If another charge is introduced into the space surrounding an electric charge, then the Coulomb force will act on it; This means that in the space surrounding electric charges, there is force field. According to the concepts of modern physics, the field really exists and, along with matter, is one of the forms of existence of matter, through which certain interactions are carried out between macroscopic bodies or particles that make up the substance. In this case, we talk about the electric field - the field through which electric charges interact. We consider electric fields that are created by stationary electric charges and are called electrostatic.

For discovery and experimental research electrostatic field used test point positive charge - such a charge that does not distort the field under study (does not cause a redistribution of charges creating the field). If in the field created by the charge Q, place a test charge Q 0, then a force acts on it F, different at different points of the field, which, according to Coulomb’s law, is proportional to the test charge Q 0 . Therefore the ratio F/ Q 0 does not depend on Q 0 and characterizes the electrostatic field at the point where the test charge is located. This quantity is called tension and is force characteristic of the electrostatic field.

Electrostatic field strength at a given point there is a physical quantity determined by the force acting on a test unit positive charge placed at this point in the field:

Field strength of a point charge in vacuum

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

The unit of electrostatic field strength is newton per coulomb (N/C): 1 N/C is the intensity of a field that acts on a point charge of 1 C with a force of 1 N; 1 N/C = 1 V/m, where V (volt) is the unit of electrostatic field potential. Graphically, the electrostatic field is represented using tension lines - lines, the tangents to which at each point coincide with the direction of vector E (Fig.).

Since at any given point in space the tension vector has only one direction, the tension lines never intersect. For uniform field(when the tension vector at any point is constant in magnitude and direction) the tension lines are parallel to the tension vector. If the field is created by a point charge, then the intensity lines are radial straight lines emerging from the charge if it is positive (Fig. A), and included in it if the charge is negative (Fig. b). Due to its great clarity, the graphical method of representing the electrostatic field is widely used in electrical engineering.

In order to use tension lines to characterize not only the direction, but also the value of the intensity of the electrostatic field, it was agreed to draw them with a certain density: the number of tension lines penetrating a unit surface area perpendicular to the tension lines must be equal to the modulus of the vector E. Then the number of tension lines , penetrating the elementary area d S, normal n which forms an angle a with the vector E, equals E d Scos a = E n d S, Where E p-vector projection E to normal n to site d S(rice.).

Value dФ E =E n dS= E dS is called tension vector flow through platform d S. Here d S= d Sn- a vector whose modulus is d S, and the direction coincides with the direction of the normal n to the site. Selecting the vector direction n(and therefore d S) is conditional, since it can be directed in any direction. The unit of flux of the electrostatic field strength vector is 1 V×m.

For an arbitrary closed surface S vector flow E through this surface

,

where the integral is taken over the closed surface S. Flow vector E is algebraic quantity: depends not only on the field configuration E, but also on the choice of direction n. For closed surfaces, the positive direction of the normal is taken to be outer normal, that is, the normal pointing outward to the area covered by the surface.

The principle of independence of force action is applied to Coulomb forces, i.e. the resulting force F acting from 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 Q i: . F = Q 0 E and F i = Q 0 E i , where E is the strength of the resulting field, and E i is the strength of the field created by the charge Q i . Substituting this into the expression above, we get . This formula expresses the principle of superposition (imposition) of electrostatic fields, according to which the strength E of the resulting field created by a system of charges is equal to the geometric sum of the field strengths created at a given point by each of the charges separately.

The principle of superposition is applicable to calculate the electrostatic field of an electric dipole. An electric dipole is a system of two opposite point charges of equal magnitude (+Q, –Q), the distance l between which is significantly less than the distance to the field points under consideration. According to the principle of superposition, the strength E of the dipole field at an arbitrary point , where E+ and E– are the field strengths created by positive and negative charges, respectively.

12. Dielectrics in an electric field. Molecules of polar and non-polar dielectrics in an electric field. Polarization of dielectrics. Types of polarization.

1. Polar dielectrics.

In the absence of a field, each of the dipoles has an electric moment, but the vectors of the electric moments of the molecules are randomly located in space and the sum of the projections of the electric moments to any direction is zero:

If the dielectric is now placed in an electric field (Fig. 18), then a pair of forces will begin to act on each dipole, which will create a moment under the influence of which the dipole will rotate around an axis perpendicular to the arm, tending to the final position when the vector of the electric moment is parallel to the voltage vector electric field. The latter will be hampered by the thermal movement of molecules, internal friction, etc. and therefore

the electric moments of the dipoles will make some angles with the direction of the external field vector, but now a larger number of molecules will have components of the projection of the electric moments in the direction that coincides, for example, with the field strength and the sum of the projections of all electric moments will already be different from zero.

A value indicating the ability of a dielectric to create greater or less polarization, that is, characterizing the compliance of the dielectric to polarization called dielectric susceptibility or dielectric polarizability ().

16. Flow of the electric induction vector (uniform and inhomogeneous induction). Flow through a closed surface. T.Gauss for el. Fields in the environment.

Similar to the flow of the tension vector, we can introduce the concept induction vector flow , leaving the same property as for tension - the induction vector is proportional to the number of lines passing through a unit surface area. You can specify the following properties:

1.Flux through a flat surface in a uniform field (Fig. 22). In this case, the induction vector is directed along the field and the induction line flux can be expressed as follows:

2. The flux of the induction vector through a surface in a non-uniform field is calculated by dividing the surface into elements so small that they can be considered flat, and the field near each element is uniform. The total flux of the induction vector will be equal to:

3. Flow of the induction vector through a closed surface.

Let us consider the flow of the induction vector crossing a closed surface (Fig. 23). Let us agree to consider the direction of external normals to be positive. Then at those points of the surface where the induction vector is directed tangentially to the induction line outward, the angle

and the flux of induction lines will be positive, and where the induction vector D will be positive, and where the vector D is directed inside the surface, the flux of induction lines will be negative, because and . Thus, the total flux of induction lines penetrating the closed surface through and through is zero.

Based on Gauss's theorem, we find that there are no uncompensated electric charges inside a closed surface conducted in a conductor. This property remains the same when the conductor is given an excess charge.

An equal but positive charge will appear on the opposite side. As a result, inside the conductor there will be induced electric field E ind , directed towards the external field, which will grow until it becomes equal to the external field and thus the resulting field inside the conductor becomes zero. This process occurs within a very short time.

The induced charges are located on the surface of the conductor in a very thin layer.

The potential at all points of the conductor remains the same, i.e. the outer surface of the conductor is equipotential.

A closed hollow conductor screens only the field of external charges. If electric charges are inside the cavity, then induction charges will arise not only on outer surface conductor, but also on the inside and the closed conducting cavity no longer screens the field of electric charges placed inside it.

. The field strength near a conductor is directly proportional to the surface charge density on it.

Charged bodies can influence each other without contact through an electric field. A field that is created by static electrical particles, is called electrostatic.

Instructions

1. If another charge Q0 is placed in the electric field created by charge Q, then it will influence it with a certain force. This collision is called the electric field strength E. It is the ratio of the force F with which the field acts on a regular electric charge Q0 at a certain point in space to the value of this charge: E = F/Q0.

2. Depending on a certain point in space, the value of field strength E can change, which is expressed by the formula E = E (x, y, z, t). Consequently, the electric field strength refers to vector physical quantities.

3. Since the field strength depends on the force acting on a point charge, the electric field strength vector E is identical to the force vector F. According to Coulomb’s law, the force with which two charged particles interact in a vacuum is directed along a straight line that connects these charges .

4. Michael Faraday proposed to visually depict the field strength of an electric charge with the support of tension lines. These lines coincide with the tension vector at all tangential points. In drawings they are usually designated by arrows.

5. If the electric field is uniform and its intensity vector is continuous in magnitude and direction, then the intensity lines are parallel to it. If the electric field is created by a properly charged body, the lines of tension are directed away from it, and in the case of a negatively charged particle, towards it.

Tip 2: How to detect electric field strength

In order to discover tension electrical fields, introduce a known test charge into it. Measure the force that acts on it from the side fields and calculate the tension value. If the electric field is created by a point charge or a capacitor, calculate it using special formulas.

You will need

• electrometer, dynamometer, voltmeter, ruler and protractor.

Instructions

1. Determination of the voltage of an arbitrary electrical fields Take a charged body, the dimensions of which are insignificant compared to the size of the body generating the electric field. A charged metal ball with low mass is ideal. Measure the amount of its charge with an electrometer and place it in an electric field. Balance the force acting on the charge from the electric fields dynamometer and take readings in Newtons. After this, divide the force value by the amount of charge in Coulombs (E=F/q). The result will be tension electrical fields in volts per meter.

2. fields point charge If the electric field is generated by a charge whose magnitude is known, to determine its intensity at a certain point in space remote from it, measure the distance between the selected point and the charge in meters. After this, divide the amount of charge in Coulombs by the measured distance raised to the second power (q/r?). Multiply the resulting total by 9*10^9.

3. Determination of electrical voltage fields capacitor Measure the potential difference (voltage) between the capacitor plates. To do this, connect a voltmeter parallel to it, record the result in volts. After this, measure the distance between these plates in meters. Divide the voltage value by the distance between the plates, the result will be tension electrical fields. If air is not placed between the plates, determine the dielectric constant of this medium and divide the total by its value.

4. Definition of electrical fields made by several fields mi If the field at a given point is the result of the superposition of several electric fields, find the vector sum of the values ​​of these fields, taking into account their direction (field superposition thesis). If you need to detect the electric field formed by two fields we, construct their vectors at a given point, measure the angle between them. After this, square each of their values ​​and find their sum. Calculate the product of the field strength values, multiply it by the cosine of the angle, the one that is equal to 180? minus the angle between the tension vectors, and multiply the total by 2. Then subtract the resulting number from the sum of the squares of the tensions (E=E1?+E2?-2E1E2*Cos(180?-?)). When constructing fields, consider that power lines leave the correct charges and enter the negative ones.

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The objects of vector algebra are line segments having a direction and a length called a modulus. In order to determine module vector, should be removed Square root from a quantity representing the sum of the squares of its projections onto the coordinate axes.

Instructions

1. Vectors are characterized by two basic properties: length and direction. Length vector called the module or norm and represents a scalar value, the distance from the start point to the end point. Both properties are used to graphically represent different quantities or actions, say, physical forces, movement elementary particles etc.

2. Location vector in two-dimensional or three-dimensional space does not affect its properties. If you move it to another place, then only the coordinates of its ends will change, however module and the direction will remain the same. This autonomy allows the use of vector algebra tools in various calculations, for example, determining angles between spatial lines and planes.

3. The entire vector can be specified by the coordinates of its ends. Let's first look at two-dimensional space: let's preface vector is located at point A (1, -3), and the end is at point B (4, -5). In order to detect their projections, lower the perpendiculars to the abscissa and ordinate axis.

4. Determine the projections of yourself vector, which can be calculated using the formula: ABx = (xb – xa) = 3; ABy = (yb – ya) = -2, where: ABx and ABy are projections vector on the Ox and Oy axis; xa and xb are the abscissas of points A and B; ya and yb are the corresponding ordinates.

5. In the graphic image you will see a right triangle formed by legs with lengths equal to the projections vector. The hypotenuse of a triangle is the quantity that needs to be calculated, i.e. module vector. Apply the Pythagorean theorem: |AB|? = ABx? +ABy? ? |AB| = ?((xb – xa)? + (yb – ya)?) = ?13.

6. Apparently, for three-dimensional space the formula becomes more complicated by adding a third coordinate - applicate zb and za for the ends vector:|AB| = ?((xb – xa)? + (yb – ya)? + (zb – za)?).

7. Let in the considered example za = 3, zb = 8, then: zb – za = 5;|AB| = ?(9 + 4 + 25) = ?38.

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In order to determine the modulus of point charges of identical magnitude, measure the force of their interaction and the distance between them and make a calculation. If it is necessary to detect the charge modulus of individual point bodies, introduce them into an electric field with a known intensity and measure the force with which the field acts on these charges.

You will need

• – torsion scales;
• - ruler;
• - calculator;
• – electrostatic field meter.

Instructions

1. If there are two charges identical in modulus, measure the force of their interaction using a Coulomb torsion balance, which is also an emotional dynamometer. Later, when the charges come into balance and the wire of the scales compensates for the force of electrical interaction, record the value of this force on the scale. Later, using a ruler, calipers, or a special scale on the scales, find the distance between these charges. Consider that unlike charges attract, and like charges repel. Measure force in Newtons and distance in meters.

2. Calculate the value of the modulus of one point charge q. To do this, divide the force F with which two charges interact by the exponent 9 10^9. Take the square root of the result. Multiply the result by the distance between charges r, q=r?(F/9 10^9). You will receive the charge in Coulombs.

3. If the charges are unequal, then one of them must be previously known. Determine the interaction force between the known and unknown charge and the distance between them using Coulomb torsion balances. Calculate the modulus of the unknown charge. To do this, divide the force of interaction of charges F by the product of the exponent 9 10^9 by the modulus of the famous charge q0. Take the square root of the resulting number and multiply the total by the distance between the charges r; q1=r ?(F/(9 10^9 q2)).

4. Determine the modulus of an unfamiliar point charge by introducing it into an electrostatic field. If its intensity at a given point is unknown in advance, insert an electrostatic field meter sensor into it. Measure voltage in volts per meter. Place a charge at a point of known tension and, with the support of an emotional dynamometer, measure the force in Newtons acting on it. Determine the charge modulus by dividing the value of the force F by the electric field strength E; q=F/E.

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Note!
The tension vector has only one direction at any point in space, therefore the tension lines never intersect.

1 .Two types of electric charges and their properties. The smallest indivisible electrical charge. Law of conservation of electric charges. Coulomb's law. Unit of charge. Electrostatic field. Field detection method. Tension as a characteristic of the electrostatic field. Tension vector, its direction. Electric field strength of a point charge. Tension units. The principle of superposition of fields.

Electric charge - the quantity is invariant, i.e. does not depend on the reference frame, and therefore does not depend on whether the charge is moving or at rest.

two kinds (types) of electric charges : positive charges and negative charges.

It has been experimentally established that like charges repel, and unlike charges attract.

An electrically neutral body must have an equal number of positive and negative charges, but their distribution throughout the volume of the body must be uniform.

Law of conservation of el. charge : algebraic sum of elec. charges of any closed system (a system that does not exchange charges with external heat) remains unchanged, no matter what processes occur within this system.

Elek. charges are not spontaneously created and do not arise, they can only be separated and transferred from one body to another.

Exists the smallest charge, it was called the elementary charge - this is the charge that an electron has and the charge on the body is a multiple of this elementary charge: e=1.6*10 -19 Cl. A negative elementary charge is associated with an electron, and a positive one is associated with a positron, whose charge and mass quantitatively coincide with the charge and mass of the electron. However, due to the fact that the positron lifetime is short, they are absent on bodies and therefore the positive or negative charge of bodies is explained by either a lack or excess of electrons on the bodies.

Coulomb's Law: the forces of interaction between two point charges located in a homogeneous and isotropic medium are directly proportional to the product of these charges and inversely proportional to the square of the distance between them, are equal to each other and are directed in a straight line passing through these charges. g is the distance between charges q 1 and q 2, k is the proportionality coefficient, depending on the choice of the system of physical units.

m/F, a =8.85*10 -12 F/m - dielectric constant

A point charge should be understood as charges concentrated on bodies whose linear dimensions are small compared to the distances between them.

In this case, charge is measured in coulombs - the amount of electricity flowing through the cross-section of a conductor in one second at a current of 1 ampere.

The force F is directed along the straight line connecting the charges, i.e. is the central force and corresponding to attraction (F<0) в случае разноименных зарядов и отталкиванию (F>0) in the case of charges of the same name. This force is called Coulomb force.

Faraday's later studies showed that the electrical interactions between charged bodies depend on the properties of the medium in which these interactions occur.

The purpose of the lesson: give the concept of electric field strength and its definition at any point in the field.

Lesson objectives:

• formation of the concept of electric field strength; give the concept of tension lines and a graphical representation of the electric field;
• teach students to apply the formula E=kq/r 2 in solving simple problems of calculating tension.

An electric field is a special form of matter, the existence of which can only be judged by its action. It has been experimentally proven that there are two types of charges around which there are electric fields characterized by lines of force.

When depicting the field graphically, it should be remembered that the electric field strength lines:

1. do not intersect with each other anywhere;
2. have a beginning on a positive charge (or at infinity) and an end on a negative charge (or at infinity), i.e. they are open lines;
3. between charges are not interrupted anywhere.

Fig.1

Positive charge lines:

Fig.2

Negative charge lines:

Fig.3

Field lines of interacting charges of the same name:

Fig.4

Field lines of unlike interacting charges:

Fig.5

The strength characteristic of the electric field is intensity, which is denoted by the letter E and has units of measurement or. Tension is a vector quantity, as it is determined by the ratio of the Coulomb force to the value of a unit positive charge

As a result of transforming the formula of Coulomb's law and the intensity formula, we have the dependence of the field strength on the distance at which it is determined relative to a given charge

Where: k– proportionality coefficient, the value of which depends on the choice of units of electric charge.

In the SI system N m 2 / Cl 2,

where ε 0 is the electrical constant equal to 8.85·10 -12 C 2 /N m 2 ;

q – electric charge (C);

r is the distance from the charge to the point at which the voltage is determined.

The direction of the tension vector coincides with the direction of the Coulomb force.

An electric field whose strength is the same at all points in space is called uniform. In a limited region of space, the electric field can be considered approximately uniform if the field strength within this region varies slightly.

The total field strength of several interacting charges will be equal to the geometric sum of the strength vectors, which is the principle of field superposition:

Let's consider several cases of determining tension.

1. Let two opposite charges interact. Let's place a point positive charge between them, then at this point there will be two voltage vectors directed in the same direction:

According to the principle of field superposition, the total field strength at a given point is equal to the geometric sum of the strength vectors E 31 and E 32.

The tension at a given point is determined by the formula:

E = kq 1 /x 2 + kq 2 /(r – x) 2

where: r – distance between the first and second charge;

x is the distance between the first and point charge.

Fig.6

2. Consider the case when it is necessary to find the voltage at a point distant at a distance a from the second charge. If we take into account that the field of the first charge is greater than the field of the second charge, then the intensity at a given point of the field is equal to the geometric difference in intensity E 31 and E 32.

The formula for tension at a given point is:

E = kq1/(r + a) 2 – kq 2 /a 2

Where: r – distance between interacting charges;

a is the distance between the second and point charge.

Fig.7

3. Let's consider an example when it is necessary to determine the field strength at a certain distance from both the first and second charge, in this case at a distance r from the first and at a distance b from the second charge. Since like charges repel, and unlike charges attract, we have two tension vectors emanating from one point, then to add them we can use the method; the opposite angle of the parallelogram will be the total tension vector. Algebraic sum we find vectors from the Pythagorean theorem:

E = (E 31 2 + E 32 2) 1/2

Hence:

E = ((kq 1 /r 2) 2 + (kq 2 /b 2) 2) 1/2

Fig.8

Based on this work, it follows that the intensity at any point in the field can be determined by knowing the magnitude of the interacting charges, the distance from each charge to a given point and the electrical constant.

4. Reinforcing the topic.

Verification work.

Option #1.

1. Continue the phrase: “electrostatics is...

2. Continue the phrase: an electric field is….

3. How are the field lines of intensity of this charge directed?

4. Determine the signs of the charges:

Homework tasks:

1. Two charges q 1 = +3·10 -7 C and q 2 = −2·10 -7 C are in a vacuum at a distance of 0.2 m from each other. Determine the field strength at point C, located on the line connecting the charges, at a distance of 0.05 m to the right of the charge q 2.

2. At a certain point in the field, a charge of 5·10 -9 C is acted upon by a force of 3·10 -4 N. Find the field strength at this point and determine the magnitude of the charge creating the field if the point is 0.1 m away from it.