Electric field strength vector flow. Let a small playground DS(Fig. 1.2) cross the lines of force of the electric field, the direction of which is with the normal n
corner to this site a. Assuming that the tension vector E
does not change within the site DS, define tension vector flow through the site DS how
DFE =E DS cos a.(1.3)
Since the density of field lines is equal to the numerical value of the tension E, then the number of lines of force crossing the areaDS, will be numerically equal to the value of the streamDFEthrough the surfaceDS. We represent the right side of expression (1.3) as a scalar product of vectors E andDS= nDS, where nis the unit normal vector to the surfaceDS. For elementary area d S expression (1.3) takes the form
dFE = E d S
across the site S the intensity vector flux is calculated as an integral over the surface
Electric induction vector flow. The flow of the electric induction vector is determined similarly to the flow 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 you can specify two normals in the opposite direction. For a closed surface, the outward normal is considered positive.
Gauss theorem. Consider point positive electric charge q, located inside an arbitrary closed surface S(Fig. 1.3). Flow of the induction vector through the 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, at the center of which there is a chargeq.
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Given that d S D/ r 2 equals elementary bodily corner dw, under which from the point where the chargeqsurface element d 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 flow of the electric induction vector through a closed surface of arbitrary shape is equal to the charge enclosed inside this surface.
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If an arbitrary closed surface S does not cover a point charge q(Fig. 1.4), then, having built a conical surface with a vertex at the point where the charge is located, we divide the surface S into two parts: S 1 and S 2. Vector flow D through the surface S we find as the algebraic sum of the flows 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. So 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 flux Ф 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 , 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 the fluxes created by each of the charges. The flow 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 charges qi do not necessarily have to be point, a necessary condition is that the charged region 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 considered 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 the closed surface S:
(1.6)
Expression (1.6) is the most general formulation Gauss theorems: 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 considered surface. The Gauss theorem can also be written for the flow 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 the Gauss theorem. Note that integral form the Gauss 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. By dividing the volume V for small volumes Vi, we get the expression
valid both in general and for each term. We 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, with unlimited division of volume V. In mathematics, this limit is called divergence vector (in this case, the vector of electric induction D):
Vector Divergence D in Cartesian coordinates:
Thus, expression (1.7) is transformed to the form:
.
Taking into account that with unlimited division, the sum on the left side of the last expression goes into a volume integral, we get
The resulting relation must hold 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 the Gauss theorem in differential form.
Note that in the process of passing to the differential form of the Gauss 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 that bounds 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 cubeq. What is the flow of the 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 is linear, surface, volume charge density.
4) Indicate the relationship between volume and surface charge density.
5) Can the field outside oppositely and uniformly charged parallel infinite planes be different from zero
6) An electric dipole is placed inside a closed surface. What is the flow through this surface
Gauss's theorem for electrical induction (electrical displacement)[
For a field in a dielectric medium, the electrostatic theorem of Gauss can be written in another way (alternatively) - 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 inside this surface:
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In differential form:
Gauss' theorem for magnetic induction
The flux of the magnetic induction vector through any closed surface is zero:
or in differential form
This is equivalent to the fact that in nature there are no "magnetic charges" (monopoles) that would create a magnetic field, just as electric charges create an electric field. In other words, the Gauss theorem for magnetic induction shows that the magnetic field is (completely) eddy.
Gauss' theorem for Newtonian gravity
For the field strength of Newtonian gravity (acceleration of free fall), the Gauss theorem practically coincides with that in electrostatics, with the exception of only constants (however, they still depend on an arbitrary choice of the system of units) and, most importantly, the sign:
where g- intensity of the gravitational field, M- gravitational charge (i.e. mass) inside the surface S, ρ - mass density, G is the Newtonian constant.
conductors in an electric field. The field inside the conductor and on its surface.
Conductors are bodies through which electric charges can pass from a charged body to an uncharged one. The ability of conductors to pass electric charges through them is explained by the presence of free charge carriers in them. Conductors - metal bodies in solid and liquid state, liquid solutions of electrolytes. The free charges of a conductor introduced into an electric field begin to move under its action. The redistribution of charges causes a change in the electric field. When the electric field strength in the conductor becomes zero, the electrons stop moving. The phenomenon of separation of opposite charges in a conductor placed in an electric field is called electrostatic induction. There is no electric field inside the conductor. This is used for electrostatic protection - protection with metal conductors from an electric field. The surface of a conducting body of any shape in an electric field is an equipotential surface.
Capacitors
To obtain devices that, at a small potential relative to the medium, would accumulate on themselves (condense) charges of noticeable magnitude, they use the fact that the electric capacitance of a conductor increases when other bodies approach it. Indeed, under the action of a field created by charged conductors, induced (on a conductor) or bound (on a dielectric) charges appear on a body brought to it (Fig. 15.5). Charges that are opposite in sign to the charge of the conductor q are located closer to the conductor than those of the same name with q, and, therefore, have a great influence on its potential.
Therefore, when a body is brought to a charged conductor, the field strength decreases, and, consequently, the potential of the conductor decreases. According to the equation, this means an increase in the capacitance of the conductor.
The capacitor consists of two conductors (plates) (Fig. 15.6), separated by a dielectric layer. When a certain potential difference is applied to a conductor, its plates are charged with equal charges of the opposite sign. The electric capacitance of a capacitor is understood as a physical quantity proportional to the charge q and inversely proportional to the potential difference between the plates
Let's determine the capacitance of a flat capacitor.
If the area of the plate is S, and the charge on it is q, then the field strength between the plates
On the other hand, the potential difference between the plates whence 
The energy of a system of point charges, a charged conductor and a capacitor.
Any system of charges has some potential energy of interaction, which is equal to the work spent on the creation of this system. Energy of a system of point charges q 1 , q 2 , q 3 ,… q N is defined as follows:
where φ 1 - the potential of the electric field created by all charges except q 1 at the point where the charge is q 1 etc. If the configuration of the system of charges changes, then the energy of the system also changes. To change the configuration of the system, work must be done.
The potential energy of a system of point charges can be calculated in another way. Potential energy of two point charges q 1 , q 2 at a distance from each other is equal. If there are several charges, then the potential energy of this system of charges can be defined as the sum of the potential energies of all pairs of charges that can be compiled for this system. So, for a system of three positive charges, the energy of the system is equal to
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Electric field of a point charge q 0 at a distance from it in a medium with permittivity ε (see figure 3.1.3).
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The potential is a scalar, its sign depends on the sign of the charge that creates the field. |
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The electric field of a uniformly charged sphere of radius at point C at a distance from its surface (Figure 3.1.4). The electric field of a sphere is similar to the field of a point charge equal to the charge of the sphere q sf and concentrated in its center. The distance to the point where the tension is determined is ( R+a) |
Out of scope:
The potential inside the sphere is constant and equal to and the tension inside the sphere is zero |
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Electric field of a uniformly charged infinite plane with surface density σ (see figure 3.1.5).
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A field whose intensity is the same at all points is called homogeneous. Surface density σ is the charge per unit surface (, where are the charge and area of the plane, respectively). The dimension of the surface charge density. |
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The electric field of a flat capacitor with equal in magnitude but opposite in sign charges on the plates (see Figure 3.1.6).
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The tension between the plates of a flat capacitor, outside the capacitor E=0. Potential difference u between the plates (plates) of the capacitor: , where d is the distance between the plates, is the permittivity of the dielectric placed between the plates of the capacitor. The surface charge density on the plates of a capacitor is equal to the ratio of the magnitude of the charge on it to the area of the plate:. |
Figure 3.1.3
; 
Figure 3.1.4.
; 
,
Figure 3.1.5.
Figure 3.1.6Energy of a charged solitary conductor and capacitor
If a solitary conductor has a charge q, then there is an electric field around it, the potential of which on the surface of the conductor is , and the capacitance is C. Let's increase the charge by dq. When transferring charge dq from infinity, work equal to 
. But the potential of the electrostatic field of a given conductor at infinity is equal to zero. Then
When the charge dq is transferred from the conductor to infinity, the same work is done by the forces of the electrostatic field. Consequently, with an increase in the charge of the conductor by dq, the potential energy of the field increases, i.e.
Integrating this expression, we find the potential energy of the electrostatic field of a charged conductor as its charge increases from zero to q:
Applying the relation , one can obtain the following expressions for the potential energy W:
For a charged capacitor, the potential difference (voltage) is therefore equal to the ratio for the total energy of its electrostatic field have the form
The purpose of the lesson: The Ostrogradsky–Gauss theorem was established by the Russian mathematician and mechanic Mikhail Vasilievich Ostrogradsky in the form of some general mathematical theorem and by the German mathematician Carl Friedrich Gauss. This theorem can be used in the study of physics at the profile level, as it allows 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 electric induction vector and the flux of this vector Ф.
It is known that the electrostatic field is often depicted using lines of force. Suppose that we determine the tension at a point lying on the interface between two media: air (=1) and water (=81). At this point, when passing 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 for calculating fields, certain inconveniences are created due to the discontinuity of the strength vector at the interface between media and on dielectrics. To avoid them, a new vector is introduced, which is called the electric induction vector:
The electric induction vector is equal to the product of the vector and the electric constant and the permittivity of the medium at a given point.
Obviously, 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 electric induction vector is measured in coulombs per square meter (C / m 2). 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 field of tension (for example, for a point charge, see Fig. 1). For a vector field, the principle of superposition takes place:
Electrical induction flux
The electric induction vector characterizes the electric field at each point in space. One more quantity can be introduced, depending 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 a surface area S, placed in a uniform electric field. The normal to the conductor plane makes an angle with the direction of the electric induction vector (Fig. 2).
The flow of electrical induction through the surface S is called a value equal to the product of the modulus of the induction vector and the area S and the cosine of the angle between the vector and the normal:
Derivation of the Ostrogradsky–Gauss theorem
This theorem allows you to find the flow of the electric induction vector through a closed surface, inside 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 induction flux 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 enclose the charge q, then the total flow Ф \u003d 0 (since each line will enter the surface, and another time it will leave it).
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 flux F does not depend on the shape of the surface. It also does not depend on the arrangement of charges inside 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 we only 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 inside the surface: .
It can be seen from the formula that the dimension of the electric flow is the same as that of the electric charge. Therefore, the unit of the flow of electrical induction is the pendant (C).
Note: if the field is inhomogeneous 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 homogeneous. 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. We 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 constant electric field in time are motionless electric charges.
Application of the Gauss theorem
Field of continuously distributed charges
Let us now determine, using the Ostrogradsky-Gauss theorem, the field strength for a number of cases.
1. Electric field of a uniformly charged spherical surface.
A 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). The surface charge density is the ratio of the 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) Let's take the point A, which is at a distance r>R from the center of the charged spherical surface. Let us mentally draw a spherical surface S of radius r through it, having a common center with a charged spherical surface. It is obvious from symmetry considerations that the lines of force are radial straight 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. So the total flow through the sphere is N = E? S; N=E. On the other hand . 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 was in its center (Fig. 5).
b) Let us find the field strength at the points lying inside the charged spherical surface. Let's take a point B separated from the center of the sphere at a distance 2. Field strength of a uniformly charged infinite plane 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 lines of tension are perpendicular to the plane and directed from it in both directions (Fig. 6). We choose a 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 bases and are parallel to the plane, and the base passes through point A (Fig. 7). Let us calculate the flux of tension through the considered cylindrical surface. The flow through the side surface is 0, because tension lines are parallel to the lateral surface. Then the total flow is the sum 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 all the fields created by each element: ; . Let us introduce the concept of the flow 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 us introduce the concept of a vector-area. Let us agree to understand the area vector as a vector directed perpendicular to the area and numerically equal to the size of the area. Figure 1 - To the definition of the vector - the site Let's call the vector flow Vector flow If the field is uniform and the surface is flat The above expression determines the number of field lines penetrating the area Flow of the electric induction vector through an arbitrary closed surface Expression (6) is the O-G theorem in integral form. Theorem 0-G operates with an integral (total) effect, i.e. if To find the located charges and their magnitude in a given field, it is necessary to have a relation connecting the electric induction vector Suppose we need to determine the presence of a charge at a point a(fig.2) Figure 2 - To the calculation of the vector divergence We apply the O-G theorem. The flow of the electric induction vector through an arbitrary surface that limits the volume in which the point is located a, is equal to The algebraic sum of charges in a volume can be written as a volume integral where To obtain a connection between the field and the charge at a point a we will decrease the volume by contracting the surface to a point a. In this case, we divide both parts of our equality by the value 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 vector divergence In this way: Consequently where With the help of this relation, the inverse problem of electrostatics is simply solved, i.e. finding distributed charges in a known field. If the vector Expression (8) represents Theorem 0-G 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. When there are many charges, some difficulties arise in calculating the fields. Gauss's theorem helps to overcome them. essence Gauss theorems reduces to the following: if an arbitrary number of charges is mentally surrounded by a closed surface S, then the electric field strength flux through the elementary area dS can be written as dФ = Есоsα۰dS where α is the angle between the normal to the plane and the field strength vector The total flow through the entire surface will be equal to the sum of flows from all charges arbitrarily distributed inside it and proportional to the magnitude of this charge Let us determine the flow of the tension vector through a spherical surface of radius r, in the center of which there is a point charge +q (Fig. 12.8). The lines of tension are perpendicular to the surface of the sphere, α = 0, hence сosα = 1. Then If the field is formed by a system of charges, then Gauss theorem:
the flow of the electrostatic field strength vector in vacuum through any closed surface is equal to the algebraic sum of the charges enclosed inside this surface, divided by the electrical constant. If there are no charges inside the sphere, then Ф = 0. The Gauss theorem makes it relatively easy to calculate electric fields for symmetrically distributed charges. Let us introduce the concept of the density of distributed charges. The linear density is denoted τ and characterizes the charge q per unit length ℓ. In general, it can be calculated by the formula With a uniform distribution of charges, the linear density is equal to The surface density is denoted σ and characterizes the charge q per unit area S. In general terms, it is determined by the formula With a uniform distribution of charges over the surface, the surface density is equal to Bulk density, denoted ρ, characterizes the charge q per unit volume V. In general terms, it is determined by the formula With a uniform distribution of charges, it is equal to Since the charge q is evenly distributed on the sphere, then σ = const. Let's apply the Gauss theorem. Let's draw a sphere with a radius through point A. The flow of the intensity vector in Fig. 12.9 through the spherical surface of the radius is cosα = 1, since α = 0. According to the Gauss theorem, From expression (12.14) it follows that the field strength outside the charged sphere is the same as the field strength of a point charge placed in the center of the sphere. On the surface of the sphere, i.e. r 1 \u003d r 0, tension Inside the sphere r 1< r 0
(рис.12.9) напряжённость Е = 0, так как сфера
радиусом r 2
внутри
никаких зарядов не содержит и, по теореме
Гаусса, поток вектора сквозь такую
сферу равен нулю. A cylinder of radius r 0 is uniformly charged with a surface density σ (Fig. 12.10). Let us determine the field strength at an arbitrarily chosen point A. Let us draw an imaginary cylindrical surface with radius R and length ℓ through point A. Due to the symmetry, the flow will exit only through the side surfaces of the cylinder, since the charges on the cylinder of radius r 0 are uniformly distributed over its surface, i.e. the lines of tension will be radial straight lines perpendicular to the side surfaces of both cylinders. Since the flow through the base of the cylinders is zero (cos α = 0), and the side surface of the cylinder is perpendicular to the lines of force (cos α = 1), then We express the value of E through σ - surface density. By definition, Substitute the value of q into the formula (12.15) By definition of line density, those. the field strength generated by an infinitely long charged cylinder is proportional to the linear charge density and inversely proportional to the distance. The intensity of the field created by an infinite uniformly charged plane
According to the Gauss theorem, Because Thus, the field strength of an infinite charged plane is proportional to the surface charge density and does not depend on the distance to the plane. Therefore, the field of the plane is homogeneous. The intensity of the field created by two oppositely uniformly charged parallel planes
Between the planes, the field strengths have the same directions, so the resulting strength is equal to Thus, the field between two oppositely uniformly charged planes is homogeneous and its intensity is twice as large as the field strength created by one plane. There is no field to the left and right of the planes. The field of finite planes has the same form, the distortion appears only near their boundaries. Using the obtained formula, you can calculate the field between the plates of a flat capacitor.
through the site 
dot product of vectors 
and 
. In this way,
through an arbitrary surface 
is found by integrating all elementary flows
(4)
located perpendicular to the field, then:
. (5)
per unit of time.Ostrogradsky-Gauss theorem. Electric field strength divergence
is equal to the algebraic sum of free electric charges 
covered by this surface
(6)
then it is not known whether this means the absence of charges at all points of the studied part of space, or whether the sum of positive and negative charges located at different points of this space is equal to zero.
at a given point with a charge at the same point.
(7)
- charge per unit volume 
;
- volume element.
. Passing to the limit, we get:
.
.
,
, (8)
is the bulk charge density. 
is given, so 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 
there are no charges. At the points where 
positive, there is a positive charge with a bulk density equal to 
, and at those points where 
will have a negative value, a negative charge is found, the density of which is also determined by the divergence value.
. (fig.12.7)
(12.9)
(12.10)
(12.11)
(12.12)
(12.13)
.
.
or
(12.14)
.
or
(12.15)
Consequently, 
(12.16)
, where 
; we substitute this expression into the formula (12.16):
(12.17)
Let us determine the strength of the field created by an infinite uniformly charged plane at point A. Let the surface charge density of the plane be σ. As a closed surface, it is convenient to choose a cylinder whose axis is perpendicular to the plane, and the right base contains point A. The plane divides the cylinder in half. It is obvious that the lines of force are perpendicular to the plane and parallel to the side surface of the cylinder, so all the flow passes only through the bases of the cylinder. On both bases, the field strength is the same, because. points A and B are symmetrical with respect to the plane. Then the flow through the bases of the cylinder is
, then 
, where
(12.18)
The resulting field created by two planes is determined by the principle of field superposition: 
(fig.12.12). The field created by each plane is homogeneous, the strengths of these fields are equal in absolute value, but opposite in direction: 
. According to the principle of superposition, the strength of the total field outside the plane is zero:
