2.2. Stream vector intensity

2.3. Ostrogradsky-Gaussian theorem

2.4. Application of the Ostrogradsky-Gaussian theorem to the calculation of electric fields

1. Field of an endless uniformly charged plane

2. Field of two uniformly charged planes

3. Field of a charged endless cylinder (thread)

4. The field of two coaxial cylinders with the same linear charge density, but different sign

5. Field of charged hollow ball

6. Field of a volume charged ball

2.5. Differential form of the Ostrogradsky-Gaussian theorem

2.1. Power lines of electrostatic field

The electrostatic field can be specified by specifying for each point the magnitude and direction of the vector . The combination of these vectors forms a field of electrostatic field strength vector. Graphic image of an electrostatic field with tension vector at various points in the field very uncomfortable. Tension vectors are superimposed on each other, and it turns out a very confusing picture. The method proposed by M. Faraday of images of electrostatic fields is more visual silest lines tension. Power lines of tension are lines tangents to which at each point coincide with the direction of the vector . Tension lines are directed as as vector fields in the point under consideration. For example, in Fig.2 of the line of tension directed from left to right. Tension lines do not intersect, because At every point of the field. Vector it has only one specified direction. Tensions begin on positive charge And end on the negative. The lines of lines are chosen so that the number of lines penetrating the unit of the surface perpendicular to the intensity lines was equal to the numerical module of the vector . Then, on the picture of tension lines, you can judge the direction and value of the vector at different points of space (Fig. 2.1).

Uniform The electrostatic field is called, in all points of which the intensity is the same in size and direction,those.

A homogeneous electrostatic field is depicted by parallel power lines at an equal distance from each other (such a field exists, for example, between the capacitor plates) (drawing).

In the case of a point charge, tension lines proceed from a positive charge and go into infinity; And from infinity included in a negative charge. Because

that and the thickness of the power lines is inversely proportional to the square of the distance from charge. However, the surface area of \u200b\u200bthe sphere through which these lines pass itself in proportion to the square of the distance, so total number Lines remain constant at any distance from charge.

For charge system, as we see, the power lines are directed from a positive charge to the negative (Figure 2.2).

Figure 2.2.

Figure 2.3 shows, just that the thickness of the power lines can serve as an indicator of the magnitude .

The thickness of the power lines should be such that the single platform, normal to the tension vector crossed such their number, which is equal to the modulus of the tension vector.

Example 1: If in Figure 2.3, highlight the platform,

then the intensity of the depicted field will be equal to

Figure 2.3

Example 2: Playground

located in a homogeneous field

How many lines crosses this platform if the angle is 30º (Figure 2.4).

We will consider the charge positive. The plane is charged with a constant surface density. The symmetry implies that the tension at any point of the field has a direction perpendicular to the plane (Fig. 2.10). Obviously, in symmetrical relatively plane points, the field strength of the field is equal in size and opposite in direction.

We highlight on the charged platter. Surrounding this platform with a closed surface. As a closed surface, they will present a cylindrical surface with forming, perpendicular to the plane and the bases of the magnitude, located relative to the plane symmetrically. Apply to this surface Theorem Gauss . The flow through the side of the surface will be absent, since at each its point is zero. For bases coincides with. Consequently, the total flow through the surface will be equal. Inside the surface is a charge. According to the Gauss Theorem, a condition must be performed: From where. (3)

The result obtained does not depend on the length of the cylinder, i.e. At any distances from the plane, the field strength is the same in size. The picture of tension lines looks like shown in fig. 2.11. For a negatively charged plane, the direction of the vectors will change to the inverse. If the plane of the final sizes, the result obtained will be valid only for points whose distance from the edge of the plate significantly exceeds the distance from the plate itself (Fig. 2.12).

Calculation electric field In many cases, the use of the theorem, which was established by M.V. Odrogradsky in the form of some common mathematical theorem and Gauss - with reference to the case of an electric field created by a point charge, through a closed ball surface.

The value of the projection of the vector of tension on the surface of the radius sphere R.:

Express the stream of the vector through any closed surface. It is equal to the number of power lines coming out, i.e. Starting on charge.

Now let's say that inside a closed surface are N. spot charges q. 1 , q. 2 , …, q N.. By virtue of the principle of superposition :.

If the line of tension crosses the surface more than once, and several, then be sure to the odd number, so that in the integral it will be taken into account only once, and the expression will keep the force for this case. Thus, for electrostatics ostrogradsky-Gaussian theoremformulated as follows: the flow of the electric field strength through a closed surface is equal to the algebraic amount of prisoners inside this surface of charges divided by ε 0 .

Apply the Ostrogradsky-Gauss theorem for calculating specific fields:

but) electric fieldgenerated by an endless uniformly charged plane

Let be surface density Charge. As a closed surface, through which we calculate the magnitude of the stream of the electric field strength of the electric field, we choose a cylindrical surface as shown in the figure.

From the consideration of symmetry (the electric field has a flat symmetry) implies that the field strength at any point has a direction perpendicular to the plane. Obviously, the tension module at points, symmetrical relative to the plane is the same: E." = E." =E.. The direction of the external normal is perpendicularly charged surface.

According to the Gauss Theorem :. Taking into account what we get :. Electric field strength is the same at any distance from the plane.

b) electric field created by two endless charged planes

Let the superficial charge density on each surface. In the case of multi-dimensionally charged endless plane-parallel plates, as can be seen from the figure, taking into account the expression for the electric field strength created by an infinitely charged plane ().

For space between the plates; For space behind the plates (to the left of the left and right of the right plate).

In the case of the same name charged endless plane-parallel plates,

as can be seen from the drawing, for space behind the plates; For space between the plates.

c) electrical field of an infinite charged cylinder

R. - radius of the cylindrical surface. As a closed surface, through which we calculate the magnitude of the stream of the electric field strength of the electric field, we choose a cylindrical surface as shown in the figure. For symmetry considerations (electric field has an axial symmetry), the electric field strength is directed perpendicular to the side surface of the closed cylindrical surface. The direction of the external normal to the side surface of the cylinder in the direction is radial :. From where. With considering E \u003d E. N. We have: when; when; at.

Or inserting a linear density by writing the Ostrogradsky-Gauss theorem. Where considering that we obtain an expression to determine the magnitude of the electric field strength: for various areas of space, we have: for (inside this area electric charges absent) :; For:; For :.

d) field of charged spherical surface

Let the superficial density of the charge, R. - Radius of a spherical surface. As a closed surface, through which the magnitude of the stream of the electric field strength is calculated, select a spherical surface, as shown in the figure. For symmetry considerations (the electric field has a central symmetry) the tension of the electric field is directed perpendicular to the spherical surface. The direction of the external normal to the spherical surface in the direction is radial. Using the Ostrogradsky-Gauss theorem, we get :; . With considering E \u003d E. N. , we have: for (inside the field of electric charges are absent); for; For :.

We also obtain an expression for the electric field strength in the case when the charge value is known q. on a spherical surface. The use of the Ostrogradsky-Gauss theorem gives the following result :. From where. With considering E \u003d E. N. , we have: for (inside the field of electric charges are absent); For:; For:. . Considering, the resulting expression can be converted :. With considering E \u003d E. N. We have :.