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
(12.45)
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
(12.46)
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 
is graphically represented by electric displacement lines in the same way as the field 
depicted by lines of force.
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 electrostatic field in 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.
(12.47)
Electric field strength vector flux. Let a small platform DS(Fig. 1.2) cross power lines electric field, 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
Flow vector electrical induction. 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 
(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 the charge is locatedq.
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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.
Electrical induction vector flow through a closed surface free form equal to the charge contained within this surface.
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If an arbitrary closed surface S does not cover 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 pointwise, necessary condition- 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:
.
From Gauss's theorem it follows important property electric field: 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:
(1.7)
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 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 physical meaning Gauss' 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
General formulation: The flow of the electric field strength vector through any arbitrarily chosen closed surface is proportional to the electric charge contained inside this surface.
In the SGSE system:
In the SI system:
is the flow of the electric field strength vector through a closed surface.
- the total charge contained in the volume that limits the surface.
- electrical constant.
This expression represents Gauss's theorem in integral form.
In differential form, Gauss's theorem corresponds to one of Maxwell's equations and is expressed as follows
in the SI system:
,
in the SGSE system:
Here is the volumetric charge density (in the case of the presence of a medium, the total density of free and bound charges), and is the nabla operator.
For Gauss's theorem, the principle of superposition is valid, that is, the flow of the intensity vector through the surface does not depend on the charge distribution inside the surface.
The physical basis of Gauss's 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 flow of the electric displacement vector (electrical induction). In this case, the formulation of the theorem is as follows: the flow of the electric displacement vector through a closed surface is proportional to the free electric charge contained inside 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 located 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
Flux of the magnetic induction vector through any closed surface equal to zero:
.
This is equivalent to the fact that in nature there are no “magnetic charges” (monopoles) that would create a magnetic field, like electric charges create an electric field. In other words, Gauss's 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:
Volumetric charge density (see above).
Surface charge density
where dS is an infinitesimal surface area.
Linear charge density
where dl is the length of an infinitesimal segment.
Let's consider the field created by an infinite uniform charged plane. Let the surface charge density of the plane be the same and equal to σ. Let us imagine a cylinder with generatrices perpendicular to the plane and a base ΔS located symmetrically relative to the plane. Due to symmetry. The flux of the tension vector is equal to . Applying Gauss's theorem, we get:
,
from which
in the SSSE 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.
Let us introduce the concept of electric induction vector flow. Let's consider an infinitesimal area. In most cases, it is necessary to know not only the size of the site, but also its orientation in space. Let us introduce the concept of vector-area. Let us agree that by area vector we mean a vector directed perpendicular to the area and numerically equal to the size of the area.
Figure 1 - Towards the definition of the vector - site
Let's call the vector flow 
through the platform 
dot product of vectors 
And 
. Thus,
Flow vector 
through an arbitrary surface 
is found by integrating all elementary flows
(4)
If the field is uniform and the surface is flat 
located perpendicular to the field, then:
. (5)
The given expression determines the number of lines of force piercing the site 
per unit of time.
Ostrogradsky-Gauss theorem. Electric field strength divergence
Electrical induction vector flow through an arbitrary closed surface 
equal to the algebraic sum of free electric charges 
, covered by this surface
(6)
Expression (6) is the O-G theorem in an integral form. Theorem 0-Г operates with the integral (total) effect, i.e. If 
it is unknown whether this means the absence of charges at all points of the studied part of space, or that the sum of positive and negative charges located at different points of this space is equal to zero.
To find the located charges and their magnitude in a given field, a relation is needed that relates the vector of electrical induction 
at a given point with a charge at the same point.
Suppose we need to determine the presence of charge at a point A(Fig.2)
Figure 2 – To calculate vector divergence
Let's apply the O-G theorem. The flow of the electrical induction vector through an arbitrary surface that limits the volume in which the point is located A, is equal
The algebraic sum of charges in a volume can be written as a volume integral
(7)
Where 
- charge per unit volume 
;
- element of volume.
To obtain the connection between the field and the charge at a point A we will reduce the volume by contracting the surface to a point A. In this case, we divide both sides of our equality by the value 
. Moving 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 electrical 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 vector divergence 
.
Thus:
,
hence
, (8)
Where 
- volumetric charge density. 
Using this relationship, the inverse problem of electrostatics is simply solved, i.e. finding distributed charges over a known field.
If the vector 
is given, which means its projections are known 
,
,
onto the coordinate axes as a function of the coordinates and to calculate the distributed density of charges that created a given field, it turns out that it is 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 
positive, there is a positive charge with a volume 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 that the sources of the electric field are free electric charges; the field lines of the electrical induction vector begin and end at positive and negative charges, respectively.
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. 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: ; .
