The Ostrograd Gauss theorem for the electric induction vector. Ostrogradsky–Gauss theorem
Objective of the lesson: The Ostrogradsky–Gauss theorem was established by the Russian mathematician and mechanic Mikhail Vasilyevich Ostrogradsky in the form of a general mathematical theorem and by the German mathematician Carl Friedrich Gauss. This theorem can be used when studying physics at a specialized level, as it allows for more rational calculations of electric fields.
Electric induction vector
To derive the Ostrogradsky–Gauss theorem, it is necessary to introduce such important auxiliary concepts as the electrical induction vector and the flux of this vector F.
It is known that the electrostatic field is often depicted using lines of force. Let's assume that we determine the tension at a point lying at the interface between two media: air (=1) and water (=81). At this point, when moving from air to water, the electric field strength according to the formula 
will decrease by 81 times. If we neglect the conductivity of water, then the number of lines of force will decrease by the same factor. When solving various problems of calculating fields, due to the discontinuity of the voltage vector at the interface between media and on dielectrics, certain inconveniences are created. To avoid them, a new vector is introduced, which is called the electrical induction vector:
The electric induction vector is equal to the product of the vector and the electric constant and the dielectric constant of the medium at a given point.
It is obvious that when passing through the boundary of two dielectrics, the number of electric induction lines does not change for the field of a point charge (1).
In the SI system, the vector of electrical induction is measured in coulombs per square meter (C/m2). Expression (1) shows that the numerical value of the vector does not depend on the properties of the medium. The vector field is graphically depicted similarly to the intensity field (for example, for a point charge, see Fig. 1). For a vector field, the principle of superposition applies:
Electrical induction flux
The electric induction vector characterizes the electric field at each point in space. You can introduce another quantity that depends on the values of the vector not at one point, but at all points of the surface bounded by a flat closed contour.
To do this, consider a flat closed conductor (circuit) with surface area S, placed in a uniform electric field. The normal to the plane of the conductor makes an angle with the direction of the electrical induction vector (Fig. 2).
The flow of electrical induction through the surface S is a quantity equal to the product of the modulus of the induction vector by the area S and the cosine of the angle between the vector and the normal:
Derivation of the Ostrogradsky–Gauss theorem
This theorem allows us to find the flow of the electric induction vector through a closed surface, inside of which there are electric charges.
Let first one point charge q be placed at the center of a sphere of arbitrary radius r 1 (Fig. 3). Then 
; . Let's calculate the total flux of induction passing through the entire surface of this sphere: ; 
(). If we take a sphere of radius , then also Ф = q. If we draw a sphere that does not cover charge q, then the total flux Ф = 0 (since each line will enter the surface and leave it another time).
Thus, Ф = q if the charge is located inside the closed surface and Ф = 0 if the charge is located outside the closed surface. The flow Ф does not depend on the shape of the surface. It is also independent of the arrangement of charges within the surface. This means that the result obtained is valid not only for one charge, but also for any number of arbitrarily located charges, if only we mean by q the algebraic sum of all charges located inside the surface.
Gauss's theorem: the flow of electrical induction through any closed surface is equal to the algebraic sum of all charges located inside the surface: .
From the formula it is clear that the dimension of the electric flow is the same as that of the electric charge. Therefore, the unit of electrical induction flux is the coulomb (C).
Note: if the field is non-uniform and the surface through which the flow is determined is not a plane, then this surface can be divided into infinitesimal elements ds and each element can be considered flat, and the field near it is uniform. Therefore, for any electric field, the flow of the electric induction vector through the surface element is: dФ=. As a result of integration, the total flux through a closed surface S in any inhomogeneous electric field is equal to: 
, where q is the algebraic sum of all charges surrounded by a closed surface S. Let us express the last equation in terms of the electric field strength (for vacuum): .
This is one of Maxwell's fundamental equations for the electromagnetic field, written in integral form. It shows that the source of a time-constant electric field is stationary electric charges.
Application of Gauss's theorem
Field of continuously distributed charges
Let us now determine the field strength for a number of cases using the Ostrogradsky-Gauss theorem.
1. Electric field of a uniformly charged spherical surface.
Sphere of radius R. Let the charge +q be uniformly distributed over a spherical surface of radius R. The charge distribution over the surface is characterized by the surface charge density (Fig. 4). Surface charge density is the ratio of charge to the surface area over which it is distributed. . In SI.
Let's determine the field strength:
a) outside the spherical surface,
b) inside a spherical surface.
a) Take point A, located at a distance r>R from the center of the charged spherical surface. Let us mentally draw through it a spherical surface S of radius r, which has a common center with the charged spherical surface. From considerations of symmetry, it is obvious that the lines of force are radial lines perpendicular to the surface S and uniformly penetrate this surface, i.e. the tension at all points of this surface is constant in magnitude. Let us apply the Ostrogradsky-Gauss theorem to this spherical surface S of radius r. Therefore the total flux through the sphere is N = E? S; N=E. On the other side . We equate: . Hence: for r>R.
Thus: the tension created by a uniformly charged spherical surface outside it is the same as if the entire charge were in its center (Fig. 5).
b) Let us find the field strength at points lying inside the charged spherical surface. Let's take point B at a distance from the center of the sphere 2. Field strength of a uniformly charged infinite plane Let's consider the electric field created by an infinite plane, charged with a density constant at all points of the plane. For reasons of symmetry, we can assume that the tension lines are perpendicular to the plane and directed from it in both directions (Fig. 6). Let's choose point A lying to the right of the plane and calculate at this point using the Ostrogradsky-Gauss theorem. As a closed surface, we choose a cylindrical surface so that the side surface of the cylinder is parallel to the lines of force, and its base is parallel to the plane and the base passes through point A (Fig. 7). Let us calculate the tension flow through the cylindrical surface under consideration. The flux through the side surface is 0, because lines of tension are parallel to the lateral surface. Then the total flow consists of the flows and passing through the bases of the cylinder and . Both of these flows are positive =+; =; =; ==; N=2. – a section of the plane lying inside the selected cylindrical surface. The charge inside this surface is q. Then ; – can be taken as a point charge) with point A. To find the total field, it is necessary to geometrically add up all the fields created by each element: ; . 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 force of interaction between 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 superposition principle apply, for example, to the gravitational field. Rice. 9. Lines of electric field strength of a point charge intersecting a closed surface X In order to formulate Gauss's theorem, let us return to the picture of the electric field lines of a stationary point charge. The field lines of a solitary point charge are symmetrically located radial straight lines (Fig. 7). You can draw any number of such lines. Let us denote their total number by Then the density of field lines at a distance from the charge, i.e., the number of lines crossing a unit surface of a sphere of radius is equal to Comparing this relationship with the expression for the field strength of a point charge (4), we see that the density of lines is proportional to the field strength. We can make these quantities numerically equal by properly choosing the total number of field lines 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 are broken and no new ones are added. Since the field lines are continuous, the same number of field lines intersects any closed surface (Fig. 9) covering the charge Lines of force have a direction. In the case of a positive charge, they come out from the closed surface surrounding the charge, as shown in Fig. 9. In the case of a negative charge, they go inside the surface. If the number of outgoing lines is considered positive and the number of incoming lines negative, then in formula (8) we can omit the sign of the modulus of the charge and write it in the form Flow of tension. Let us now introduce the concept of field strength vector flow through a 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's 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 is the same, and the density of the lines, according to the accepted condition, is equal to the modulus of the field strength E, then The value a is the projection of the vector E onto the direction of the normal to the site Therefore, the number of power lines crossing the area is equal to The product is called the field strength flux through the surface. Formula (10) shows that the flux of vector E through the surface is equal to the number of field lines crossing this surface. Note that the intensity vector flux, like the number of field lines passing through the surface, is a scalar. Rice. 11. Flow of the tension vector 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 the fluxes through the elementary areas into which this surface can be divided. By virtue of relations (9) and (10), it can be stated that the flow of the field strength of a point charge through any closed surface 2 enveloping the charge (see Fig. 9), as the number of field lines emerging from this surface is equal to. In this case, the normal vector to the elementary areas closed surface should be directed outward. If the charge inside the surface is negative, then the field lines enter inside 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 flows of their field strengths will add up. The total flux will be equal to where by should be understood as the algebraic sum of all charges located inside the surface. If there are no electric charges inside a closed surface or their algebraic sum is zero, then the total flux of field strength through this surface is zero: as many lines of force enter the volume bounded by the surface, the same number go out. Now we can finally formulate Gauss’s theorem: the flow of the electric field strength vector E in a vacuum through any closed surface is proportional to the total charge located inside this surface. Mathematically, Gauss's theorem is expressed by the same formula (9), where by is meant the algebraic sum of charges. In absolute electrostatic in the SGSE system of units, the coefficient and Gauss’s theorem are written in the form In SI and the flux 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 fields created by symmetrically located charges. Fields of symmetrical sources. Let us apply Gauss's theorem to calculate the intensity of the electric field uniformly charged over the surface of a ball of radius . For definiteness, we will assume its charge to be positive. The distribution of charges creating the field has spherical symmetry. Therefore, the field also has the same symmetry. The lines of force of such a field are directed along the radii, and the intensity modulus 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, let us 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 absolute value, the intensity flow is simply equal to the product of the field strength and the surface area of the sphere: But this quantity can also be expressed using Gauss’s theorem. If we are interested in the field outside the ball, i.e., then, for example, in SI and, comparing with (13), we find In the system of units SGSE, obviously, Thus, outside the ball the field strength is the same as that of a point charge placed at the center of the ball. If we are interested in the field inside the ball, i.e., then since the entire charge distributed over the surface of the ball is located outside the sphere we have mentally drawn. Therefore, there is no field inside the ball: Similarly, using Gauss's theorem, one can calculate the electrostatic field created by an infinitely 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 field lines at different points were different, then moving a 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 applying Gauss's 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 side surface is zero, so the total flux through the closed surface is equal to the sum of the fluxes through the bases of the cylinder: Rice. 12. Towards 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 it is equal to Comparing these expressions for the flux, we find In the SGSE 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 field lines will be bent. At very large distances compared to the size of the plate, the field decreases with distance in the same way as the field of a point charge. Other examples of fields created by symmetrically distributed sources include the field of a uniformly charged along the length of an infinite rectilinear thread, the field of a uniformly charged infinite circular cylinder, the field of a ball, uniformly charged throughout the volume, etc. Gauss's theorem makes it possible to easily calculate the field strength in all these cases. Gauss's theorem gives a relationship between the field and its sources, in some sense the opposite of that given by Coulomb's law, which allows one to determine the electric field from given charges. Using Gauss's theorem, you can 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 interactions? What is electric field strength? What do they mean when it is called the force characteristic of the electric field? How can one judge the direction and magnitude of the field strength at a certain point from the pattern of field lines? Can electric field lines intersect? Give reasons for your answer. Draw a qualitative picture of the electrostatic field lines of two charges such that . The flow of electric field strength through a closed surface is expressed by different formulas (11) and (12) in the GSE and SI units. How can this be reconciled with the geometric meaning of flow, determined by the number of lines of force crossing the surface? How to use Gauss's theorem to find the electric field strength when the charges creating it are symmetrically distributed? How to apply formulas (14) and (15) to calculate the field strength of a ball with a negative charge? Gauss's theorem and the geometry of physical space. Let's look at the proof of Gauss's 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 a 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 it arose due to the fact that the force of interaction between charges, described by Coulomb’s law, is inversely proportional to the square of the distance between the charges. On the left side, the appearance is related to geometry: the surface area of a sphere is proportional to the square of its radius. The proportionality of surface area to the square of linear dimensions is a hallmark of Euclidean geometry in three-dimensional space. Indeed, the proportionality of areas precisely to the squares of linear dimensions, and not to any other integer degree, is characteristic of space three dimensions. The fact that this exponent is exactly equal to two, and does not differ from two, even by a negligibly small amount, indicates that this three-dimensional space is not curved, i.e., 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 these laws themselves were established. Thus, I. Kant, three decades before the discovery of Coulomb’s law, wrote about the properties of space: “Three-dimensionality occurs, apparently, because 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 comes from the idea of a force field filling space, i.e., 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 visually imagined as a set of field lines, the immutability of the flow is manifested in the continuity of these lines. Gauss's theorem, based on the inverse proportionality of 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 true for the gravitational field. It is clear that this is not just a coincidence, but a reflection of the fact that both electrical and gravitational interactions play out in three-dimensional Euclidean physical space. What feature of the law of interaction of electric charges is the Gauss theorem based on? Prove, based on Gauss's theorem, that the electric field strength of a point charge is inversely proportional to the square of the distance. What properties of space symmetry 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 indicates the Euclidean nature of geometry and the three-dimensionality of physical space? The main applied task of electrostatics is the calculation of electric fields created in various devices and devices. In general, this problem is solved using Coulomb's law and the principle of superposition. However, this task becomes very complicated when considering a large number of point or spatially distributed charges. Even greater difficulties arise when there are dielectrics or conductors in space, when under the influence of an external field E 0 a redistribution of microscopic charges occurs, creating their own additional field E. Therefore, to practically solve these problems, auxiliary methods and techniques are used that use complex mathematical apparatus. We will consider the simplest method based on the application of the Ostrogradsky–Gauss theorem. To formulate this theorem, we introduce several new concepts: If the charged body is large, then you need to know the distribution of charges inside the body. Volume charge density– measured by the charge per unit volume: Surface charge density– measured by the charge per unit surface of a body (when the charge is distributed over the surface): Linear charge density(charge distribution along the conductor): b) electrostatic induction vector Vector of electrostatic induction
Vector Let's check the dimension D in SI units: then the dimensions D and E do not coincide, and their numerical values are also different. From the definition Field To understand the meaning of the introduction At the boundary of the cavity with the dielectric, associated negative charges are concentrated and For the same case: D = Eεε 0 Thus– continuity of induction lines greatly facilitates the calculation V) electrostatic induction vector flux Consider the surface S in an electric field and choose the direction of the normal 1. If the field is uniform, then the number of field lines through the surface S: 2. If the field is non-uniform, then the surface is divided into infinitesimal elements dS, which are considered flat and the field around them is uniform. Therefore, the flux through the surface element is: dN = D n dS, and the total flow through any surface is: Induction flux N is a scalar quantity; depending on can be > 0 or< 0, или = 0. Electric field strength vector flux. Let a small platform DS(Fig. 1.2) intersect the electric field lines, the direction of which is with the normal n
angle to this site a. Assuming that the tension vector E
does not change within the site DS, let's define tension vector flow through the platform DS How DFE
=E DS cos a.(1.3) Since the density of the power lines is equal to the numerical value of the tension E, then the number of power lines crossing the areaDS, will be numerically equal to the flow valueDFEthrough the surfaceDS. Let us represent the right side of expression (1.3) as a scalar product of vectors E AndDS=
nDS, Where n– unit vector normal to the surfaceDS. For an elementary area d S expression (1.3) takes the form dFE =
E d S
Across the entire site S the flux of the tension vector is calculated as an integral over the surface Electrical induction vector flow. The flux of the electric induction vector is determined similarly to the flux of the electric field strength vector dFD
= D d S
There is some ambiguity in the definitions of flows due to the fact that for each surface two
normals in the opposite direction. For a closed surface, the outer normal is considered positive. Gauss's theorem. Let's consider point positive electric charge q, located inside an arbitrary closed surface S(Fig. 1.3). Induction vector flux through surface element d S equals 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 the charge is locatedq.
Considering that d S D/ r 2 is equal elementary bodily corner dw, under which from the point where the charge is locatedqsurface element d visible S, we transform expression (1.4) to the form d FD =
q
d w / 4
p, from where, after integration over the entire space surrounding the charge, i.e. within the solid angle from 0 to 4p, we get FD = q. The flow of the electrical 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 a point charge q(Fig. 1.4), then, having constructed a conical surface with the vertex at the point where the charge is located, we divide the surface S into two parts: S 1 and S 2. Flow vector D
through the surface S we find as the algebraic sum of fluxes through the surfaces S 1 and S 2: Both surfaces from the point where the charge is located q visible from one solid angle w. Therefore the flows are equal Since when calculating the flow through a closed surface, we use outer normal to the surface, it is easy to see that the flow F 1D
< 0, тогда как поток Ф2D> 0. Total flow Ф D= 0. This means that the flow 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
,
qn, 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 determined as the sum of the fluxes created by each of the charges. The flow of the electrical 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 a 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 side of expression (1.5), the algebraic summation of charges is replaced by integration over the volume enclosed inside a closed surface S: (1.6) Expression (1.6) is the most general formulation Gauss' theorem: the flow of the electric induction vector through a closed surface of arbitrary shape is equal to the total charge in the volume covered by this surface and does not depend on the charges located outside the surface under consideration. Gauss's theorem can also be written for the flow of the electric field strength vector:
An important property of the electric field follows from Gauss’s theorem: lines of force begin or end only on electric charges or go to infinity. Let us emphasize once again that, despite the fact that the electric field strength E
and electrical induction D
depend on the location in space of all charges, the flows of these vectors through an arbitrary closed surface S are 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 (tension or induction) in the volume V arbitrary, but sufficient for the formation of integral relations, magnitude. By dividing the volume V for small volumes V i, we get the expression valid both as a whole and for each term. Let us transform the resulting expression as follows: and consider the limit to which the expression on the right side of the equality, enclosed in curly brackets, tends for an unlimited division of the volume V. In mathematics this limit is called divergence vector (in this case, the vector of electrical induction D): Vector divergence D in Cartesian coordinates: Thus, expression (1.7) is transformed to the form: Considering that with unlimited division the sum on the left side of the last expression goes into a volume integral, we obtain The resulting relationship must be satisfied for any arbitrarily chosen volume V. This is possible only if the values of the integrands at each point in space are the same. Therefore, the divergence of the vector D is related to the charge density at the same point by the equality or for the electrostatic field strength vector These equalities express Gauss's theorem in differential form. Note that in the process of transition to the differential form of Gauss's theorem, a relation is obtained that has a general character: The expression is called the Gauss-Ostrogradsky formula and connects the volume integral 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 Gauss's theorem for the electrostatic field in vacuum 2)
There is a point charge in the center of the cubeq. What is the flux of a vector? E:
a) through the full surface of the cube; b) through one of the faces of the cube. 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 are linear, surface, volume charge densities. 4)
Indicate the relationship between volume and surface charge densities. 5)
Can the field outside oppositely and uniformly charged parallel infinite planes be non-zero? 6)
An electric dipole is placed inside a closed surface. What is the flow through this surface The most difficult thing is to study electrical phenomena in a non-uniform electrical environment. In such a medium, ε has different values, changing abruptly at the dielectric boundary. Let's assume 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, during 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 electrostatic field strength vector at the interface between two media creates certain difficulties when calculating fields. As for Gauss's theorem, under these conditions it generally loses its meaning. Since the polarizability and voltage of dissimilar dielectrics are different, the number of field lines 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 E 1 = ε 2 E 2 =E 0 =const Multiplying all parts of these equalities by the electric constant ε 0 we obtain ε 0 ε 1 E 1 = ε 0 ε 2 E 2 =ε 0 E 0 =const Let us introduce the notation ε 0 εE=D then the penultimate relation will take the form D 1 = D 2 = D 0 = const Vector D, equal to the product of the electric field strength in the dielectric and its absolute dielectric constant, is calledelectric displacement vector
Electrical displacement unit – pendant per square meter(C/m2). Electrical displacement is a vector quantity and can also be expressed as D = εε 0 E =(1+χ)ε 0 E = ε 0 E + χε 0 E = ε 0 E+P In contrast to the voltage E, the electrical displacement D is constant in all dielectrics (Fig. 12.25, b). Therefore, it is convenient to characterize the electric field in an inhomogeneous dielectric medium not by the intensity E, but by the displacement vector D. Vector D describes the electrostatic field created by free charges (i.e. in a vacuum), but with their distribution in space as in the presence of a dielectric, since bound charges arising in dielectrics can cause a redistribution of free charges creating the field. Vector field Electrical displacement line
- these are lines whose tangents at each point coincide in direction with the electric displacement vector. The lines of vector E can begin and end on any charges - free and bound, while the lines of vectorD- only on free charges. Vector linesDUnlike tension lines, they are continuous. Since the electric displacement vector does not experience a discontinuity at the interface between two media, all induction lines emanating from charges surrounded by some closed surface will penetrate it. Therefore, for the electric displacement vector, Gauss's theorem completely retains its meaning for an inhomogeneous dielectric medium. Gauss's theorem for the electrostatic field in a dielectric
: the flow of the electric displacement vector through an arbitrary closed surface is equal to the algebraic sum of the charges contained inside this surface.
A) charge density
(electric displacement vector) is a vector quantity characterizing the electric field.
equal to the product of the vector 
on the absolute dielectric constant of the medium at a given point:
, because 
,
it follows that for the vector field 
the same superposition principle applies as for the field 
:
graphically represented by induction lines, just like the field
. The induction lines are drawn so that the tangent at each point coincides with the direction 
, and the number of lines is equal to the numerical value of D at a given location.
Let's look at an example.
ε> 1
The field decreases by a factor of and the density decreases abruptly.
, then: lines 
go on continuously. Lines 
start on free charges (at 
on any - bound or free), and at the dielectric boundary their density remains unchanged.
, and, knowing the connection 
With 
you can find the vector 
.
(6)
(1.4)
.
.
(1.7)
.
.
(12.45)
(12.46)
is graphically represented by electric displacement lines in the same way as the field 
depicted by lines of force.
(12.47)
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