The electrostatic field can be clearly depicted using lines of force (tension lines). Power lines are called curves whose tangents at each point coincide with the tension vector E.

Lines of force are a conventional concept and do not really exist. The field lines of a single negative and single positive charge are shown in Fig. 5 are radial lines coming from a positive charge or going to a negative charge.

If the density and direction of the field lines throughout the entire volume of the field remain unchanged, such an electrostatic field is considered homogeneous (the number of lines should be numerically equal to the field strength E).

The number of field lines marked ">dS, perpendicular to them, determines flow of the electrostatic field strength vector:

formula" src="http://hi-edu.ru/e-books/xbook785/files/17-1.gif" border="0" align="absmiddle" alt="- projection of the vector E onto the direction of the normal n to the site dS (Fig. 6).

Accordingly, the flow of vector E through an arbitrary closed surface S

mark">S not only the magnitude, but also the sign of the flow can change:

1) with formula" src="http://hi-edu.ru/e-books/xbook785/files/17-4.gif" border="0" align="absmiddle" alt="

3) when selecting "> Let's find the flow of vector E through a spherical surface S, in the center of which there is a point charge q.

In this case, the mark ">E and n coincide at all points of the spherical surface.

Taking into account the field strength of a point charge, the formula" src="http://hi-edu.ru/e-books/xbook785/files/18-2.gif" border="0" align="absmiddle" alt="(! LANG:we get

formula" src="http://hi-edu.ru/e-books/xbook785/files/Fe.gif" border="0" align="absmiddle" alt="- an algebraic quantity depending on the sign of the charge. For example, when q<0 линии Е направлены к заряду и противоположны направлению внешней нормали n ..gif" border="0" align="absmiddle" alt="around the charge q has an arbitrary shape. Obviously, the surface is marked ">E, as is the surface S. Therefore, the flow of vector E through an arbitrary surface is the formula" src="http://hi-edu.ru/e-books/xbook785/files/Fe.gif" border ="0" align="absmiddle" alt=".

If the charge is located outside the closed surface, then, obviously, how many lines enter the closed area, the same number will leave it. As a result, the flux of vector E will be equal to zero.

If the electric field is created by a system of point charges formula" src="http://hi-edu.ru/e-books/xbook785/files/18-4.gif" border="0" align="absmiddle" alt="

This formula is the mathematical expression of Gauss's theorem: the flow of the electric field strength vector E in a vacuum through an arbitrary closed surface is equal to the algebraic sum of the charges that it covers, divided by formula" src="http://hi-edu.ru/e-books/xbook785/files/18-6.gif" border="0" align="absmiddle" alt="

To complete the description, let us also present Gauss’s theorem in a local form, relying not on integral relations, but on the field parameters at a given point in space. To do this, it is convenient to use the differential operator - vector divergence, -

formula" src="http://hi-edu.ru/e-books/xbook785/files/nabla.gif" border="0" align="absmiddle" alt="(“nabla”) -

formula" src="http://hi-edu.ru/e-books/xbook785/files/19-1.gif" border="0" align="absmiddle" alt="

In mathematical analysis, the Gauss-Ostrogradsky theorem is known: the flow of a vector through a closed surface is equal to the integral of its divergence over the volume limited by this surface -

formula" src="http://hi-edu.ru/e-books/xbook785/files/ro.gif" border="0" align="absmiddle" alt=":

formula" src="http://hi-edu.ru/e-books/xbook785/files/19-4.gif" border="0" align="absmiddle" alt="

This expression is Gauss's theorem in local (differential) form.

Gauss's theorem (2.2) allows us to determine the strengths of various electrostatic fields. Let's look at several examples of the application of Gauss's theorem.

1. Let's calculate E electrostatic field created by a uniformly charged spherical surface.

Let us assume that a spherical surface of radius R carries a uniformly distributed charge q, i.e. surface charge density is the same mark everywhere ">r >R from the center of the sphere, we mentally construct a new spherical surface S, symmetrical to the charged sphere. In accordance with Gauss’s theorem

formula" src="http://hi-edu.ru/e-books/xbook785/files/20-1.gif" border="0" align="absmiddle" alt="

For points located on the surface of a charged sphere of radius R, by analogy we can write:

selection">inside a charged sphere, does not contain electrical charges within itself, so the flux mark">E = 0.

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

The flux of the magnetic induction vector through any closed surface is zero:

.

This is equivalent to the fact that in nature there are no “magnetic charges” (monopoles) that would create a magnetic field, just as 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.

In a number of cases, Gauss's theorem makes it possible to find the electric field strength of extended charged bodies without resorting to calculating cumbersome integrals. This usually applies to bodies whose geometric shape has certain elements of symmetry (ball, cylinder, plane). Let's look at some examples of using Gauss's theorem to calculate the strength of electric fields.

Example 1. Field of a uniformly charged plane.

The electric field created by an infinitely extended uniformly charged plane is uniform - at every point in space outside the plane its intensity is the same everywhere. This field is directed perpendicular to the plane in both directions (Fig. 2.5). Therefore, for the flow of the field strength vector through an arbitrarily chosen cylindrical surface resting on an element of the plane ΔS, we can write: , from where , where is the surface charge density. Dimension in SI: .

Thus, the desired electric field strength uniformly charged plane .

Example 2. Field of a uniformly charged thread (cylinder).

In this case, the electric field has axial symmetry - it does not depend on the azimuthal angle φ and coordinate z and is directed along the radius vector (Fig. 2.6). Therefore, for the vector flow through a selected cylindrical surface with an axis coinciding with the charged thread, we have: , where is an element of a cylindrical surface; l– length of an arbitrary section of thread.

On the other hand, according to Gauss's theorem, this flow is equal to: where , is the linear charge density of the thread. From here we find: .

The required electric field strength uniformly charged thread : .

Example 3. Field of a uniformly charged sphere.

A) Metal ball. In equilibrium, the charges are evenly distributed over the outer surface of the charged ball (Fig. 2.7). Therefore, when< (внутри шара) электрическое поле отсутствует: .

Outside the ball (>), the electric field created by charges uniformly distributed over its surface has spherical symmetry (directed along radial lines), therefore, according to Gauss’s theorem:

.

We see that the electric field of a uniformly charged metal ball does not depend on the radius of the ball and coincides with the field point charge .

b) Dielectric ball .

Consider a ball with a conditional dielectric constant ε = 1, uniformly charged throughout the volume with charge density (Fig. 2.8).

Dimension of volumetric charge density in SI: .

The total charge of the ball is obviously: .

By Gauss' theorem we have:

1) Inside the ball(r< R) : , where Δq = is the charge of the internal region of the ball, limited by the selected spherical surface of radius r. From here we find: .

2) Outside the ball (r > R): , from where = ,

that is outside charged dielectric ball electric field the same , as in the case metal ball.

Figure 2.9 shows the qualitative progression of the dependencies E(r) for metal and dielectric balls.

metal Fig.2.9. Addiction E(r). dielectric

1.4 Gauss's theorem. Electric induction vector.

Gauss's theorem.

Calculating the field strength of a system of electric charges using the principle of superposition of electrostatic fields can be significantly simplified using Gauss's theorem, which determines the flow of the electric field strength vector through an arbitrary closed surface.

Consider the flow of the tension vector through a spherical surface of radius r, covering a point charge q, located at its center

This result is true for any closed surface of arbitrary shape, covering the charge.

If the closed surface does not cover the charge, then the flow through it is zero, since the number of tension lines entering the surface is equal to the number of tension lines leaving it.

Let's consider general case arbitrary surface surrounding n charges.According to the principle of superposition, the field strength created by all charges is equal to the sum of the intensities created by each charge separately. That's why

Gauss's theorem for the electrostatic field in vacuum:the flux vector of the electrostatic field strength in vacuum through an arbitrary closed surface is equal to algebraic the sum of the charges contained inside this surface divided by ε 0 .

In general, electric charges can be distributed with a certain volume density, different in different places in space. Then the total charge of the volume V covered by the closed surface S is equal to and Gauss's theorem should be written in the form .

LECTURE No. 7. OSTROGRADSKY-GAUSS THEOREM FOR ELECTROSTATIC FIELD

INTRODUCTION

In this lecture we continue to get acquainted with the most important characteristics of the electrostatic field.

The introduction of the concept of electrical induction is associated, first of all, with the convenience of describing the electrostatic field and simplifying the solution of many problems of electrostatics, mainly related to the electrostatic field in dielectrics.

The fact is that another quantity characterizing the electrostatic field is the flow of the induction vector of the electrostatic field through any surface is determined only by free charges, and not by all the charges inside the volume limited by a given surface.

In further study of electric and magnetic fields, we will encounter similar concepts more than once - magnetic field induction, magnetic induction flux. The physical meaning of these concepts is of course different, but their mathematical nature is completely equivalent.

1. FLOW OF THE ELECTROSTATIC FIELD INDUCTION VECTOR

As is known, the electrostatic field strength depends on the properties of the medium: in a homogeneous isotropic medium, the field strength is inversely proportional to the dielectric constant .

Therefore, when moving from one medium to another, the electrostatic field strength undergoes abrupt changes, thereby creating inconvenience when calculating electrostatic fields. That is why it turned out to be necessary, in addition to the intensity vector, to characterize the field with another vector quantity - the electric displacement vector or the electrostatic field induction vector.

Definition. Electrical displacement (electrical induction) is a vector physical quantity equal to the product of the absolute dielectric constant of the medium and the electric field strength.

, (1)

where the quantity is called the absolute dielectric constant of the medium.

From the formula (1) it follows that the vector of electrical induction and the vector of electrostatic field strength for isotropic media, i.e. media whose properties are the same in all directions are always collinear, since the absolute dielectric constant is a strictly positive value.

Let's find the induction of the electric field of a point charge.

Fig.1

(2)

From the formula (2) it is clear that, indeed, the value does not depend on the properties of the medium. The value is the same in all media (water, kerosene, etc.).

Dimension of electrical induction in the SI system:

Electrical displacement lines can be used to graphically represent the electrostatic field.

Definition. Electric field induction lines are imaginary lines, the tangents to which at each point coincide with the electric field induction vector at a given point.

Let's consider an electric field characterized by the electric displacement vector . Let this field contain some elementary flat surface with area - (Fig.2) .

Fig.2

Let's construct a unit normal to the surface and direct it “outward”. Then we introduce the vector of the oriented area equal to the product of the area of ​​this elementary surface and the unit normal vector:

It is obvious that and , since .

Definition Elementary stream vector of electrical induction through the site dS is a scalar physical quantity equal to the scalar product of a vector on a vector-oriented area.

Where - angle between the induction vector and normal to the surface, - projection of the electrical induction vector onto the normal direction.

The total flux of a vector through any surface is equal to the sum of the elementary fluxesthrough elementary surfaces into which a given surface of arbitrary shape can be divided, that is:

(4)

The dimension of the flow of electrical induction in the SI system is coulomb:

.

Comment.

1) For closed surfaces S the vector flux through this surface is:

()

The positive direction of the normal is taken to be the direction of the external normal, i.e. the normal pointing outward to the area covered by the surface.

In this part of the lecture, we studied new physical quantities that characterize the electric field - the induction of the electric field and the flow of the electric field induction vector. The electrical induction vector is an auxiliary quantity, but nevertheless plays an important role in the process of studying the electric field. Similar quantities will be introduced when studying the magnetic field.

2. OSTROGRADSKY-GAUSS THEOREM

It is known that the field strength created by a system of charges can be calculated using the principle of superposition of electrostatic fields. But this in most cases is associated with cumbersome calculations.

These calculations can be greatly simplified by using the fundamental theorem of electrostatics, the Ostrogradsky-Gauss theorem, which determines the flow of the electrical induction vector through any closed surface.

The Ostrogradsky-Gauss theorem is formulated as follows:

“The induction flux of an electrostatic field through any closed surface is equal to the algebraic sum of the charges contained within this surface.”

Mathematically, the Ostrogradsky-Gauss theorem for electrostatic fields is written as follows:

= (5)

Notes.

1) The surface must be closed; the shape of the surface does not matter and can be anything.

2) If the surface S does not cover charges, then the flow of electrical induction through it is zero (Fig. 3):

Fig.3

3) If the algebraic sum of charges is 0, then the flux is zero.

The significance of the Ostrogradsky-Gauss theorem is enormous - it allows you to find the induction and strength of an electric field of a complex configuration.

Algorithm (scheme) for using Theorem OcTrogradsky-Gausswhen calculating the strength of the electrostatic field created by an arbitrary configuration of charges, consists of the following points:

1) We select the point at which we will determine and

2) Through this point we draw a closed surface covering all charges;

3) Calculating the flow electrical induction through this surface by definition, that is, according to the formula:

4) We calculate the same flow, but according to the Ostrogradsky-Gauss theorem:

(5)

5) We equate the expressions obtained in the third and fourth paragraphs and find the value of electrical induction at a given point:

6) Knowing the electrical induction, it is easy to determine the magnitude of the electrostatic field strength at a given point:

As mentioned above, the Ostrogradsky-Gauss theorem is one of the main theorems of electrostatics, with the help of which it is easy to calculate the strength and electrical induction of electrostatic fields of various configurations. Every student should know the algorithm for applying the Ostrogradsky-Gauss theorem by heart.

3. APPLICATION OF THE OSTROGRADSKY-GAUSS THEOREM TO CALCULATE THE STRENGTH OF ELECTROSTATIC FIELDS

Often when solving problems it is convenient to assume that the charges distributed continuously in a charged body – along a certain line (for example, in the case of a charged thin rod), surface (for example, in the case of a charged plate), or volume. They use the concepts accordingly linear, surface and volume charge densities.

The volumetric density of electric charges is a scalar physical quantity equal to the ratio of the charge of a body to the volume of the body over which the charge is distributed:

If the charge is distributed uniformly throughout the volume of the body, then the volumetric charge density is a constant value and it can be easily calculated using the formula:

The dimension of the volumetric charge density is determined from the indicated formulas and in the international system of units is equal to: .

The surface density of electric charges is determined in a similar way - it is a scalar physical quantity equal to the ratio of the charge of the entire surface to the area of ​​​​this surface:

Surface charge density is measured in SI units in coulombs divided per square meter:

The linear density of electric charges is a scalar physical quantity equal to the ratio of the charge of an extended body to the length of this body:

The dimension of linear charge density in the International System of Units is coulomb divided by meter:

3.1. The strength of the electrostatic field created by a uniformly charged spherical surface.

Since the sphere is charged uniformly, the surface charge density is a constant value:

Let the radius of the sphere be known to us and equal to . Then from the formula given above, the total charge of the entire sphere can be easily expressed:

We will assume that the sphere is positively charged. Due to the uniform distribution of charge over the surface of the sphere, the field created by these charges has spherical symmetry. Therefore, the electric induction lines (and the electrostatic field strength lines) are directed radially from the sphere (Fig. 4).

Fig.4

In accordance with the above algorithm for applying the Ostrogradsky-Gauss theorem, we perform the following actions:

1. Select an arbitrary point A, located at a distance from the center of the sphere and determine the strength of the electrostatic field at this point;

2. Let us draw a closed surface through the point . Considering the spherical symmetry of the problem, it is convenient to construct a sphere of radius with a center at the point where the center of the charged sphere is located;

3. We calculate the flow of electrical induction through the surface by definition:

since the problem has spherical symmetry, the magnitude of the electric induction vector at any point located at the same distance from the center of the charged sphere will be constant, therefore we have the right to remove this value from under the integral sign. In addition, the angle - the angle between the vector of electrical induction and the normal vector to the spherical surface at any point of the spherical surface along which integration is carried out is equal to zero.

The integral of the form is equal to the area of ​​the surface over which the integration is carried out, so we can finally write:

;

4. We calculate the same flow, but according to the Ostrogradsky-Gauss theorem:

5. We compare the results obtained in paragraphs 3 and 4:

Or ,

and find the value of electrical induction at point A:

Or

6. Determine the electrostatic field strength at point:

or

Notes:

1) If point A is located inside a charged sphere, that is, then the electrical induction and the electrostatic field strength at such a point are identically equal to zero, and since there are no charges inside the charged sphere, the flow of electrical induction through any closed surface located inside the charged sphere will be equal to zero. In other words, there is no electric field inside a charged sphere.

2) If point A is on the surface of a charged sphere, that is, then the electric induction and the electric field strength on the surface of the charged sphere are respectively equal:

Or

Or

Graph of the dependence of the electrostatic field strength on the distance to the center of the sphere (Fig. 5):

Rice. 5

3.2. Field strength of a uniformly charged infinite plane

Let there be a uniformly charged infinite plane with a constant surface charge density (Fig. 6).

Rice. 6

We will consider a plane to be infinite if the distance from the plane to the point where is determined is much less than the linear dimensions of the plane. Electric displacement lines, as well as vector force lines, in this case are directed perpendicular to the plane and run symmetrically in both directions

We will use the Ostrogradsky-Gauss theorem according to the well-known algorithm:

1. Select a point at a distance from the plane.

2. Let us draw through this point a closed surface in the form of a cylinder, the axis of which is perpendicular to the charged surface. The point lies at the base of the cylinder.

3. Let us calculate the induction flux through the constructed cylindrical surface by definition.

,

where is the induction flux through the side surface of the cylinder, and is the induction flux through the base of the cylinder.

The induction flux through the side surface is zero, since the angle between the normal to the side surface and the induction vector is equal to . Flow through the base of the cylinder:

4. Let us calculate the induction flux using the Ostrogradsky–Gauss theorem.

,

where is the electric charge located inside the closed surface we have constructed - the cylinder.

5. Let’s equate the results obtained in points 3 and 4 and find:

, from here

6. Let us calculate the strength of the electric field created by a uniformly charged infinite plane:

.

Rice. 7

Thus, the induction and field strength of a uniformly charged plane do not depend on the distance to the plane and are constant at any point in the field: the field of a charged surface is uniform. For a negatively charged surface, the result will be the same, only the direction of the vectors will change to the opposite. The dependence graph for such a field is shown in Fig. 7.

From these formulas it is clear that the electric field of an infinite uniformly charged plane is uniform and does not depend on distance.

Using the principle of superpositions for the electrostatic field, one can easily obtain expressions for the strength and electrical induction of the electric field of a flat capacitor:

Conclusion

The Ostrogradsky-Gauss theorem was derived mathematically for a vector field of any nature by the Russian mathematician M.V. Ostrogradsky, and then independently of him, Gauss obtained this theorem as applied to the electrostatic field.

When proving this theorem, Gauss relied on Coulomb's law and therefore the Ostrogradsky-Gauss theorem for the electrostatic field is a consequence of Coulomb's law.

At its core, Gauss's theorem mathematically expresses the fact that it is electric charges that are the sources of the electrostatic field, therefore Gauss's theorem is the main theorem of electrostatics.

4. METHODOLOGICAL INSTRUCTIONS AND EXAMPLES OF SOLVING PROBLEMS

TASK No. 1 . Two isolated metal concentrically located spheres with radii of 5 centimeters and 10 centimeters are assigned charges of 10 nanocoulombs and 20 nanocoulombs, respectively. The space between the spheres is filled with a dielectric with dielectric constant. Determine the strength of the electrostatic field and the magnitude of electrical induction at a distance of 2 centimeters, 7 centimeters and 12 centimeters from the center of both spheres.

GIVEN:

FIND:

SOLUTION: this problem is solved using the Ostrogradsky-Gauss theorem. Let's find electrical induction and electrostatic field strength at a point located at a distance of 2 centimeters from the common center of these spheres, for this we construct a spherical surface with a radius of 2 centimeters, the center of which coincides with the center of the metal spheres. After this, we will find the flow of electrical induction through this spherical surface in two ways - using the Ostrogradsky-Gauss theorem and by determining the flow of electrical induction . The first method gives a trivial value - the electrical induction flux must be equal to zero - , since there is no electric charge inside a spherical surface with a radius of 2 centimeters. The second method gives the following result:

,

since the angle at any point on the spherical surface through which we are looking for the flow of electrical induction. In addition, here we took into account that the integral over a closed surface equal to the area of ​​a spherical surface with a radius of 2 centimeters.

Let us equate the two results obtained: . It follows that electrical induction is zero at a distance of 2 centimeters from the center of the metal spheres and, in general, at any point located inside both spheres Let us now find the strength of the electrostatic field. To do this, we use the definition of electrical induction . From this equality it follows that . Thus, the electrostatic field strength will also be zero at a distance of 2 centimeters from the center of the spheres and at any point inside the metal charged spheres .

Let's move on to a point located between the charged metal spheres at a distance of 7 centimeters from their common center. We will follow the same algorithm. First, we draw a spherical surface with a radius of 7 centimeters, the center of which coincides with the center of the metal spheres. Then we calculate the flow of electrical induction through this surface in two ways. From the Ostrogradsky-Gauss theorem it follows that . Using the definition of electrical induction flux gives a different result:

.

Here we took into account the same considerations that were used in the first case:

And

Equating these expressions, we get:

.

Thus, the electrical induction at a point located between charged spheres at a distance of 7 centimeters from their common center depends only on the charge of the inner sphere , the outer sphere does not in any way affect the electric field that exists inside it.

The electrostatic field strength at the point of interest to us will be equal to

,

Where – dielectric constant of the substance filling the space between charged spheres.

Let's check the dimension of the resulting working formulas:

And

The dimension corresponds to reality, so you can start calculating the final result:

,

Let's move on to the third stage of the task. To find the value of electrical induction and electrostatic field strength outside both charged spheres, at a point located at a distance of 12 centimeters from their common center, we draw a spherical surface with a radius of 12 centimeters, the center of which coincides with the center of the charged spheres.

Let us determine the flow of electrical induction through this surface in two ways. The Ostrogradsky-Gauss theorem gives the following result:

Determining the flow of electrical induction leads to a different result:

The left sides of these two equalities are the same, which means that the right sides of these equalities must be equal to each other, that is: .

Let us express the required quantities:

And

Thus, both spheres participate in the creation of an electric field outside the charged spheres. Since the space surrounding the outer charged sphere is not filled with anything (it is a vacuum), then .

The dimension of these formulas need not be checked, since this operation has already been carried out above.

,

The minus sign gives us information about the direction of the electrical induction vector and the electrostatic field strength vector at a point located at a distance of 12 centimeters from the center of the charged spheres. Indeed, at any point lying outside the charged spheres, the induction vector and the electrostatic field strength vector will be directed radially to the outer charged sphere.

TASK No. 2 . Two infinitely extended uniformly charged plates are located at some distance from each other. The electrostatic field strength between the plates is 3000 volts per meter, and outside the plates it is 1000 volts per meter. Find the surface charge density on each plate.

GIVEN:

FIND:

SOLUTION: when solving this problem, we will use the results of applying the Ostrogradsky-Gauss theorem to calculate the strength and electrical induction of the electrostatic field created by an infinite uniformly charged plane. It turns out that the electrostatic field that exists near such a plane is uniform in nature, the lines of force of such an electrostatic field are directed perpendicular to the plane. If the charge on the plane is positive, then the lines of force are directed from the plane in both directions, but if the charge on the plane is negative, then the lines of force are directed on both sides of the plane. The magnitude of the tension at any point in space near an infinite uniformly charged plane is equal to .

The fact that the electrostatic field strength between the plates is greater than the field strength outside the plates indicates that the plates are charged with opposite charges - one positively, the other negatively. Since outside the vector plates directed in opposite directions , and between the plates - in one direction, that is .

Rice. 2

If the plates are charged with the same charges, let’s say positively, it will be the other way around - the electrostatic field strength between the plates will be less than the intensity outside the plates, since

PROBLEM No. 3. With what force does the electric field of a flat capacitor act on the electric charge of 1 nanocoulomb located in it? Find the force of interaction between the capacitor plates. The surface charge density on the capacitor plates is 0.1 nanocoulombs per square meter, and the area of ​​the capacitor plates is 100 square centimeters.

GIVEN:

FIND:

SOLUTION: The electrostatic field inside a parallel-plate capacitor consists of the electric field created by a positively charged plate and a negatively charged plate. The strength of the resulting field will be equal to the vector sum of the electric field strengths created by the first and second plates:

The magnitude of the voltage of an infinite uniformly charged plate can be found using the Ostrogradsky-Gauss theorem. As is known, its value is equal to:

Summarizing all of the above, we can find the electrostatic field strength inside a parallel-plate capacitor:

This result tells us that the electric field inside the parallel-plate capacitor is uniform.

If you place a charged particle inside a flat capacitor, it will be in an electrostatic field, which will act on it with a certain force:

Let's check the dimension of the resulting working formula:

The dimension is correct, since force is actually measured in newtons.

Mathematical calculations give the following result:

The interaction force, namely the force of attraction between the plates of a flat capacitor, can be found as follows: consider one charged capacitor plate located in an electrostatic field created by another charged plate. The amount of charge on the entire capacitor plate is equal to , where is the area of ​​one plate of a flat capacitor. The strength of the electrostatic field in which this capacitor plate is located is equal to . Consequently, the force that will act on one plate of the capacitor from the electrostatic field created by the other plate will be described by the following formula:

So, we answered the second question of the problem - we found the interaction force (the force with which they attract) the plates of a flat-plate capacitor.

Let's check the dimension of this formula:

The dimension corresponds to reality, let's proceed to mathematical calculations:

Let us determine the flow of the electrostatic field strength of charges q 1 ,q 2 ,...q n in a vacuum (e=1) through an arbitrary closed surface surrounding these charges.

Let us first consider the case of a spherical surface of radius R surrounding one charge +q located at its center (Fig. 1.7).

, where is the integral over the closed surface of the sphere. At all points of the sphere, the magnitude of the vector is the same, and it itself is directed perpendicular to the surface. Hence . The surface area of ​​the sphere is . It follows that

.

The result obtained will also be valid for a surface S¢ of arbitrary shape, since it is penetrated by the same number of lines of force.

Figure 1.8 shows an arbitrary closed surface covering a charge q>0. Some lines of tension either leave the surface or enter it. For all tension lines, the number of intersections with the surface is odd.

As noted in the previous paragraph, tension lines emerging from a volume bounded by a closed surface create a positive flow F e; lines entering the volume create a negative flow -F e. The flows of lines at the entrance and exit are compensated. Thus, when calculating the total flow through the entire surface, only one (uncompensated) intersection of the closed surface by each tension line should be taken into account.

If the charge q is not covered by a closed surface S, then the number of lines of force entering and exiting this surface is the same (Fig. 1.9). The total vector flux through such a surface is zero: Ф E =0.

Let us consider the most general case of a surface of arbitrary shape covering n charges. According to the principle of superposition of electrostatic fields, the intensity created by charges q 1 , q 2 ,...q n is equal to the vector sum of the intensities created by each charge separately: . The projection of the vector - the resulting field strength onto the direction of the normal to the site dS is equal to the algebraic sum of the projections of all vectors onto this direction: ,

The flow of the electrostatic field strength vector in a vacuum through an arbitrary closed surface is equal to the algebraic sum of the charges covered by this surface, divided by the electric constant e 0 . This formulation is a theorem of K. Gauss.

In general, electric charges can be distributed with a certain volume density, different in different places in space. Then the total charge of the volume V covered by the closed surface S is equal to and Gauss's theorem should be written in the form .

Gauss's theorem is of significant practical interest: it can be used to determine the field strengths created by charged bodies of various shapes.