. The distribution of charges that create the field has spherical symmetry. Therefore, the field also possesses the same symmetry. The lines of force of such a field are directed along the radii, and the modulus of intensity is the same at all points equidistant from the center of the ball.

In order to find the field strength at a distance from the center of the ball, mentally draw a spherical surface of radius concentric with the ball.Since at all points of this sphere the field strength is directed perpendicular to its surface and is the same in magnitude, the intensity flux is simply equal to the product of the field strength by the surface area of ​​the sphere:

But this quantity can also be expressed using the Gauss theorem. If we are interested in the field outside the ball, that is, for then, for example, in SI and, comparing with (13), we find

In the CGSE system of units, it is obvious that

Thus, outside the ball, the field strength is the same as that of the field of a point charge placed in the center of the ball. If we are interested in the field inside the ball, that is, when, then since the entire charge distributed over the surface of the ball is outside the sphere we mentally draw. Therefore, there is no field inside the ball:

Similarly, using the Gauss theorem, you can calculate the electrostatic field created by an infinite charged

plane with a constant density at all points of the plane. For reasons of symmetry, we can assume that the lines of force are perpendicular to the plane, directed from it in both directions, and have the same density everywhere. Indeed, if the density of the lines of force at different points were different, then the movement of the charged plane along itself would lead to a change in the field at these points, which contradicts the symmetry of the system - such a shift should not change the field. In other words, the field of an infinite uniformly charged plane is uniform.

As a closed surface for the application of the Gauss theorem, we choose the surface of a cylinder constructed as follows: the generatrix of the cylinder is parallel to the lines of force, and the bases have areas parallel to the charged plane and lie on opposite sides of it (Fig. 12). The field strength flux through the lateral surface is zero, therefore the total flux through the closed surface is equal to the sum of the flux through the cylinder base:

Rice. 12. To the calculation of the field strength of a uniformly charged plane

According to Gauss's theorem, the same flux is determined by the charge of that part of the plane that lies inside the cylinder, and in SI is equal. Comparing these expressions for the flux, we find

In the CGSE system, the field strength of a uniformly charged infinite plane is given by the formula

For a uniformly charged plate of finite dimensions, the obtained expressions are approximately valid in a region located sufficiently far from the edges of the plate and not too far from its surface. Near the edges of the plate, the field will no longer be uniform and its lines of force are bent. At very large distances in comparison with the dimensions of the plate, the field decreases with distance in the same way as the field of a point charge.

As other examples of fields created by symmetrically distributed sources, one can cite the field of an infinite rectilinear filament uniformly charged along the length, the field of a uniformly charged infinite circular cylinder, the field of a ball,

uniformly charged over the volume, etc. Gauss's theorem makes it possible to easily calculate the field strength in all these cases.

Gauss's theorem gives a connection between the field and its sources, in a sense, the opposite of the one that gives the Coulomb's law, which allows you to determine the electric field for given charges. Using Gauss's theorem, it is possible to determine the total charge in any region of space in which the distribution of the electric field is known.

What is the difference between the concepts of long-range and short-range action when describing the interaction of electric charges? To what extent can these concepts be applied to gravitational interaction?

What is electric field strength? What do they mean when they call it the strength characteristic of an electric field?

How can one judge the direction and modulus of the field strength at a certain point from the pattern of the field lines?

Can the lines of force of an electric field intersect? Give reasons for your answer.

Draw a qualitative picture of the lines of force of the electrostatic field of two charges such that.

The flux of electric field strength through a closed surface is expressed by different formulas (11) and (12) in the systems of units of the GSE and in SI. How can this be reconciled with the geometric meaning of the flow, determined by the number of force lines crossing the surface?

How to use the Gauss theorem to find the strength of the electric field with a symmetric distribution of the charges that create it?

How to apply formulas (14) and (15) to the calculation of the field strength of a sphere with a negative charge?

Gauss's theorem and the geometry of physical space. Let's look at the proof of Gauss' theorem from a slightly different point of view. Let us return to formula (7), from which it was concluded that the same number of lines of force passes through any spherical surface surrounding the charge. This conclusion is due to the fact that there is a reduction in the denominators of both sides of the equality.

On the right side arose due to the fact that the force of interaction of charges, described by Coulomb's law, is inversely proportional to the square of the distance between the charges. On the left, the appearance is associated with geometry: the surface area of ​​a sphere is proportional to the square of its radius.

The proportionality of surface area to a square of linear dimensions is a hallmark of Euclidean geometry in three-dimensional space. Indeed, the proportionality of areas precisely to squares of linear dimensions, and not to any other integer degree, is characteristic of space

three dimensions. The fact that this exponent is exactly two, and does not differ from two, even by a negligible amount, testifies to the non-curvature of this three-dimensional space, that is, to the fact that its geometry is precisely Euclidean.

Thus, Gauss's theorem is a manifestation of the properties of physical space in the fundamental law of interaction of electric charges.

The idea of ​​a close connection between the fundamental laws of physics and the properties of space was expressed by many outstanding minds long before the establishment of these laws themselves. So, I. Kant three decades before the discovery of Coulomb's law wrote about the properties of space: "Three-dimensionality occurs, apparently, because the substances in the existing world act on one another in such a way that the force of action is inversely proportional to the square of the distance."

Coulomb's law and Gauss's theorem actually represent the same law of nature, expressed in different forms. Coulomb's law reflects the concept of long-range action, while Gauss's theorem proceeds from the concept of a force field that fills space, that is, from the concept of short-range action. In electrostatics, the source of the force field is a charge, and the characteristic of the field associated with the source - the flow of intensity - cannot change in empty space, where there are no other charges. Since the flow can be visualized as a set of field lines of force, the invariability of the flow manifests itself in the continuity of these lines.

Gauss's theorem, based on the inverse proportionality of the interaction to the square of the distance and on the principle of superposition (additivity of interaction), is applicable to any physical field in which the inverse square law operates. In particular, it is also valid for the gravitational field. It is clear that this is not just an accidental coincidence, but a reflection of the fact that both electrical and gravitational interactions are playing out in three-dimensional Euclidean physical space.

On what feature of the law of interaction of electric charges is Gauss's theorem based?

Prove, based on the Gauss theorem, that the electric field strength of a point charge is inversely proportional to the square of the distance. What properties of symmetry of space are used in this proof?

How is the geometry of physical space reflected in Coulomb's law and Gauss's theorem? What feature of these laws testifies to the Euclidean nature of the geometry and three-dimensionality of physical space?