Electric field strength vector flow. Let a small playground DS(Fig. 1.2) cross the lines of force electric field, whose direction is equal to 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 spherical surface 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 algebraic sum charges covered by this surface:
It should be noted that charges q i 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 flux 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:
.
From the Gauss theorem it follows important property electric field: 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 physical meaning Gauss's theorems 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
Law of interaction electric charges- Coulomb's law - can be formulated differently, in the form of the so-called Gauss theorem. Gauss's theorem is obtained as a consequence of Coulomb's law and the principle of superposition. The proof is based on the inverse proportionality of the interaction force of two point charges to the square of the distance between them. Therefore, the Gauss theorem is applicable to any physical field where the inverse square law and the principle of superposition operate, for example, to the gravitational field.
Rice. 9. Lines of electric field strength of a point charge crossing a closed surface X
In order to formulate the Gauss theorem, let us return to the picture of the lines of force of the electric field of an immobile point charge. The lines of force of a solitary point charge are symmetrically located radial straight lines (Fig. 7). Any number of such lines can be drawn. Let us denote their total number through Then the density of field lines at a distance from the charge, i.e., the number of lines crossing the unit surface of a sphere of radius is equal to Comparing this ratio 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 appropriately choosing the total number N of field lines:
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 gap between any two concentric spheres of different radii, none of the lines breaks and no new ones are added. Since the lines of force are continuous, the same number of lines of force intersect any closed surface (Fig. 9) enclosing the charge
The lines of force have a direction. In the case of a positive charge, they come out of the closed surface surrounding the charge, as shown in Fig. 9. In the case of a negative charge, they enter inside the surface. If the number of outgoing lines is considered positive, and the number of incoming lines is negative, then in formula (8) one can omit the sign of the modulus of the charge and write it in the form
The flow of tension. Let us now introduce the concept of the flux of the field strength vector through the surface. An arbitrary field can be mentally divided into small regions in which the intensity varies in magnitude and direction so little that within this region the field can be considered uniform. In each such region, the lines of force are parallel straight lines and have a constant density.
Rice. 10. To determine the flow of the field strength vector through the area
Consider how many lines of force permeate 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 intersecting is the same, and the density of 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 on the direction of the normal to the site
Therefore, the number of lines of force crossing the area is
The product is called the flux of the field strength through the surface. Formula (10) shows that the flux of the vector E through the surface is equal to the number lines of force crossing this surface. Note that the intensity vector flux, as well as the number of lines of force passing through the surface, is a scalar.
Rice. 11. The flow of the intensity vector E through the site
The dependence of the flow on the orientation of the site relative to the field lines is illustrated in Fig.
The flux of the field strength 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 argued that the flux of the field strength of a point charge through any closed surface 2 enclosing the charge (see Fig. 9), as the number of lines of force emerging from this surface, is equal 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 lines of force 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 fluxes of their field strengths will be added. The total flux will be equal to where by should be understood 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 equal to zero, then the total flux of the field strength through this surface zero: how many lines of force enter the volume bounded by the surface, the same number goes out.
Now we can finally formulate the Gauss theorem: the flow of the electric field strength vector E in vacuum through any closed surface is proportional to the total charge inside this surface. Mathematically, the Gauss theorem is expressed by the same formula (9), where by is understood the algebraic sum of charges. In absolute electrostatic
system of CGSE units, the coefficient and the Gauss theorem are written in the form
In SI and the intensity flux through a closed surface is expressed by the formula
Gauss's theorem is widely used in electrostatics. In some cases, with its help, the fields created by symmetrically located charges are easily calculated.
Fields of symmetrical sources. Let us apply the Gauss theorem to calculate the strength of the electric field of a ball of radius uniformly charged over the surface. For definiteness, we assume that its charge is positive. The distribution of charges that create the field has spherical symmetry. Therefore, the field has the same symmetry. The lines of force of such a field are directed along the radii, and the modulus of tension 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, we draw a spherical surface of radius mentally 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 strength flux 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 the Gauss theorem. If we are interested in the field outside the ball, i.e., for then, for example, in SI and, comparing with (13), we find
In the CGSE system of units, obviously,
Thus, outside the ball, the field strength is the same as that of the field of a point charge placed in the center of the ball. If, however, we are interested in the field inside the ball, i.e., at that, since the entire charge distributed over the surface of the ball is outside the sphere mentally drawn by us. Therefore, there is no field inside the ball:
Similarly, using the Gauss theorem, one can calculate the electrostatic field created by an infinite charged
plane with 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 in different points was different, then moving the charged plane along itself would lead to a change in the field at these points, which contradicts the symmetry of the system - such a shift should not change the field. In other words, the field of an infinite uniformly charged plane is uniform.
As a closed surface for applying the Gauss theorem, we choose the surface of a cylinder constructed as follows: the generatrix of the cylinder is parallel to the lines of force, and the bases have areas parallel to the charged plane and lie on opposite sides of it (Fig. 12). Field strength flow through side surface is zero, so the total flow through a closed surface is equal to the sum of the flows through the bases of the cylinder:
Rice. 12. To the calculation of the field strength of a uniformly charged plane
According to the Gauss theorem, the same flow is determined by the charge of that part of the plane that lies inside the cylinder, and in SI it is equal. Comparing these expressions for the flow, we find
In the CGSE system, the field strength of a uniformly charged infinite plane is given by the formula
For a uniformly charged plate of finite size, the expressions obtained are approximately valid in a region that is far enough from the edges of the plate and not too far from its surface. Near the edges of the plate, the field will no longer be uniform and its lines of force will be bent. At very large distances compared to the dimensions of the plate, the field decreases with distance in the same way as the field of a point charge.
As other examples of fields created by symmetrically distributed sources, one can cite the field of an infinite rectilinear filament uniformly charged along the length, the field of a uniformly charged infinite circular cylinder, the field of a ball,
uniformly charged in volume, etc. The Gauss theorem makes it easy to calculate the field strength in all these cases.
The Gauss theorem gives a connection between the field and its sources, in a sense the opposite of that given by Coulomb's law, which allows you to determine the electric field from given charges. Using the Gauss theorem, one 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 in describing the interaction of electric charges? To what extent can these concepts be applied to gravitational interaction?
What is electric field strength? What do they mean when it is called the force characteristic of the electric field?
How can one judge the direction and modulus of the field strength at a certain point from the pattern of field lines?
Can electric field lines intersect? Justify 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 systems of units of GSE and in SI. How to connect it with geometric sense flow determined by the number of lines of force crossing the surface?
How to use the Gauss theorem to find the strength of an electric field with a symmetrical distribution of the charges that create it?
How to apply formulas (14) and (15) to calculate the field strength of a ball with a negative charge?
Gauss' 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 the charge. This conclusion is due to the fact that there is a reduction in the denominators of both parts of the equality.
On the right side, it arose due to the fact that the force of interaction of charges, described by Coulomb's law, is inversely proportional to the square of the distance between the charges. On the left 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 the space
three dimensions. The fact that this exponent is equal to exactly two, and does not differ from two, even if by a negligible amount, indicates that this three-dimensional space is not curved, that is, that its geometry is precisely Euclidean.
Thus, the Gauss theorem is a manifestation of the properties of physical space in the fundamental law of the 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 prominent minds long before the establishment of these laws themselves. So, I. Kant, three decades before the discovery of Coulomb's law, wrote about the properties of space: “Three-dimensionality occurs, apparently, because substances in existing world act on each other in such a way that the force of the action is inversely proportional to the square of the distance.
Coulomb's law and Gauss's theorem actually represent the same law of nature, expressed in different forms. Coulomb's law reflects the concept of long-range action, while Gauss's theorem proceeds from the idea of a force field filling space, i.e., from the concept of short-range action. In electrostatics, the source of a force field is a charge, and the characteristic of the field associated with the source - the intensity flux - cannot change in empty space, where there are no other charges. Since the flow can be visualized as a set of field lines of force, the invariance of the flow is manifested in the continuity of these lines.
The Gauss theorem, based on the inverse proportionality of the interaction to the square of the distance and on the principle of superposition (additivity of the interaction), is applicable to any physical field in which the inverse square law operates. In particular, it is also valid for the gravitational field. It is clear that this is not just 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 the Gauss 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 testifies to the Euclidean nature of the geometry and three-dimensionality of the physical space?
Consider how the value of the vector E changes at the interface between two media, for example, air (ε 1) and water (ε = 81). The field strength in water decreases abruptly by a factor of 81. This vector behavior E creates certain inconveniences when calculating fields in various environments. To avoid this inconvenience, a new vector is introduced D is the vector of induction or electric displacement of the field. Communication of vectors D And E has the form
D = ε ε 0 E.
Obviously, for the field of a point charge, the electrical displacement will be equal to
It is easy to see that the electrical displacement is measured in C/m 2 , does not depend on properties, and is graphically represented by lines similar to lines of tension.
The direction of field lines characterizes the direction of the field in space (of course, lines of force do not exist, they are introduced for the convenience of illustration) or the direction of the field strength vector. With the help of lines of tension, it is possible to characterize not only the direction, but also the magnitude of the field strength. To do this, we agreed to carry them out with a certain density, so that the number of lines of tension penetrating a unit surface, perpendicular to the lines of tension, was proportional to the modulus of the vector E(Fig. 78). Then the number of lines penetrating the elementary area dS, the normal to which n forms an angle α with the vector E, is equal to E dScos α = E n dS,
where E n - vector component E in the direction of the normal n. The value dФ Е = E n dS = E d S called tension vector flow through the site d S(d S= dS n).
For an arbitrary closed surface S, the flow of the vector E through this surface is
A similar expression has the flow of the electric displacement vector Ф D
.
Ostrogradsky-Gauss theorem
This theorem allows you to determine the flow of vectors E and D from any number of charges. Take a point charge Q and define the flow of the vector E through a spherical surface of radius r, in the center of which it is located.
For a spherical surface α = 0, cos α = 1, E n = E, S = 4 πr 2 and
Ф E = E · 4 πr 2 .
Substituting the expression for E we get
Thus, from each point charge comes the flux Ф E of the vector E equal to Q/ ε 0 . Generalizing this conclusion to the general case of an arbitrary number of point charges, we give the formulation of the theorem: the total flow of the vector E through a closed surface of arbitrary shape is numerically equal to the algebraic sum of electric charges enclosed inside this surface, divided by ε 0 , i.e.
For the electric displacement vector flux D you can get a similar formula
the flow of the induction vector through a closed surface is equal to the algebraic sum of the electric charges covered by this surface.
If we take a closed surface that does not enclose the charge, then each line E And D will cross this surface twice - at the inlet and outlet, so the total flow turns out to be zero. Here it is necessary to take into account the algebraic sum of lines, incoming and outgoing.
Application of the Ostrogradsky-Gauss theorem to calculate electric fields generated by planes, a sphere and a cylinder
A spherical surface of radius R carries a charge Q uniformly distributed over the surface with surface density σ
Let's take a point A outside the sphere at a distance r from the center and mentally draw a sphere of radius r symmetrical to the charged one (Fig. 79). Its area is S = 4 πr 2 . The flow of the vector E will be equal to
According to the Ostrogradsky-Gauss theorem 
, hence, 
taking into account that Q = σ 4 πr 2 , we obtain 
For points located on the surface of a sphere (R = r)
D 
For points inside a hollow sphere (there is no charge inside the sphere), E = 0.
2
. Hollow cylindrical surface with radius R and length l charged with a constant surface charge density 
(Fig. 80). Let us draw a coaxial cylindrical surface of radius r > R.
Vector flow E through this surface
According to the Gauss theorem
Equating the right parts of the given equalities, we obtain
.
If the linear charge density of a cylinder (or a thin thread) is given 
That
3. Field of infinite planes with surface charge density σ (Fig. 81).
Consider the field created by an infinite plane. From considerations of symmetry it follows that the intensity at any point of the field has a direction perpendicular to the plane.
At symmetrical points, E will be the same in magnitude and opposite in direction.
Let us construct mentally the surface of a cylinder with base ΔS. Then, through each of the bases of the cylinder, a stream will exit
F E = E ∆S, and the total flow through the cylindrical surface will be equal to F E = 2E ∆S.
Inside the surface there is a charge Q = σ · ΔS. According to the Gauss theorem,
where 
The result obtained does not depend on the height of the selected cylinder. Thus, the field strength E at any distance is the same in magnitude.
For two oppositely charged planes with the same surface charge density σ, according to the superposition principle, outside the space between the planes, the field strength is equal to zero E = 0, and in the space between the planes 
(Fig. 82a). If the planes are charged with like charges with the same surface charge density, the reverse picture is observed (Fig. 82b). In the space between the planes E=0, and in the space outside the planes 
.
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 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 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 side . 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: ; .
