ELECTROMAGNETISM

Chapter 9. Electric field in vacuum

Field tension

It was experimentally found that particles may experience the interaction much stronger than gravity. To explain this to the mass m. Particles added another particle characteristic - electric charge q.measured by coulins (CL).

Let's call a charged particle, i.e., a particle having a charge q., point charge q. (Unlike the charged body, the sizes of which cannot be neglected under the conditions of this task). Each fixed charge point charge q. Creates in the surrounding space electric field (More precisely electrostatic field). On any other point charge in this field will act electrical power :

where vector is called tension electric field At the point where the charge is. Power can be directed or charged q. or from him. In this regard, two types of charge introduced: positive and negative. Multimame charges are attracted, and the same names are repelled from each other (Fig. 31.1).

Tensions are determined as force acting on a single positive point charge at this point point:

where\u003e 0. From the expression (31.2) it can be seen that the dimension is Newton to the pendant (N / CL).

Experience shows that moving point charge q. Creates at a distance r. From him tension

(31.3)

where ε 0 is the electrical constant (ε 0 \u003d 8.85 · 10 -12 KL 2 / (N · m 2)), - a single vector of the radius-vector spent on the center of the field placed at the beginning of the coordinate in which the point charge is located q., Before you are interested in the point of the field.

From the expression (31.1) it can be seen that the electrical force acting on the charge is directed as well as the vector if the charge is positive, and the oppositely vector if the charge is negative (Fig. 31.2).

From experience it follows that system field strength N. stationary spot charges

where - the field strength in the point you are interested in, created i.- point charge in the absence of other point charges. The ratio (31.4) expresses the principle of superposition of electric fields.

Example 31.1. Two balls with masses of 0.3 kg are at such a distance that the interaction of their charges is balanced by the force of gravitational attraction. Find the radii of the balls if the surface density of the ball of balls

Given: m. 1 = m. 2 = m.\u003d 0.3 kg F. E \u003d F. G. R. 1 = R. 2 = R.	Decision . F. E \u003d F. c. .
R. – ?

Answer:R.\u003d 4 cm.

Example 31.2. Point charges q. 1 = 2q. and q. 2 = – q. Located, as shown in Fig. 31.4. Distance between charges is equal d.. Determine at what distance x. 1 from charge q. 1 Electric field strength is zero.

Given: q. 1 = 2q. q. 2 = – q. d. E ( x. 1) = 0	Decision Fig. 31.3.
x. 1 – ?

According to the principle of the superposition of electric fields at the point where the condition should be performed

where and - the tensions of the charges created by charges q. 1 I. q. 2 At this point. Obviously, this condition will not be performed outside the axis x. (vectors and directed at an angle to each other), as well as on the axis x. Left from charge q. 1, where always E. 1 > E. 2 (see formula (31.3) and the condition of the task). Between charges on the axis x. It cannot be zero, since the vectors are directed in one direction. It remains to assume that the desired point lies on the axis x. Right from charge q. 2 (see Fig. 31.3). Distance x. 1 from charge q. 1 Find from the condition

Spere on and extract the root square from the left and right parts of equality:

Answer: x. 1 = 3,5 d..

§ 32. Vector stream

Viscable field vector depicted using lines vector , which are followed as follows:

1) tangent to them at each point coincides with the direction of the vector;

2) The number of lines penetrating the unit of the surface area, perpendicular lines (lines thickness), is equal to the module of the vector (Fig. 32.1).

Electric field Call uniformIf at every point of the field vector \u003d const. The vector lines of such a field are parallel and the distance between them are the same (Fig. 32.2).

Fig. 32.1 Fig. 32.2.

Lines vector electrostatic field Start on positive charges and end on negative charges.

Take the elementary platform ds. In the field of the vector (Fig. 32.3). Let - a single vector of normal to the site ds., α is the angle between the vectors and. Then the number of lines vector penetrating ds.well

where - vector, whose module is equal ds.and direction coincides with a single vector of normal to the site ds..

Name flow f vectorthrough an arbitrary surface S. The number of vector lines that permeate this surface. Obviously

integral on the surface S. From the scalar product of vectors and. Stream - the value of algebraic. The flow sign depends on the choice of the normal direction to ds.. For closed surfaces, it is customary to take an external normal.

§ 33. Gauss theorem for a vector field

Theorem. Vector stream Through any closed surface S. Raven q. vn. / ε 0, where q. vn. - Algebraic amount of charges inside this surface:

(33.1)

where the circle of the integral means that the integration is carried out along a closed surface.

Proof of Theorem. Consider the electrical field of one fixed point charge q.. Let be q. \u003e 0. Mentally surrounding the charge q. arbitrary closed surface S. (Fig. 33.1).

We find the flow d.F vector through element ds. Surfaces. Obviously

where - elementary body (spatial) angle inside a cone based on ds., with a vertex at the charge point q..

Vector stream Through the entire closed surface S.

where is a full body angle. We got

which coincides with the expression (33.1).

Now consider the electric field created by the system N. Fixed point charges mentally surrounding this charge system of an arbitrary closed surface S.. Using the principle of superposition of electric fields, we can write

where q. - Algebraic amount N. charges, which coincides with the expression (33.1).

The Gauss Theorem allows in some cases it is very easy to determine the tension at any point of the electric field.

Example 33.1. We have an infinite uniformly charged plane with surface density Charge σ. Determine intensive aboutst E. x. From the plane.

We will spend through the Gaussian point of interest to us S. In the form of a symmetric relative to the plane of the cylinder so that the point is on the base of the cylinder (Fig. 32.2). We will find the flow of the vector through Gaussian surface:

where S. OSN. - The base area of \u200b\u200bthe cylinder. When integrating, we learned that the flow of the vector through the side surface of the cylinder is zero (vector lines do not permeate this surface) and for all points of the base of the cylinder α \u003d 0 and E. \u003d const.

According to Theorem Gauss

where - the charge of the plane focused inside the cylinder. Find him. By definition, the surface density of charge

In the case of a uniformly charged plane (σ \u003d const) we can write

(from fig. 33.2 it can be seen that the charge is concentrated on the part of the plane with an area S. OWN), from where

(33.4)

Substituting the expressions (33.2) and (33.4) to the ratio (33.3), we get

From the expression (33.5) it can be seen that E. depends on distance x. from a charged plane, i.e.

Consequently, the electric field created by an endlessly charged plane is uniform.

Example 33.2. We have a uniformly charged sphere with the surface density of the charge σ. Radius of sphere R.. Determine intensive aboutst E. electric field at a distance r. from the center of the sphere.

(From fig. 33.3 can be seen, then there is no charge inside the Gaussian surface), whence it follows that

Consequently, inside the charged sphere tension E. Electric field is zero.

Now define E. At a point outside the charged sphere ( r.> R.). Let the sphere charged positively. Due to symmetry. Vector E. The fields created by the sphere, in the point you are interested in radially from the center of the sphere.

E. \u003d const.

According to Theorem Gauss

From fig. 33.3 It can be seen that the charged sphere is inside the Gaussian surface and so the charge q. vn. equal to charging q. SF. spheres. In the case of a uniformly charged sphere (σ \u003d const) we can record

(33.8)

Substituting the expressions (33.6) and (33.8) to the ratio (33.7), we get

Consequently, tension E. Fields outside the charged sphere decreases with distance r.. Graphically addiction E.(r.) The electrical field of a uniformly charged sphere is presented in Fig. 33.4.

Example 33.3. We have a uniformly charged ball with the bulk density of the charge ρ. Radius of a ball R.. Determine intensive aboutst E. electric field at a distance r. From the center of the ball.

When integrating, we took into account that for all points of the Gaussian sphere α \u003d 0 and E. \u003d const.

According to Theorem Gauss

where - the charge of the part of the ball, focused inside the Gaussian sphere. Find him. By definition, the bulk density of charge

In the case of a uniformly charged bowl (ρ \u003d const) we can write

where - the volume of the ball inside the Gaussian sphere. From the expression (33.12) we find

(33.13)

Substituting expressions (33.10) and (33.13) to the ratio (33.11), we get

Each electrical charge surrounds the electric field. As a result of long-term research, physician scientists came to the conclusion that the interaction of charged bodies is due to the electric fields, others. They are a special form of matter, which is inextricably linked with every electric charge.

The study of the electric field is carried out by introducing minor charged bodies into it. These bodies call "trial charges." For example, often as a trial charge is used by a charged cork ball.

When making a test charge into an electric body field having a positive charge, a light positively charged cork ball under its action will deviate the more, the closer we will bring it to the body.

When moving a test charge in an electrical field of an arbitrary charged body, it can be easily detected that the force acting on it will be different in different places.

Thus, when placed sequentially in one point of the field of various in the magnitude of the test positive charges Q1, Q2, Q3, ..., Qn can be found that the forces acting on them, F1, F2, F3, ..., Fn are different, but the ratio of force to size A certain charge for such a point of the field is consistently:

F1 / Q1 \u003d F2 / Q2 \u003d F3 / Q3 \u003d ... \u003d Fn / Qn.

If we investigate different points of the field in this way, then we obtain the following conclusion: for each individual point in the electric field, the ratio of the amount of force acting on the test charge, to the magnitude of such charge is invariably and regardless of the value of the trial charge.

It follows from this that the magnitude of this relationship characterizes the electric field in the arbitrary point. The value that is measured by the ratio of force acting on a positive charge, located at this point of the field, to the charge size and is the electric field strength:

It is, as seen from its definition, is equal to force that acts per unit of positive charge placed at a specific point of the field.

Behind the unit of electrical strength takes the size of a single electrostatic unit with a force in one diena. Such a unit is called an absolute electrostatic unit of tension.

To determine the strength of the electric field of any point charge Q in an arbitrary point of the field A of this charge, which is distinguished from it at a distance R1, it is necessary to put the test charge Q1 into this arbitrary point and calculate the force of Fa, which acts on it (for vacuum).

Fa \u003d (Q1Q) / R²₁.

If we take the ratio of the amount of force that affects the charge, to its value Q1, then the calculation of the tension of the electro-tube at the point A:

In addition, you can find tensions at an arbitrary point in; It will be equal to:

Therefore, the tension of the electric field of a point charge at a certain point of the field (in vacuum) will be directly proportional to the size of this charge and inversely proportional to the square of the distance between this charge and the point.

The field strength acts as its strength characteristic. Knowing it in an arbitrary point of the field E, it is easy to calculate and strength F, affecting Q at this point:

Fields - the direction of tension at each specific point of the field will be combined with the direction of force acting on a positive charge, placed in the point.

When the field is formed by several charges: Q1 and Q2 - the intensity E at any point and this field will be equal to the geometric sum of the E1 and E2 tension created at this point separately by charges Q1 and Q2.

The electric field strength at an arbitrary point can be displayed graphically using a directional segment, which comes from this point, similar to the image of strength and other vector quantities.

If in the space surrounding the electric charge, make another charge, then the Coulomb force will act on it; So, in space surrounding electrical charges, exists force field. According to the ideas of modern physics, the field really exists and along with a substance is one of the forms of the existence of matter, through which certain interactions are carried out between macroscopic bodies or particles that are part of the substance. In this case, they say the electric field - the field by which electrical charges interact. We consider electrical fields that are created by fixed electric charges And called electrostatic.

To detect and experienced the electrostatic field is used test point positive charge - Such a charge that does not distort the field under study (does not cause the redistribution of charges that create the field). If in the charge field Q, Place a trial charge Q. 0, then there is power F., different in different points of the field, which, according to the law of the coulon, is proportional to the trial charge Q. 0. Therefore, the ratio F / Q. 0 does not depend on Q. 0 and characterizes the electrostatic field at that point where the trial charge is located. This value is called tension and is silence characteristic electrostatic field.

Electrostatic field tension At this point there are physical quantitydetermined by force acting on a trial single positive charge placed in this field:

Point Dot Charge Field Tension in Vacuum

The direction of the vector E coincides with the direction of force acting on a positive charge. If the field is created by a positive charge, the vector E is directed along the radius-vector from charge in external space (repulsion of a test positive charge); If the field is created negative charge, then the vector E is directed to the charge (Fig.).

The unit of tension of the electrostatic field - Newton on the pendant (N / CL): 1 N / CL - the intensity of this field, which to the point charge 1 CL acts with force in 1 H; 1 N / CL \u003d 1 V / m, where in (volts) is the unit of the potential of the electrostatic field. Graphically electrostatic field are depicted using tension lines - Lines tangents to which at each point coincide with the direction of the vector E (Fig.).

Since in each given point of space, the vector of tension has only one direction, then the line of tension never intersect. For uniform field (When the vector of tension at any point is permanent in size and direction) intensity line parallel to the tension vector. If the field is created by a point charge, the line of tension is radial straight, emerging from charge, if it is positive (Fig. but), and included in it if the charge is negative (Fig. b.). Due to great clarity, the graphic method of representing the electrostatic field is widely used in electrical engineering.

To use tension lines, not only direction, but also the value of the electrostatic field strength, it was agreed to carry them out with a certain dense: the number of tension lines that permeate the unit of the surface area, perpendicular to the intensity lines, should be equal to the vector of E. Then the number of tension lines penetrating the elementary platform D S, normal n. which forms angle a with a vector E.well E.d. S cos.a. \u003d E N.d. S, Where E P.-Products vector E. On Normal n. To the court D. S.(Fig.).

The value of DF E \u003d E N DS \u003d E.dS is called stream vector intensity through the platform D. S. Here D. S. \u003d D. S.n. - vector, the module of which is D S, And the direction coincides with the direction of normal n. To the site. Selection of direction vector n. (and therefore S.It is conditional, since it can be directed to any side. Unit of the flow of the electrostatic field intensity - 1 V × m.

For an arbitrary closed surface S. Stream vector E. Through this surface

,

where the integral is taken on a closed surface S. Stream vector E. is an algebraic value: Depends not only on the field configuration E., but also from the choice of direction n.. For closed surfaces for the positive direction of the Normal external normal i.e. Normal directed by the outside area covered by the surface.

The principle of independence of the forces is applicable to the Coulomb forces, that is, the resulting force F, acting on the side of the field on the test charge Q 0, is equal to the vector sum of the forces Fi applied to it from each of the charges of Q I :. F \u003d Q 0 E and F I \u003d Q 0 E I, where the e-tension of the resulting field, and E I is the field strength created by the charge Q i. Substituting this in the expression above, we get. This formula expresses the principle of superposition (imposition) of electrostatic fields, according to which the intensity of the resulting field created by the charge system is equal to the geometric sum of field tensions created at this point by each of the charges separately.

The principle of superposition is applicable to calculate the electrostatic field of an electric dipole. Electric dipole is a system of two equal in the module of multi-dimensional point charges (+ q, -q), the distance L between which is significantly less than the distance to the points under consideration. According to the principle of superposition, the tension of the field of the dipole at an arbitrary point where E + and the elastic tensions of fields created by respectively positive and negative charges.